Answer: A
\nSolution: e m ratio of cathode rays is independent to the nature of gas.
Answer: B
\nSolution: e m e m e e m mA A B B \u239b \u239d \u239c \u239e \u23a0 \u239f \u239b \u239d \u239c \u239e \u23a0 \u239f = \u00d7 A B \u21d2 2 3 3 2 = \u00d7 e e A B \u21d2 e e A B = 4 9
Answer: B
\nSolution: eV = 1 2 2 m v \u21d2 v e m = \u00d7 \u00d7 2 V = \u00d7 \u00d7 \u00d7 = \u00d7 2 1 764 10 200 8 2 10 11 6 . . m/s
Answer: C
\nSolution: e m e m \u239b \u239d \u239c \u239e \u23a0 \u239f \u239b \u239d \u239c \u239e \u23a0 \u239f = \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f = mesor -particle \u03b1 1 1 1836 208 2 4 17 65 1 .
Answer: D
\nSolution: eV = = 1 2 2 2 2 mv p m \u21d2 p m = 2 eV \u2234 p p p e = = 1836 1 42 85 1 .
Answer: A
\nSolution: \u03c5 \u03c5 = = \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f \u00d7 \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f = \u00d7 \u2212 c s z 3 10 2 10 50 1 2 10 10 6 1 15 cm cm H .
Answer: A
\nSolution: p E t n hc t = = \u22c5 \u22c5 = \u00d7 \u00d7 \u00d7 \u00d7 \u00d7 \u00d7 \u00d7 \u2212 \u2212 \u03bb 6 10 6 626 10 3 10 1 662 6 10 15 34 8 9 . . = 1.8 \u00d7 10 \u20133 J/s \u2013 m 2
Answer: D
\nSolution: p E t n hc t = = \u22c5 \u22c5 \u03bb \u21d2 14 100 200 6 626 10 3 10 1 1987 8 7 10 34 8 9 \u00d7 = \u00d7 \u00d7 \u00d7 \u00d7 \u00d7 \u00d7 \u2212 \u2212 n . . \u2234 n = 4 \u00d7 10 19 s \u20131
Answer: D
\nSolution: E n hc hc = \u22c5 = \u00d7 \u00d7 \u00d7 \u00d7 = \u2212 \u2212 \u03bb ( . N ) A 1 75 10 2500 10 84 4 10 J
Answer: B
\nSolution: E x E abs \u00d7 = 100 emit \u21d2 n hc x n hc 1 1 2 2 100 \u22c5 \u22c5 = \u22c5 \u03bb \u03bb \u2234 x n n = \u00d7 = \u00d7 = 2 1 1 2 53 100 4530 5080 47 3 \u03bb \u03bb .
Answer: B
\nSolution: \u03bb = = 1240 5 248 nm
Answer: A
\nSolution: E = n \u22c5 hc \u03c5 = 1 \u00d7 6.626 \u00d7 10 \u201334 \u00d7 3 \u00d7 10 8 \u00d7 1650763.73 = 3.28 \u00d7 10 \u201319 J/quanta.
Answer: C
\nSolution: E n h c = \u03bb \u21d2 0 36 6 626 10 3 10 662 6 10 34 8 9 . . . = \u00d7 \u00d7 \u00d7 \u00d7 \u00d7 \u2212 \u2212 n \u2234 n = 1.2 \u00d7 10 18
Answer: D
\nSolution: Energy needed for photochemical dissociation = + = + \u239b \u239d \u239c \u239e \u23a0 \u239f = 482 5 1 2 482 5 96 5 1 2 6 2 . . . . . . KJ mol eV eV eV \u2234 \u03bb \u2248 = 1240 6 2 200 . nm
Answer: A
\nSolution: \u03bb \u2248 \u239b \u239d \u239c \u239e \u23a0 \u239f = 1240 289 5 96 5 413 33 . . . nm
Answer: B
\nSolution: Energy absorbed per mole of H 2 = \u00d7 \u00d7 \u00d7 \u00d7 \u00d7 \u00d7 \u00d7 = \u2212 \u2212 \u2212 6 10 6 6 10 3 10 270 10 10 440 23 34 8 9 3 . KJ \u2234 Percentage of absorbed energy corrected into K.E. = 440 429 440 100 \u2212 \u00d7 = 2.5 %
Answer: D
\nSolution: E = \u00d7 \u00d7 = 9 12400 6900 23 372 kcal/mole \u2234 Energy conversion efficiency = \u00d7 = 111 6 372 100 30 . % EXERCISE II (JEE ADVANCE)\n13.45 Atomic Structure HINTS AND EXPLANATIONS
Answer: C
\nSolution: E n hc = \u22c5 \u03bb \u21d2 6 626 6 626 10 3 10 360 10 34 8 9 . . = \u00d7 \u00d7 \u00d7 \u00d7 \u00d7 \u2212 \u2212 n \u2234 Mole of photons absorbed = = \u00d7 \u00d7 = \u00d7 \u2212 n A ~ . 1 2 10 6 10 2 10 19 23 5 \u2234 Quantum efficiency = \u00d7 \u00d7 \u2212 \u2212 1 10 2 10 0 5 5 5 .
Answer: C
\nSolution: Theoretical
Answer: C
\nSolution: Theoretical
Answer: D
\nSolution: h h E \u03c5 \u03c5 1 0 = + and h h \u03c5 \u03c5 2 0 = + \u22c5 E K \u2234 \u03c5 \u03c5 \u03c5 0 1 2 1 = \u2212 \u2212 K K
Answer: B
\nSolution: Theoretical
Answer: C
\nSolution: 1 2 2 0 mv hc hc max = \u2212 \u03bb \u03bb \u21d2 v hc m max = \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f \u23a1 \u23a3 \u23a2 \u23a4 \u23a6 \u23a5 2 0 0 1 2 \u03bb \u03bb \u03bb \u03bb
Answer: B
\nSolution: h h \u03c5 \u03c5 = + K.E. 0 \u21d2 \u03c5 \u03c5 = \u22c5( ) + 1 0 h K.E.
Answer: C
\nSolution: c a z \u03bb = \u2212 ( ) 1 and c a z 4 1 \u03bb = \u2032 \u2212 ( ) \u2234 \u2032 = + z z z 1
Answer: B
\nSolution: Number of atoms in the disc = \u00d7 \u00d7 = \u00d7 1 12 6 10 5 10 23 22 Now, F K q q r = \u22c5 1 2 2 \u21d2 10 9 10 10 5 9 2 2 2 \u2212 \u2212 = \u00d7 \u00d7 q ( ) \u21d2 q = \u2212 10 3 10 C \u2234 Number of excess electron on negatively charged disc = \u00d7 = \u2212 \u2212 10 3 1 6 10 10 4 8 10 19 9 / . . Hence, Number of excess electron Number of atoms = \u00d7 = \u2212 10 4 8 5 10 10 9 22 / . 1 14 2 4 .
Answer: A
\nSolution: r n = r 1 \u00d7 n 2 \u21d2 21.2 \u00d7 10 \u201311 = 5.3 \u00d7 10 \u201311 \u00d7 n 2 \u21d2 n = 2
Answer: C
\nSolution: 2 p r n = 26.5 \u00c5 \u21d2 2 p \u00d7 0.529 \u00d7 n 2 2 = 26.5 \u21d2 n = 4
Answer: D
\nSolution: r n = 0.529 \u00d7 n z 2 \u00c5
Answer: C
\nSolution: r n = r 1 \u00d7 n 2 \u2234 r n \u2013 r n \u2013 1 = r 1 \u00d7 n 2 \u2013 r 1 \u00d7 ( n \u2013 1) 2 = (2 n \u2013 1) \u22c5 r 1 Where n is the higher orbit.
Answer: D
\nSolution: A A r r r r r r 2 1 2 2 2 1 2 1 2 1 2 1 2 2 16 1 = = \u239b \u239d \u239c \u239e \u23a0 \u239f = \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f = ( ) ( ) \u03c0 \u03c0
Answer: C
\nSolution: r 4 \u2013 r 2 = 2.116 \u00c5 \u21d2 0.529 \u00d7 4 2 z \u2013 0.529 \u00d7 2 2 z = 2.116 \u2234 z = 3 \u21d2 Li 2+ ion
Answer: A
\nSolution: d = 2 p r \u00d7 100 = 2 p \u00d7 0 529 2 4 10 2 10 . \u00d7 \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f \u2212 m \u00d7 100 = 3.32 \u00d7 10 \u20138 m
Answer: B
\nSolution: Circumference = Z p r = 2 p \u22c5 r 0 \u00d7 n 2 1 = 2 p r 0 n 2 and n = 1, 2, 3,.\u2026
Answer: C
\nSolution: V n = 2.188 \u00d7 10 6 z n m/s \u21d2 0.547 \u00d7 10 6 = 2.188 \u00d7 10 6 \u00d7 1 n \u2234 n = 4 Now, r n = 0.529 \u00d7 n z 2 = 0.529 \u00d7 4 1 2 = 8.464 \u00c5
Answer: D
\nSolution: m r ze r \u03bd \u03c0\u03b5 2 0 2 2 1 4 = \u22c5 \u21d2 \u03bd \u03c0\u03b5 = ( ) ze m r 2 0 4
Answer: D
\nSolution: v c ze nh c = ( ) \u22c5 2 4 2 0 \u03c0 \u03c0\u03b5
Answer: A
\nSolution: Solution of Q.36
Answer: A
\nSolution: F ze r = \u22c5 = \u00d7 \u00d7 \u00d7 \u00d7 \u00d7 = \u00d7 \u2212 \u2212 \u2212 1 4 9 10 2 1 6 10 4 10 2 88 10 0 2 2 9 19 2 10 2 9 \u03c0\u03b5 ( . ) ( ) . N
Answer: C
\nSolution: T n z n z , . sec = \u00d7 \u2212 1 5 10 16 3 2 T T 2 3 3 2 3 2 2 2 3 3 , , / / H Li e + 2+ = \u21d2 T x 2 2 3 , sec H e + =\n13.46 Chapter 13 HINTS AND EXPLANATIONS
Answer: C
\nSolution: r r n n 1 2 1 2 2 2 = \u21d2 r r n n 4 2 1 2 2 = \u21d2 n n 1 2 1 2 = \u2234 T T n n 1 1 3 2 3 1 8 2 = =
Answer: C
\nSolution: T n a n 3 and n a r n \u21d2 T n a r n 3 2 /
Answer: B
\nSolution: N T n z = = \u00d7 \u00d7 = \u00d7 \u2212 \u2212 \u2212 10 10 1 5 10 2 1 8 33 10 8 8 16 3 2 6 sec , . .
Answer: C
\nSolution: \u03bb \u03bd = = \u22c5 = \u00d7 \u00d7 \u00d7 \u00d7 = \u00d7 \u2212 \u2212 c c T n z , . . 3 10 1 5 10 1 1 4 5 10 8 16 3 2 8 m
Answer: A
\nSolution: K.E. J V = = \u22c5 ( )\u22c5 = \u22c5 1 2 1 2 2 2 mv mvr r r \u03bd
Answer: C
\nSolution: K.E. = = \u22c5 \u239b \u239d \u239c \u239e \u23a0 \u239f = 1 2 1 2 2 8 2 2 2 2 2 2 mv m nh mr n h mr \u03c0 \u03c0 = \u22c5 \u22c5 = \u22c5 n h m a n h ma n 2 2 2 0 2 4 2 2 0 2 2 8 8 \u03c0 \u03c0 ( )
Answer: A
\nSolution: \u0394 = \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f = \u00d7 \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f = E n n ( ) . . I.E. eV 1 1 14 4 1 1 1 4 13 5 1 2 2 2 2 2
Answer: B
\nSolution: Reduced mass effect: \u2032 = \u22c5 + \u239b \u239d \u239c \u239e \u23a0 \u239f r r m m e n 1 On increasing the nuclear mass, radius decreases.
Answer: C
\nSolution: Reduced mass effect: (I.E.) \u2032 = ( )\u22c5 + \u239b \u239d \u239c \u239e \u23a0 \u239f I.E. 1 1 m m e n On increasing the nuclear mass, ionisation energy increases.
Answer: A
\nSolution: u m m m m m m m m m m p p p p p e = + = \u22c5 + = = \u00d7 1 2 1 2 2 1836 2 \u2234 r pm = = 0 529 918 0 058 . . \u00c5
Answer: C
\nSolution: n = 2 but z = 3 \u2013 2 = 1 \u2234 E 2 2 2 13 6 1 2 3 4 = \u2212 \u00d7 = \u2212 . . eV and I.E. = 3.4 eV
Answer: A
\nSolution: K.E. of emitted electron = 0.5 \u00d7 13.6 eV Now, K.E. mV = 1 2 2 or, 6 8 1 6 10 1 2 9 1 10 19 31 2 . . . \u00d7 \u00d7 = \u00d7 \u00d7 \u00d7 \u2212 \u2212 v \u21d2 v = 1.55 \u00d7 10 6 m/s
Answer: D
\nSolution: \u0394 = = = \u00d7 \u2212 E 1240 589 6 2 1 19 . . eV 3.37 10 J = 48.5 kcal/mol
Answer: A
\nSolution: \u03bb = = \u00d7 1240 0 0141 8 8 10 4 . . nm = 8.8 \u00d7 10 \u20135 m = 88 nm
Answer: A
\nSolution: Minimum is 1 (4 \u2192 1 transition in both atoms) and maximum is 4 (4 \u2192 3 \u2192 2 \u2192 1 in one atom and any other transition in other atom).
Answer: B
\nSolution: Number of available orbits is only
Answer: D
\nSolution: Hence, maximum number of spectral lines = 4 6 2 C = .
Answer: A
\nSolution: 12.1 10.2 1.9 At least two atoms are needed for these three transitions.
Answer: D
\nSolution: 1 1 1 2 1 2 1 4 2 1 2 2 2 2 2 2 \u03bb = \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f = \u00d7 \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f Rz n n R \u21d2 l = 1223 \u00c5 \u2234 UV region.
Answer: C
\nSolution: n n ( ) \u2212 = 1 2 15 \u21d2 n = 6 Now, for shortest wavelength, required transition is 6 \u2192
Answer: C
\nSolution: \u2234 1 1 1 1 1 6 35 36 2 2 2 \u03bb = \u00d7 \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f = R R \u21d2 \u03bb = 36 35 R
Answer: C
\nSolution: 1 1 1 2 1 4 4 2 2 2 2 2 \u03bb = \u00d7 \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f = \u22c5 \u2212 R n R n n ( ) \u2234 \u03bb = \u2212 = \u22c5 \u2212 4 4 4 2 2 2 2 n R n K n n ( ) \u21d2 K R = 4
Answer: D
\nSolution: \u03bb\u03b1 1 2 z \u21d2 \u03bb \u03bb \u03bb H He Li : : : : : : + + = = 2 1 1 1 2 1 3 36 9 4 2 2 2
Answer: B
\nSolution: 1 1 1 1 1 2 2 2 \u03bb = \u00d7 \u00d7 \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f R n \u21d2 n R R = \u2212 \u03bb \u03bb 1\n13.47 Atomic Structure HINTS AND EXPLANATIONS
Answer: A
\nSolution: Required transition is 4 \u2192 2 1 1 1 2 1 4 3 16 2 2 2 \u03bb = \u00d7 \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f = R R \u21d2 \u03bb = 16 3 R
Answer: C
\nSolution: \u03bb\u03b1 1 2 z \u21d2 \u03bb \u03bb Na H 10 1 10 2 2 + = \u21d2 \u03bb Na 10 12 16 + = . \u00c5
Answer: C
\nSolution: Excited state is n = 6 [3 \u2192 2, 4 \u2192 2, 5 \u2192 2, 6 \u2192 2] \u2234 Number of spectral lines in 1 R region =
Answer: D
\nSolution: 66. Required transition is 3 \u2192
Answer: A
\nSolution: Modified Rydberg constant is given by, \u2032 = \u00d7 = R R R me h c 2 2 4 2 4 0 2 3 as \u03c0 \u03c0\u03b5 ( ) Now, 1 1 1 2 1 3 2 5 36 2 2 2 \u03bb = \u2032 \u00d7 \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f = \u22c5 R R \u2234 \u03bb = 18 5 R
Answer: D
\nSolution: \u03c5 = \u22c5 \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f = \u2212 R n R n n 1 2 1 4 4 2 2 2 2 ( )
Answer: B
\nSolution: \u0394 = \u00d7 \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f = E 1312 1 2 1 3 182 22 2 2 . KJ
Answer: A
\nSolution: All are visible radiations. Next line is from transition 7 \u2192
Answer: B
\nSolution: 70. 410.2 nm n2 n1 2 486.1 nm ? Both are visible radiations. For required series, we get only n 1 . Now, 1 486 1 10 1 09 10 1 1 2 1 9 7 2 2 1 2 . . \u00d7 = \u00d7 \u00d7 \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f \u2212 n \u21d2 n 1 =
Answer: A
\nSolution: Hence, the series is Brackett series.
Answer: C
\nSolution: 1 2 1 2 1 3 1 2 2 2 \u03bb = \u00d7 \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f R \u21d2 \u03bb 1 9 5 = R 1 2 1 1 1 2 2 2 2 2 \u03bb = \u00d7 \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f R \u21d2 \u03bb 2 1 5 = R From question, l 1 \u2013 l 2 = 132 nm or, 9 5 1 3 132 10 9 R R \u2212 = \u00d7 \u2212 m \u21d2 R = 1.11 \u00d7 10 9 m \u20131
Answer: D
\nSolution: \u221e n + 1 n 1 \u03bb 2 \u03bb \u03c5 = \u00d7 = \u00d7 \u00d7 + \u2212 \u221e \u239b \u239d \u239c \u239e \u23a0 \u239f 2 725 10 1 09 10 1 1 1 1 6 7 2 2 2 . . ( ) n \u2234 n = 3 Now, 1 1 09 10 2 1 3 1 4 7 2 2 2 \u03bb req = \u00d7 \u00d7 \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f . \u21d2 l req = 471.8 nm
Answer: B
\nSolution: 1240 108 5 1 2 1 5 2 2 . = \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f B.E. \u21d2 B.E. = 54.4 eV
Answer: A
\nSolution: K.E. of electron = 13 6 2 1 1 1 2 13 6 27 2 2 2 2 . . . \u00d7 \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f \u2212 = eV Now, 27.2 \u00d7 1.6 \u00d7 10 \u201319 = 1 2 9 1 10 31 2 \u00d7 \u00d7 \u00d7 \u2212 . v \u2234 v \u2248 3.1 \u00d7 10 6 m/s
Answer: D
\nSolution: X E n Y E n = = + 2 2 3 and ( ) \u2234 X Y n = + 1 3
Answer: A
\nSolution: 1 1 1 1 1 3 2 2 2 \u03bb = \u00d7 \u00d7 \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f R \u21d2 \u03bb = 9 8 R
Answer: B
\nSolution: n r \u03bb \u03c0 = 2 \u21d2 2 3 3 6 2 \u03c0 \u03c0 \u00d7 \u00d7 = x x
Answer: C
\nSolution: \u03bb = h m 2 E \u21d2 E m \u03b1 \u03bb 1 ( ) for same
Answer: B
\nSolution: \u03bb = h m 2 E \u21d2 \u03bb \u03bb \u03b1 p = \u00d7 \u00d7 = 4 2 1 1 2 2 1
Answer: C
\nSolution: p mv mv v = = = 1 2 1 2 2 const \u21d2 l = Constant
Answer: B
\nSolution: \u03bb min . . = \u00d7 \u00d7 = \u00d7 \u2212 \u2212 1 24 10 5 10 2 48 10 6 4 11 m
Answer: B
\nSolution: m h c h c R = \u22c5 = \u00d7 \u00d7 \u2212 \u221e \u239b \u239d \u239c \u239e \u23a0 \u239f = \u00d7 \u2212 \u03bb 2 1 2 1 2 4 10 2 2 2 35 . kg\n13.48 Chapter 13 HINTS AND EXPLANATIONS
Answer: A
\nSolution: \u03bb = h m 2 E \u21d2 E E 2 1 1 2 2 2 100 99 1 02 = \u239b \u239d \u239c \u239e \u23a0 \u239f = \u239b \u239d \u239c \u239e \u23a0 \u239f \u2248 \u03bb \u03bb . \u2234 E 2 is about 2 % greater than E 1 .
Answer: C
\nSolution: \u0394 = \u22c5 \u0394 = \u22c5 = v h m x h m h mv v min 4 4 4 \u03c0 \u03c0 \u03c0
Answer: C
\nSolution: \u03bb = h m 2 E \u21d2 \u0394 = \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f E h m 2 2 2 1 2 2 1 1 \u03bb \u03bb = \u00d7 \u00d7 \u00d7 \u00d7 \u2212 \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f \u2212 \u2212 \u2212 \u2212 ( . ) . ( ) ( ) 6 626 10 2 9 1 10 1 50 10 1 100 10 34 2 31 9 2 9 2 = 7.24 \u00d7 10 \u201323 J = 4.5 \u00d7 10 \u20134 eV
Answer: C
\nSolution: \u03bb = = = \u00d7 \u00d7 \u00d7 \u00d7 \u00d7 \u00d7 \u00d7 \u00d7 \u2212 \u2212 \u2212 h m h m 2 2 2 6 626 10 2 4 1 66 10 2 1 6 10 6 34 27 19 E V . . . = 4.15 \u00d7 10 \u201312 m
Answer: A
\nSolution: \u03bb = \u22c5 3 32 . n z \u00c5 \u21d2 3 32 3 32 2 . . = \u00d7 n \u21d2 n = 2 Energy of photon liberated in 2 \u2192 1 transition, \u0394 = \u00d7 \u00d7 \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f = E 13 6 2 1 1 1 2 40 8 2 2 2 . . eV \u2234 K.E. of emitted electron from H-atom = 40.8 \u2013 13.6 = 27.2 eV Hence, its de Broglie wavelength is given by, \u03bb = = 150 27 2 2 348 . . \u00c5
Answer: C
\nSolution: K.E. of electrons = 12400 3000 12400 4000 1 03 \u2212 = . eV \u2234 \u03bb = = 150 1 03 12 05 . . \u00c5
Answer: D
\nSolution: \u0394 \u22c5 \u0394 \u2265 x \u03bb \u03bb \u03c0 2 4 and \u03bb = = 150 6 5\u00c5 \u2234 \u0394 = \u22c5 \u0394 = \u00d7 \u00d7 \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f = \u00d7 \u2212 \u2212 \u2212 \u03bb \u03bb \u03c0 \u03c0 \u03c0 min ( ) . 2 10 2 9 11 4 5 10 4 1 10 6 25 10 x m
Answer: A
\nSolution: \u0394 E = 2.55 eV = 13 6 1 1 1 2 1 2 2 2 . \u00d7 \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f n n eV \u2234 n 1 = 2 and n 2 = 4 Now, \u0394 = \u00d7 \u2212 \u00d7 = \u03bb 3 32 4 1 3 32 2 1 6 64 . . . \u00c5
Answer: B
\nSolution: Orbital angular momentum = + \u22c5 l l h ( ) 1 2 \u03c0
Answer: B
\nSolution: Electron of 1 s level can never emit photon.
Answer: B
\nSolution: Maximum permissible value of l = ( n \u2013 1)
Answer: B
\nSolution: m = \u20131 \u21d2 l \u2265 1 \u21d2 can not be s -orbital.
Answer: A
\nSolution: Theoretical
Answer: A
\nSolution: Theoretical
Answer: A
\nSolution: Energy 2 s < 2 p < 3 s < 3 p < 4 s < 3 d
Answer: D
\nSolution: m = \u20133, \u20132, \u20131, 0, +1, +2, +3
Answer: B
\nSolution: Number of radial nodes = n \u2013 l \u2013 1 = 3 \u2013 2 \u2013 1 = 0
Answer: B
\nSolution: Theoretical
Answer: C
\nSolution: Probability of fi nding electron at the nucleus = 0
Answer: D
\nSolution: Theoretical
Answer: D
\nSolution: Mg( z = 12) 1 s 2 2 s 2 2 p 6 3 s 2
Answer: C
\nSolution: Theoretical
Answer: B
\nSolution: 2(1 s ) + 2(2 s ) + 2(2 p ) + 1(3 s ) = 7
Answer: D
\nSolution: L l l h h = + + \u22c5 = \u22c5 ( ) 1 2 5 \u03c0 \u03c0 \u21d2 l = 4 Number of orbitals = 2 l + 1 = 9
Answer: A, B, C
\nSolution: (a) (K.E.) Initial = (P.E.) at distance of closest approach or 4.0 MeV = K. q q r 1 2 or, 4 \u00d7 10 6 \u00d7 1.6 \u00d7 10 \u201319 = 9 \u00d7 10 9 \u00d7 ( . ) ( . ) 2 1 6 10 50 1 6 10 19 19 \u00d7 \u00d7 \u00d7 \u00d7 \u00d7 \u2212 \u2212 r \u2234 Distance of closest approach, r = 3.6 \u00d7 10 \u201314 m (b) P.E. = K \u22c5 q q r 1 2 = 9 \u00d7 10 9 \u00d7 ( . ) ( . ) 2 1 6 10 50 1 6 10 9 10 19 19 14 \u00d7 \u00d7 \u00d7 \u00d7 \u00d7 \u00d7 \u2212 \u2212 \u2212 = 10 25 \u00d7 (1.6 \u00d7 10 \u201319 ) 2 J = 10 1 6 10 1 6 10 25 19 2 19 \u00d7 \u00d7 \u00d7 \u2212 \u2212 ( . ) . e V = 1.6 MeV (c) P.E. = K q q r \u22c5 = \u00d7 \u00d7 \u00d7 \u00d7 \u00d7 \u00d7 \u00d7 \u00d7 \u00d7 \u00d7 \u2212 \u2212 \u2212 1 2 9 19 19 14 9 10 2 1 6 10 50 1 6 10 4 5 10 1 6 1 ( . ) ( . ) . ( . 0 0 10 19 6 \u2212 \u00d7 ) K q q r \u22c5 = \u00d7 \u00d7 \u00d7 \u00d7 \u00d7 \u00d7 \u00d7 \u00d7 \u00d7 \u00d7 \u2212 \u2212 \u2212 1 2 9 19 19 14 9 10 2 1 6 10 50 1 6 10 4 5 10 1 6 1 ( . ) ( . ) . ( . 0 0 10 19 6 \u2212 \u00d7 ) = 3.2 MeV \u2234 K.E. of a -particle at this distance = 4.0 \u2013 3.2 = 0.8 MeV
Answer: A, C, D
\nSolution: \u03b5 \u03b5 n n = 1 2
Answer: A, C, D
\nSolution: Theoretical
Answer: A, B, C, D
\nSolution: Theoretical
Answer: B, C
\nSolution: Theoretical
Answer: A, C, D
\nSolution: c d a 4 3 2 1
Answer: A, B, C
\nSolution: 1 1 1 1 1028 10 1 09 10 1 1 1 1 2 1 2 2 2 10 7 2 2 2 \u03bb = \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f \u21d2 \u00d7 = \u00d7 \u00d7 \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f \u2212 RZ n n n . \u2234 n = 3 1 3 2 3 2 1028 \u00c5 1 Induced radiations \u03bb 1 = 1028 \u00c5 \u03bb \u03bb \u03bb 2 1 2 2 2 2 2 1 1 1 3 1 2 1 3 6579 2 = \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f \u21d2 = . \u00c5 \u03bb \u03bb \u03bb 3 1 1 1 3 1 1 1 2 1218 4 1 2 2 2 2 3 = \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f \u21d2 = . \u00c5
Answer: A, B
\nSolution: Per atom only one photon is emitted out and hence, the concerned transition is 2 \u2192
Answer: B, C, D
\nSolution: \u0394 E Z Z = \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f = 13 6 1 1 1 2 10 2 2 2 2 2 . . eV
Answer: A, B, C
\nSolution: (a) r 1 : r 2 : r 3 = 1 2 : 2 2 : 3 2 = 1 : 4 : 9 (c) \u03bb = = \u00d7 \u00d7 = \u00d7 = \u2212 c v 3 10 6 10 5 10 500 8 14 7 m nm (d) \u03bb \u03bb \u03b1 = \u21d2 h mE m 2 1 \u2234 \u03bb \u03bb \u03bb H : : : : : : H e cn 4 1 1 1 4 1 16 4 2 1 = =
Answer: A, B
\nSolution: h n = \u03d5 + (K.E.) max For A: 4.25 = \u03d5 A + T A and \u03bb A A 2 = h mT For B: 4.20 = \u03d5 B + T B and \u03bb B B = h mT 2 As T B = T A \u2013 1.50 and \u03bb B = 2 \u03bb A \u03d5 A = 2.25 eV; \u03d5 B = 3.70 eV; T A = 2.0 eV; T B = 0.5 eV
Answer: A, C
\nSolution: l = 3.32 n z \u00c5
Answer: A, B, C
\nSolution: Theoretical\n13.50 Chapter 13 HINTS AND EXPLANATIONS
Answer: B, C
\nSolution: Theoretical
Answer: B, C
\nSolution: Theoretical
Answer: C
\nSolution: Theoretical
Answer: A, B, C
\nSolution: 1 S 2 S 2 P
Answer: A, B, C
\nSolution: Theoretical
Answer: B, C, D
\nSolution: Na (11) 1 s 2 2 s 2 2 p 6 3 s 1
Answer: A, B, D
\nSolution: Theoretical
Answer: B
\nSolution: For minimum l , K.E. of photoelectron should be maximum. For it, the power should be maximum and number of photons is minimum. E max for photon = 5 4 10 1 25 10 18 18 \u00d7 = \u00d7 \u2212 . J \u2234 (K.E.) max of photoelectron = 1.25 \u00d7 10 \u201318 \u2013 4.5 \u00d7 10 \u201319 = 8.0 \u00d7 10 \u201319 J = 5 eV \u2234 \u03bb min = 150 5 30 = \u00c5
Answer: D
\nSolution: i min = 4 \u00d7 10 18 \u00d7 1.6 \u00d7 10 \u201319 = 0.64 A
Answer: C
\nSolution: i i max min . . = \u00d7 \u00d7 \u00d7 \u00d7 \u00d7 \u00d7 = \u2212 \u2212 9 10 1 6 10 4 10 1 6 10 9 4 18 19 18 19 Comprehension II
Answer: B
\nSolution: F du dr K r MV r V K mr = \u2212 = = \u21d2 = 4 4 5 2 4 2 (1) From Bohr\u2019s quantization, V n h m r 2 2 2 2 2 2 4 = \u03c0 (2) \u2234 4 4 16 4 4 2 2 2 2 2 2 2 2 K mr n h m r r mK n h nh mK = \u21d2 = = \u03c0 \u03c0 \u03c0 .
Answer: B
\nSolution: V nh mr nh m nh mK n h m mK = = = 2 2 4 8 2 2 2 \u03c0 \u03c0 \u03c0 \u03c0 .
Answer: C
\nSolution: E = K.E. + P.E. = 1 2 2 4 mv K r + \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f = \u22c5 \u2212 = \u22c5 \u239b \u239d \u239c \u239e \u23a0 \u239f 1 2 4 4 4 4 4 m K mr K r K nh mK \u03c0 \u2234 E n h m K = 4 4 4 2 256 \u03c0 Comprehension III \u0394 E z n n = \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f 13 6 1 1 2 2 2 1 2 . eV 10.2 + 17.0 = 13.6 z 2 1 2 1 2 2 \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f n (1) 4.25 + 5.95 = 13.6 z 2 1 3 1 2 2 \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f n (2)
Answer: D
\nSolution: n = 6
Answer: B
\nSolution: z = 3
Answer: C
\nSolution: \u0394 E = \u00d7 \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f = 13 6 3 1 1 1 7 119 9 2 2 2 . . eV\n13.51 Atomic Structure HINTS AND EXPLANATIONS Comprehension IV
Answer: D
\nSolution: After excitation, n =
Answer: C
\nSolution: Hence, initial excited state is n =
Answer: A
\nSolution: 11. n = 3
Answer: B
\nSolution: 1 1 1 1 1654 10 1 09 10 1 2 1 3 2 1 2 2 2 10 7 2 2 2 \u03bb = \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f \u21d2 \u00d7 = \u00d7 \u00d7 \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f \u2212 RZ n n z . \u2234 z = 2 \u00de He + ion
Answer: C
\nSolution: \u0394 E z n n = \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f = \u00d7 \u2212 \u221e \u239b \u239d \u239c \u239e \u23a0 \u239f = 13 6 1 1 13 6 2 1 3 1 6 04 2 1 2 2 2 2 2 2 . . . eV Comprehension V
Answer: C
\nSolution: Final excited state, after absorption of 2.7 eV, is
Answer: C
\nSolution: On de-excitation, the sample emit radiations equal to less than or more than 2.7 eV and hence, the initial excited state must be
Answer: B
\nSolution: 4 2.7 eV 2.7 eV Less than 2.7 eV More than 2.7 eV 3 2 1
Answer: C
\nSolution: \u0394 E I E n n I E = \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f \u21d2 = \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f . . . . . 1 1 2 7 1 2 1 4 1 2 2 2 2 2 \u2234 I . E . = 14.4 eV
Answer: A
\nSolution: \u0394 E I E n n min . . . = \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f = \u00d7 \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f 1 1 14 4 1 3 1 4 1 2 2 2 2 2 = 0.7 eV Comprehension VI
Answer: B
\nSolution: r n h mze n z = = \u00d7 ( ) . 4 4 0 529 0 2 2 2 2 2 \u03c0\u03b5 \u03c0 \u00c5 (for H-like atom) For this system, r = \u00d7 = \u00d7 = \u2212 0 529 1 207 2 56 10 0 256 3 . . . \u00c5 pm
Answer: C
\nSolution: I . E . = 2 4 13 6 2 2 4 0 2 2 2 2 2 \u03c0 \u03c0\u03b5 mz e n h z n ( ) . = \u00d7 eV ( for H-like atom) For this system, I . E . = 13.6 \u00d7 207 = 2835.9 eV
Answer: B
\nSolution: Rydberg constant for this system = 1.09 \u00d7 10 7 \u00d7 207 m \u20131 \u2234 1 1 09 10 207 1 1 1 2 5 91 10 7 2 2 10 \u03bb \u03bb = \u00d7 \u00d7 \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f \u21d2 = \u00d7 \u2212 ( . ) . m Comprehension VII
Answer: C
\nSolution: n n n ( ) \u2212 = \u21d2 = 1 2 6 4 Now, 1 1 1 1 10 1 09 10 1 1 1 1 4 2 1 2 2 2 10 7 2 2 2 \u03bb = \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f \u21d2 \u00d7 = \u00d7 \u00d7 \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f \u2212 RZ n n x . \u2234 x = 978.6
Answer: C
\nSolution: n = 4
Answer: A
\nSolution: For max l , transition : n = 4 to n = 3 \u2234 1 1 09 10 1 1 3 1 4 1 887 10 7 2 2 2 6 \u03bb \u03bb max max . . = \u00d7 \u00d7 \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f \u21d2 = \u00d7 \u2212 m\n13.52 Chapter 13 HINTS AND EXPLANATIONS
Answer: B
\nSolution: For max n , transition : n = 4 to n = 1 \u2234 Hz \u03bd \u03bb max . . = = \u00d7 \u00d7 = \u00d7 \u2212 c 3 10 978 6 10 3 066 10 8 10 15
Answer: C
\nSolution: 1R radiations involve transition : n = 4 to n = 3 only.
Answer: B
\nSolution: Visible radiation involve transitions : n = 4 to n = 2 (489.3 nm) and n = 3 to n = 2 (660.5 nm). Comprehension VIII
Answer: B
\nSolution: 5 4 3 2 1 1 2 3 4 \u23af \u2192 \u23af \u23af \u2192 \u23af \u23af \u2192 \u23af \u23af \u2192 \u23af
Answer: C
\nSolution: 5 3 2 1 \u23af \u2192 \u23af \u23af \u2192 \u23af \u23af \u2192 \u23af and 5 \u23af \u2192 \u23af \u23af \u2192 \u23af \u23af \u2192 \u23af 4 3 1
Answer: D
\nSolution: 6 + 1 (any possibility after than Q.27)
Answer: A
\nSolution: 5 5 1 2 10 ( ) \u2212 =
Answer: C
\nSolution: 4 5 3 2 1 1 1 1 1 3 4 2 5 6 2 Comprehension IX \u0394 E Z n n eV Z n = \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f = \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f 13 6 1 1 204 13 6 1 1 1 2 2 1 2 2 2 2 2 2 . . ( ) (1) 40 8 13 6 1 1 2 2 2 2 . . ( ) = \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f Z n n (2)
Answer: B
\nSolution: n = 2 \u21d2 2 n = 4
Answer: D
\nSolution: Z = 4
Answer: C
\nSolution: E 1 2 2 13 6 4 1 217 6 = \u2212 \u00d7 = \u2212 . . eV
Answer: A
\nSolution: \u0394 E min . . = \u00d7 \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f = 13 6 4 1 3 1 4 10 58 2 2 2 eV Comprehension X
Answer: B
\nSolution: 1 1 1 2 1 4 1 4 16 16 2 1 2 2 2 2 2 2 2 2 \u03bb = \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f = \u00d7 \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f = \u2212 R Z n n R n R n n ( ) \u2234 \u03bb = \u2212 = \u2212 \u21d2 = = 4 16 16 4 366 97 2 2 2 2 n R n cn n C R ( ) . nm
Answer: C
\nSolution: For series limit, n = \u221e \u2234 l = C = 366.97 nm
Answer: D
\nSolution: For n = 5, l = 1019.36 nm For n = \u221e , l = 366.97 nm Comprehension XI
Answer: A
\nSolution: S 1 = 2 s
Answer: C
\nSolution: E E S H 1 13 6 3 2 2 25 2 2 = \u2212 \u00d7 = \u00d7 . .
Answer: D
\nSolution: S 2 = 3 p \u21d2 l = 1\n13.53 Atomic Structure HINTS AND EXPLANATIONS Comprehension XII
Answer: A
\nSolution: n = 2 l = 1 j = 3 2 1 2 or m = = \u2212 \u2212 + \u2212 + \u2212 3 2 1 2 1 2 3 2 , , , for j = 3 2 = \u2212 + 1 2 1 2 , for j = 1 2
Answer: B
\nSolution: n = 3 l = 0 \u21d2 j = 1 2 \u21d2 m = \u2212 + 1 2 1 2 , = 2 \u21d2 j = 3 2 \u21d2 m = \u2212 \u2212 + + 3 2 1 2 1 2 3 2 , , , = \u21d2 = \u2212 \u2212 \u2212 + + + 5 2 5 2 3 2 1 2 1 2 3 2 5 2 m , , , , , Comprehension XIII
Answer: A
\nSolution: Radial nodes = n \u2013 l \u2013 1 = 3 Angular nodes = 1
Answer: A
\nSolution: Orbital angular momentum = + \u22c5 l l h ( ) 1 2 \u03c0
Answer: B
\nSolution: Theory based
Answer: D
\nSolution: Theory based
Answer: C
\nSolution: Spin quantum number is independent from wave function.
Answer: D
\nSolution: Theory based
Answer: C
\nSolution: Be is the reactive element.
Answer: A
\nSolution: 2 p \u21d2 2 p x + 2 p y + 2 p z \u21d2 Total 3 angular nodes.
Answer: D
\nSolution: dz 2 has two conical nodes.
Answer: D
\nSolution: 3 p x and 3 p y diff ers in angular function.
Answer: C
\nSolution: 4s energy level is lower than 3 d .
Answer: A \u2192 R, S; B \u2192 Q, S; C \u2192 P, Q; D \u2192 P, R
\nSolution: Radial nodes = n \u2013 l \u2013 1 Angular nodes = l
Answer: A \u2192 R, S; B \u2192 P, S; C \u2192 Q; D \u2192 Q
\nSolution: Theory based
Answer: A \u2192 R; B \u2192 Q; C \u2192 P; D \u2192 S
\nSolution: (A) V K P E K E n n = = \u2212 = \u2212 . . . . mV mV 2 1 2 2 2 (B) \u03b5 n n r \u221d \u2212 ( ) 1 (C) Lowest energy level is 1 s . (D) r z n \u221d 1
Answer: A \u2192 S; B \u2192 P; C \u2192 R; D \u2192 Q
\nSolution: Theory based
Answer: A \u2192 P; B \u2192 S; C \u2192 Q; D \u2192 R
\nSolution: Graph of s -orbital status with some value but for other orbitals, it starts from zero. Radial nodes: 3 s = 2, 4 s = 3, 2 p = 0, 3 p = 1
Answer: A \u2192 P, U; B \u2192 Q, T; C \u2192 S, W; D \u2192 R, V
\nSolution: (A) r n z \u221d 2 (B) V z n \u221d (C) F mv r z n = \u221d 2 3 4 (D) f v r z n = \u221d 2 2 3 \u03c0\n13.54 Chapter 13 HINTS AND EXPLANATIONS
Answer: A \u2192 P; B \u2192 P, Q, S; C \u2192 P, R; D \u2192 Q, S
\nSolution: (A) 3 radial nodes \u21d2 4 s , 5 p , 6 d but graph does not start from origin and hence, only 4 s . (B) 3 radial nodes \u21d2 4 s , 5 p , 6 d (C) Only s- orbital (D) l \u2265 1
Answer: A \u2192 Q, R; B \u2192 P, Q, R, S; C \u2192 P, Q, R; D \u2192 P, Q
\nSolution: Theory based
Answer: A \u2192 R, S; B \u2192 Q, P; C \u2192 P
\nSolution: 10. (A) V V 6 4 4 6 2 3 = = v \u221d \u239b \u239d \u239c \u239e \u23a0 \u239f 1 n (B) \u03bb \u03bb 3 2 1 1 1 1 2 1 4 1 2 2 2 2 = \u2212 \u221e \u239b \u239d \u239c \u239e \u23a0 \u239f \u2212 \u221e \u239b \u239d \u239c \u239e \u23a0 \u239f = (C) \u03bb \u03bb c p = \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f = 1 3 1 6 1 1 1 3 3 3 2 2 2 2 2 (D) \u0394 \u0394 E E n H e + = = 1 2 1 4 2 2
Answer: A \u2192 Q; B \u2192 P; C \u2192 S; D \u2192 R
" } } ] }, { "title": "Chem Sec 6", "originalName": "Section F - Integer", "questions": [ { "question_id": "atomic-structure-chem-sec-6-1-164", "marks": 4.0, "negMarks": 1.0, "partialMarks": null, "subject": "chemistry", "chapter": "atomic-structure", "chapterTitle": "Atomic Structure", "type": "sa", "rawChapterType": "NAT", "originalNumber": 164, "displayNumber": 1, "image": "./assets/chapter-packs/atomic-structure/Chemistry Section 1__--__164__--__1.png", "solutionImage": null, "question": { "content": "Answer: 4
\nSolution: \u03b5 = = 1240 300 4 13 . eV For photoelectric effect, e \u2265 f \u21d2 N 0 = 4
Answer: 6
\nSolution: Frequency of reduction \u221d z n 2 3 \u2234 T T 3 2 2 3 8 2 2 7 1 3 4 8 10 1 2 1 28 10 1 6 = \u00d7 \u00d7 \u00d7 \u00d7 = \u2212 \u2212 . .
Answer: 5
\nSolution: 1 1 1 2 1 2 2 2 \u03bb = \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f R Z n n \u21d2 1 108 5 10 1 30 4 10 1 09 10 2 1 1 1 7 7 7 2 2 2 . . . \u00d7 + \u00d7 = \u00d7 \u00d7 \u00d7 \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f \u2212 \u2212 n \u2234 n = 5
Answer: 3
\nSolution: \u03bb 3 \u2013 \u03bb 2 = 59.3 nm or, 1 1 2 1 3 1 1 1 1 2 59 3 2 2 2 2 2 2 R Z R Z \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f \u2212 \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f = . nm \u21d2 z = 3
Answer: 3
\nSolution: Final excited state = 5th orbit As only the wavelengths are longer than absorbed radiation initial excited state = 3rd orbit
Answer: 4
\nSolution: \u0394 E = 12.75 = 13.6 \u00d7 1 2 4 2 1 4 2 n m n n \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f \u21d2 =
Answer: 5
\nSolution: \u0394 \u0394 x h m V min . . = = \u00d7 \u00d7 \u00d7 \u00d7 = \u00d7 \u2212 \u2212 \u2212 \u2212 4 6 626 10 4 10 3 313 10 5 10 34 6 3 26 \u03c0 \u03c0 \u03c0 m
Answer: 1
\nSolution: m h x v min . . = \u22c5 \u22c5 = \u00d7 \u00d7 \u00d7 \u00d7 = \u2212 \u2212 \u2212 4 6 626 10 4 10 5 27 10 1 34 11 24 \u03c0 \u03c0 \u0394 \u0394 kg
Answer: 2
\nSolution: 2 p r = nl \u21d2 l = 4 2 nm = 2 nm
Answer: 2
\nSolution: For radial node, y 23 = 0 \u21d2 r 0 = 2 a 0 Four-digit Integer Type
Answer: 0025
\nSolution: Initial K. E. = P. E. at distance of closest approach or, p q q r 2 0 1 0 2 1 4 m = \u22c5 \u03c0\u03b5 . or ( . ) . . . 3 2 10 2 4 10 6 10 9 10 2 1 6 10 1 6 10 1 20 2 3 23 9 19 19 \u00d7 \u00d7 \u00d7 \u00d7 = \u00d7 \u00d7 \u00d7 \u00d7 \u00d7 \u00d7 \u00d7 \u2212 \u2212 \u2212 \u2212 z 5 5 10 13 \u00d7 \u2212 \u2234 z = 25 Orbital Radial nodes Angular nodes 3 d 0 2 2 p 0 1 3 p 1 1 5 d 2 2\n13.55 Atomic Structure HINTS AND EXPLANATIONS
Answer: 0084
\nSolution: t = = \u00d7 \u00d7 \u00d7 = Distance Speed sec 2 12600 10 3 10 0 084 3 8 .
Answer: 0030
\nSolution: c a \u03bb = \u2212 (z ) 6 c a 180 27 1 = \u2212 ( ) c a z z 144 1 30 = \u2212 \u21d2 = ( )
Answer: 8400
\nSolution: nh n = ms \u2219 \u0394 T or, n \u00d7 6.626 \u00d7 10 \u201334 \u00d7 2.45 \u00d7 10 10 = 245 \u00d7 4.2 \u00d7 (99.5 \u2013 19.5) \u2234 Number of photons = 5.04 \u00d7 10 27 \u2234 Moles of photon = 5 04 10 6 10 8400 27 23 . \u00d7 \u00d7 =
Answer: 0917
\nSolution: \u0394 E = \u0394 E 1 + \u0394 E 2 = \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f \u00d7 + \u00d7 \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f \u00d7 = 1310 1 1 1 3 45 100 1310 1 1 1 2 40 100 917 2 2 2 2 kJ
Answer: 0091
\nSolution: \u03bb = = 1240 13 6 91 17 . . nm
Answer: 0100
\nSolution: Moles of H 2 = PV RT x = \u00d7 \u00d7 = 1 1 0 08 300 . \u0394 E x x = \u00d7 + \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f \u00d7 = 436 1312 1 1 1 2 2 100 16 2 2 . kJ
Answer: 0090
\nSolution: 2 1 ? 360 nm 120 nm \u221e 1 1 120 1 360 90 \u03bb \u03bb = + \u21d2 = nm
Answer: 0024
\nSolution: \u03bb = \u21d2 = \u21d2 = 150 2 5 150 1 2 V V 24 \u00af V V .
Answer: 0005
\nSolution: \u03bb \u03bb\u03b1 = = \u22c5 \u21d2 \u22c5 h h m KT m T 2 2 3 2 1 mE \u2234 \u03bb \u03bb H N e e = \u00d7 \u00d7 = 20 1000 4 200 5
Answer: C
\nSolution: K eq for the reaction in backward direction = = \u00d7 \u00d7 = \u2212 \u2212 \u2212 \u2212 \u2212 \u2212 K K s b f 2 1 10 3 9 10 53 846 3 1 1 5 1 1 . . . L mol s L mol
Answer: C
\nSolution: Stability constant, K K K f b = = \u00d7 \u00d7 = \u00d7 \u2212 1 45 10 1 22 10 1 1885 10 13 9 17 . . .
Answer: C
\nSolution: K K K B A f b eq = = [ ] [ ] 2 \u21d2 1 5 10 100 10 10 10 3 2 5 . ( / ) ( / ) \u00d7 = \u2212 \u2212 K b \u21d2 K b = 1.5 \u00d7 10 \u201311 M \u20131 S \u20131
Answer: A
\nSolution: For pent hydrate to be efflorescent, Q < K p or, P H O 2 atm 2 4 2 10 < \u2212 \u21d2 P H O 2 atm mm < = \u2212 10 7 6 2 .
Answer: C
\nSolution: Q P P K P P = \u00d7 = \u00d7 = < NH 2 CO 3 atm 2 10 20 2000 2 3 Hence, the reaction should shift forward. But as solid NH 2 COONH 4 is not present initially, the pressure will remain at 30 atm.
Answer: D
\nSolution: Q K P K P P p < \u21d2 < H O 2 2 \u21d2 40 760 100 1 21 10 2 4 \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f < \u00d7 \u2212 R H . . \u2234 R.H. < 20.9 %
Answer: B
\nSolution: H 2 (g) + I 2 (s) \u001f 2HI(g) ; K p = 6.4 \u00d7 10 \u20134 atm I 2 (s) \u001f I 2 (g) ; K p = 1.6 \u00d7 10 \u20134 atm \u2234 H 2 (g) + I 2 (g) \u001f 2HI(g) ; K P = \u00d7 \u00d7 = \u2212 \u2212 6 4 10 1 6 10 4 4 4 . .
Answer: A
\nSolution: Net rate of reaction of HI, \u2212 \u22c5 = \u2212 = \u2212 \u2212 1 2 1 2 1 2 2 d dt r r b f [ ] [ ] [ ][I ] HI K HI K H
Answer: D
\nSolution: \u03b1 = \u2212 \u2212 \u22c5 = \u2212 \u2212 \u00d7 = M M n M 0 1 208 5 124 2 1 124 0 681 ( ) . ( ) .
Answer: A
\nSolution: PCl g PCl g 3 2 5 ( ) Cl ( ) ( ) + g \u001f \u21c0 \u001f \u21bd \u001f \u001f Initial partial pressure P 0 P 0 0 Equilibrium partial pressure P 0 \u2013 0.75 P 0 P 0 \u2013 0.75 P 0 0.75 P 0 \u20130.25 P 0 \u20130.25 P 0 Now, K P P P P = \u00d7 PCl PCl Cl 5 3 2 \u21d2 2 0 75 0 25 0 25 0 0 0 = \u00d7 . . . P P P \u21d2 P 0 = 6 atm \u2234 Initial total pressure of mixture = 2 P 0 = 12 atm
Answer: A
\nSolution: N g 3H g NH g 2 2 3 2 ( ) ( ) ( ) + \u001e \u21c0 \u001e \u21bd \u001e \u001e Initial moles 1 3 0 Moles at equilibrium 1 \u2013 x 3 \u2013 3 x 2 x Total moles of gases = (1 \u2013 x ) + (3 \u2013 3 x ) + 2 x = 4 \u2013 2 x Equilibrium partial pressure 1 4 2 3 3 4 2 2 4 2 \u2212 \u2212 \u00d7 \u2212 \u2212 \u00d7 \u2212 \u00d7 x x P x x P x x P Now, K x x P x x P x x P x P = \u2212 \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f \u2212 \u2212 \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f \u2212 \u2212 \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f = 2 4 2 1 4 2 3 3 4 2 4 27 1 2 3 2 ( \u2212 \u2212 \u00d7 \u2212 \u2248 \u00d7 \u00d7 x x P x P ) ( ) 4 2 2 2 2 4 2 4 16 27 \u2234 x K P P K P P = \u22c5 = \u22c5 27 64 3 3 8 2
Answer: D
\nSolution: X 2 + Y 2 \u001f 2 XY Initial moles 2 3 0 Final moles 2 \u2013 x 3 \u2013 x 2 x\n6.37 Chemical Equilibrium HINTS AND EXPLANATIONS [ ] . XY = = 2 5 0 7 x \u21d2 x = 1.75 \u2234 [X ] . 2 2 5 0 05 = \u2212 = x M and [Y ] . 2 3 5 0 25 = \u2212 = x M
Answer: A
\nSolution: N 2 + O 2 \u001f 2NO Equilibrium moles 1 \u2013 x 1 \u2013 x 2 x 0 09 2 1 1 2 . ( ) ( )( ) = \u2212 \u2212 x x x \u21d2 x = 0.13
Answer: B
\nSolution: N 2 + O 2 \u001f 2NO Initial moles 4 a a 0 Equilibrium moles 4 a \u2013 x a \u2013 x 2 x Now, 0 0004 2 4 4 4 2 2 . ( ) ( )( ) = \u2212 \u2212 \u2248 \u22c5 x a x a x x a a \u21d2 x a = 0 02 . \u2234 Per cent of NO = 2 5 100 0 8 x a \u00d7 = . %
Answer: A
\nSolution: N 2 + 3H 2 \u001f 2NH 3 Initial moles 1 5 0 Moles at equilibrium 1 \u2013 x 5 \u2013 3 x 2 x Total moles = (1 \u2013 x ) + (5 \u2013 3 x ) + 2 x = 6 \u2013 2 x From question, 2 6 2 0 4 x x \u2212 = . \u21d2 x = 6 7 K x x x x P P = \u2212 \u00d7 \u2212 \u00d7 \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f = \u00d7 \u2212 \u2212 ( ) ( ) ( ) . 2 1 5 3 6 2 2 6 10 2 3 2 4 2 atm
Answer: B
\nSolution: N 2 + 3H 2 \u001f 2NH 3 Initial moles 4 16 0 Moles at equilibrium 4 \u2013 x 16 \u2013 3 x 2 x Total moles = (4 \u2013 x ) + (16 \u2013 3 x ) + 2 x = 20 \u2013 2 x From question, 20 9 10 20 2 \u00d7 =\u2212 x \u21d2 x = 1 Now, K x x x V C = \u2212 \u2212 \u22c5 = \u00d7 \u2212 \u2212 ( ) ( )( ) . 2 4 16 3 6 07 10 2 3 2 4 2 M
Answer: B
\nSolution: If reactants are taken in stoichiometric amount, then their mass ratio does not change at any stage of reaction. For 3 mole N 2 , there should be 9 mole H 2 . Hence, at any stage, m m m N H NH 2 3 gm 2 3 28 9 2 102 + + = \u00d7 + \u00d7 = .
Answer: A
\nSolution: 2SO 2 + O 2 \u001f 2SO 3 Initial moles 2 1 0 Moles at equilibrium 2 \u2013 2 x 1\u2013 x 2 x From question n eq SO 2 = n eq MnO 4 \u2212 . or, (2 \u2013 2 x ) \u00d7 2 = 0.4 \u00d7 5 \u21d2 x = 0.5 \u2234 K x x x C = \u2212 \u00d7 \u2212 = \u2212 ( ) ( ) ( ) 2 2 2 1 2 2 2 1 M .
Answer: D
\nSolution: CH 3 COOH + C 2 H 5 OH \u001f CH 3 C00C 2 H 5 + H 2 O Case I 60 60 1 = mole 46 46 1 = mole 0 0 Moles at Equ. 1 \u2013 x 1 \u2013 x x = = 44 88 0 5 . x Case II 120 60 2 = mole 46 46 1 = mole 0 0 Moles at Equ. 2 \u2013 y 1 \u2013 y y y K x x x x y y y y eq = \u22c5 \u2212 \u22c5 \u2212 = \u22c5 \u2212 \u22c5 \u2212 ( ) ( ) ( ) ( ) 1 1 2 1 \u21d2 y = 2 3 \u2234 Mass of CH 3 COOC 2 H 5 at equilibrium = 2 3 88 \u00d7 =
Answer: D
\nSolution: R 1 OH + CH 3 COOH \u001f CH 3 COOR 1 + H 2 O Initial moles 1 1 0 0 Equ. moles 1 \u2013 x 1 \u2013 ( x + y ) x x + y\n6.38 Chapter 6 HINTS AND EXPLANATIONS R 2 OH + CH 3 COOH \u001f CH 3 COOR 2 + H 2 O Initial moles 1 1 0 0 Equ. moles 1 \u2013 y 1 \u2013 ( x + y ) y x + y From question, x + y = 0.8 and x y = 3 2 \u2234 x = 0.48 and y = 0.32 Now, K x x y x x y 1 1 1 0 48 0 8 0 52 0 2 3 69 = \u22c5 + \u2212 \u2212 + = \u00d7 \u00d7 = ( ) ( )[ ( )] . . . . .
Answer: B
\nSolution: 2NO( g ) + Cl 2 ( g ) \u001f 2NOCl( g ) Initial partial pressure 2P 0 P 0 0 Equ. partial pressure 2P 0 \u2013 2x P 0 \u2013 x 2x From question, (2 P 0 \u2013 2 x ) + ( P 0 \u2013 x ) + 2 x = 1 \u21d2 3 P 0 \u2013 x = 1 (1) and 2 1 4 0 x P x = \u2212 ( ) (2) From (1) and (2), P 0 = 9 x and x = 1 26 \u2234 K x P x P x P = \u2212 \u2212 = \u2212 ( ) ( ) ( ) 2 2 2 13 256 2 0 2 0 1 atm
Answer: B
\nSolution: S 8 ( g ) \u001f 4 S 2 ( g ) Initial partial pressure 1 atm 0 Equ. partial pressure 1 \u2013 0.3 4 \u00d7 0.3 = 0.7 atm = 1 .2 atm \u2234 K P = = ( . ) . . 1 2 0 7 2 96 4 3 atm
Answer: D
\nSolution: HCl( ) O Cl H O 2 g g g g + + 1 4 1 2 1 2 2 2 ( ) ( ) ( ) \u001e \u21c0 \u001e \u21bd \u001e \u001e Initial partial pressure 730 8 100 \u00d7 730 92 100 \u00d7 = 58.4 mm = 671.6 mm Equilibrium partial pressure 58.4 \u2013 58.4 \u00d7 0.08 671 6 58 4 0 08 4 . . . \u2212 \u00d7 = 670.432 mm
Answer: B
\nSolution: H 3 BO 3 + Glycerin \u001f complex Initial concent. 0.1 a M 0 Equ. Concert 0.1 \u2013 0.06 ( a \u2013 0.06) M 0.06 M = 0.04 M Now, K a eq = = \u00d7 \u2212 0 9 0 06 0 04 0 06 . . ( . ) ( . ) \u21d2 a = 1.73 M
Answer: B
\nSolution: 2A( g ) \u001f A 2 ( g ); K P = 8 \u00d7 10 8 atm \u20131 Initial partial pressure 1 atm 0 Partial pressure on complete reaction 0 0.5 atm Equilibrium partial pressure 2 x atm 0.5 \u2013 x \u2248 0.5 atm Now, 8 10 0 5 8 2 \u00d7 = . P A \u21d2 P A = 2.5 \u00d7 10 \u20135 atm
Answer: C
\nSolution: K eq = 3.8 \u00d7 10 \u20137 10 6 3 2 \u2212 \u2212 \u00d7 [ ] [ ] HCO CO \u21d2 [ ] [ ] . HCO CO 3 2 0 38 \u2212 =
Answer: A
\nSolution: A(g) \u001f nB(g) Initial mole 1(say) 0 Equilibrium mole 1 \u2013 a n a Total moles = 1 \u2013 a + n a = 1 + a ( n \u2013 1) Now K P P n n P n P n P P B n A n n n = = + \u2212 \u22c5 \u23a1 \u23a3 \u23a2 \u23a4 \u23a6 \u23a5 \u2212 + \u2212 \u22c5 \u23a1 \u23a3 \u23a2 \u23a4 \u23a6 \u23a5 = \u22c5 \u2212 \u03b1 \u03b1 \u03b1 \u03b1 \u03b1 1 1 1 1 1 1 ( ) ( ) ( ) ( ( ) [ ( )] 1 1 1 1 \u2212 \u22c5 + \u2212 \u2212 \u03b1 \u03b1 n n
Answer: B
\nSolution: A + B \u001f C + D Initial moles a a 0 0 Equilibrium moles a \u2013 x a \u2013 x x x From question, [ A ] = 2[ C ] \u21d2 a \u2013 x = 2 x \u21d2 a = 3 x Now, K K K x x a x a x f b eq = = \u22c5 \u2212 \u22c5 \u2212 ( ) ( ) \u21d2 2 10 2 2 3 \u00d7 = \u22c5 \u22c5 \u2212 K x x x x b \u2234 K b = 8 \u00d7 10 \u20133 mol \u20131 L S \u20131\n6.39 Chemical Equilibrium HINTS AND EXPLANATIONS
Answer: D
\nSolution: (Cl 2 CHCOOH) 2 \u001f 2Cl 2 CHOOH Initial Conc. 0 0129 258 100 1000 . / / 0 = 5 \u00d7 10 \u20134 M Equ. Conc. 5 \u00d7 10 \u20134 \u2013 x 2 x Now, K eq = 5 \u00d7 10 \u20134 = ( ) ( ) 2 5 10 2 4 x x \u00d7 \u2212 \u2212 \u21d2 x = 1.95 \u00d7 10 \u20134 \u2234 [Cl 2 CHOOH] = 3.90 \u00d7 10 \u20134 M
Answer: C
\nSolution: NH 2 CONH 4 ( s ) \u001f 2NH 3 ( g ) + CO 2 ( g ) Initial moles 1 0 0 Equ. moles 1 \u2013 a 2 a a From question, 3 \u03b1 = \u22c5 P V RT \u2234 Percentage dissociation of solid = 100 a % = \u22c5 100 3 PV RT %
Answer: A
\nSolution: P H O eq 2 , = ( K P ) 1/4 = (8.1 \u00d7 10 \u20137 ) = 0.03 atm P H O eq actual 0.04 atm 2 30 4 760 , , . = = \u2234 Mass of water vapour absorbed = ( . . ) . . 0 09 0 03 1 642 0 0821 300 18 \u2212 \u00d7 \u00d7 \u00d7 = 0.012 gm
Answer: C
\nSolution: Addition of CO will shift second reaction backward. Decrease in Cl 2 will shift the fi rst reaction forward.
Answer: B
\nSolution: P g P P g P H O HCl(g) H O HCl(g) 2 2 new ( ) ( ) , 2 2 2 = \u00d7 \u21d2 P P HCl (g), new HCl (g) = \u00d7 2
Answer: C
\nSolution: Ionic form of the reaction is NH H O NH OH H 4 2 4 + + + + \u001e \u21c0 \u001e \u21bd \u001e \u001e
Answer: D
\nSolution: K A B AB K AB AB B 1 2 2 = = + \u2212 \u2212 \u2212 [ ][ ] [ ] [ ] [ ][ ] and Now, [ ] [ ] [ ] A AB K K B + \u2212 \u2212 = \u22c5 2 1 2 2
Answer: B
\nSolution: NH 4 HS(s) \u001f NH 3 ( g ) + H 2 S( g ) Equ. partial pressure P 2 atm P 2 atm New Equ.partial pressure P atm P \u2032 atm Now, P P P P 2 2 \u00d7 = \u00d7 \u2032 \u21d2 P \u2032 = 0.25 P
Answer: A
\nSolution: N 2 + 3H 2 \u001f 2NH 3 Equ. partial pressure 100 mm 400 mm 1000 mm New equ. partial pressure 100 \u2013 a + x 400 + 3 x = 700 mm 1000 \u2013 2 x = 800 mm \u2234 x = 100 mm Now, K P P N = \u00d7 = \u00d7 1000 100 400 800 700 2 3 2 3 2 \u21d2 P N 2 11 94 = . mm K B A K C A 1 2 = = [ ] [ ] , [ ] [ ] Now, X A A B C A A K A K A K K A = + + = + + = + + [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] 1 2 1 2 1 1
Answer: B
\nSolution: CO and H 2 are initially in 1 : 3 mole ratio, as they are formed by 2nd reaction. CO + 2H 2 \u001f CH 3 OH Initial moles 1 3 0 Equilibrium moles 1 \u2013 0.25 3 \u2013 0.25 \u00d7 2 0.25 = 0.75 = 2.5 Total moles = 0.75 + 2.5 + 0.25 = 3.5 Now, K P P = \u00d7 \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f = \u00d7 \u2212 \u2212 0 25 0 75 2 5 3 5 6 23 10 2 2 3 . . ( . ) . . \u2234 P = 10.24 bar
Answer: C
\nSolution: A \u001f B + C; K 1 = 10 6 Initial moles 1 0 0 Equilibrium moles 1 \u2013 x + y x \u2013 y x B + D \u001f A; K 2 = 10 \u20136 x 1 1 Equilibrium moles x \u2013 y 1 \u2013 y 1 + y \u2013 x\n6.40 Chapter 6 HINTS AND EXPLANATIONS As K 1 >> 1, we may assume x \u2248 1 Now, K y x x y y y y y 2 1 1 1 1 = + \u2212 \u2212 \u2212 \u2248 \u2212 \u22c5 \u2212 ( ) ( )( ) ( ) ( ) As K 2 << 1, we may assume y << 1 K y y y y 2 1 1 = \u2212 \u2212 ( )( ) \u001a \u2234 [A] = 1 \u2013 x + y \u2248 y = 10 \u20136 M
Answer: A
\nSolution: A \u001f B Initial a M b M Equilibrium ( a \u2013 x ) M ( b + x ) M Now, K K K b x a x eq = = + \u2212 1 2 \u2234 x K a K b K K = \u2212 + 1 2 1 2
Answer: A
\nSolution: r b = K b \u22c5 P C(g)
Answer: A
\nSolution: \u0394 G \u00b0 = \u20132.303 RT \u22c5 ln K p \u00b0 \u0394 H \u00b0 \u2013 T \u22c5 \u0394 S \u00b0 or \u20132.303 \u00d7 8.314 \u00d7 T \u00d7 ln 1.0 = 240 \u00d7 10 3 \u2013 T \u00d7 50 \u2234 T = 4800 K
Answer: B
\nSolution: \u0394 G \u00b0 = \u20132.303 RT \u22c5 ln K eq \u21d2 \u20132.303 \u00d7 10 3 = \u20132.303 \u00d7 2 \u00d7 500 \u00d7 ln K eq \u2234 K eq = 10 Now, K P P P eq = \u00d7 HI H I 2 2 1 2 1 2 / / \u21d2 10 0 001 2 1 2 1 2 P H / / ( . ) \u00d7 \u2234 P H atm 2 1000 =
Answer: A
\nSolution: \u0394 \u00b0 = \u2212 \u22c5 = \u2212 \u22c5 = \u2212 \u22c5 G RT B RT RT ln [ ] [ ] ln ln . \u03b1 64 36 1 78
Answer: C
\nSolution: There is no net change at equilibrium.
Answer: D
\nSolution: K eq at 27\u00b0C, K 1 3 4 2 10 2 10 4 = \u00d7 \u00d7 = \u2212 \u2212 and K eq at 127\u00b0C, K 2 2 3 8 10 4 10 20 = \u00d7 \u00d7 = \u2212 \u2212 Now, ln K K H R T T 2 1 1 2 1 1 = \u0394 \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f or, 2 303 20 4 1 300 1 400 . log = \u0394 \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f H R \u2234 \u0394 H = 2.303 \u00d7 8.314 \u00d7 1200 log(5) J/mol
Answer: D
\nSolution: n A \u001f A n Initial moles 1 0 Equilibrium moles 1 \u2013 x x n Now, K x n V x V x V x n x V n x C n n n n = \u239b \u239d \u239c \u239e \u23a0 \u239f \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f = \u22c5 \u2212 \u22c5 \u2248 \u22c5 << \u2212 \u2212 / ( ) ( ) 1 1 1 1 1 as Now, total moles = (1 \u2013 x ) + x n = 1 + x \u22c5 1 1 n \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f = + \u22c5 \u22c5 \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f = + \u2212 \u22c5 \u2212 \u2212 1 1 1 1 1 1 n KC V n n n K V n C n ( )
Answer: A
\nSolution: \u03b1 = \u2212 \u2212 \u22c5 = \u2212 \u2212 \u22c5 = \u2212 M M M M M M M M mix mix mix mix mix ( ) ( ) n 1 2 1 1 = \u239b \u239d \u239c \u239e \u23a0 \u239f \u2212 = \u2212 M dRT P PM dRT 1 1
Answer: B
\nSolution: n RT RT NO = \u00d7 = 0 4 250 100 . n RT RT O 2 0 8 100 80 = \u00d7 = . 2 NO + O 2 \u2192 2 NO 2 \u001f N 2 O 4 Initial moles 100 RT 80 RT 0 0 Final moles 0 30 RT 100 RT x \u2212 x 2 From question, 30 100 2 0 3 350 RT RT x x RT + \u2212 + = \u00d7 . \u2234 x RT = 50 Now, K P of second reaction = P P x RT x RT N O 2 4 2 2 2 1 2 100 0 3 0 3 350 NO = \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f \u00d7 \u00d7 \u239b \u239d \u239c \u239c \u239e \u23a0 \u239f \u239f \u2212 . . = 3.5 atm
Answer: A
\nSolution: N 2 + 3H 2 \u001f 2NH 3 Initial moles 1 3 0 Equilibrium moles 1 \u2013 x 3 \u2013 3 x 2 x From question, 2 4 2 x x a \u2212 =\n6.41 Chemical Equilibrium HINTS AND EXPLANATIONS \u21d2 x a a = + 2 1 Now, K x x x P x x x x P P = \u2212 \u2212 \u22c5 \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f = \u22c5 \u2212 \u2212 \u22c5 \u2212 ( ) ( )( ) ( ) ( ) 2 1 3 3 4 2 4 4 2 27 1 2 3 2 2 2 4 2 or, K x x x P a a a a a P = \u22c5 \u2212 \u2212 \u22c5 = \u22c5 + \u239b \u239d \u239c \u239e \u23a0 \u239f \u22c5 \u2212 + \u239b \u239d \u239c \u239e \u23a0 \u239f \u22c5 \u2212 2 4 2 27 1 2 2 1 4 4 1 27 1 2 1 2 ( ) ( ) + + \u239b \u239d \u239c \u239e \u23a0 \u239f \u22c5 a P 2 = \u22c5 \u2212 \u22c5 32 27 1 2 a a P ( ) \u2234 a a P ( ) 1 2 \u2212 \u03b1
Answer: B
\nSolution: K P P = \u22c5 \u2212 \u03b1 \u03b1 2 2 1 \u21d2 ( . ) ( ) . 0 3 1 1 0 3 0 1 1 2 2 2 2 \u00d7 \u2212 \u2212 = \u00d7 \u2212 \u03b1 \u03b1 \u21d2 a = 0.973
Answer: A
\nSolution: K P = P 2 Now, ln K K H R T T 2 1 1 2 1 1 = \u0394 \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f or, ln ( ) P 2 2 3 2 7 10 3360 2 1 300 1 400 \u00d7 = \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f \u2212 \u21d2 P 2 = 1.4 \u00d7 10 \u20132 atm
Answer: B
\nSolution: CO(g) + H 2 (g) \u001f CO 2 (g) + H 2 (g) Initial 1 5 0 1 Equilibrium 1 \u2013 x 5 \u2013 x x 1 \u2013 x Now, K x x x x eq = = \u22c5 + \u2212 \u22c5 \u2212 1 3 1 1 5 ( ) ( ) ( ) \u21d2 x = 1 2
Answer: D
\nSolution: NH 2 COONH 4 (s) \u001f N 2 + 3H 2 + CO + 1 2 2 O Equilibrium partial pressure 22 5 5 4 . = 3 22 5 5 12 \u00d7 = . 22 5 5 4 . = 22 2 5 5 2 \u00d7 = . Now, K p = \u00d7 \u00d7 \u00d7 = \u00d7 4 12 4 2 27 2 3 1 2 10 5 ( ) ( ) ( ) / .
Answer: A
\nSolution: NH 2 COONH 4 (s) \u001f 2NH 3 (g) + CO 2 (g) Equ. partial pressure 2 P 0 P 0 New Equ. partial pressure 3 P 0 P 0 Now, K P P P P P = \u22c5 = \u22c5 ( ) ( ) 2 3 0 2 0 0 2 \u21d2 P P = 4 9 0 Now, 3 3 31 27 0 0 P P P + =
Answer: C
\nSolution: \u0394 H \u00b0 = \u0394 E \u00b0 + \u0394 n g \u22c5 RT = (+30) + (3 \u2013 2) \u00d7 2 1000 300 30 6 \u00d7 = + . K cal Now, \u0394 G \u00b0 \u2013 RT \u22c5 ln K eq = \u0394 H \u00b0 \u2013 T \u22c5 \u0394 S \u00b0 or, \u2013 2 \u00d7 300 \u00d7 ln K eq = 30.6 \u00d7 10 3 \u2013 300 \u00d7 100 \u21d2 ln K eq = \u2234 K eq = 1 e
Answer: B
\nSolution: trans \u001f Cis ; \u0394 G \u00b0 = 22.112 \u2013 30.426 = \u2013 8.314 KJ Now, \u0394 \u00b0 = \u2212 \u22c5 G RT Cis trans ln [ ] [ ] \u21d2 \u2013 8.314 \u00d7 10 3 = \u2013 8.314 \u00d7 300 \u00d7 ln [ ] [ ] Cis trans \u2234 [ ] [ ] Cis trans = 28 1
Answer: D
\nSolution: CO( g ) + H 2 O( g ) \u001f CO 2 ( g ) + H 2 ( g ) Initial moles 2 5 0 2 Equilibrium moles 2 \u2013 x 5 \u2013 x x 2 + x Now, K x x x x eq = = \u22c5 + \u2212 \u2212 3 0 2 2 5 . ( ) ( )( ) \u21d2 x = 1.5 \u2234 [ ] . H M 2 2 2 1 75 = + = x
Answer: A
\nSolution: NH 4 HS(s) \u001f NH 3 (g) + H 2 S(g) X 2 bar X 2 bar \u2234 \u0394 G \u00b0 = \u2013 RT \u22c5 ln K P \u00b0 = \u2013 RT ln X X RT 2 2 2 2 , (ln ln ) \u239b \u239d \u239c \u239e \u23a0 \u239f = \u2212 \u00d7 \u2212\n6.42 Chapter 6 HINTS AND EXPLANATIONS
Answer: C
" } } ] }, { "title": "Chem Sec 2", "originalName": "Section B - Multi Correct", "questions": [ { "question_id": "chemical-equilibrium-chem-sec-2-1-61", "marks": 4.0, "negMarks": 1.0, "partialMarks": 1, "subject": "chemistry", "chapter": "chemical-equilibrium", "chapterTitle": "Chemical Equilibrium", "type": "mcqm", "rawChapterType": "MSQ", "originalNumber": 61, "displayNumber": 1, "image": "./assets/chapter-packs/chemical-equilibrium/Chemistry Section 1__--__61__--__1.png", "solutionImage": null, "question": { "content": "Answer: A, B, C
\nSolution: NH 4 HS(s) \u001f NH 3 (g) + H 2 S(g) Equ. partial pressure 0.2 atm 0.2 atm Second Equ. partial pressure 0.5 atm P atm Now, K P = 0.2 \u00d7 0.2 = 0.5 \u00d7 P \u21d2 P = 0.08
Answer: A, B
\nSolution: \u0394 ng = 0
Answer: A, B, C
\nSolution: N 2 O 5 (g) \u001f 2NO 2 (g) + 1 2 2 O ( ) g Initial partial pressure P 0 0 0 Equ. partial pressure P 0 (1 \u2013 a ) 2 a P 0 \u03b1 P 0 2 and \u03b1 = \u2212 \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f \u22c5 = \u2212 M D D M D D 2 5 2 1 2 2 3 Total equilibrium pressure = P 0 (1 \u2013 a ) + 2 a P 0 + \u03b1 \u03b1 P P 0 0 2 1 3 2 = + \u239b \u239d \u239c \u239e \u23a0 \u239f
Answer: A, B
\nSolution: Le Chatelier\u2019s principle
Answer: A, B, D
\nSolution: Theory based
Answer: A, B, C
\nSolution: Theory based
Answer: D
\nSolution: Vapour pressure of a particular liquid system depends only on temperature.
Answer: B, C
\nSolution: Cl 2 (g) \u001f 2Cl(g) T \u2191 P \u2193
Answer: B, C
\nSolution: Addition of insert gas at constant pressure shifts the equilibrium in the direction of increase in moles of gases.
Answer: A, B, D
\nSolution: Decrease in pressure favors the reaction is the direction of increase in moles of gas and hence, B should be monomer.
Answer: A, B
\nSolution: \u0394 \u00b0 > > f G : NO N O NO 2 2 5
Answer: A, D
\nSolution: At 300 K : \u0394 G \u00b0 = (\u201341) \u2013 300 \u00d7 (\u20130.04) = \u201329 KJ/mol Hence, the reaction is spontaneous in forward direction. At 1200 K : \u0394 G \u00b0 = (\u201333) \u2013 1200 \u00d7 (\u20130.03) = + 3 KJ/mol Hence, the reaction is spontaneous in backward direction.
Answer: B, C
\nSolution: Theory based.
Answer: A, B
\nSolution: Theory based.
Answer: D
\nSolution: S A e H RT = \u22c5 \u2212\u0394 / \u21d2 ln s = ln A H RT \u2212 \u0394 Positive slope represents that \u0394 H = negative.
Answer: A, B, C
\nSolution: Theory based.
Answer: A, B, C
\nSolution: K P P g 2 2 0 2 = = Cl atm ( ) . K P P P g 1 2 8 8 25 9 0 2 0 001 2 10 = \u22c5 = \u00d7 = \u00d7 \u2212 Cl H O(g) 2 atm ( ) . ( . ) P H O(g) 2 = Vapour pressure of ice.
Answer: A, C
\nSolution: PCl 5 (g) \u001f PCl 3 (g) + Cl 2 (g) Initial moles 5 0 0 Moles at equilibrium 5 \u2013 x x x From question, ( ) . . 5 4 4 8 112 0 0821 546 \u2212 + + + = \u00d7 \u00d7 x x x \u21d2 x = 3 \u2234 \u03b1 = = x 5 0 6 . and K x x x P = \u22c5 \u2212 \u22c5 \u239b \u239d \u239c \u239e \u23a0 \u239f = 5 4 8 12 1 8 . . atm
Answer: A, B, C, D
\nSolution: 4HCl(g) + O 2 (g) \u001f 2Cl 2 (g) + 2H 2 O(g) Initial partial pressure 1.0 atm 0.25 atm 0 0.4 atm On completion 0 0 0.5 atm 0.4 atm Equ. partial pressure 4 x atm x atm 0.5 atm 0.4 atm K x x P = \u00d7 = \u00d7 5 10 0 5 0 4 4 12 2 2 4 ( . ) ( . ) ( ) \u21d2 x = 5 \u00d7 10 \u20134
Answer: A, B
\nSolution: \u0394 \u00b0 = \u00d7 \u0394 \u00b0 \u2212 \u0394 \u00b0 = G G G f g f g 2 0 2 4 NO N O 2 ( ) ( ) \u21d2 K P \u00b0 = 1 Now, \u0394 = \u0394 \u00b0 + \u22c5 = + \u22c5 G G Q RT P P RT ln NO N O 2 4 0 2 2 ln = \u22c5 RT ln 10 10 2 = positive.\n6.43 Chemical Equilibrium HINTS AND EXPLANATIONS
Answer: B, C
\nSolution: AB 2 (g) + A( s ) \u001f 2 AB(g) Initial partial pressure 0.7 bar 0 Equ. partial pressure (0.7 \u2013 x ) bar 2 x bar Second equ. partial pressure y bar (0.4 \u2013 y ) bar From question, (0.7 \u2013 x ) + 2 x = 0.95 \u21d2 x = 0.25 \u2234 K x x P = \u2212 = = ( ) ( . ) ( . ) . 2 0 7 0 5 0 45 5 9 2 2 Now, 5 9 0 4 2 = \u2212 ( . ) y y \u21d2 y = 0.13 \u2234 At second equilibrium, the volume per cent of AB 2 0 13 0 4 100 32 5 = \u00d7 = . . . %
Answer: A, D
\nSolution: PCl 5 ( g ) \u001f PCl 3 ( g ) + Cl 2 ( g ) Initial moles 1 1 0 Equ. moles 1 \u2013 x 1 + x x \u2248 1 \u2248 1 = 0.004 \u2234 K C = \u00d7 \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f = 1 0 004 1 1 10 0 0004 . . M
Answer: C
\nSolution: \u0394 \u00b0 = \u2212 \u22c5 \u00b0 G RT K P ln \u21d2 \u20131743 = \u2013 8.3 \u00d7 300 \u00d7 ln K P \u00b0 \u2234 K P \u00b0 = 2
Answer: A, C, D
\nSolution: K K K 1 2 3 1 1 0 24 = \u00d7 = . As \u0394 n g = 0, [ A ] + [ B ] + [ C ] = 1 M
Answer: C, D
\nSolution: Addition of water will shift the reaction in the direction of increase in mole of aq species.
Answer: C
\nSolution: For SrCl 2 \u22c5 2H 2 O(s), P H O 2 = (2.56 \u00d7 10 \u201310 ) 1/4 = 0.004 atm For Na 2 HPO 4 \u22c5 7H 2 O P H O 2 = (2.43 \u00d7 10 \u201313 ) 1/5 = 0.003 atm For Na 2 SO 4 (s), P H O 2 = (1.024 \u00d7 10 \u201327 ) 1/10 = 0.002 atm As P H O 2 is minimum for Na 2 SO 4 (s), it is the best dehydrating agent.
Answer: D
\nSolution: For Na 2 SO 4 (s), 10H 2 O(s) to be efflorescent, P H O 2 < 0.002 atm or 0 04 100 0 002 . . . . \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f < R H \u21d2 R . H . < 5 %
Answer: C
\nSolution: Na 2 HPO 4 \u22c5 7H 2 O(s) to be deliquescent, P H O 2 > 0.003 atm or, 0 04 100 0 003 . . . . \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f > R H \u21d2 R . H . > 7.5 % Comprehension II
Answer: C
\nSolution: CO( g ) + 2H 2 ( g ) \u001f CH 3 OH( g ) Initial moles 0.2 a (say) 0 Moles at equ. 0.2 \u2013 x a \u2013 2 x x = 0.1 = a \u2013 0.2 = 0.1 Total moles = 0.1 + ( a \u2013 0.2) + 0.1 = 7 5 2 463 0 0821 750 . . . \u00d7 \u00d7 \u21d2 a = 0.3 Now, K P = \u00d7 \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f = \u2212 \u2212 0 1 0 1 0 1 7 5 0 3 0 16 2 2 2 . . ( . ) . . . atm
Answer: B
\nSolution: K K RT C P n g = = \u00d7 = \u0394 \u2212 \u2212 ( ) . ( . ) 0 16 0 0821 750 607 2 2 M
Answer: D
\nSolution: P = + \u00d7 \u00d7 = ( . . ) . . . 0 2 0 3 0 0821 750 2 463 12 5 atm\n6.44 Chapter 6 HINTS AND EXPLANATIONS Comprehension III
Answer: C
\nSolution: C 2 H 5 OH + CH 3 COOH \u001f CH 3 COOC 2 H 5 + H 2 O Initial moles a a 0 0 Equilibrium moles a \u2013 x a \u2013 x x x From question, x a 2 0 333 1 3 = = . \u21d2 x a = 2 3 Now, K x x a x eq a x = \u22c5 \u2212 = \u22c5 \u2212 ( ) ( ) 4
Answer: A
\nSolution: C 2 H 5 OH + CH 3 COOH \u001f CH 3 COOC 2 H 5 + H 2 O Initial moles a 3 2 3 a 0 0 Equilibrium moles a x 3 \u2212 2 3 a x x x Now, K x x a x a x eq = = \u22c5 \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f 4 3 2 3 \u21d2 x = 0.2833 a \u2234 Fraction of alcohol reacted = x a / . 3 0 85 =
Answer: B
\nSolution: Solution of 0.7 = x a \u00d7 = 100 66 67 . % Comprehension IV
Answer: D
\nSolution: 2HI( g ) \u001f H 2 ( g ) + I 2 ( g ) Initial moles 1(say) 0 0 Equilibrium moles 1 \u2013 0.2222 = 0.7778 0.1111 0.1111 \u2234 K eq = \u00d7 = \u2248 0 1111 0 1111 0 7778 1 49 0 02 2 . . ( . ) .
Answer: C
\nSolution: In the presence of I 2 ( g ), the extent of dissociation of HI will decrease.
Answer: D
\nSolution: Addition of He( g ) will not affect thequilibrium. Comprehension V
Answer: D
\nSolution: NH 4 HS ( s ) \u001f NH 3 ( g ) + H 2 S ( g ) Initial partial pressure P mm 0 Equilibrium partial pressure ( P + x )mm x mm From question, P + x = 625 and ( P + x ) + x = 725 \u2234 x = 100 and P = 525
Answer: B
\nSolution: K P = ( P + x ) \u22c5 x = 625 \u00d7 100 mm 2 \u2234 \u2032 K P (required) = 1 1 6 10 5 K P = \u00d7 \u2212 . mm \u20132 .
Answer: B
\nSolution: P P K P NH H S mm 3 2 250 = = =
Answer: D
\nSolution: Minimum mass of NH 4 HS ( s ) needed. = \u00d7 \u00d7 \u00d7 250 760 5 0 0 0821 300 51 . . gm\n6.45 Chemical Equilibrium HINTS AND EXPLANATIONS Comprehension VI
Answer: C
\nSolution: K A A e e e eq f b H RT = \u22c5 = \u2212 \u2212 \u00d7 \u00d7 = \u2212\u0394 / ( . ) . 24 942 10 8 314 300 3 10
Answer: A
\nSolution: K K K e b f eq = = \u2212 1 10 \u0394 H = Ea f \u2013 Ea b = Ea f \u2013 3 2 Ea f \u21d2 Ea f = 2 \u22c5 (\u2013 \u0394 H) Now, K A e e e f f Ea RT f = \u22c5 = \u00d7 = \u2212 \u2212 \u00d7 \u00d7 \u00d7 \u2212 / . . 1 2 24 942 10 8 314 300 20 3 and K b = e \u201330 Comprehension VII
Answer: C
\nSolution: 2SO 3 \u001f 2SO 2 + O 2 Equilibrium moles 1 \u2013 a a \u03b1 2 Now. K K P P = \u22c5 \u2212 \u22c5 + \u239b \u239d \u239c \u239c \u239e \u23a0 \u239f \u239f \u2032 \u03b1 \u03b1 \u03b1 \u03b1 2 2 2 1 1 2 ( ) \u21d2 \u03b1 = 2 3
Answer: B
\nSolution: 2NH 3 \u001f N 2 + 3H 2 Equilibrium moles 1 \u2212 \u03b1 \u03b1 2 3 2 \u03b1 = 1 3 = 1 3 = 1 Now, K P = \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f \u239b \u239d \u239c \u239c \u239e \u23a0 \u239f \u239f = 1 3 1 1 3 50 5 3 2700 3 2 2 2 atm
Answer: A
\nSolution: n initial \u00d7 (17 + 28 + 2 + 20) = 134 \u21d2 n initial = 2 2NH 3 \u001f N 2 + 3H 2 Initial moles 2 2 2 Moles at equilibrium 2 \u2013 2 x 2 + x 2 + 3 x = 1.0 = \u00d7 134 0 5224 28 . = 3.5 \u2234 x = 0.5 Now, K P P = = \u00d7 \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f 2700 2 5 3 5 1 0 9 3 2 2 . ( . ) ( . ) \u21d2 P = \u00d7 \u00d7 2700 81 2 5 3 5 3 . ( . ) atm Comprehension VIII N 2 + 3H 2 \u001f 2NH 3 ; K P 1 Initial partial pressure 9 P 13 P 0 Equilibrium partial pressure 9 P \u2013 x \u2013 y 13 P \u2013 3 x \u2013 2 y 2 x N 2 + 2H 2 \u001f N 2 H 4 ; K P 2 Initial partial pressure 9 P 13 P 0 Equilibrium partial pressure 9 P \u2013 y \u2013 x 13 P \u2013 2 y \u2013 3 x y From question, P x P NH 3 2 0 = = \u21d2 x P = 0 2 P P x y P H 2 13 3 2 2 0 = \u2212 \u2212 = and P total = (9 P \u2013 x \u2013 y ) + (13 P \u2013 3 x \u2013 2 y ) + 2 x + y = 7 P 0 \u2234 y P = 3 2 0 and P P = 0 2 K P P P P P P P P 1 3 2 2 2 3 0 2 0 0 3 0 2 5 2 2 1 20 = \u22c5 = \u00d7 = NH N H ( ) \u2234 K P (required) = 20 0 2 P
Answer: C
\nSolution: K P P P P P P P P 2 2 4 2 2 2 0 0 0 2 0 2 3 2 5 2 2 3 20 = \u22c5 = \u239b \u239d \u239c \u239e \u23a0 \u239f \u00d7 = N H N H ( )\n6.46 Chapter 6 HINTS AND EXPLANATIONS
Answer: A
" } }, { "question_id": "chemical-equilibrium-chem-sec-3-24-93", "marks": 4.0, "negMarks": 1.0, "partialMarks": 1, "subject": "chemistry", "chapter": "chemical-equilibrium", "chapterTitle": "Chemical Equilibrium", "type": "mcq", "rawChapterType": "MCQ", "originalNumber": 93, "displayNumber": 24, "image": "./assets/chapter-packs/chemical-equilibrium/Chemistry Section 1__--__93__--__1.png", "solutionImage": null, "question": { "content": "Answer: D
" } } ] }, { "title": "Chem Sec 4", "originalName": "Section D - Assertion Reason", "questions": [ { "question_id": "chemical-equilibrium-chem-sec-4-1-94", "marks": 4.0, "negMarks": 1.0, "partialMarks": 1, "subject": "chemistry", "chapter": "chemical-equilibrium", "chapterTitle": "Chemical Equilibrium", "type": "mcq", "rawChapterType": "MCQ", "originalNumber": 94, "displayNumber": 1, "image": "./assets/chapter-packs/chemical-equilibrium/Chemistry Section 1__--__94__--__1.png", "solutionImage": null, "question": { "content": "Answer: D
\nSolution: Direction of shifting of equilibrium will depend on relative values of a and b .
Answer: A
\nSolution: Equilibrium opposes the changes.
Answer: A
\nSolution: Theory based
Answer: A
\nSolution: Theory based
Answer: A
\nSolution: Theory based
Answer: A
\nSolution: K K RT P C n g = \u22c5 \u0394 ( )
Answer: B
\nSolution: Theory based
Answer: C
\nSolution: Exothermic direction is favoured on lowering temperature.
Answer: A
\nSolution: NaCl( s ) \u001f Na + ( aq ) + Cl \u2013 ( aq )
Answer: C
\nSolution: On decreasing the volume, moles of A( g ) as well as B( s ) will increase.
Answer: A \u2192 S; B \u2192 Q, R; C \u2192 Q; D \u2192 P
\nSolution: Theory based
Answer: A \u2192 S; B \u2192 R; C \u2192 P; D \u2192 Q
\nSolution: K K RT P C n g = \u22c5 \u0394 ( )
Answer: A \u2192 P, R; B \u2192 S; C \u2192 Q; D \u2192 S
\nSolution: Le Chatelier\u2019s principle
Answer: A \u2192 P, Q, S; B \u2192 P, Q; C \u2192 R
\nSolution: Le Chatelier\u2019s principle
Answer: A \u2192 P, Q, R, S; B \u2192 Q, R, S; C \u2192 P, Q, R, S, T; D \u2192 Q, R
\nSolution: Le Chatelier\u2019s principle
Answer: A \u2192 S, T; B \u2192 R; C \u2192 Q; D \u2192 P
\nSolution: Le Chatelier\u2019s principle
Answer: A \u2192 P, S, T; B \u2192 Q, R, S; C \u2192 S; D \u2192 Q, R, S
\nSolution: Le Chatelier\u2019s principle
Answer: A \u2192 P, S, T; B \u2192 Q, R; C \u2192 Q, R; D \u2192 P, S
\nSolution: Le Chatelier\u2019s principle
Answer: A \u2192 P, R; B \u2192 Q, R; C \u2192 Q, S
\nSolution: P K P CO atm 2 2 463 = = . \u2234 n CO 2 at equilibrium = \u00d7 \u00d7 = 2 463 15 0 0821 900 0 5 . . . (A) % of CaCO 3 decomposed = \u00d7 = 0 5 1 0 100 50 . . % ( ) Eqn (B) % of CaCO 3 decomposed = \u00d7 = 0 5 0 5 100 100 . . % ( ) Eqn (C) % of CaCO 3 decomposed = 100% (non ) -Eqn
Answer: A \u2192 Q, R, S; B \u2192 P, R, S; C \u2192 P, R, S; D \u2192 Q, R, S
\nSolution: Le Chatelier\u2019s principle
Answer: 2
\nSolution: 2H 2 S(g) \u001f 2H 2 (g) + S(g); K e = 10 \u20136 Initial moles 0.1 0 0 Equilibrium moles 0.1 \u2013 x x x 2 \u001a 0 1 . Now, K x x C = = \u00d7 \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f \u2212 10 2 0 1 1 0 4 6 2 2 ( . ) . \u21d2 x = 2 \u00d7 10 \u20133 \u2234 Percentage dissociation = \u00d7 \u00d7 = \u2212 2 10 0 1 100 2 3 . %
Answer: 4
\nSolution: V V CF CO ml 4 2 500 300 200 = \u2212 = = \u2234 V COF ml 2 500 2 200 100 = \u2212 \u00d7 = Hence, P P CF CO atm 4 2 200 500 10 4 = = \u00d7 = P COF atm 2 100 500 10 2 = \u00d7 = K p = \u00d7 = 4 4 2 4 2
Answer: 7
\nSolution: Initial: n PCl 5 62 55 208 5 0 3 = = . . . and n Cl 2 4 48 22 4 0 2 = = . . . PCl 5 \u001f PCl 3 + Cl 2 Initial moles 0.3 0 0.2 Equilibrium moles 0.3 \u2013 x x 0.2 + x\n6.47 Chemical Equilibrium HINTS AND EXPLANATIONS Now, K p = x x x p x x x x RT V \u22c5 + \u2212 \u22c5 + \u239b \u239d \u239c \u239e \u23a0 \u239f = + \u2212 \u00d7 ( . ) ( . ) . ( . ) ( . ) 0 2 0 3 0 5 0 2 0 3 or, 8 0 2 0 3 0 0821 546 4 48 = + \u2212 \u00d7 \u00d7 x x x ( . ) . . . \u21d2 x = 0.2 \u2234 Final pressure = + \u00d7 \u00d7 = ( . ) . . 0 5 0 0821 546 4 48 7 x atm
Answer: 1
\nSolution: NaOH is used to neutralize acetic acid. From the given data, half of the acid taken is neutralize. CH 3 COOH + C 2 H 5 OH \u001f CH 3 COOC 2 H 5 + H 2 O Initial moles a a 0 0 Equilibrium moles a a \u2212 2 a a \u2212 2 a 2 a 2 \u2234 K a a a a eq = \u00d7 \u00d7 = 2 2 2 2 1
Answer: 2
\nSolution: 2HI \u001f H 2 + I 2 Equilibrium moles 1 \u2013 0.8 0.4 0.4 = 0.2 \u2234 K eq = \u00d7 = 0 4 0 4 0 2 4 2 . . ( . ) Now, H 2 + I 2 \u001f 2HI Initial moles 2 2 0 Equilibrium moles 2 \u2013 x 2 \u2013 x 2 x K x x x eq = = \u2212 \u2212 1 4 2 2 2 2 ( ) ( )( ) \u21d2 x = 0.4 Now, n eq of I 2 = n eq of Na 2 S 2 O 3 or, (2 \u2013 x ) \u00d7 2 = V \u00d7 (1.6 \u00d7 1) \u21d2 V = 2 L
Answer: 5
\nSolution: Cl 2 CHCOOH + C 5 H 10 \u001f Cl 2 CHCOOC 5 H 11 I: Initial moles 1 4 0 Equilibrium moles 1 \u2013 x 4 \u2013 x x = 0.5 II: Initial moles 1 a 0 Equilibrium moles 1 \u2013 y a \u2013 y y = 0.6 Now, K a eq = \u00d7 \u00d7 = \u00d7 \u2212 \u00d7 0 5 0 5 3 5 0 7 0 6 0 4 0 6 0 72 . . . . . . ( . ) . \u21d2 a = 5\n6.48 Chapter 6 HINTS AND EXPLANATIONS
Answer: 5
\nSolution: n CO2 at equilibrium = 0.05 \u2234 Minimum mass of CaCO 3 needed = 0.05 \u00d7 100 = 5 gm
Answer: 5
\nSolution: Ag + (aq) + Fe 2+ (aq) \u001f Fe 3+ (aq) + Ag(s) Initial moles 500 0 9 1000 0 45 \u00d7 = . . 500 1 0 1000 0 50 \u00d7 = . . 0 0 Equilibrium moles 0.45 \u2013 x 0.50 \u2013 x x x Now, n eq Fe 2+ = n eq MnO 4 \u2212 or, ( . ) . 0 50 1000 30 1 25 0 06 1000 5 \u2212 \u00d7 \u00d7 = \u00d7 \u00d7 x \u21d2 x = 0.25 \u2234 K x x x eq M = \u2212 \u2212 = \u2212 ( . )( . ) 0 45 0 5 5 1
Answer: 1
\nSolution: Sb2S3(s) + 3H2(g) \u001f 2Sb(s) + 3H2S(g) Initial moles 0.01 0.01 0 0 Equ. moles 0.01 \u2013 x 0.01 \u2013 3 x 2 x 3 1 19 238 x = . = 5 \u00d7 10 \u20133 = 5 \u00d7 10 \u20133 Now, K c = \u00d7 \u00d7 = \u2212 \u2212 ( ) ( ) 5 10 5 10 1 3 3 3 3
Answer: 4
\nSolution: H 2 O + D 2 O \u001f 2HDO Initial moles 28 28 0 Equ. moles 28 \u2013 14 = 14 28 \u2013 14 = 14 2 \u00d7 14 = 28 K C = \u00d7 = ( ) 28 14 14 4 2
Answer: 6
\nSolution: 3A 2 (g) \u001f A 6 (g), K p 1 1 6 2 = \u2212 . atm Initial partial pressure 2 P 0 0 Equilibrium partial pressure 2 P 0 \u2013 3 a \u2013 b a A 2 (g) + C (g) \u001f A 2 C (g), K x p 2 1 = \u2212 atm Initial partial pressure 2 P 0 P 0 0 Equilibrium partial pressure 2 P 0 \u2013 b \u2013 3 a P 0 \u2013 b b From question, a = 0.2, P P P A A A 6 2 2 3 3 1 6 0 2 1 6 = \u21d2 = . . . \u21d2 P P a b A 2 0 5 2 3 0 = = \u2212 \u2212 . and (2 P 0 \u2013 3 a \u2013 b ) + a + ( P 0 \u2013 b ) + b = 1.4 \u21d2 P 0 = 0.7 and b = 0.3 Now, K b P a b P b p 2 2 3 0 3 0 5 4 1 5 0 0 1 = \u2212 \u2212 \u2212 = \u00d7 = \u2212 ( )( ) . . . . atm
Answer: 4
\nSolution: Initial partial pressure of IBr (g) = \u00d7 \u00d7 = 8 28 207 0 0821 500 0 1642 10 . . . 2IBr (g) \u001f I 2 (g) + Br 2 (g) Initial partial pressure 10 0 0 Equilibrium partial pressure 10 \u2013 2 x x x = 4 \u2234 K p = \u00d7 = 4 4 2 4 2 ( )
Answer: 4
\nSolution: H 2 (g) + I 2 (g) \u001f 2HI (g) I: Initial moles 1 3 0 Equilibrium moles 1 2 \u2212 x 3 2 \u2212 x x II: Initial moles 3 3 0 Equilibrium moles 3 \u2013 x 3 \u2013 x 2 x\n6.49 Chemical Equilibrium HINTS AND EXPLANATIONS Now, K x x x x x x eq = \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f = \u2212 \u2212 ( ) ( ) ( )( ) 2 2 1 2 3 2 2 3 3 \u21d2 x = 3 2 \u2234 K eq = 4
Answer: 1
\nSolution: N 2 O 5 (g) \u001f N 2 O 3 (g) + O 2 (g); K C 1 2 5 = . M Initial moles 4 0 0 Equilibrium moles 4 \u2013 x x \u2013 y x + y N 2 O 3 (g) \u001f N 2 O(g) + O 2 (g); K C 2 x 0 0 Equilibrium moles x \u2013 y y x + y From question, [ ] . O 2 2 2 5 = + = x y \u21d2 x + y = 5 And K x y x y y C 1 2 5 5 4 1 2 5 2 5 1 1 2 = = \u2212 \u00d7 \u2212 \u00d7 = \u2212 \u00d7 \u2212 \u00d7 . ( ) ( ) ( ) \u21d2 y = 2 \u2234 [N O] 2 2 1 = = y M
Answer: 1
\nSolution: In left chamber, P H e = 2 atm \u2234 P P NH H 3 2 atm = = \u2212 = 3 4 2 2 1 \u2234 K p = \u00d7 = 1 1 1 2 atm Four-digit Integer Type
Answer: 0480
\nSolution: P-xyloquinone + M.W \u001f P-xylohydroquinone + M.B. Initial conc. 0.012 M 0 0.24 M 10 \u20133 M Equ. con. 0.012 + 4 \u00d7 10 \u20135 4 \u00d7 10 \u20135 M 0.24 \u2013 4 \u00d7 10 \u20135 10 4 100 10 3 3 \u2212 \u2212 \u2212 \u00d7 \u2248 0.012 \u2248 0.24 M 0.96 \u00d7 10 \u20133 m \u2234 K eq = 0 24 0 96 10 0 012 4 10 480 3 5 . . . \u00d7 \u00d7 \u00d7 \u00d7 = \u2212 \u2212
Answer: 0015
\nSolution: 6HCHO \u001f C 6 H 12 0 6 ; K eq = 6.4 \u00d7 10 19 Initial conc. 0 1 M Equ. con. 6 x 1 \u2013 x = 1 M Now, K eq = 6.4 \u00d7 10 19 = 1 6 [ ] HCHO \u21d2 [HCHO] = 5 \u00d7 10 \u20134 M = 5 \u00d7 10 \u20134 \u00d7 30 g/L = 15 mg/L
Answer: 0600
\nSolution: PCl 5 \u001f PCl 3 + Cl 2 Initial equ. moles 2 2 2 Moles on adding Cl 2 2 2 2 + x Final Equ. moles 2 + y 2 \u2013 y 2 + x \u2013 y From question, (2 + y) + (2 \u2013 y ) + 2 + ( x \u2013 y ) = 2 \u00d7 6 or, x \u2013 y = 6 and K c = 2 2 2 1 2 2 2 1 2 \u00d7 \u00d7 = \u2212 \u00d7 + \u2212 + \u00d7 V y x y y V ( ) ( ) ( ) Or, 4 = ( ) ( ) 8 8 4 \u2212 \u00d7 \u2212 x x \u21d2 x = 20 3
Answer: 1784
\nSolution: K \u00b0 = eq 1 \u21d2 \u2206 G \u00b0 = 0 \u21d2 T = \u0394 \u0394 H S \u00b0 \u00b0 = \u00d7 \u2212 \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f 223 10 223 33 1520 10 3 3 = 1784 K
Answer: 0180
\nSolution: Graphite \u001f Diamond; \u2206 G\u00b0 = (3.0 \u2013 0) kJ/mol; P 1 = 1 bar \u2206 G = 0 P 2 = P Now, \u2206 ( \u2206 G ) = \u2206 V \u2219 \u2206 P or, ( \u2206 G \u2013 \u2206 G \u00b0) = ( V Dia \u2013 V Gra ) ( P 2 \u2013 P 1 ) or, (0 \u2013 3.0 \u00d7 10 3 ) = 12 3 6 12 2 4 10 10 6 5 2 . . \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f \u00d7 \u23a1 \u23a3 \u23a2 \u23a4 \u23a6 \u23a5 \u2212 \u23a1 \u23a3 \u23a4 \u23a6 \u2212 P \u2234 P 2 = 1.8 \u00d7 10 9 Pa = 1.8 \u00d7 10 4 bar
Answer: 0532
\nSolution: K P P P \u00b0 = = CO CO 2 10 400 6 Now, \u2206 G \u00b0 = \u20135320 \u2013 5.6 T = \u2013RT ln K P \u00b0\n6.50 Chapter 6 HINTS AND EXPLANATIONS = \u20132 \u00d7 T \u00d7 ln 10 400 6 \u2234 T = 532 k
Answer: 0032
\nSolution: 4HNO 3 (g) \u001f 4NO 2 (g) + 2H 2 O (g) + O 2 (g) Initial partial pressure P 0 0 0 0 Equ. partial pressure P 0 \u2013 4 x 4 x 2 x x From question, P 0 \u2013 4 x = 2 atm and ( P 0 \u2013 4 x ) + 4 x + 2 x + x = 30 atm \u2234 P 0 = 18 atm and x = 4 atm Now, K x x x P x p o = \u00d7 \u00d7 \u2212 = ( ) ( ) ( ) 4 2 4 2 4 2 4 20 3 atm and K K RT c p n g = ) = \u00d7 = \u0394 ( ( . ) 2 0 08 400 32 20 3 M 3
Answer: 0032
\nSolution: Initial: P NOcl = P 0 bar and P N 2 = (1 \u2013 P 0 ) bar 2NOCl \u001f 2NO + Cl 2 Initial partial pressure P 0 0 0 Eqn. partial pressure P 0 \u2013 2 x 2 x x = 1.2\u20131.0 = 0.2 Par. pre. on adding Cl 2 P 0 \u2013 2 x 2 x x + (8.3 \u2013 1.2) New Equ.partial pre. P 0 \u2013 2 x + 2 y 2 x \u2013 2 y 7.1 + x \u2013 y From question, y = 8.3 \u2013 8.2 = 0.1 Now, K x x P x x y x y P x y p o = \u00d7 \u2212 = \u2212 \u00d7 + \u2212 \u2212 + ( ) ( ) ( ) ( . ) ( ) 2 2 2 2 7 1 2 2 2 2 2 0 2 or, 0 4 0 2 0 4 0 2 7 2 0 2 0 5 2 0 2 2 0 2 0 . . ( . ) . . ( . ) . \u00d7 \u2212 = \u00d7 \u2212 \u21d2 = P P P and K p = 3.2
Answer: 0170
\nSolution: Initial equilibrium: A + 2B \u001f C Initial partial pressure P 0 2 P 0 O Equ. partial pressure P 0 \u2013 x 2 P 0 \u2013 2 x x From question, ( P 0 \u2013 x ) + (2 P 0 \u2013 2 x ) + x + 3 P 0 = 5 6 6 0 \u00d7 P or, x = P 0 2 Second equilibrium: A + 2B \u001f C Initial partial pressure 2 P 0 4 P 0 0 Equ. partial pressure 2 P 0 \u2013 y 4 P 0 \u2013 2 y y \u2013 2 z 2C + D \u001f 2F y 6 P 0 0 Equ. partial pressure y \u2013 2 z 6 P 0 \u2013 z 2 z From question, 2 P 0 \u2013 y = y \u2013 2 z Now for the first reaction, K x P x P x y z P y P y p o = \u2212 \u2212 = \u2212 \u2212 \u2212 ( )( ) ( )( ) 2 2 2 2 4 2 0 2 0 0 2 or, 1 1 4 2 3 2 2 0 2 0 2 0 0 P P y y P z P = \u2212 \u21d2 = = ( ) and Total equilibrium pressure: First equilibrium = 5 P 0 Second equilibrium = (2 P 0 \u2013 y ) + (4 P 0 \u2013 2y) + ( y \u2013 2 z ) + (6 P 0 \u2013 z ) + 2 z = 3.5 P 0
Answer: 0018
\nSolution: A 2 + B 2 \u001f 2AB Initial moles 0.2 0.2 0 Equ. moles 0.2 \u2013 x 0.2 \u2013 x 2 x = 0.3 = 0.05 = 0.05 K eq = K 1 = ( . ) . . 0 3 0 05 0 05 36 2 \u00d7 = After adding C 2 : A 2 + B 2 \u001f 2AB Initial moles 0.2 0.2 0 Equ. moles 0.2 \u2013 y \u2013 z 0.2 \u2013 y 2 y = 0.24 = 0.08 \u2013 z = 0.08 A 2 + C 2 \u001f 2AC Initial moles 0.2 0.1 0 Equ. moles 0.2 \u2013 z \u2013 y 0.1 \u2013 z 2 z Now, K z z 1 2 36 0 24 0 08 0 08 0 06 = = \u2212 \u00d7 \u21d2 = ( . ) ( . ) . . \u2234 K eq = K 2 = ( . ) . . 0 12 0 02 0 04 18 2 \u00d7 =
Answer: 2400
\nSolution: \u0394 G = \u0394 G \u00b0 + RT. ln Q = \u2013RT. ln K K f b + RT [Product] [Reactants] .ln\n6.51 Chemical Equilibrium HINTS AND EXPLANATIONS = RT. ln K K b f [Product] [Reactants] = RT. ln r r b f = 2 \u00d7 300 \u00d7 ln 1 4 e = \u20132400 cal
Answer: 0250
\nSolution: A + B \u001f C; K 1 = 4 \u00d7 10 10 Initial moles 5 5 0 Equ. moles 5 \u2013 ( x + y ) 5 \u2013 x x A + D \u001f C; K 2 = 10 10 Initial moles 5 5 0 Equ. moles 5 \u2013 ( x + y ) 5 \u2013 y y As K 1 and K 2 are very large, ( x + y ) = 5 (1) and K K x x y y x y 1 2 4 5 5 2 = = \u2212 \u00d7 \u2212 \u21d2 = (2) From (1) and (2), x = 10 3 \u2234 Moles of B at equilibrium = 5 \u2013 x = 5 3
Answer: 0800
\nSolution: Br 2 (l) + Cl 2 (g) \u001f 2Br Cl (g); K p = 1 atm Initial moles x 10 0 Equ. moles \u2248 0 10 \u2013 x 2 x Br 2 (l) \u001f Br 2 (g); K p = 0.25 atm Initial moles y 0 Equ. moles \u2248 0 y From question: y y \u00d7 \u00d7 = \u21d2 = 0 082 300 164 0 25 5 3 . . and ( ) . . 10 2 0 082 300 164 2 00 10 3 \u2212 + \u00d7 \u00d7 = \u21d2 = x x x \u2234 Minimum mass of Br 2 (l) = ( x + y ) \u00d7 160 gm = 800 gm
Answer: 0016
\nSolution: 2SO 3 \u001f 2SO 2 + O 2 Initial moles 1 (say) 0 0 Equ. moles 1 \u2013 0.4 = 0.6 0.4 0.2 \u2234 M av = 1 80 1 2 \u00d7 . Now, d = PM RT p p \u21d2 = \u00d7 \u00d7 \u21d2 = 16 80 1 2 0 0821 920 0 0821 216 . . . atm \u2234 K p = \u00d7 \u00d7 = ( . ) . ( . ) . 0 4 0 2 0 6 216 1 2 16 2 2 atm
Answer: 0008
\nSolution: A 2 \u001f 2A; K 1 = x atm Initial partial pressure 1 atm 0 Equ. partial pressure 1 \u2013 ( x + z ) 2 x B 2 \u001f 2B; K 2 = y atm Initial partial pressure 1 atm 0 Equ.partial pressure 1 \u2013 ( y + z ) 2 y A 2 + B 2 \u001f 2AB; K 3 = 2 Initial partial pressure 1 1 0 Equ. partial pressure 1 \u2013 ( x + z ) 1 \u2013 ( y + z ) 2 z = 0.5 (1) From question, [1 \u2013 ( x + z )] + 2 x + [1 \u2013 ( y + z )] + 2 y + 2 z = 2.75 \u2234 x + y = 0.75 (2) Now, K 3 = ( . ) ( . )( . ) . . 0 5 0 75 0 75 2 0 25 0 50 2 \u2212 \u2212 = \u21d2 = x y x or y = 0.50 or 0.25 \u2234 K K y y z x x z y x x y 2 1 2 2 2 2 2 1 2 1 2 0 75 2 0 75 = \u2212 + \u2212 + = \u00d7 \u2212 \u00d7 \u2212 ( ) ( ) ( ) ( ) ( ) ( . ) ( ) ( . ) = = 1 8 1 or 8
Answer: B
\nSolution: Negative sign is for reactants and positive for products.
Answer: B
\nSolution: r d dt K K rxn = \u2212 \u22c5 = \u2212 1 2 2 1 2 2 2 2 4 [ ] [ ] [ ] NO NO N O \u2234 Rate of disappearance of NO 2 is given by, \u2212 = \u2212 d dt K K [ ] [ ] [ ] NO NO N O 2 1 2 2 2 2 4 2 2
Answer: D
\nSolution: P n RT V A A = \u21d2 dP dt RT V dn dt A A = \u22c5 or, ( ) ( ) \u2212 \u22c5 = \u2212 \u22c5 K P RT K C A n A n 1 2 \u2234 K K RT P C K RT RT RT A A n n n 2 1 1 1 = \u22c5 \u239b \u239d \u239c \u239e \u23a0 \u239f = \u22c5 = \u2212 ( ) ( )
Answer: D
\nSolution: \u2212 \u22c5 = \u2212 1 2 2 2 1 2 2 d dt K K [ ] [ ] [ ][ ] HI HI H I
Answer: C
\nSolution: Order = 1 \u2190\u23af with respect to A \u21d2 a = 1 Order = 2 \u2190\u23af with respect to B \u21d2 b = 2 Hence, reaction is A + 2B P. r d A dt d B dt = \u2212 = \u2212 [ ] [ ] 1 2
Answer: C
\nSolution: A + 2B C + D t = 0 0.6 atm 0.8 atm t = t 0.6 \u2013 x 0.8 \u2013 2 x = 0.3 atm = 0.2 atm \u21d2 x = 0.3 \u2234 r r K K t 0 2 2 0 3 0 2 0 6 0 8 1 32 = \u00d7 \u00d7 \u00d7 \u00d7 = . ( . ) . ( . )\n11.45 Chemical Kinetics HINTS AND EXPLANATIONS
Answer: B
\nSolution: For no change in temperature, \u0394 H net = 0 and hence, for 3 moles of B reacted, 4 moles of Q should form.
Answer: A
\nSolution: 2.82 = (2) x \u21d2 x = 3 2 9 = (3) y \u21d2 y = 2 \u2234 overall order = x + y = 7 2
Answer: D
\nSolution: r 1 = 0.0068/65 = 1.046 \u00d7 10 \u20134 gm/min r 2 = 0.0031/120 = 2.583 \u00d7 10 \u20135 gm/min r 3 = 0.0032/60 = 5.333 \u00d7 10 \u20135 gm/min From (1) and (2) : order with respect to K 2 C 2 O 4 = 2 From (1) and (3) : order with respect to HgCl 2 = 1
Answer: D
\nSolution: From (2) and (3) : order with respect to I \u2013 = 1 From (1) and (3) : order with respect to ClO \u2013 = 1 From (3) and (4) : order with respect to OH \u2013 = 1
Answer: B
\nSolution: (1 + K 2 \u22c5 C A ) \u2248 1
Answer: A
\nSolution: For steady state, + = d R dt [ ] 0 or, K 1 [ A ] \u2013 K 2 [ R ][ B ] \u2013 K 3 [ R ][ C ] = 0 \u2234 [ ] [ ] [ ] [ ] R K A K B K C = + 1 2 3 Now, dx dt K R C K K A C K B K C = = + 3 3 1 2 3 [ ][ ] [ ][ ] [ ] [ ]
Answer: C
\nSolution: r K = 3 3 [ ][ ] O O For 1st step, K K 1 2 2 3 = [ ][ ] [ ] O O O \u2234 r = K 3 [O][O 3 ] = K K K 3 1 3 2 2 2 [ ] [ ] O O
Answer: D
\nSolution: r d X dt d Y dt = \u2212 = + [ ] [ ] and rate decreases with time.
Answer: C
\nSolution: A 2B t = 0 0.1 M 0 t = 1 min 0.1 \u2013 x 2 x For zero order reaction: [ A 0 ] \u2013 [ A ] = Kt or, 0.1 \u2013 (0.1 \u2013 x ) = 0.01 \u00d7 1 \u21d2 x = 0.01 \u2234 [ B ] = 2 x = 0.02 M
Answer: B
\nSolution: 2NH 3 N 2 + 3H 2 r r r r rxn = = = = = NH N H atm/s Constant 3 2 2 2 1 3 0 1 . Hence, after 10 seconds: P NH atm 3 3 2 0 1 10 1 = \u2212 \u00d7 \u00d7= . P N atm 2 0 1 10 1 = \u00d7 = . P H atm 2 3 0 1 10 3 = \u00d7 \u00d7 = . \u2234 P total = 1 + 1 + 3 = 5 atm
Answer: B
\nSolution: For \u2212 = d A dt K A n [ ] [ ] and n \u2260 1 [ A ] 1 \u2013 n = [ A 0 ] 1 \u2013 n \u2013 K (1 \u2013 n ) \u22c5 t For given graph, 1 \u2013 n = \u20133 \u21d2 n = 4 and \u2013 K (1 \u2013 n ) = tan 45\u00b0 \u21d2 K (4 \u2013 1) = 1 \u2234 K = \u2212 \u2212 1 3 3 1 M min Now, r d A dt K A rxn = \u2212 \u22c5 = \u22c5 \u22c5 1 3 1 3 4 [ ] [ ] = \u00d7 \u00d7 = \u00d7 \u2212 \u2212 1 3 1 3 0 2 16 9 10 4 4 1 ( . ) min M
Answer: C
\nSolution: 1 0 25 2 8 0 1 2 M M hr t t = \u00d7 \u23af \u2192 \u23af\u23af\u23af / . . \u21d2 t 1/2 = 4.0 hr 0.6M M hr t t = \u23af \u2192 \u23af\u23af 1 2 4 0 0 3 / . .
Answer: C
\nSolution: The successive t 1/2 are double of previous one and hence, order =
Answer: C
\nSolution: 20. r = K [ A ] n 10 = K (0.8) n (1) 0.625 = K (0.2) n (2) \u2234 n = 2
Answer: A
\nSolution: C 2 H 6 C 2 H 4 + H 2 t = 0 3 bar 0 0 t = ? (3 \u2013 x ) bar x bar x bar From question: (3 \u2013 x ) + x + x = 5 \u21d2 x = 2 From the unit of rate constant, the order of reaction is 2, hence, t K P P x = \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f = \u2212 \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f \u00d7 \u00b0 1 1 1 1 0 0015 1 3 1 3 1 10 2 6 2 6 5 C H C H . = 4.44 \u00d7 10 \u20133 hr = 16 seconds
Answer: B
\nSolution: Let the reaction be first-order. K 1 1 8 100 20 0 201 = \u22c5 = ln . K 2 1 18 100 10 0 128 = \u22c5 = ln .\n11.46 Chapter 11 HINTS AND EXPLANATIONS As K 1 \u2260 K 2 , the reaction is not first-order. Let the reaction be second-order. K 1 1 8 1 0 2 1 1 0 5 = \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f = . . K 2 1 18 1 0 1 1 1 0 5 = \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f = . . As K 1 = K 2 , order = 2
Answer: D
\nSolution: Here, t 1/2 is independent from sugar concentration and hence, the order with respect to sugar is
Answer: D
\nSolution: Now, r = K [Sugar][H + ] n = K \u2032 \u22c5 [Sugar] t K K n 1 2 2 2 / ln ln [ ] = \u2032 = + H 500 2 10 5 = \u22c5 \u2212 ln ( ) K n and 50 2 10 6 = \u22c5 \u2212 ln ( ) K n \u2234 n = \u20131
Answer: C
\nSolution: r = K [ester] [ H + ] = K \u2032 \u22c5 [ester] \u2234 t K K 1 2 2 2 0 693 0 1 0 01 693 / ln ln [ ] . . . = \u2032 = = \u00d7 = + H hr
Answer: C
\nSolution: For n th order reaction ( n \u2260 1) Kt A A n n n = \u2212 \u2212 \u2212 \u2212 [ ] [ ] 0 1 1 1 For n = 0.5, Kt A A = \u2212 [ ] [ ] / / 0 1 2 1 2 1 2 Now, t T A K A K 100 0 1 2 1 2 0 1 2 2 0 2 % / / / ([ ] ) [ ] = = \u2212 = and t t A A K A 50 1 2 0 1 2 0 1 2 0 1 2 2 2 2 1 1 2 % / / / / [ ] [ ] [ ] = = \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f \u239b \u239d \u239c \u239e \u23a0 \u239f = \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f \u239f K \u2234 T t 1 2 1 1 1 2 1 0 3 / . = \u2212 =
Answer: A
\nSolution: For A : t t A A B A = \u22c5 = \u22c5 1 2 0 0 2 10 2 8 / log log [ ] [ ] log log [ ] [ ] For B : t t B B B A = \u22c5 = \u22c5 1 2 0 0 2 20 2 / log log [ ] [ ] log log [ ] [ ] From question, 10 2 8 20 2 0 0 log log [ ] [ ] log log [ ] [ ] \u22c5 = \u22c5 B A B A \u2234 [ ] [ ] B A 0 8 = \u21d2 t B A = \u22c5 = 20 2 60 0 log log [ ] [ ] min Alternate method: A B B B B B B : [ ] [ ] [ ] [ ] [ ] [ ] 8 4 2 2 4 0 10 0 10 0 10 0 10 0 10 0 10 \u23af \u2192 \u23af \u23af \u2192 \u23af \u23af \u2192 \u23af \u23af \u2192 \u23af \u23af \u2192 \u23af \u23af \u23af \u2192 \u23af [ ] B 0 8 B B B B B : [ ] [ ] [ ] [ ] 0 20 0 20 0 20 0 2 4 8 \u23af \u2192 \u23af \u23af \u2192 \u23af \u23af \u2192 \u23af
Answer: A
\nSolution: 0 1 0 025 2 40 1 2 . . / min M M t t = \u23af \u2192 \u23af\u23af \u21d2 t 1/2 = 20 min Now, r K A t A = = = \u00d7 [ ] ln [ ] . min . / 2 0 693 20 0 01 1 2 M = 3.465 \u00d7 10 \u20134 M min \u20131
Answer: A
\nSolution: + = \u2212 = d B dt d A dt K A [ ] [ ] [ ] / 1 3 or, \u2212 = \u22c5 \u222b \u222b d A B K dt A A t [ ] [ ] / [ ] [ ]/ / 1 3 2 0 0 0 1 2 t A K 1 2 0 2 3 2 3 5 3 3 2 1 2 / / / / [ ] ( ) = \u2212 \u22c5
Answer: C
\nSolution: N 2 O 5 2NO 2 + 1 2 O 2 t = 0 a mole 0 t = t ( a \u2013 x ) mole x V t 2 mole \u03b1 t = \u221e \u001f 0 a V 2 mole \u03b1 \u221e K t t a a x t V V V t = \u22c5 = \u22c5 \u2212 = \u22c5 \u2212 \u221e \u221e 1 1 1 2 5 0 2 5 ln [ ] [ ] ln ln N O N O Now, 1 20 9 6 9 6 4 8 1 40 9 6 9 6 \u22c5 \u2212 = \u22c5 \u2212 ln . . . ln . . V t \u21d2 V t = 7.2 ml
Answer: A
\nSolution: For zero order reaction, t P 1 2 3 / \u03b1 NH \u00b0 \u2234 315 70 150 1 2 t / = \u21d2 t 1/2 = 675 sec
Answer: A
\nSolution: A n B t = 0 P 0 0 t = t P 0 \u2013 x n.x Now, ( P 0 \u2013 x ) = P 0 \u22c5 e \u2013 Kt \u21d2 x = P 0 (1 \u2013 e \u2013 Kt )\n11.47 Chemical Kinetics HINTS AND EXPLANATIONS Now, P total = ( P 0 \u2013 x ) + nx = P 0 \u22c5 e \u2013 Kt + n \u22c5 P 0 (1 \u2013 e \u2013 Kt ) = P 0 [ n + (1 \u2013 n ) e \u2013 Kt ]
Answer: B
\nSolution: t K n K n n n n n 1 2 1 1 1 1 2 1 1 2 1 / ( ( ) ( ) [ ] ( ) = \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f \u2212 = \u2212 \u2212 \u2212 \u2212 \u2212 \u2212 C ) C C 0 0 0 and t K n K n n n n n 3 4 1 1 1 2 1 4 1 1 2 1 / ( ) ( ( ) ( ) [ ] ( ) = \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f \u2212 = \u2212 \u2212 \u2212 \u2212 \u2212 \u2212 C ) C C 0 0 0 \u2234 t t n n n 3 4 1 2 2 1 1 1 1 2 1 2 1 2 / / ( ) ( ) ( ) = \u2212 \u2212 = + \u2212 \u2212 \u2212
Answer: D
\nSolution: r r A A n 2 1 2 1 = \u239b \u239d \u239c \u239e \u23a0 \u239f [ ] [ ] \u21d2 2 = (4) n \u21d2 n = 1 2 Now, t 1/2 a [ A 0 ] 1 \u2013 n \u21d2 t 1/2 a [ A 0 ] 1/2 100 50 25 16 16 2 t t = = \u23af \u2192 \u23af\u23af\u23af \u23af \u2192 \u23af\u23af\u23af\u23af min / min \u2234 Time for 75 % reaction = 16 16 2 27 3 + = . min
Answer: D
\nSolution: Kt a a x x a = \u2212 = \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f ln ln 1 1 or, 2.5 \u00d7 10 \u20135 \u00d7 (100 \u00d7 60) = \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f ln 1 1 x a \u21d2 x a = 0 138 . \u2234 Percentage decomposition = \u00d7 = x a 100 13 8 . %
Answer: A
\nSolution: ( )( ) / / t t 1 2 1 1 2 2 = \u21d2 0 693 1 1 2 0 . [ ] K K A = \u2234 [ ] . . . . . A K K 0 1 2 2 0 693 6 93 10 0 693 0 2 0 5 = = \u00d7 \u00d7 = \u2212 M
Answer: A
\nSolution: t t t t t t 1 2 1 2 1 1 2 2 1 2 1 1 2 2 100 75 2 100 25 = \u22c5 \u22c5 = ( ) log log ( ) log log ( ) ( / / / / ) ) log log 2 4 3 4 \u22c5 \u239b \u239d \u239c \u239e \u23a0 \u239f = \u00d7 \u2212 = 3 2 0 6 0 48 0 6 3 10 . . .
Answer: B
\nSolution: r = K [ A ] n 1 100 0 02 60 0 02 \u00d7 = . ( . ) K n (1) 1 100 0 04 15 0 04 \u00d7 = . ( . ) K n (2) \u2234 n = 3
Answer: B
\nSolution: t T A A = \u22c5 gen ln ln [ ] [ ] 2 0 \u21d2 60 75 2 0 = \u22c5 ln ln [ ] [ ] A A \u2234 [ ] [ ] . A A e 0 0 56 =
Answer: A
\nSolution: ( ) ( ) [ ] [ ] / / t t A A n 1 2 1 1 2 2 0 1 0 2 1 = \u239b \u239d \u239c \u239e \u23a0 \u239f \u2212 \u21d2 37 82 18 95 0 05 0 10 1 . . . . = \u239b \u239d \u239c \u239e \u23a0 \u239f \u2212 n \u21d2 n = 2 Now, ( ) . . . / t 1 2 1 37 82 0 15 0 05 = \u239b \u239d \u239c \u239e \u23a0 \u239f \u2212 \u21d2 t 1/2 = 12.6 hr
Answer: C
\nSolution: Now, T N a 0 1000 2 \u22c5 = \u00d7 (1) and T N a x x x t \u22c5 = \u2212 \u00d7 + \u00d7 + \u00d7 1000 2 2 1 ( ) (2) \u2234 a a x T T T t \u2212 = \u2212 0 0 3 2
Answer: A
\nSolution: B n + B ( n + 4)+ t = 0 a mole 0 t = 10 min ( a \u2013 x ) mol x mol Now, 25 1000 2 \u00d7 = \u00d7 N a (1) and, 32 5 1000 2 5 . ( ) \u00d7 = \u2212 \u00d7 + \u00d7 N a x x (2) Now, K t a a x = \u22c5 \u2212 = \u22c5 \u2212 = \u2212 1 1 10 12 5 12 5 2 5 0 02 1 ln min ln . . . . min CH (Br) COOH CH (Br) COOH a mole 0 x mole 0 x mole ( a \u2013 x ) mole t = 0 t = t CHCOOH C Br COOH + H Br \u2192\n11.48 Chapter 11 HINTS AND EXPLANATIONS
Answer: B
\nSolution: A 2B + C t = 0 a mol 0 0 t = 10 sec ( a \u2013 x ) mol 2 x mol x mol Now, r r P P M M A B A B B A = \u22c5 \u21d2 1 2 2 4 16 = \u2212 \u22c5 a x x \u21d2 x a = 3 Now, K t a a x a a a = \u22c5 \u2212 = \u22c5 \u2212 = \u2212 1 1 10 2 3 0 04 1 ln sec ln . sec
Answer: A
\nSolution: r = K [ A ] 2 [ B ] = K \u2032 \u22c5 [ A ] 2 as [ B 0 ] >> [ A 0 ] \u2234 t K A K B A 1 2 0 0 0 1 1 1 0 5 0 002 2 0 500 / [ ] [ ][ ] . . . min = \u2032 \u22c5 = = \u00d7 \u00d7 =
Answer: D
\nSolution: r = K [ester][H + ] \u2234 r r HA HX HA H = = + 1 100 1 0 [ ] . \u21d2 [H + ] HA = 0.01 M Now, Ka A HA ( ) [ ][ ] [ ] . . ( . ) HA H = = \u00d7 \u2212 \u2248 + \u2212 \u2212 0 01 0 01 1 0 01 10 4
Answer: A
\nSolution: As [ A 0 ] = [ B 0 ] and the stoichiometric coefficients of both A and B are 1, at any time [ A ] = [ B ]. Hence, r = K [ A ] 1/2 [ B ] 1/2 = K [ A ]. Required time = 2 2 0 693 2 31 10 600 1 2 3 \u00d7 = \u00d7 \u00d7 = \u2212 t / . . sec
Answer: C
\nSolution: \u2212 = + dC dt C C \u03b1 \u03b2 1 \u21d2 \u2212 + \u239b \u239d \u239c \u239e \u23a0 \u239f \u22c5 = \u22c5 \u222b \u222b 1 0 2 1 2 C dC dt t Co Co \u03b2 \u03b1 / / \u2234 t C 1 2 0 1 2 2 / ln = + \u22c5 \u239b \u239d \u239c \u239e \u23a0 \u239f \u03b1 \u03b2
Answer: C
\nSolution: r = K \u2032 [CH 3 COOH][C 2 H 5 OH] = K \u2032 \u22c5 [CH 3 COOH] 2 \u2234 t K 1 2 0 1 / [ = \u2032 \u22c5 CH COOH] 3 \u21d2 50 1 10 0 2 3 = \u00d7 \u00d7 \u2212 ( ) . K \u2234 K = 100 M \u20132 min \u20131
Answer: C
\nSolution: r = K [ A ] x [ B ] y For case-I : r = K [ A ] x [ B ] y = K \u2032 \u22c5 [ B ] y where K \u2032 = K [ A 0 ] x In equal time interval, the concentrations of B are in G.P. and hence, y = 1 and \u2032 = \u22c5 = \u2212 K 1 10 0 01 0 008 0 02 1 ln . . . min For case-II: r = K [ A ] x [ B ] y = K \u2033 [ A ] x where K \u2033 = K [ B 0 ] y In equal time interval, the concentration of A are in G.P. and hence, x = 1 and \u2032\u2032 = \u22c5 = \u2212 K 1 10 0 02 0 018 0 01 1 ln . . . min Now, r = K [ A ][ B ] \u2234 K K A K B = \u2032 \u2032\u2032 = = \u2212 \u2212 [ ] , [ ] . . . . . min 0 0 1 1 0 02 2 0 0 01 1 0 0 01 or or M
Answer: B
\nSolution: K K C C K C app = \u22c5 + \u22c5 = + 1 1 1 1 \u03b1 \u03b1 lim C K K \u2192\u221e = app 1 \u03b1 From question, K C C K 1 1 1 90 100 \u22c5 + \u22c5 = \u00d7 \u03b1 \u03b1 or, C C 1 9 10 90 100 1 9 10 5 5 + \u00d7 = \u00d7 \u00d7 \u21d2 C = 10 \u20135 M
Answer: A
\nSolution: A 2B + C t = 0 1 0 0 t = 12 hr 1 \u2013 x 2 x x t = 24 hr 1 \u2013 y 2 y y V.P. of solution, P = X 2 \u22c5 P o or, 20 180 18 180 18 1 2 24 = + + \u00d7 / ( ) x \u21d2 x = 0.5 \u2234 t = 12 hr = t 1/2 Now, t = 24 hr = 2 \u00d7 t 1/2 \u21d2 y = 0.75 Now, V.P. of solution, P X P y = \u22c5 \u00b0 = + + \u00d7 2 10 10 1 2 24 ( ) = 19.2 mm Hg
Answer: B
\nSolution: [ ] [ ] . . B C K K = = \u00d7 \u00d7 = \u2212 \u2212 1 2 4 5 1 26 10 3 15 10 4 1 \u2234 Percentage of B = \u00d7 = 4 5 100 80%\n11.49 Chemical Kinetics HINTS AND EXPLANATIONS
Answer: A
\nSolution: A R t = t a \u2013 x x \u2234 r = K ( a \u2013 x ) \u22c5 x For maximum rate, dr dx = 0 \u21d2 x a = 2 \u21d2 C A = C R
Answer: A
\nSolution: K dt by y dy t \u22c5 = + \u2212 \u22c5 \u222b \u222b 0 0 2 1 2 1 / / Co Co \u21d2 t K b b 1 2 1 1 2 2 / ( ) ln = + \u22c5 \u22c5 \u2212 \u22c5 \u23a1 \u23a3 \u23a2 \u23a4 \u23a6 \u23a5 Co Co
Answer: A
\nSolution: t t B A 50 94 %, %, = or, 1 100 50 1 100 6 2 1 K K \u22c5 = \u22c5 ln ln \u21d2 K K 1 2 4 067 1 = .
Answer: A
\nSolution: Percentage product by S N 2 mechanism = \u00d7 \u00d7 + \u00d7 \u00d7 \u2212 \u2212 \u2212 ( . )[ ]( . ) ( . )[ ]( . ) . [ ] 4 8 10 0 01 4 8 10 0 01 2 4 10 100 5 5 6 RX RX RX = = 16 67 . %
Answer: B
\nSolution: [H 2 ] : [O 2 ] : [OH] : [H 2 O] : [O] = K 1 : K 1 : 2 K 2 : K 3 : K 3 = 0.60 : 0.60 : 2 \u00d7 0.30 : 0.10 : 0.10 = 6 : 6 : 6 : 1 : 1
Answer: A
\nSolution: [ A ] + [ B ] + [ C ] = [ A 0 ] when [ A ] = [ B ] = [ C ], [ A ] = [ ] [ ] A A e Kt 0 0 3 = \u22c5 \u2212 or, 1 3 3 3 = \u2212 + \u22c5 e t (ln ln ) \u21d2 t = 0.5 min
Answer: D
\nSolution: As K 1 = K 2 = K (Say), t K max . min = = = 1 1 0 02 50 and [ ] [ ] . max B A e e = = 0 0 2 M
Answer: B
\nSolution: After long time, r A = r B \u21d2 K 1 [ A ] = K 2 [ B ] \u2234 [ ] [ ] A B K K = = 2 1 40
Answer: A
\nSolution: [ ] [ ] [ ] ( ) [ ] ( ( ) ( ) C A K A K K e A e K K K e K K t K K t = + \u2212 \u22c5 = + \u2212 + \u2212 + \u22c5 2 0 1 2 0 2 1 2 1 1 2 1 2 ( ( ) ) K K t 1 2 1 + \u22c5 \u2212 = \u2212 = \u2212 \u22c5 \u00d7 \u00d7 \u00d7 \u2212 9 10 1 9 10 1 1 1 10 10 1 25 10 3600 1 5 K K e e K t ( ) [ ] . = 0.5112
Answer: C
\nSolution: At steady state, K 1 [ A ] = K 2 [ B ] \u2234 K K A B 2 1 4 3 1 2 5 10 0 2 0 01 5 10 = = \u00d7 \u00d7 = \u00d7 \u2212 \u2212 \u2212 [ ] [ ] . . . min
Answer: B
\nSolution: Reaction may be considered as A C K 1 \u23af \u2192 \u23af \u2234 [ C ] = [ A 0 ] ( ) 1 1 \u2212 \u2212 e K t
Answer: A
\nSolution: A B K 1 2 \u23af \u2192 \u23af A C K 2 \u23af \u2192 \u23af t = 0 1 atm 0 1 atm 0 t = 10 min (1 \u2013 x \u2013 y ) 2 x 1 \u2013 x \u2013 y y t = \u221e (1 \u2013 a \u2013 b ) 2 a (1 \u2013 a \u2013 b ) b \u2248 0 \u2248 0 From question, a + b = 1 and 2 a + b = 1.5 \u2234 a = b = 0.5 Now, P P K K a b x y B C = = = 2 2 2 1 2 \u21d2 K K x y 1 2 1 = = 0 Now, P x y x y 10 1 2 1 4 min ( ) . = \u2212 \u2212 + + = \u21d2 x = y = 0.4 \u2234 P x y A = \u2212 \u2212 = 1 0 2 . atm at t = 10 min Now, K 1 + K 2 = 1 1 10 1 0 2 0 16 1 t P P A A \u22c5 \u00b0 = \u22c5 = \u2212 ln ln . . min \u2234 K 1 = K 2 = 0.08 min \u20131
Answer: A
\nSolution: r z u N av max * = = \u22c5 \u22c5 11 2 2 1 2 \u03c0\u03c3 = \u00d7 \u00d7 \u00d7 \u00d7 \u00d7 \u00d7 \u2212 \u2212 \u2212 1 2 4 10 2 10 2 10 8 2 4 1 19 3 2 \u03c0 ( ( ) ( ) cm) cm s cm = 2.842 \u00d7 10 28 cm \u20133 s \u20131 = 4.74 \u00d7 10 7 mol l \u20131 s \u20131
Answer: B
\nSolution: d K dT E RT a (ln ) = 2 or, 0 2 2 + + = \u03b2 \u03b3 T T E RT a \u21d2 E a = ( b T + \u03b3 ) R
Answer: A
\nSolution: K K B C 1 2 40 60 2 3 = = = [ ] [ ] Now, E K E K E K K a a a ( ) overall = \u22c5 + \u22c5 + 1 1 2 2 1 2 = 32 kcal/mol\n11.50 Chapter 11 HINTS AND EXPLANATIONS
Answer: C
\nSolution: r r uncat cat = \u00d7 1 2 \u21d2 K K uncat cat = \u00d7 1 2 or, A e A e E RT E RT T a a \u22c5 = \u00d7 \u22c5 \u2212 \u2212 \u00d7 ( ) ( ) / / uncat cat 0.5 1 2 or, ln . ( ) ( ) 2 20 0 5 \u2212 = \u2212 \u2212 E RT E RT a a uncat uncat \u2234 E a (uncat) = 38.58 kcal/mol
Answer: B
\nSolution: For A B; K 1 = 8 min \u20131 at T = 300 K \u2032 = K 1 ? at T = ? ln \u2032 = \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f K R T 1 8 20 1 300 1 KJ (1) For A C; K 2 = 2 min \u20131 at T = 300 K \u2032 = K 2 ? at T = ? ln . \u2032 = \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f K R T 2 2 28 314 1 300 1 KJ (2) From (1) and (2), ln / / . . \u2032 \u2032 = \u00d7 \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f K K T 2 1 3 2 8 8 314 10 8 314 1 300 1 or, ln 1 2 8 2 1 300 1 10 3 \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f = \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f \u00d7 T \u21d2 T = 379.75 K
Answer: C
\nSolution: Given: K K 1 310 1 300 2 ( ) ( ) = , K 1 310 2 30 ( ) ln min = K K 1 310 1 310 2 ( ) ( ) = and E E a a 2 1 1 2 = For reaction 1: ln ( ) ( ) K K E R a 1 310 1 300 1 1 300 1 310 \u23a1 \u23a3 \u23a2 \u23a2 \u23a4 \u23a6 \u23a5 \u23a5 = \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f (1) For reaction 2: ln ( ) ( ) K K E R a 2 310 2 300 2 1 300 1 310 \u23a1 \u23a3 \u23a2 \u23a2 \u23a4 \u23a6 \u23a5 \u23a5 = \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f (2) From (1) \u00f7 (2) : ln ln ( ) ( ) 2 2 2 310 2 300 K K \u23a1 \u23a3 \u23a2 \u23a4 \u23a6 \u23a5 = or, K K 2 310 2 300 2 ( ) ( ) \u23a1 \u23a3 \u23a2 \u23a2 \u23a4 \u23a6 \u23a5 \u23a5 = \u21d2 K 2 (300) = K 2 310 2 2 2 30 2 ( ) = \u00d7 ln = 0.0327 min \u20131
Answer: B
\nSolution: At 27\u00b0C, K 1 1 1 21 6 100 25 2 10 8 = \u22c5 = \u2212 . ln ln . min Now, ln . . K K E R T T a 2 1 1 2 3 1 1 9 6 10 2 1 300 1 320 1 0 = \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f = \u00d7 \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f = \u2234 K K e 2 1 2 7 = = . \u21d2 K 2 2 7 2 10 8 2 4 = \u00d7 = . ln . ln \u21d2 ( t 1/2 ) 2 = 4 min \u2234 Percentage decomposition in 8.0 min = 75 %
Answer: C
\nSolution: K A e A e A e A E Rt RT RT a = \u22c5 = \u22c5 = \u2212 \u2212 / / . \u001f 0 37
Answer: A
\nSolution: ln 2 1 280 1 290 = \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f E R a (1) and ln x E R a = \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f 1 290 1 300 (2) From (2) \u00f7 (1), ln ln x 2 280 300 = \u21d2 x = 1.91
Answer: B
\nSolution: Informative
Answer: D
\nSolution: Theory based
Answer: A, B, C, D
\nSolution: Theory based
Answer: C
\nSolution: Theory based
Answer: B, C
\nSolution: Informative
Answer: A, C, D
\nSolution: r K A n = \u22c5 [ ] \u21d2 n r K A = ln( / ) ln [ ] Now, r r A A n 2 1 2 1 = \u239b \u239d \u239c \u239e \u23a0 \u239f [ ] [ ] \u21d2 n r r A A = \u2212 \u2212 ln ln ln[ ] ln[ ] 2 1 2 1\n11.51 Chemical Kinetics HINTS AND EXPLANATIONS And, t 1/2 a [ A 0 ] 1\u2013 n \u21d2 ( ) ( ) [ ] [ ] / / t t A A n 1 2 2 1 2 1 0 2 0 1 1 = \u239b \u239d \u239c \u239e \u23a0 \u239f \u2212 \u2234 n A A t t = \u2212 \u2212 \u2212 1 0 2 0 1 1 2 2 1 2 1 ln[ ] ln[ ] ln( )ln( ) / /
Answer: A, D
\nSolution: [ A ] = [ A 0 ] (1 \u2013 a ) = [ A 0 ] \u22c5 e \u2013 Kt \u21d2 \u03b1 = 1 \u2013 e \u2013 Kt
Answer: A, C
\nSolution: (a) \u2212 = \u22c5 d A dt K A n [ ] [ ] f A A A d A A = \u2212 = \u2212 [ ] [ ] [ ] [ ] [ ] 1 2 1 From question, f d A A = \u2212 [ ] [ ] \u2234 f A t K A n [ ] [ ] = \u21d2 f t K A n = \u22c5 \u2212 [ ] 1 or, log log ( ) log[ ] f t K n A \u239b \u239d \u239c \u239e \u23a0 \u239f = + \u2212 \u22c5 1 (b) [ ] [ ] A A n Kt n n 0 1 1 1 \u2212 \u2212 \u2212 \u2212 = \u21d2 [ A ] 1 \u2013 n = [ A 0 ] 1 \u2013 n + ( n \u2013 1) \u22c5 Kt (c) t t A A A A n n n n 3 4 1 2 0 1 0 1 0 1 0 1 4 2 1 2 / / [ ] [ ] [ ] [ ] ( = \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f = \u2212 \u2212 \u2212 \u2212 \u2212 2 2 1 1 1 1 2 1 2 ) n n n \u2212 \u2212 \u2212 \u2212 = +
Answer: B, C, D
\nSolution: t 1/2 = C \u22c5 (C 0 ) 1 \u2013 n \u21d2 ln t 1/2 = ln C + (1 \u2013 n ) \u22c5 ln C 0
Answer: B
\nSolution: K \u2032 = K \u22c5 [H + ] On doubling [H + ], K \u2032 will double but K will remain unchanged.
Answer: D
\nSolution: Theory based
Answer: A, D
\nSolution: (a) For steady state, 6 93 10 80 0 693 100 6 3 . . [ ] \u00d7 = \u00d7 \u2212 SO \u2234 [SO 3 ] = 1.25 \u00d7 10 \u20135 M (b) n eq SO 3 = n eq NaOH \u21d2 1.25 \u00d7 10 \u20135 \u00d7 10 3 \u00d7 2 = V NaOH \u00d7 1 \u2234 V NaOH = 2.5 \u00d7 10 \u20132 L = 25 ml (c) Mole of SO 3 needed = 980 10 98 10 3 4 \u00d7 = \u2234 Air needed = \u00d7 = \u00d7 \u2212 10 1 25 10 8 10 4 5 8 . L (d) 1000 days = 10 t 1/2 \u2234 [ ] . . SO M 3 5 10 8 1 25 10 2 1 25 10 = \u00d7 \u2248 \u00d7 \u2212 \u2212
Answer: C, D
\nSolution: 3A( g ) 2B( g ) + 2C( s ) t = 0 6 atm 0 \u2013 t = 20 min (6 \u2013 x ) atm 2 3 x atm 0.05 atm t = \u221e \u2248 0 4 atm 0.05 atm But from question, P \u221e = 4.05 atm and hence, (4.05 \u2013 4) = 0.05 atm is the vapour pressure of C( s ). Now, P x x 20 6 2 3 0 05 5 05 = \u2212+ + = ( ) . . \u21d2 x = 2 \u2234 t = 20 min = t 1/2
Answer: A, B
\nSolution: \u2212 = \u22c5 d dt K \u03b8 \u03b8 \u21d2 Kt = ln \u03b8 \u03b8 0 (a) t K = \u22c5 = \u22c5 = 1 1 0 04 596 298 17 5 0 ln . ln . sec \u03b8 \u03b8 (b) t = \u22c5 = 1 0 04 1192 298 35 . ln sec
Answer: A, C
\nSolution: (a) A + B 2C t = 0 2 a a t = t 2 a \u2013 x a \u2013 x As [A] \u2260 [B] throughout, the overall reaction is not fi rst-order. (b) r = K [A] \u20131 [B] 2 = K \u2032 \u22c5 [B] 2 \u21d2 t K B 1 2 0 1 / [ ] = \u2032 (c) r = K [A] \u20131 [B] 2 = K \u2033 [A] \u20131 (d) As [A] = [B] = stoichiometric ratio, then the mole ratio will remain constant throughout.
Answer: B, D
\nSolution: A P K 1 \u23af \u2192 \u23af ; t K 1 2 1 0 693 / . = B Q K 2 \u23af \u2192 \u23af ; t K B K 1 2 2 0 2 1 1 / [ ] = = From question, 0 693 1 1 2 . K K = \u21d2 K 2 > K 1
Answer: A, B, D
\nSolution: A 4 4A t = 0 a M 0 t = 30 min ( a \u2013 x ) M 4 x M As a \u2013 x = 4 x \u21d2 x a = 5 \u2234 Percentage reaction at t = 30 min = \u00d7 = x a 100 20% Now, 30 2 1 2 = \u22c5 \u2212 t a a x / log log \u21d2 t 1/2 = 90 min\n11.52 Chapter 11 HINTS AND EXPLANATIONS
Answer: A, C, D
\nSolution: (a) \u0394 r H = \u2211 \u0394 f H Products \u2013 \u2211 \u0394 f H Reactants = 2 \u00d7 (\u20131263) \u2013 [(\u20132238) + (\u2013285)] = \u20133 KJ/mol (b) Can not confirm because in aqueous medium, there is no combustion. (d) Concentration in G.P. in equal time interval.
Answer: B, D
\nSolution: n = 1 \u21d2 t 100% = 1 0 0 K A \u22c5 = ln [ ] Infinite n \u2260 1 \u21d2 t 100 % = [ ] ( ) ( ) [ ] ( ) A K n A K n n n n n 0 1 1 0 1 0 1 1 \u2212 \u2212 \u2212 \u2212 \u2212 = \u2212 if < 1 = Infi nite if n > 1
Answer: A, D
\nSolution: [ ] [ ] B C K K = = 1 2 1 2 \u21d2 [ C ] > [ B ] Hence, after long time, the solution will be dextrorotatory. Now, K = K 1 + K 2 = 6.93 \u00d7 10 \u20132 + 13.86 \u00d7 10 \u20132 = 3 \u00d7 6.93 \u00d7 10 \u20132 min \u20131 \u2234 t K 1 2 2 2 0 693 3 6 93 10 10 3 / ln . . min = = \u00d7 \u00d7 = \u2212 A B A C t = 0 2M 0 2M 0 t = t 2 \u2013 ( x + y )M x M 2 \u2013 ( x + y )M y M From question, x + y = 1.5 and x y = 1 2 \u2234 x = 0.5, y = 1.0 Hence, total rotation = 0.5 \u00d7 60\u00b0 + 0.5 \u00d7 (\u201372\u00b0) + 1.0 \u00d7 42\u00b0 = 36\u00b0
Answer: A, B, C
\nSolution: [ B ] : [ C ] : [ D ] = 1 \u00d7 3 K : 2 \u00d7 2 K : 3 \u00d7 K = 3 : 4 : 3 A B A 2C A 3D t = 0 1M 0 1M 0 1M 0 t = t 1 \u2013 ( x + y + z ) x M 1 \u2013 ( x + y + z ) 2 y M 1 \u2013 ( x + y + z ) 3 z M t = \u221e 1 \u2013 ( a + b + c ) a M 1 \u2013 ( a + b + c ) 2 b M 1 \u2013 ( a + b + c ) 3 c M As a : 2 b : 3 c = 3 : 4 : 3 and a + b + c = 1 [ C ] = 2 b = 0.67 M As [ A 0 ] = 1 M, [ B ] \u2260 1M
Answer: A, B, C, D
\nSolution: For S N 1 path : r 1 = (3 \u00d7 10 \u20134 s \u20131 ) [RX] For S N 2 path : r 2 = (5 \u00d7 10 \u20134 M \u20131 s \u20131 ) [RX] [ ] \u001f\u001f Nu (a) [ ] \u001f\u001f Nu = 0.1 M, then r 1 > r 2 (b) [ ] \u001f\u001f Nu = 1.0 M, then r 1 < r 2 (c) [ ] \u001f\u001f Nu = 0.6 M, then r 1 = r 2 (d) [ ] \u001f\u001f Nu = 0.4 M, then r r 1 2 2 3 = \u2234 Percentage product by S N 1 = 2 2 3 100 40 + \u00d7 = %
Answer: A, B, C, D
\nSolution: \u2212 = + d A dt d B dt [ ] [ ] always
Answer: A, B, C, D
\nSolution: As mole is not changing, C A + C B + C C = C A 0 Now, C C C C C C K K K B A A B B C 0 1 1 2 \u2212 = + = +
Answer: A, C, D
\nSolution: Informative
Answer: B, D
\nSolution: Theoretical
Answer: A, C
\nSolution: Increase in temperature will result in greater increase in the rate of reaction A \u2192 B than B \u2192 C.
Answer: B
\nSolution: Informative
Answer: C
\nSolution: Informative
Answer: C, D
\nSolution: K 1 = K 2 \u21d2 \u2212 + = \u2212 + 14000 5 20000 10 RT RT \u21d2 T K = 1200 8 314 . Now, P P e e A B K t K t 2 3 1 2 1 1 1 1 = \u00d7 \u00d7 = \u2212 \u2212 Now, initial pressure P 0 1 1 0 0821 1200 8 314 100 0 237 = + \u00d7 \u00d7 = ( ) . . . atm As number of moles will increase on reaction, the total pressure can never be less than 0.2 atm Now, P P K K A B = = 2 3 2 3 1 2
Answer: A, B, C
\nSolution: 1 1 25 10 6 3 2 K dK dT d K dT T E RT a \u22c5 = = \u00d7 = (ln ) . \u2234 E R T a = \u00d7 = \u00d7 \u00d7 = 1 25 10 1 25 10 2 250 10 6 6 4 . . cal/mol
Answer: A, B, C
\nSolution: \u0394 = \u2212 H E E a a f b \u21d2 \u2212 = \u2212 2 8 E a f \u21d2 E a f = 6 kcal/mol Now, the fraction of molecules crossing energy barrier = \u2212 e E RT a / and K e H Rt eq = \u2212\u0394 /\n11.53 Chemical Kinetics HINTS AND EXPLANATIONS
Answer: A
\nSolution: The overall reaction is first-order.
Answer: B
\nSolution: K K K 1 2 3 2 4 1 = = \u21d2 2 K 1 = K 2 = 4 K 3
Answer: A
\nSolution: 2N 2 O 5 4NO 2 + O 2 2 \u00d7 108 gm 4 \u00d7 46 gm 32 gm 108 gm 92 gm 16 gm Comprehension II
Answer: D
\nSolution: CO(g) + Cl 2 (g) COCl 2 (g) r d dt K COCl 2 COCl COCl][Cl = + = \u22c5 [ ] [ ] 2 5 2 (1) Now, for steady state of COCl, + = d dt [ ] COCl 0 or K 3 [Cl][CO] \u2013 K 4 [COCl] \u2013 K 5 [COCl][Cl 2 ] = 0 \u2234 [ [ [ ] COCl] Cl][CO] Cl = + K K K 3 4 5 2 (2) \u2234 For steady state of Cl, d dt [ ] Cl = 0 or 2 K 1 [Cl 2 ] \u2013 2 K 2 [Cl] 2 \u2013 K 3 [Cl][CO] + K 4 [COCl] + K 5 [COCl][Cl 2 ] = 0 \u2234 [ [ ] / Cl]= Cl K K 1 2 2 1 2 \u239b \u239d \u239c \u239e \u23a0 \u239f (3) From (1), (2), (3), r K K K K K K COCl CO Cl 2 1 1 2 5 3 2 3 2 2 1 2 4 5 2 = + / / / [ ][Cl ] ( [ ])
Answer: C
\nSolution: r 4 >> r 5 or K 4 [COCl] >> K 5 [COCl][Cl 2 ] or K 4 >> K 5 [Cl 2 ] \u2234 r K K K K K K K K K COCl CO Cl CO 2 1 1 2 3 5 2 3 2 2 1 2 4 5 2 1 1 2 3 5 = + \u2248 / / / / [ ][Cl ] ( [ ]) [ ] ][Cl ] / / 2 3 2 2 1 2 4 K K
Answer: A
\nSolution: A overall = A A A A A 1 1 2 3 5 2 1 2 4 / / \u22c5 \u22c5 \u22c5
Answer: A
\nSolution: E E E E E E a a a a a a overall = + + \u2212 \u2212 1 2 1 2 1 3 5 2 4 Comprehension III For steady state of Br, + = d dt [Br] 0 or, 2 K 1 [Br 2 ] \u2013 K 2 [Br][H 2 ] + K 3 [H][Br 2 ] + K 4 [H][HBr] \u2013 2 K 5 [Br] 2 = 0 (1) For steady state of H, + = d dt [H] 0 or, K 2 [Br][H 2 ] \u2212 K 3 [H][Br 2 ] \u2013 K 4 [H][HBr] = 0 (2)
Answer: C
\nSolution: From (1) and (2), [ [ ] / Br] Br = \u239b \u239d \u239c \u239e \u23a0 \u239f K K 1 2 5 1 2
Answer: D
\nSolution: [ [Br][H ] [Br ] [HBr]) [Br ] [H ] / / / H] = + = \u22c5 \u22c5 \u22c5 K K K K K K 2 2 3 2 4 2 1 1 2 2 1 2 2 5 1 2 ( ( [ ] [ K K 3 2 4 Br HBr]) +
Answer: A
\nSolution: + d dt [HBr] = K 2 [Br][H 2 ] + K 3 [H][Br 2 ] \u2013 K 4 [H][HBr] = = + 2 2 3 2 3 2 1 1 2 2 3 2 2 5 1 2 3 2 4 K K K K K K K [ ] [Br ] [H ] ( [ ] [ / / / H][Br Br HBr])
Answer: B
\nSolution: At t = 0, [HBr] = 0 and hence, initial rate is given by, r K K K 0 2 1 1 2 2 1 2 2 5 1 2 2 = / / / [Br ] [H ]\n11.54 Chapter 11 HINTS AND EXPLANATIONS Comprehension IV
Answer: A
\nSolution: K t t V V t t = \u22c5 = \u22c5 1 1 0 0 ln ln [H O ] [H O ] 2 2 2 2 For t = 10 min, K 1 1 1 10 25 6 16 1 6 10 = \u22c5 = \u2212 ln . ln . min For t = 20 min, K 2 1 1 20 25 6 10 1 6 10 = \u22c5 = \u2212 ln . ln . min As K 1 = K 2 , order of reaction = 1
Answer: C
\nSolution: t K 1 2 2 2 1 6 10 15 / ln log log . min = = \u239b \u239d \u239c \u239e \u23a0 \u239f =
Answer: D
\nSolution: Kt a a x x a t = = \u2212 = \u2212 ln ln ln [H O ] [H O ] 2 2 2 2 0 1 1 or, ln . ln 1 6 10 25 1 1 \u00d7 = \u2212 x a \u21d2 x a = 11 16
Answer: A
\nSolution: Order = 1, but molecularity = 2 (as per reaction). Comprehension V C 8 H 18 O 2 ( g ) \u2192 2CH 3 COCH 3 ( g ) + C 2 H 6 ( g ) t = 0 800 torr 0 0 t = t (800 \u2013 x ) torr 2 x torr x torr
Answer: D
\nSolution: P t t C H O 8 18 2 torr torr = \u23af \u2192 \u23af\u23af\u23af = \u00d7 800 100 3 1 2 / \u2234 t = 3 \u00d7 80 = 240 min
Answer: B
\nSolution: P acetone = 2 x = 1200 \u21d2 x = 600 \u2234 P t t C H O 8 18 2 torr torr = \u23af \u2192 \u23af\u23af\u23af = \u00d7 800 200 2 1 2 / \u2234 t = 2 \u00d7 80 = 160 min
Answer: A
\nSolution: 800 \u2013 x = 700 \u21d2 x = 100 \u2234 P total = (800 \u2013 x ) + 2 x + x = 1000 torr Comprehension VI
Answer: C
\nSolution: From (1) and (2) data : order w.r.t OH = 1 From (2) and (3) data : order w.r.t H 2 S = 1 \u2234 r = K [H 2 S][OH]
Answer: C
\nSolution: K r = = \u00d7 \u00d7 \u00d7 \u00d7 \u2212 \u2212 \u2212 \u2212 [ . ( . ) ( . ) H S][OH] M s M M 2 1 4 10 2 1 10 1 3 10 6 1 8 8 = 5.1 \u00d7 10 9 M \u20131 s \u20131
Answer: D
\nSolution: r = K [H 2 S][OH] = 5.1 \u00d7 10 9 \u00d7 (1.0 \u00d7 10 \u20138 ) \u00d7 (1.7 \u00d7 10 \u20138 ) = 8.67 \u00d7 10 \u20137 M s \u20131
Answer: A
\nSolution: r = 8.67 \u00d7 10 \u20137 \u00d7 0.1 = 8.67 \u00d7 10 \u20138 mol s \u20131 Comprehension VII
Answer: C
\nSolution: K t P P x x = \u22c5 \u00b0 1 ln For t = 100 min, K 1 1 1 100 800 400 2 100 = \u22c5 = \u2212 ln ln min For t = 200 min, K 2 1 1 200 800 200 2 100 = \u22c5 = \u2212 ln ln min As K 1 = K 2 , order of reaction = 1
Answer: B
\nSolution: K = = \u00d7 \u2212 \u2212 ln . min 2 100 6 93 10 3 1 \u2234 K K rxn = = \u00d7 \u2212 \u2212 2 3 465 10 3 1 . min
Answer: B
\nSolution: Time for 87.5 % reaction = 3 3 2 6 93 10 1 2 3 \u00d7 = \u00d7 \u00d7 \u2212 t / ln . = 300 min
Answer: B
\nSolution: 2X( g ) 3Y( g ) + 2Z( g ) t = 0 800 0 0 t = t 800 \u2013 x 3 2 x x = 700 \u2234 P total = 800 + 3 2 x = 950 torr\n11.55 Chemical Kinetics HINTS AND EXPLANATIONS Comprehension VIII
Answer: B
\nSolution: A + 2B C + D t = 0 a M b M 0 0 t = t ( a \u2013 x ) M ( b \u2013 2 x ) M Now, r = K \u22c5 C B \u21d2 \u2212 = \u2212 d dt K b x [A] ( ) 2 \u21d2 dx dt K b x = \u2212 ( ) 2 or, dx b x K dt x t \u2212 = \u22c5 \u222b \u222b 2 0 0 \u21d2 x b e Kt = \u2212 \u2212 2 1 2 ( ) \u2234 C A = a \u2013 x = a b e Kt \u2212 \u2212 \u2212 2 1 2 ( )
Answer: B
\nSolution: For C a a a b e A Kt = = \u2212 \u2212 \u2212 2 2 2 1 2 , ( ) \u2234 ( ) ln / t K b b a A 1 2 1 2 = \u22c5 \u2212
Answer: D
\nSolution: For ( ) ( ) , [ ] [ ] / / t t A B a b A B 1 2 1 2 1 2 = = = Comprehension IX
Answer: B
\nSolution: r n dn dt K n rxn A A = \u2212 \u22c5 = \u22c5 1 1 \u21d2 n A = n A \u00b0 \u22c5 e \u2013 n , kt
Answer: B
\nSolution: n 1 A n 2 A t = 0 a mole 0 t = t ( a \u2013 x )mole n n x 2 1 \u22c5 mole = a \u22c5 e \u2013 n , kt \u2234 x = a (1 \u2013 e \u2013 n , kt ) Now, V V n n a x n n x a 2 1 2 1 = = \u2212 + \u22c5 final initial ( ) or, V V a x n n a n n e n kt 2 0 2 1 2 1 1 1 1 1 = + \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f = + \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f \u22c5 \u2212 \u2212 ( ) , \u2234 V V n n n n e n kt 2 0 2 1 2 1 1 = + \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f \u23a1 \u23a3 \u23a2 \u23a4 \u23a6 \u23a5 \u2212 ,
Answer: D
\nSolution: If n 1 = 1, n 2 = 2, then V 2 = V 0 (2 \u2013 e \u2013 kt ) Now, [ ] ( ) [ ] A n V n e V e A e e A A kt kt kt kt = = \u00b0 \u22c5 \u22c5 \u2212 = \u22c5 \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f \u2212 \u2212 \u2212 \u2212 2 0 0 2 2 Comprehension X
Answer: B
\nSolution: df f K dt f t 1 0 0 \u2212 = \u22c5 \u222b \u222b \u21d2 t f K = \u2212 \u2212 ln( ) 1 Now, K = \u2212 \u2212 = \u2212 ( ) 3 200 3 200 1 hr \u2234 t K 1 2 2 0 693 3 200 46 2 / ln . . = = \u239b \u239d \u239c \u239e \u23a0 \u239f = hr
Answer: D
\nSolution: t f K = \u2212 \u2212 ln( ) 1 \u21d2 f = 1 \u2013 e \u2013 Kt = 1 \u2013 e \u20133 t /200 Comprehension XI
Answer: B
\nSolution: Unit of K = s \u20131 \u21d2 order = 1
Answer: C
\nSolution: K K B = \u00d7 = \u00d7 \u00d7 = \u00d7 \u2212 \u2212 \u2212 10 100 10 100 1 5 10 1 5 10 4 5 1 . . s\n11.56 Chapter 11 HINTS AND EXPLANATIONS Comprehension XII
Answer: A
\nSolution: [ ] [ ] ( . ) . M . A A e e K t = \u22c5 = \u00d7 = \u2212 \u2212 \u00d7 0 0 04 25 1 1 0 0 368 M
Answer: B
\nSolution: [ ] [ ] ( ) B K A K K e e K t K t = \u2212 \u2212 \u2212 \u2212 1 0 2 1 1 2 = \u00d7 \u2212 \u2212 \u2212 \u00d7 \u2212 \u00d7 0 04 1 0 0 06 0 04 0 04 25 0 06 25 . ( . . . ( ) . . M) e e = 0.29 M
Answer: C
\nSolution: [ C ] = [ A 0 ] \u2013 [ A ] \u2013 [ B ] = 1.0 \u2013 0.368 \u2013 0.29 = 0.342 M
Answer: A
\nSolution: t K K K K max ln ln . . min = \u239b \u239d \u239c \u239e \u23a0 \u239f \u2212 = \u239b \u239d \u239c \u239e \u23a0 \u239f \u2212 = 2 1 2 1 3 2 0 06 0 04 20
Answer: C
\nSolution: [ ] [ ] ( . ) max . . . B A K K K K K = \u239b \u239d \u239c \u239e \u23a0 \u239f = \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f \u2212 \u2212 \u2212 \u2212 0 2 1 0 06 0 06 0 0 2 2 1 1 0 3 2 M 4 4 = 0.3 M
Answer: C
\nSolution: [ ] [ ] [ ] [ ] A B C A = = = 0 3 Now, t K A A K = \u22c5 = = = 1 3 1 1 0 04 27 5 1 0 1 ln [ ] [ ] ln . . . min
Answer: A
\nSolution: \u2212 = + d A dt d C dt [ ] [ ] \u21d2 K 1 [ A ] = K 2 [ B ] \u21d2 t = 20 min \u2234 [ ] [ ] ( . M) e . . A A e K t = \u22c5 = \u22c5 = \u2212 \u2212 \u00d7 0 0 04 20 1 1 0 0 45M
Answer: B
\nSolution: ( r C ) max = K 2 [ B ] max = 0.06 \u00d7 0.3 = 1.8 \u00d7 10 \u20132 M/ min Comprehension XIII A B t = 0 0.15 M 0 t = 10 (0.15 \u2013 x ) M x M = 0.125 M = 0.025 M t = t eq (0.15 \u2013 x eq ) M x eq M = 0.10 M = 0.05 M Now, K K K eq f b = = = 0 05 0 10 1 2 . . (1) and t K K f b 1 2 2 / ln = + \u21d2 10 0 693 min . = + K K f b (2)
Answer: C
\nSolution: From (1) and (2), K f = 2.31 \u00d7 10 \u20132 min \u20131
Answer: D
\nSolution: From (1), K b = 4.62 \u00d7 10 \u20132 min \u20131
Answer: B
\nSolution: K eq = 0.5
Answer: C
\nSolution: t 1/2 = 10 min Comprehension XIV
Answer: B
\nSolution: E K E K E K K a a a ( ) overall = \u22c5 + \u22c5 + 1 2 1 2 1 1 or, 10 5 12 9 1 2 1 2 . = \u00d7 + \u00d7 + K K K K \u21d2 K 1 = K 2 or, A e A e E RT E RT a a 1 2 1 2 \u22c5 = \u22c5 \u2212 \u2212 / / or, E E RT A A a a 1 2 1 2 \u2212 = ln \u21d2 ( ) ln 12 9 10 2 2 10 2 10 3 14 14 2 \u2212 \u00d7 \u00d7 = \u00d7 \u00d7 \u00d7 \u2212 T e \u2234 T = 750 K
Answer: C
\nSolution: Above 750 K, Y will be the major product and below 750 K, Z will be the major product as E E a a 1 2 > .
Answer: B
\nSolution: Reactions with higher E a are more sensitive towards temperature change.\n11.57 Chemical Kinetics HINTS AND EXPLANATIONS Comprehension XV Energy (kcal/mol) 27.5 19.9 30.1 16.9 67.9 10.2 2.4 4.3 13.2 7.6 42.8 A B C D Reaction coordinates
Answer: A
\nSolution: C \u2192 D [ : ] . . . E a A B C D 27 5 30 1 4 3 \u23af \u2192 \u23af\u23af \u23af \u2192 \u23af\u23af \u23af \u2192 \u23af
Answer: B
\nSolution: C \u2192 B [ : D A] . . . E a 67 9 16 9 19 9 \u23af \u2192 \u23af\u23af \u23af \u2192 \u23af\u23af \u23af \u2192 \u23af\u23af C B
Answer: A
\nSolution: C \u2192 D [Lowest E a ]
Answer: B
\nSolution: B \u2192 C [Highest E a ]
Answer: A
\nSolution: D \u2192 C [Highest E a ]
Answer: C
\nSolution: Molecularity can never be fractional.
Answer: C
\nSolution: t A K n n 100 0 1 1 % [ ] ( ) = \u2212 \u2212 when n < 1 = Infi nite when n \u2265 1
Answer: D
\nSolution: For a particular step, rates always increase with increase in temperature.
Answer: A
\nSolution: Relative increase in rate constant with increase in temperature is higher for the reaction with higher activation energy.
Answer: C
\nSolution: \u0394 H = E E a a f b \u2212
Answer: A
\nSolution: Theoretical
Answer: C
\nSolution: For zero order reaction : t A K t A K 1 2 0 100 0 2 / % [ ] , [ ] = =
Answer: D
\nSolution: Order is in dependent from stoichiometry of reaction.
Answer: A
\nSolution: t A K 1 2 0 2 / [ ] =
Answer: A
\nSolution: Theoretical
Answer: A \u2192 P, Q; B \u2192 R, S; C \u2192 P, Q; D \u2192 R, S
\nSolution: Informative
Answer: A \u2192 Q; B \u2192 P; C \u2192 P; D \u2192 R, S
\nSolution: Informative
Answer: A \u2192 Q; B \u2192 P, R; C \u2192 P; D \u2192 R, S
\nSolution: Theoretical
Answer: A \u2192 R; B \u2192 P; C \u2192 Q, S; D \u2192 Q, S
\nSolution: Theoretical
Answer: A \u2192 P; B \u2192 S; C \u2192 R; D \u2192 Q
\nSolution: Theoretical
Answer: A \u2192 R, U; B \u2192 P, X; C \u2192 S, V; D \u2192 Q, W
\nSolution: Theoretical
Answer: A \u2192 Q; B \u2192 P, R; C \u2192 S; D \u2192 T
\nSolution: Theoretical
Answer: A \u2192 P; B \u2192 Q, S; C \u2192 R, T; D \u2192 U
\nSolution: Theoretical
Answer: A \u2192 S; B \u2192 P; C \u2192 R, T; D \u2192 Q
\nSolution: (P) 2 a a t 1/3 t 19/27 = 54 sec 3 = 18 sec t 1/3 = 18 sec t 1/3 = 18 sec 4 a 9 8 a 27 (Q) 3 a a t 1/4 t 7/16 = 32 sec 4 = 16 sec t 1/4 = 16 sec 9 a 16 (R) K a a a x a = \u2212 \u23a1 \u23a3 \u23a2 \u23a4 \u23a6 \u23a5 = \u2212 \u2212 \u23a1 \u23a3 \u23a2 \u23a4 \u23a6 \u23a5 1 4 1 2 3 1 1 56 1 1 / \u21d2 x a = 7 8\n11.58 Chapter 11 HINTS AND EXPLANATIONS (S) K a a x = \u2212 = 2 3 18 30 \u21d2 x a = 5 9 (T) K a a x = \u2212 = 2 16 28 \u21d2 x a = 7 8
Answer: A \u2192 P, Q; B \u2192 P, R; C \u2192 S
\nSolution: (A) d C dt K B [ ] [ ] = 2 For d C dt [ ] max \u239b \u239d \u239c \u239e \u23a0 \u239f , [ B ] should be maximum and hence t K K K K K = \u239b \u239d \u239c \u239e \u23a0 \u239f \u2212 = ln ln 2 1 2 1 1 2 (when K 2 = 2 K 1 ) Now, ( ) ln / t K A 1 2 1 2 = (B) Rate of formation of B is maximum at t = 0, at which [ B ] = [ C ] = 0 Now, [ B ] = [ C ] K A K K e e A K e K e K K K t K t K t K t 1 0 2 1 0 2 1 2 1 1 2 1 2 1 [ ] ( ) [ ] \u2212 \u2212 = \u2212 \u22c5 \u2212 \u22c5 \u2212 \u23a1 \u23a3 \u23a2 \u23a4 \u23a6 \u2212 \u2212 \u2212 \u2212 \u23a5 \u23a5 or, e e K e K e K K t K t K t K t \u2212 \u2212 \u2212 \u2212 \u2212 = \u2212 \u22c5 \u2212 \u22c5 1 1 1 1 2 1 1 2 1 1 2 (when K 2 = 2 K 1 ) or, K e K e K K e K e K t K t K t K t 1 1 2 1 1 1 2 1 1 1 1 2 \u22c5 \u2212 \u22c5 = \u2212 \u22c5 \u2212 \u22c5 \u2212 \u2212 \u2212 \u2212 \u2234 t K = ln 2 1 (C) [A] = [B] [ ] [ ] ( ) A e K A K K e e K t K t K t 0 1 0 2 1 1 1 2 \u22c5 = \u2212 \u2212 \u2212 \u2212 \u2212 K K K e K K t 2 1 1 1 1 2 \u2212 = \u2212 \u2212 ( ) \u2234 t K K K K K = \u2212 \u22c5 \u2212 1 2 1 2 1 2 1 ln
Answer: 3
\nSolution: ii, iv, v
Answer: 2
\nSolution: Theoretical
Answer: 6
\nSolution: K K K BrO BrO Br \u2212 \u2212 \u2212 = = 3 1 2 3 \u2234 K a BrO M s 3 0 06 3 0 02 1 1 \u2212 = = = \u2212 \u2212 . . and K b Br M s \u2212 = = \u00d7 = \u2212 \u2212 2 3 0 06 0 04 1 1 . .
Answer: 5
\nSolution: 7 2 10 3600 2 10 15 1 8 2 . ( \u00d7 = \u00d7 \u00d7 \u2212 \u2212 \u2212 M s M) K K = \u2212 \u2212 1 200 1 1 M s = 5 ml mol \u20131 s \u20131
Answer: 3
\nSolution: \u2212 = \u22c5 \u22c5 dP dt K P P a b NO H 2 1 5 0 25 372 152 . . = \u239b \u239d \u239c \u239e \u23a0 \u239f a \u21d2 a = 2 and 1 60 0 79 289 144 . . = \u239b \u239d \u239c \u239e \u23a0 \u239f b \u21d2 b = 1
Answer: 1
\nSolution: 0 1 0 4 0 20 . . . = x \u21d2 x = 0.8 0 1 0 8 0 2 0 05 0 4 . . . . . = \u00d7 \u00d7 y \u21d2 y = 0.2
Answer: 5
\nSolution: t K = \u22c5 + + 1 3 0 3 ln [ ] [ ] Cr Cr = \u00d7 \u22c5 \u2212 \u2212 \u2212 1 9 10 100 100 80 5 1 s ln = 1.8 \u00d7 10 4 sec = 5 hrs
Answer: 3
\nSolution: (i) Addition of NaOH will decrease [H 3 O + ]. (ii) Addition of water will decrease the concentration of both. (iii) Acetic acid is a weak acid and hence, [H 3 O + ] will decrease. (iv) Increase in temperature increases the reaction rate.\n11.59 Chemical Kinetics HINTS AND EXPLANATIONS
Answer: 3
\nSolution: Time for certain progress of reaction, t a [ A 0 ] 1 \u2013 n 1 10 0 25 10 0 02 0 04 3 3 1 \u00d7 \u00d7 = \u239b \u239d \u239c \u239e \u23a0 \u239f \u2212 \u2212 \u2212 . . . n \u21d2 n = 3
Answer: 2
\nSolution: C 4 H 8 2C 2 H 4 t = 0 a mole 0 t = t ( a \u2013 x ) mole 2 x mole As, a \u2013 x = 2 x \u21d2 x a = 3 Now, t t K a a x a a a = \u22c5 \u2212 = \u00d7 \u00d7 \u2212 = \u2212 \u2212 ln ln 1 25 18 10 3 2 5 1 s hrs
Answer: 8
\nSolution: For 2 % reaction, we may assume that rate is almost constant. r = K [ A ] \u21d2 2 100 1 \u00d7 \u2212 [ ] min A = K [ A ] \u21d2 K = 0.02 min \u20131
Answer: 3
\nSolution: H 2 O 2 ( aq ) H 2 O( l ) + 1 2 O 2 ( g ) \u0394 H = (\u2013287) \u2013 (\u2013 187) = \u2013100 KJ/mol Moles of H 2 O 2 reacted per sec = 7.5 \u00d7 10 \u20134 \u00d7 0.02 \u00d7 2 = 3 \u00d7 10 \u20135 \u2234 Heat produced per sec = 3 \u00d7 10 \u20135 \u00d7 (100 \u00d7 10 3 ) = 3 J
Answer: 2
\nSolution: t K a a x = \u22c5 \u2212 = \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f \u22c5 \u2212 \u2212 1 1 4 5 3 1536 10 100 40 8 1 ln . . ln s = \u00d7 \u00d7 \u00d7 \u00d7 = 0 9 3 1536 10 4 5 3 1536 10 2 8 7 . . . . Year Years
Answer: 5
\nSolution: t 1 2 4 0 693 6 93 10 1000 / . . sec = \u00d7 = \u2212 A n B t = 0 a mole 0 t = 1000 sec a 2 mole n a \u22c5 2 mole Now, a n a a 2 2 3 + \u22c5 = \u21d2 n = 5
Answer: 2
\nSolution: r = K [ester][H + ] x = k 1 [ester] K 1 = K \u22c5 [H + ] x 1 0 10 10 10 3 2 . = \u239b \u239d \u239c \u239e \u23a0 \u239f \u2212 \u2212 x \u21d2 x = 1 and K K y 1 1 3 3 1 0 10 10 1 = = \u00d7 = = + \u2212 \u2212 [ ] . H
Answer: 2
\nSolution: t 3/4 = 2 \u00d7 t 1/2 and hence, a =
Answer: 1
\nSolution: Now, t K K b 1 2 2 2 / ln ln [ ] = = + H 1 0 0 5 0 02 0 01 . . . . = \u239b \u239d \u239c \u239e \u23a0 \u239f b \u21d2 b = 1
Answer: 4
\nSolution: Concentrations are in G.P. and hence, order =
Answer: 8
\nSolution: 18. A 2 B 3 ( aq ) 2A 3+ ( aq ) + 3 B 2\u2013 ( aq ) t = 0 a mole 0 0 t = 10 min a \u2013 x 2 x 3 x Now, p = CRT = r gh \u21d2 total mole a h \u2234 a a x + = 4 2 6 \u21d2 t = 10 min = t 1/2 Now, at t = t 3/4 = 2 \u00d7 t 1/2 = 20 min, x a = 3 4 \u2234 a a a h + \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f = 4 3 4 2 \u21d2 h = 8 mm p = x = r gh = 1 0 1000 0 8 3 2 . ( . gm cm cm s cm) \u239b \u239d \u239c \u239e \u23a0 \u239f \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f \u00d7 = = 800 80 2 dyne cm pascal Now, x y = = 80 20 4
Answer: 0
\nSolution: [ ] A t 4 1 1 = + \u21d2 4 1 1 3 2 8 [ ] [ ] ( ) [ ] A d A dt t A \u22c5 = \u2212 + = \u2212 \u2234 \u2212 = = = \u00d7 \u2212 \u2212 d A dt A [ ] [ ] ( . ) 5 5 5 1 4 0 2 4 8 10 M s
Answer: 8
\nSolution: A 2B t = 0 a mole 0 t = t ( a \u2013 x ) mole 2 x mole From mass conservation, a \u00d7 M 0 = ( a + x ) \u00d7 M t \u2234 x a M M M t t = \u2212 ( ) 0 If the reaction is zero order, then K a a x t x t a M M t M t t = \u2212 \u2212 = = \u2212 \u22c5 ( ) ( ) 0\n11.60 Chapter 11 HINTS AND EXPLANATIONS For t = 10 min, K a a = \u2212 \u00d7 = ( ) 42 35 10 35 50 For t = 20 min, K a a = \u2212 \u00d7 = ( ) 42 30 20 30 50 As K values are same, the reaction is of zero-order.
Answer: 2
\nSolution: r = K [ A ] n and 2 r = K (4[ A ]) n \u21d2 n = 1 2 \u2234 t 1/2 a [ A 0 ] 1 \u2013 n = [ A 0 ] 1/2 Next t 1/2 will be 1 2 times of previous one and hence, t = = 8 2 2 8 hr.
Answer: 1
\nSolution: P K P K K K e B B A A B C K K K t A B C = \u22c5 \u00b0 + + \u22c5 \u2212 \u2212 + + \u22c5 [ ] ( ) 1 = \u00d7 \u00d7 \u00d7 \u2212 = \u2212 \u2212 \u2212 \u00d7 \u00d7 \u2212 2 10 13 86 6 93 10 1 2 3 3 6 93 10 100 3 . . [ ] . e atm
Answer: 7
\nSolution: K K A e A e A A e E RT E RT E E RT a a a a I II I II I II I II I II = \u22c5 \u22c5 = \u22c5 \u2212 \u2212 \u2212 \u2212 / / ( )/ = = \u00d7 = \u2212 \u00d7 \u00d7 100 1 4 606 10 2 500 3 e . /
Answer: 4
\nSolution: Fraction of molecules having sufficient energy = = = \u00d7 \u2212 \u2212 \u00d7 \u00d7 \u2212 e e E RT a / . / . 83 14 10 8 314 500 9 3 2 10
Answer: 7200
\nSolution: ln ln K K t t E R T T a 2 1 1 2 1 2 1 1 = = \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f or, ln 1 3 1 300 1 280 = \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f E R a (1) and ln 16 1 300 1 330 t E R a = \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f (2) From (1) and (2), t = 4 hrs Four Digit Integer Type
Answer: 0025
\nSolution: r = K [O 3 ] 2 = 5 \u00d7 10 \u20134 \u00d7 (2 \u00d7 10 \u20138 ) 2 = 2 \u00d7 10 \u201319 mol l \u20131 s \u20131 = 2 \u00d7 10 \u201319 \u00d7 6 \u00d7 10 23 \u00d7 10 \u20133 \u00d7 60 = 7200 molecules ml \u20131 min \u20131
Answer: 0060
\nSolution: At t = \u221e , P total should be 400 mm, but as it is only 390 mm, some unreactive gas should also be present in the vessel. Let P P \u00b0 = C H Br 2 5 0 mm then P unreactive gas = (200 \u2013 P 0 ) mm. C 2 H 5 Br(g) C 2 H 4 (g) + HBr(g) t = 0 P 0 0 0 t = t P 0 \u2013 x x x t = \u221e 0 P 0 P 0 From question, P 0 + P 0 + (200 \u2013 P 0 ) = 390 \u21d2 P 0 = 190 and ( P 0 \u2013 x ) + x + x + (200 \u2013 P 0 ) = 342.5 \u21d2 x = 142.5 \u2234 Percentage C 2 H 5 Br undecomposed = P x P 0 0 25 \u2212 = %
Answer: 1250
\nSolution: t t A A = \u22c5 gen log log [ ] [ ] 2 0 \u21d2 96 0 30 3 = \u22c5 t gen . log \u21d2 t gen = 60 hrs
Answer: 1000
\nSolution: r dP dt K P P = \u2212 = \u2032 \u22c5 \u22c5 NO NO O 2 2 and \u2032 = \u00d7 \u00d7 \u2212 \u2212 K 1 6 10 0 08 600 5 2 2 1 . ( . ) atm s \u2234 r = \u00d7 \u00d7 \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f 1 6 10 48 48 190 760 288 760 5 2 . = = \u2212 \u2212 1250 760 1250 1 1 atm s mm s
Answer: 0200
\nSolution: From the unit of rate constant, the process is zero order. \u2234 t K 100 0 % [ ] = + H = \u00d7 \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f \u00d7 = \u00d7 = \u2212 \u2212 3 10 0 05 1000 1 0 10 6 10 1000 7 7 4 . . sec min
Answer: 0040
\nSolution: K t A A = = 1 0 ln [ ] [ ] Constant \u2234 1 100 3 1 9 0 0 0 0 \u22c5 = \u22c5 ln [ ] [ ] / ln [ ] [ ] / A A t A A \u21d2 t = 200 min
Answer: 0120
\nSolution: K t A A = \u22c5 = 1 0 ln [ ] [ ] Constant\n11.61 Chemical Kinetics HINTS AND EXPLANATIONS \u2234 1 20 500 420 1 100 70 \u22c5 = \u22c5 ln ln t \u21d2 t = 40 min
Answer: 0535
\nSolution: K t V V V t = \u22c5 \u2212 = \u221e \u221e 1 ln Constant \u2234 1 40 80 80 40 1 80 80 70 \u22c5 \u2212 = \u22c5 \u2212 ln ln t \u21d2 t = 120 min
Answer: 0123
\nSolution: 2P 4Q + R + S(l) t = 0 P 0 0 0 t = 30 min P 0 \u2013 x 2 x x 2 V.P. = 25 t = 60 min P 0 \u2013 y 2 y y 2 V.P. = 25 t = \u221e 0 2 P 0 P 0 2 V.P. = 25 From question, 2 2 25 625 0 0 P P + + = \u21d2 P 0 = 240 and ( ) P x x x 0 2 2 25 445 \u2212 + + + = \u21d2 x = 120 Now, 1 30 1 60 0 0 0 0 \u22c5 \u2212 = \u22c5 \u2212 ln ln P P x P P y \u21d2 y = 180 \u2234 P P y y y 60 0 2 2 25 535 = \u2212 + + + = ( ) mm
Answer: 0375
\nSolution: Initial moles of NH 4 NO 2 = 200 0 02 1000 0 004 \u00d7 = . . and moles of N 2 O formed = ( ) . . . 785 25 760 49 26 1000 0 0821 300 0 002 \u2212 \u00d7 \u00d7 = \u2234 t req = t 1/2 = 123 min
Answer: 0011
\nSolution: For set 1 and 2, r = K \u2032 [ B ] as [ A 0 ] >> [ B 0 ] and t K K A 1 2 0 2 2 2 / ln ln [ ] = \u2032 = \u21d2 x = 62.5 For set 3 and 4, r = K \u2033 [ A ] 2 as [ B 0 ] >> [ A 0 ] and t K A K B A 1 2 0 0 0 1 1 / [ ] [ ][ ] = \u2032\u2032 = \u21d2 y = = 625 2 312 5 . \u2234 x + y = 62.5 + 312.5 = 375
Answer: 1680
\nSolution: t = 43.5 min = 3 t 1/2 Hence, P ether atm = = 4 2 0 5 3 . \u21d2 \u0394 P ether = 3.5 atm \u2234 P fi nal = 0.5 + 3.5 \u00d7 3 = 11 atm
Answer: 4003
\nSolution: \u0394 = \u22c5 t t r r 1 2 1 2 2 / ln ln \u21d2 12 2 0 04 0 03 1 2 = \u22c5 \u239b \u239d \u239c \u239e \u23a0 \u239f t / ln ln . . \u2234 t 1/2 = 28 min = 1680 sec
Answer: 0025
\nSolution: t t V V V V t = \u22c5 \u2212 \u2212 \u221e \u221e 1 2 0 2 / ln ln \u21d2 120 2 60 20 60 55 1 2 = \u22c5 \u2212 \u2212 t / ln ln \u2234 t 1/2 = 40 min ab = 40 Now, [ [ester] HCl] = \u2212 \u221e V V V 0 0 \u21d2 [ . HCl] 6 0 20 60 20 = \u2212 \u21d2 [HCl] = 3.0 M \u2234 cd = 03
Answer: 0500
\nSolution: t 1 2 3 0 693 1 386 10 500 / . . sec = \u00d7 = \u2212 Let the initial moles of A = x , then after 500 sec, A 2B + C x x \u2212 2 2 2 \u00d7 x x 2 = x 2 = x = x 2 Total moles becomes x x x x 2 2 2 + + \u239b \u239d \u239c \u239e \u23a0 \u239f = . As moles becomes double, volume becomes double and hence, [ ] . . A req M = \u00d7 = 0 1 2 2 0 025 = 25 millimole per litre
Answer: 0260
\nSolution: A 2B + C t = 0 4 a 0 3 a t = t 4 a \u2013 x 2 x 3 a + x From question (4 a \u2013 x )(40\u00b0) + 2 x (10\u00b0) + (3 a + x ) (\u201330\u00b0) = 0\u00b0 \u2234 x a = 7 5 Now, t K a a a = \u22c5 \u2212 = \u22c5 = 1 4 4 7 5 1 0 001 20 13 500 ln . ln min\n11.62 Chapter 11 HINTS AND EXPLANATIONS
Answer: 0060
\nSolution: Exp (1): r = K \u2032 [ B ] y as [ A 0 ] >> [ B 0 ] \u2235 t t 7 8 1 2 3 / / = \u00d7 \u21d2 y = 1 Exp (2): r = K \u2033 [ A ] x as [ A 0 ] << [ B 0 ] \u2235 t t 7 8 1 2 7 / / = \u00d7 \u21d2 x = 2 Now, for exp (2) and (3), t K A K B A 1 2 0 0 0 1 1 / [ ] [ ][ ] = \u2032\u2032 = \u2234 a = \u00d7 = 10 2 2 2 5 . and b = 7 \u00d7 2.5 = 17.5 And for exp (1) and (4), t K K A 1 2 0 2 2 / ln ln [ ] = \u2032 = \u2234 c = 30 \u00d7 2 = 60 and d = 3 \u00d7 60 = 180
Answer: 1595
\nSolution: Percentage yield = K K K 2 1 2 100 4 8 3 2 4 8 100 60 + \u00d7 = + \u00d7 = . . . %
Answer: 1475
\nSolution: A + 2B + 3C D t = 0 1.0 M 1.0 M 1.0 M 0 t = t 1 \u2013 x 1 \u2013 2 x 1 \u2013 3 x x = 0.9 = 0.8 = 0.7 = 0.1 (given) \u2234 r = 2 \u00d7 10 \u20136 \u00d7 (0.9) 2 \u2013 1 4 10 0 1 0 8 0 7 1 595 10 6 2 6 . ( . ) . . . \u00d7 \u00d7 \u00d7 = \u00d7 \u2212 \u2212
Answer: 0100
\nSolution: [ C ] = 0.875 + 0.6 = 1.475 M
Answer: 0055
\nSolution: K K K B A eq f b = = [ ] [ ] \u21d2 1 38 300 0 1 0 2 . / . . K b = \u21d2 K b = \u2212 1 38 150 1 . min Now, t K K x x x f b e B e B B = + \u22c5 \u2212 = + \u22c5 \u2212 \u00d7 1 1 1 38 300 2 76 300 0 1 0 1 0 3 25 100 ln . . ln . . . , , = \u00d7 \u22c5 = 300 6 2 4 100 ln ln min
Answer: 0727
\nSolution: For completion in 30 min, the rate should be increased by 4 60 30 8 \u00d7 = times. Assuming temperature coefficient constant, the approximate temperature is 25 10 8 2 55 + \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f = \u00b0 C .
Answer: 0500
\nSolution: K 1 = K 2 \u21d2 A e A e E RT E RT a a 1 2 1 2 \u22c5 = \u22c5 \u2212 \u2212 / / \u2234 ln A A E E RT a a 2 1 2 1 = \u2212 ln ( . . ) . 10 10 171 39 152 30 10 8 3 14 13 3 = \u2212 \u00d7 \u00d7 T or, T = 1000 K = 727\u00b0 C
Answer: 0100
\nSolution: t K A e E RT a 1 2 2 2 / / ln ln = = \u22c5 \u2212 or, 1 60 0 7 5 10 13 149 4 10 8 3 3 \u00d7 = \u00d7 \u00d7 \u2212 \u00d7 \u00d7 . . / . e T \u2234 T = 500 K
Answer: C
\nSolution: (I) Cu cannot reduce Pb (II) Pb can reduce Ag (III) Ag cannot reduce Cu. Hence, reducing power: Pb > Cu > Ag
Answer: D
\nSolution: E E n = \u00b0 \u2212 \u22c5 0 06 . log [R] [O] \u21d2 0 24 0 36 0 06 1 . . . log [ [ = \u2212 R] O] \u2234 [ [ O] R] = 1 100
Answer: B
\nSolution: For the complex ion to get oxidised, its reduction potential should be low.
Answer: B
\nSolution: Ag NH Ag(NH) E + + + \u00b0 = \u2212 = 2 0 79 0 37 0 42 3 3 \u001e \u21c0 \u001e \u21bd \u001e \u001e ; . . . V Now, E n K eq \u00b0 \u2212 \u22c5 0 06 . log \u21d2 0 42 0 06 1 . . log = \u22c5 K f \u21d2 K f = 10 7
Answer: A
\nSolution: Given: Co 3+ + e \u2013 Co 2+ ; E\u00b0 = 1.81 V; \u0394 \u00b0 = \u2212 \u00d7 \u00d7 G F 1.81 1 1 Co(CN) e Co(CN) 6 3 6 4 + \u2212 \u2212 + \u23af \u2192 \u23af E\u00b0 = \u20130.83 V; \u0394 \u00b0 = \u2212 \u00d7 \u00d7 \u2212 G F 2 1 0 83 ( . ) Co 6CN Co(CN) 2 6 4 + \u2212 \u2212 + \u23af \u2192 \u23af ; K f = 10 19 ; \u0394 \u00b0 = \u2212 \u00d7 G RT ln10 3 19 Required Co CN Co(CN) 3 6 3 6 + \u2212 \u2212 + \u23af \u2192 \u23af K f = ?; \u0394 \u00b0 = \u2212 \u00d7 G RT lnK f Now, \u0394 \u00b0 = \u0394 \u00b0 \u2212 \u0394 \u00b0 + \u0394 \u00b0 G G G G 1 2 3 or, \u2212 \u22c5 = \u2212 \u2212 + \u2212 RT lnK F) F RTln10 f ( . . ( ) 1 81 0 83 19 or, RT ln K F \u22c5 = \u2212 10 2 64 19 f . \u21d2 log . . . . 10 2 64 2 303 2 64 0 06 19 K F RT f = \u2212 = \u2212 \u2234 K f = 10 63
Answer: A
\nSolution: E E Cu E Cu Cu Cu Cu Cu Cu Cu 2 2 2 0 06 2 1 0 03 2 2 + + + = \u00b0 \u2212 \u22c5 = \u00b0 + + + | | | . log [ ] . log[ ] ] = 0.34 + 0.03 \u00d7 log(0.1) = 0.31 V \u2234 E Cu/Cu 2+ = \u2212 0.31 V
Answer: D
\nSolution: E E RT F Cd Ag cell cell = \u00b0 \u2212 \u22c5 + + 2 2 2 ln [ ] [ ] As CN\u2013 will form complex with Ag+ ion in the cathodic compartment, E cell will decrease.
Answer: D
\nSolution: \u2212\u0394 = \u22c5 = \u00b0 \u2212 \u22c5 = \u00b0 \u2212 + + G nF E nF E RT nF Zn Cu nFE cell cell Cell Cell [ ln [ ] [ ] 2 2 RT T C C \u22c5 ln 1 2
Answer: B
\nSolution: E E RT F K X Ag X|Ag Ag |Ag sp \u00b0 = \u00b0 \u2212 \u22c5 \u2212 + | ln 1
Answer: C
\nSolution: Cell reaction: H + (cathode, P H = 3) H + (anode, P H = ?) E H anode H P P cell anode H catho = \u2212 \u22c5 = \u2212 + + 0 0 059 1 0 059 . log [ ] [ ] cathode . d de H \u23a1 \u23a3 \u23a4 \u23a6 or, 0 272 0 059 3 . . [P ] = \u2212 anode H \u21d2 P anode H = 7.6
Answer: C
\nSolution: Cl 2 + H 2 O \u001f Cl \u2013 + ClO \u2013 + 2H + ; E V cell \u00b0 = \u2212 = \u2212 1 36 1 63 0 27 . . . Now, E E H cell cell = \u00b0 \u2212 + 0 06 1 2 . log[ ] or, 0 0 27 0 06 1 2 2 25 = \u2212 + \u00d7= . . . P H
Answer: A
\nSolution: E E E cell quinohydrone calomel = \u2212 0.210 = E quinohydrone \u2013 0.279 \u21d2 E quinohydrone = 0.489 V EXERCISE II (JEE ADVANCED)\n8.41 Electrochemistry HINTS AND EXPLANATIONS Quinohydrone electrode is + 2H + + 2e \u2013 (Quinone, Q) (Hydroquinone, H2Q) O O OH OH E E H E P H = \u00b0 \u2212 \u22c5 = \u00b0 \u2212 + 0 06 2 1 0 06 . log [ ] . or, 0.489 = 0.699 \u2013 0.06.P H \u21d2 P H = 3.5
Answer: A
\nSolution: Anode: Ag(s) Ag + (aq) + e \u2013 1 \u00d7 6 Cathode: Cr O aq H aq Cr aq H O(l) 2 7 2 3 2 14 6 2 7 \u2212 + \u2212 + + + \u23af \u2192 \u23af + ( ) ( ) e ( ) Net: 6 14 6 2 7 2 3 2 Ag(s) Cr O aq H aq Ag aq) 2Cr aq)+7H O(l) + + \u23af \u2192 \u23af + \u2212 + + + ( ) ( ) ( ( E E n Ag Cr Cr O H cell cell = \u00b0 \u2212 \u22c5 + + \u2212 + 0 06 6 3 2 2 7 2 14 . log [ ] [ ] [ ][ ] = (1.33 \u2013 0.80) \u2013 0 06 6 0 1 0 4 1 6 0 1 0 46 6 2 14 . log ( . ) ( . ) . ( . ) . \u22c5 \u00d7 \u00d7 = V
Answer: B
\nSolution: Net cell reaction: Zn \u2013 Hg(C 1 M) Zn \u2013 Hg(C 2 M) E C C V cell = \u2212 = \u2212 = 0 0 059 2 0 059 2 1 10 0 0295 2 1 . log . log .
Answer: B
\nSolution: E K cell eq \u00b0 = \u22c5 0 06 2 . log \u21d2 0 75 1 50 3 1 68 1 3 1 0 03 . . . . log \u2212 \u00d7 \u2212 \u00d7 \u2212 = K eq \u2234 K eq = 10 \u201322
Answer: B
\nSolution: [ ] . . H Ka C C M left 1 + \u2212 = \u22c5 = \u00d7 \u00d7 = 1 8 10 0 1 5 [ ] . . H K Kb C C M right w + \u2212 \u2212 = \u22c5 = \u00d7 \u00d7 = 10 1 8 10 0 01 14 5 2 Net cell reaction, assuming as concentration cell: H + (C 2 M) H + (C 1 M) E C C V cell = \u2212 \u22c5 = \u2212 0 0 06 1 0 465 1 2 . log .
Answer: A
\nSolution: E V V V \u00b0 = \u00d7 + \u00d7 \u2212 \u00d7 = + + 2 3 1 0 616 1 0 439 1 0 799 0 256 | . . . . \u2234 E V V V \u00b0 = \u2212 + + 3 2 0 256 | .
Answer: B
\nSolution: E E n H P cell cell H = \u00b0 \u2212 \u22c5 + 0 06 2 2 . log [ ] or, 0 70 0 28 0 0 06 2 1 2 . ( . ) . log [ ] = \u2212 \u2212 \u22c5 + H \u21d2 P H = 7.0
Answer: D
\nSolution: Net cell reaction: Ag C M) Ag C K M sp + + = \u23af \u2192 \u23af = \u239b \u239d \u239c \u239e \u23a0 \u239f \u239b \u239d \u239c \u239e \u23a0 \u239f ( . / 1 2 1 3 0 1 2 4 Now, E C C cell = \u2212 \u22c5 0 0 06 1 2 1 . log \u21d2 0.162 = \u2212 \u22c5 0 06 2 0 1 1 3 . log ( ) . / K sp \u2234 K sp = 4 \u00d7 10 \u201312
Answer: A
\nSolution: E E V Tl Tl Pb Pb + + \u2212 = \u2212 | | . 2 0 444 or, E E Pb Tl Tl Tl Pb Pb + + \u2212 ( ) \u2212 \u22c5 = \u2212 + + | | . log [ ] [ ] . 2 0 06 2 0 444 2 2 or, [( . ) ( . )] . log . . . \u2212 \u2212 \u2212 \u2212 \u22c5 \u239b \u239d \u239c \u239e \u23a0 \u239f = \u2212 0 336 0 126 0 03 0 1 0 1 0 444 2 K sp \u2234 K sp = 4 \u00d7 10 \u20136
Answer: B
\nSolution: E cell = E Cu \u2013 E Zn = (E Cu \u2013 E calomel ) \u2013 (E Zn \u2013 E calomel ) From question E calomel \u2013 E Zn = 1.083 V and E calomel \u2013 E Cu = \u2013 0.018 V
Answer: C
\nSolution: The potential of hydrogen electrode at H 2 (1 bar) may be expressed as E = \u2013 0.059 P H Now, E P Ka 1 0 059 = \u2212 + \u23a1 \u23a3 \u23a2 \u23a4 \u23a6 \u23a5 . log x y and E P Ka 2 0 059 = \u2212 + \u23a1 \u23a3 \u23a2 \u23a4 \u23a6 \u23a5 . log y x \u2234 P E E ) Ka 1 2 = \u2212 + ( . 0 118\n8.42 Chapter 8 HINTS AND EXPLANATIONS
Answer: A
\nSolution: For the given reaction, E E E cell given hydrogen \u00b0 = \u00b0 \u2212 \u00b0 \u2234 (\u20130.84) \u2013 0 = \u22c5 0 06 1 . log K eq \u21d2 K eq = 10 \u201314
Answer: A
\nSolution: Net cell reaction may be written as H 2 + Zn 2+ \u001f 2H + + Zn E E n H P cell cell H = \u00b0 \u2212 \u22c5 + + 0 06 2 2 2 . log [ ] [Zn ] or, ( . ) ( . ) . log [ ] . \u2212 = \u2212 \u2212 \u22c5 \u00d7 + 0 61 0 76 0 06 2 1 0 4 2 H \u2234 [H + ] = 2 \u00d7 10 \u20133 M Now, K H SO HSO a 2 3 2 3 3 2 2 10 6 4 10 0 4 = = \u00d7 \u00d7 \u00d7 + \u2212 \u2212 \u2212 \u2212 [ ][ ] [ ] ( ) ( . ) . = 3.2 \u00d7 10 \u20134
Answer: B
\nSolution: Cu NH Cu(NH K f 2 3 3 4 2 12 4 10 + + + = \u001e \u21c0 \u001e \u21bd \u001e \u001e ) , 1.0 M excess 100 % 0 1.0 M Equ. x 2.0 M 1.0 M 10 1 0 2 0 12 4 = \u00d7 . ( . ) x \u21d2 x = = \u00d7 \u2212 \u2212 10 16 6 25 10 12 14 . Now, cell reaction: Zn + Cu 2+ \u001f Zn 2+ + Cu E E Zn Cu cell cell = \u00b0 \u2212 \u22c5 + + 0 06 2 2 2 . log [ ] [ ] = \u2212 \u2212 \u22c5 \u00d7 = \u2212 [ . ( . )] . log . . . 0 34 0 76 0 06 2 1 0 6 25 10 0 704 14 V
Answer: C
\nSolution: Net cell reaction: Zn + 2H + \u001f Zn 2+ + H 2 E E Zn P H cell cell H = \u00b0 \u2212 \u22c5 + + 0 06 2 2 2 2 . log [ ] [ ] or, 0 70 0 0 76 0 06 2 0 01 1 2 . [ ( . )] . log . [ ] = \u2212 \u2212 \u2212 \u22c5 \u00d7 + H \u21d2 [H + ] = 0.01 M Moles of HCl in RHS = 500 0 01 1000 5 10 3 \u00d7 = \u00d7 \u2212 . \u2234 Mass of NaOH needed = 5 \u00d7 10 \u20133 \u00d7 40 = 0.2 gm
Answer: D
\nSolution: On assuming concentration cell, the net cell reaction is Ag + (C 1 M, Right) Ag + (C 2 M, left) Now, C M 1 0 1 40 100 0 04 = \u00d7 = . . and C K Cl K K M sp sp sp 2 0 1 50 100 0 05 = = \u00d7 = \u2212 [ ] . . Now, E C C cell = \u2212 0 0 06 1 2 1 . log or, 0 42 0 06 1 0 05 0 04 . . log / . . = \u2212 K sp \u21d2 K sp = 2 \u00d7 10 \u201310
Answer: B
\nSolution: \u0394 G cell = \u2013 nFE cell \u21d2 \u2013 965 \u00d7 3 \u00d7 10 3 = \u201312 \u00d7 96500 \u00d7 E cell \u2234 E cell = 2.5 V
Answer: B
\nSolution: Theoretical efficiency = \u2212\u0394 \u00b0 \u2212\u0394 \u00b0 G H \u21d2 0 84 285 . = \u2212\u0394 \u00b0 G \u2234 \u0394 G\u00b0 = \u2013 0.84 \u00d7 285 KJ = \u2013nF \u22c5 E\u00b0 cell or, 0.84 \u00d7 285 \u00d7 10 3 = 2 \u00d7 96500 \u00d7 E\u00b0 cell \u21d2 E\u00b0 cell = 1.24 V
Answer: B
\nSolution: Net cell reaction: Ag(s) + H + + Cl \u2013 \u001f AgCl(s) + 1 2 H 2 (g) But for E cell calculation, reaction may be written as Ag(s) + H + \u001f Ag + + 1 2 H 2 Now, E E Ag P H cell cell H = \u00b0 \u2212 \u22c5 \u22c5 + + 0 06 1 2 1 2 . log [ ] [ ] / = \u2212 \u2212 \u22c5 \u239b \u239d \u239c \u239e \u23a0 \u239f \u00d7 = \u2212 \u2212 [ . ] . log . . . / 0 0 80 0 06 1 10 0 1 1 0 1 0 32 10 1 2 V It means that actual reaction is in reverse direction and E cell = 0.32 V
Answer: B
\nSolution: Informative
Answer: A
\nSolution: x y y x Ag NH Ag(NH + + + 3 3 \u001e \u21c0 \u001e \u21bd \u001e \u001e ) aM bM 0 b >> a Eqn. ? bM a x M K / Ag f x y a x b = \u22c5 + [ ] \u21d2 [Ag ] / + = \u22c5 \u22c5 \u239b \u239d \u239c \u239e \u23a0 \u239f a x b f y x K 1 Cell reaction: Ag + (Right) Ag + (Left)\n8.43 Electrochemistry HINTS AND EXPLANATIONS \u2234 E cell = 0 0 059 1 \u2212 \u22c5 + + . log [ ] [ ] Right Ag Left Ag Case-I: 0 118 0 059 4 10 4 10 4 2 1 . . log / = \u2212 \u22c5 \u00d7 \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f \u2212 \u2212 x \u21d2 x = 1 Case-II: 0 118 0 059 0 1 1 . . log . y/ = \u2212 \u22c5 \u239b \u239d \u239c \u239e \u23a0 \u239f x \u21d2 y = 2
Answer: C
\nSolution: E K V cell eq \u00b0 = \u22c5 = \u00d7 = \u2212 0 06 1 0 06 1 1 667 10 0 3732 6 . log . log . . Now, E E V Cu Cu Cu Cu \u00b0 \u2212 \u00b0 = \u2212 + + + 2 0 3732 | | . (1) and E E E V Cu Cu Cu Cu Cu Cu \u00b0 = \u00d7 \u00b0 + \u00d7 \u00b0 + = + + + + 2 2 1 1 1 1 0 3376 | | | . or, E E Cu Cu Cu Cu \u00b0 + \u00b0 = + + + 2 0 6752 | | . V (2) From (1) and (2), E V Cu Cu \u00b0 = + | . 0 5242
Answer: A
\nSolution: Zn + Ni 2+ \u001f Zn 2+ + Ni E K cell eq \u00b0 = 0 06 2 . log \u21d2 (\u20130.24) \u2013 ( \u20130.75) = 0.03 log K eq \u2234 K eq = 10 17 \u21d2 It means that Ni 2+ will react almost completely and [Zn 2+ ] \u2248 1.0 M Now, 10 1 0 17 2 = + . [Ni ] \u21d2 [Ni 2+ ] = 10 \u201317 M
Answer: B
\nSolution: Assuming the cell as concentration cell, the cell reaction may be written as Ag C M Ag C M + \u2212 \u2212 + \u2212 \u2212 = \u00d7 = \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f \u23af \u2192 \u23af = \u00d7 = 1 13 10 2 10 9 4 10 0 001 4 10 2 10 0 2 10 . . \u239b \u239b \u239d \u239c \u239e \u23a0 \u239f Now, E V cell = \u2212 \u22c5 \u00d7 = \u2212 \u2212 \u2212 0 0 06 1 10 4 10 0 024 9 10 . log .
Answer: A
\nSolution: 0 03 0 06 2 0 3 0 5 2 2 2 . . log [ ] [ ] . log . [ ] lower = \u22c5 = \u22c5 + + + Cu Cu Cu higher lower r \u2234 [Cu 2+ ] lower = 0.05 M
Answer: D
\nSolution: Cell reaction: H + (C 1 , HA 1 ) H + (C 2 , HA 2 ) E C C Ka Ka cell = \u2212 \u22c5 \u239b \u239d \u239c \u239e \u23a0 \u239f = \u22c5 0 0 059 1 0 059 2 1 2 1 . log . log = \u2212 = 0 059 2 0 059 1 2 . ( ) . V P P K K a a
Answer: A
\nSolution: Au + + 2CN \u2013 \u001f Au ( ) CN 2 \u2212 , \u0394 G\u00b0 1 = \u2013RT \u22c5 ln x O 2 + 2H 2 \u039f + 4e \u2013 \u001f 40H \u2013 ; \u0394 G\u00b0 2 = \u20134 \u00d7 F \u00d7 0.41 Au 3+ + 3e \u2013 \u001f Au; \u0394 G\u00b0 3 = \u20133 \u00d7 F \u00d7 1.50 Au 3+ + 2e \u2013 \u001f Au + ; \u0394 G\u00b0 4 = \u20132 \u00d7 F \u00d7 1.40 From \u0394 \u00b0 + \u0394 \u00b0 \u2212 \u0394 \u00b0 + \u0394 \u00b0 \u0394 \u00b0 = \u2212 + G G G G G RT F required 1 2 3 4 1 4 1 29 , ln . x
Answer: A
\nSolution: Informative
Answer: A
\nSolution: P H of right electrode will increase due to formation of OH \u2013 ion.
Answer: A
\nSolution: Informative
Answer: C
\nSolution: Theoretical
Answer: A
\nSolution: At low [Cl \u2013 ], O 2 becomes anode product
Answer: A
\nSolution: Informative
Answer: A
\nSolution: Theoretical
Answer: B
\nSolution: In the electrolysis of aq. KNO 3 , neither K + are NO 3 \u2212 participate in electrode reaction.
Answer: C
\nSolution: Na will react with water. S will not conduct electricity.
Answer: B
\nSolution: w E Q F = \u21d2 0 635 63 5 2 0 965 3600 96500 . . . \u00d7 = \u00d7 \u00d7 i \u21d2 i = 2 3 6 . A \u2234 % . . . % error = \u2212 = 2 3 6 0 5 2 3 6 10
Answer: B
\nSolution: Detonating gas is mixture of H 2 and O 2 n eq S n = n eq H 2 = n eq O 2 \u21d2 1 2 120 2 2 4 2 2 . \u00d7 = \u00d7 = \u00d7 n n H O \u2234 n and n H O 2 1 0 01 0 005 = = . .\n8.44 Chapter 8 HINTS AND EXPLANATIONS \u2234 Vol. of mixture of H 2 and O 2 = 0.015 \u00d7 22400 = 336 ml
Answer: B
\nSolution: n eq of Li OH = Q F \u21d2 w 24 1 2 5 4825 0 8 96500 \u00d7 = \u00d7 \u00d7 . .
Answer: B
\nSolution: n eq Cu deposited at cathode = Q F or, w 63 5 2 12 4 4825 96500 . . \u00d7 = \u00d7 \u21d2 w = 19.685 gm But the increase in mass of cathode is only 19.05 gm It represents that 20 gm of sample contains only 19.05 gm Cu. \u2234 % of Cu = 19 05 20 100 95 25 . . % \u00d7 = Now, Q F n Cu n Fe eq eq = + (oxidised at anode) or, 12 4 4825 96500 19 05 63 5 2 56 2 . . . \u00d7 = \u00d7 + \u00d7 w \u21d2 w = 0.56 gm \u2234 Percentage of Fe = 0 56 20 100 2 8 . . % \u00d7 =
Answer: B
\nSolution: Theoretical n eq of NaOH formed = n eq Cu = 3 18 63 6 2 0 1 . . . \u00d7 = Actual n eq of NaOH formed = \u00d7 = 60 1 1000 0 06 . \u2234 Percentage yeild = 0 06 0 1 100 60 . . % \u00d7 =
Answer: A
\nSolution: Cell reaction during discharge: Pb + PbO 2 + 2H 2 SO 4 2PbSO 4 + 2H 2 O n Pb taken = 200 208 and n PbO taken 2 200 240 = Hence, PbO 2 is L.R. Now, n eq PbO Q F 2 = \u21d2 200 240 2 10 96500 \u00d7 = \u00d7 t \u21d2 t = 16083.33 sec
Answer: A
\nSolution: Q F of I N a S O eq eq 2 2 3 = = \u2212 n n or, i \u00d7 \u00d7 = \u00d7 \u00d7 2 3600 96500 72 1 0 1000 1 . \u21d2 i = 0.965 A
Answer: A
\nSolution: C 14 H 10 + 2H 2 O C 14 H 8 O 2 + 6H + + 6e \u2013 n eq C 14 H 8 O 2 = Q F \u21d2 w 208 6 1 40 60 0 965 96500 \u00d7 = \u00d7 \u00d7 \u00d7 . \u2234 w = 0.832 gm
Answer: B
\nSolution: Number of coulombs required = 1000 0 00033 . per Kg Cu Energy required = \u00d7 = 1000 0 00033 0 33 10 6 . . J \u2234 Cost of electricity = \u00d7 \u00d7 = 4 10 3600 10 1 11 3 6 . Rupee
Answer: D
\nSolution: n eq CH 3 Coo \u2013 oxidised = Q F or, n \u00d7 = \u00d7 \u00d7 \u00d7 1 0 5 482 5 60 0 8 96500 . . . \u21d2 n = 0.12 \u2234 Moles of (C 2 H 6 + CO 2 ) produced = 3 2 0 12 0 18 \u00d7 = . . and total volume = 0.18 \u00d7 22.4 = 4.032 L
Answer: C
\nSolution: \u0394 \u00b0 = \u22c5 = E V 0 059 2 10 0 177 6 . log .
Answer: A
\nSolution: Back EMF = \u22c5 = \u00d7 \u2212 0 06 2 0 12 0 08 5 4 10 3 . log . . . V
Answer: B
\nSolution: Equivalent of charge used = 0 4825 10 3600 96500 0 18 . . \u00d7 \u00d7 = F Cell reaction during charge: Cu Zn Cu Zn + \u23af \u2192 \u23af + + + 2 2 100 1 1000 0 1 \u00d7 = . mole 100 1 1000 0 1 \u00d7 = . mole = 0.2 eq = 0.2 eq Final 0.2 \u2013 0.1 8 = 0.2 + 0.18 = 0.02 eq = 0.38 eq = 0.01 mole = 0.19 mole \u2234 Final [ ] . . Zn M 2 0 01 100 1000 0 1 + = \u00d7 = and [ ] . . Cu M 2 0 19 100 1000 1 9 + + \u00d7 = Now, cell reaction as galvanic cell:\n8.45 Electrochemistry HINTS AND EXPLANATIONS Zn + Cu 2+ Zn 2+ + Cu and E E n Zn Cu cell cell = \u00b0 \u2212 \u22c5 + + 0 06 2 2 . log [ ] [ ] = \u2212 \u2212 \u2212 \u22c5 = [ . ( . )] . log . . . 0 34 0 76 0 06 2 0 1 1 9 1 1084 V
Answer: C
\nSolution: Reactions involved are 2 2 2 2 H O H O 2 electrolysis \u23af \u2192 \u23af\u23af\u23af\u23af + N 2 + 3H 2 2NH 3 NH 3 + 2O 2 HNO 3 + H 2 O NH 3 + HNO 3 NH 4 NO 3 For 1 mole NH 4 NO 3 , 3 moles of H 2 O should be electrolyzed. Hence for 1152 Kg NH 4 NO 3 , moles of H 2 needed = \u00d7 \u00d7 = \u00d7 3 1152 10 80 4 32 10 3 4 . Now, n eq H Q F 2 = \u21d2 4.32 \u00d7 10 4 \u00d7 2 = i \u00d7 \u00d7 24 3600 96500 \u2234 i = 96500 A/day
Answer: A
\nSolution: n eq HC 3 H 5 O 3 = n OH Q F \u2212 = or, w 90 1 50 10 1158 96500 3 \u00d7 = \u00d7 \u00d7 \u2212 \u21d2 w = 0.054 gm \u2234 % of lactice acid = 0 054 1 100 5 4 . . % \u00d7 =
Answer: A
\nSolution: n n n H O NH S O formed 2 2 4 2 2 8 = = ( ) and n eq NH S O Q F ( ) 4 2 2 8 = \u21d2 n i \u00d7 = \u00d7 = \u00d7 \u00d7 2 102 34 2 3600 0 5 96500 . [ ] 2 2 4 2 2 8 2 SO S O e \u2212 \u2212 \u2212 \u23af \u2192 \u23af + \u21d2 i = 321.67 A
Answer: B
\nSolution: n n n As H AsO = = 3 3 and n eq H 3 AsO 3 = n eq I 2 = Q F or, w 75 2 1 68 10 96 5 96500 3 \u00d7 = \u00d7 \u00d7 \u2212 . . \u21d2 w = 6.3 \u00d7 10 \u20135 gm
Answer: D
\nSolution: w E Q F = \u21d2 52 2 87 2 19 3 2 3600 96500 . . \u00d7 = \u00d7 \u00d7 \u00d7 \u03b7 \u21d2 \u03b7 = 0.8333 or 83.33 %
Answer: B
\nSolution: n eq Cu 2+ reduced = Q F \u21d2 n \u00d7 2 = 2 10 19 3 60 96500 3 \u00d7 \u00d7 \u00d7 \u2212 . \u2234 n = 1.2 \u00d7 10 \u20135 \u2234 [ ] ( . ) . CuSO M 4 0 5 5 1 2 10 2 250 1000 9 6 10 = \u00d7 \u00d7 \u00d7 = \u00d7 \u2212 \u2212
Answer: A
\nSolution: w E Q F = \u21d2 12 3 123 6 . \u00d7 = \u00d7 Q 0.5 F \u21d2 Q = 1.2 F + 6H + + 6e \u2013 + 2H2O NH2 NO2 and Energy consumed = 1.2 F \u00d7 3.0 V = 347.4 KJ
Answer: C
\nSolution: n eq H 2 (at cathode) = n eq O 2 + n eq H 2 S 2 O 8 (at anode) or, 9 08 22 7 2 2 27 22 7 4 194 2 . . . . \u00d7 = \u00d7 + \u00d7 w \u2234 w = 38.8 gm
Answer: C
\nSolution: Initial mass of H 2 SO 4 , w 1 = 3600 \u00d7 1.5 \u00d7 40 100 2160 = gm Final mass of H 2 SO 4 , w 2 = 3600 \u00d7 1.1 \u00d7 10 100 396 = gm \u2234 Moles of H 2 SO 4 consumed = w w 1 2 98 18 \u2212 = Now, n eq H 2 SO 4 = Q F \u21d2 18 \u00d7 1 = ( ) amp-hr \u00d7 3600 96500 \u2234 Number of ampere-hr = 482.5
Answer: B
\nSolution: \u2227 \u2227 = = \u22c5 \u22c5 \u22c5 \u2217 \u2217 m m NaCl KCl NaCl NaCl KCl NaCl KCl /C) /C) R G C R G ( ) ( ) ( ( \u03ba \u03ba 1 1 1 \u22c5 \u22c5 = \u22c5 \u22c5 1 C R C) R C) KCl KCl NaCl ( ( or, \u2227 = \u00d7 \u00d7 m NaCl) ( . . 120 200 0 1 6400 0 003 \u21d2 \u2227 m(NaCl) = 125 \u03a9 \u20131 cm \u20131 mol \u20131 2 2 4 2 2 8 2 SO S O e \u2212 \u2212 \u2212 \u2192 + \u23a1 \u23a3 \u23a4 \u23a6\n8.46 Chapter 8 HINTS AND EXPLANATIONS
Answer: C
\nSolution: \u2227\u00b0 = \u00d7 \u00b0 + + \u2212 m m m NH CrO NH CrO [( ) ] ( ) ( ) 4 2 4 4 4 2 2 \u03bb \u03bb = (2 \u00d7 6.6 \u00d7 10 \u20138 + 5.4 \u00d7 10 \u20138 ) \u00d7 96500 = 0.01795 \u03a9 \u20131 m 2 mol \u20131
Answer: A
\nSolution: \u2227 = \u22c5 \u2227\u00b0 = m m C \u03b1 \u03ba or, 0.9 \u00d7 4.25 \u00d7 10 \u20132 = 382 5 3 . C 10 \u00d7 \u21d2 C = 0.1 M
Answer: C
\nSolution: \u2227 = m K C \u21d2 1.5 \u00d7 10 \u20132 = 3 06 10 2 56 10 3 3 . . \u00d7 \u2212 \u00d7 \u2212 \u2212 C \u2234 C mol m mol l V = = \u00d7 \u22c5 = \u2212 \u2212 \u2212 1 30 1 30 10 585 58 5 3 3 1 / . \u2234 V = 3 \u00d7 10 5 L
Answer: D
\nSolution: \u2227 = = \u22c5 \u2217 m C G G C \u03ba \u21d2 100 1 0 5 1 5 0 1 10 3 = \u00d7 \u00d7 \u2212 R . . . \u21d2 R ohm = 100 3 Now, V = IR \u21d2 I V R A = = = 5 100 3 0 15 / .
Answer: C
\nSolution: Ionic mobility, \u03bc \u03bb \u00b0 = = \u00b0 speed of ion Pot. gradient F m or, speed 19 3 5 50 96500 . \u239b \u239d \u239c \u239e \u23a0 \u239f = \u21d2 speed = 2 \u00d7 10 \u20133 cm/s
Answer: A
\nSolution: \u2227 eq = C \u03ba \u21d2 150 3 4 10 1 6 10 5 6 6 \u00d7 \u00d7 \u2212 \u00d7 \u2212 \u2212 . . \u2234 S = 1.2 \u00d7 10 \u20138 mol cm \u20133 = 1.2 \u00d7 10 \u20135 M
Answer: B
\nSolution: K = G \u22c5 G* \u21d2 \u03ba 1 4 1 280 1 50 . / / = \u21d2 \u03ba = 0.25 s m \u20131 for 0.5 M Now, \u2227 = = \u00d7 = \u00d7 \u2212 \u2212 m C s m mol \u03ba 0 25 0 5 10 5 10 3 4 2 1 . .
Answer: D
\nSolution: Ag A S Ag A ( ) ( )M \u001e \u21c0 \u001e \u21bd \u001e \u001e + + \u2212 + x y x M Ag B S Ag B M ( ) ( )M \u001e \u21c0 \u001e \u21bd \u001e \u001e + + \u2212 + x y y Now, ( x + y ) \u22c5 x = 3 \u00d7 10 \u201314 and ( x + y ) \u22c5 y = 1 \u00d7 10 \u201314 \u2234 [Ag ] , [A ] . ; [B ] . + \u2212 \u2212 \u2212 \u2212 \u2212 = + = \u00d7 = = \u00d7 = = \u00d7 x y x y 2 10 1 5 10 0 5 10 7 7 7 M M M Now, \u03ba solution = \u00d7 = \u00d7 \u00d7 \u00d7 + \u00d7 \u00d7 \u00d7 + \u00d7 \u2212 \u2212 \u2212 \u2212 \u2212 3 75 10 2 10 10 60 1 5 10 10 80 0 5 10 8 7 3 7 3 . . . \u2212 \u2212 \u2212 \u00b0 \u00d7 7 3 10 \u03bb B \u2234 \u03bb B \u00b0 \u2212 \u2212 = 135 ohm cm mol 1 2 1
Answer: A
\nSolution: \u2227 \u00b0 = \u2227\u00b0 + \u2227\u00b0 \u2212 \u2227 \u00b0 eq eq eq eq [Be (Po ) ] [BeCl ] [K Po ] [K ] 3 4 2 2 3 4 Cl = + \u2212 = \u2212 \u2212 160 140 100 200 1 2 1 ohm cm eq Now, n eq = \u21d2 = \u00d7 \u2212 \u03ba C C 200 1 2 10 5 . \u21d2 C = 6 \u00d7 10 \u20138 eq cm \u20133 = 6 \u00d7 10 \u20135 N = 10 \u20135 M Now, K sp = = \u00d7 = \u00d7 \u2212 \u2212 108 108 10 1 08 10 5 5 5 23 S ( ) .
Answer: B
" } } ] }, { "title": "Chem Sec 2", "originalName": "Section B - Multi Correct", "questions": [ { "question_id": "electrochemistry-chem-sec-2-1-81", "marks": 4.0, "negMarks": 1.0, "partialMarks": null, "subject": "chemistry", "chapter": "electrochemistry", "chapterTitle": "Electrochemistry", "type": "mcqm", "rawChapterType": "MSQ", "originalNumber": 81, "displayNumber": 1, "image": "./assets/chapter-packs/electrochemistry/Chemistry Section 1__--__81__--__1.png", "solutionImage": null, "question": { "content": "Answer: A, C
\nSolution: Net cell reaction is spontaneous in electrochemical cell but non-spontaneous in electrolytic cell. Cathode is +ve in electrochemical cell but \u2013ve in electrolytic cell.
Answer: A, C, D
\nSolution: For the cell: Ag(s) | Ag Cl (s) | Cl\u2212 || Ag + | Ag(s), the net cell reaction is Ag + + Cl\u2212 \u001f Ag Cl (s).
Answer: C, D
\nSolution: Left electrode: Ag(s) + Cl \u2212 (aq) \u2192 Ag Cl (s) + e \u2212 1 \u00d7 2 Right electrode : Hg 2 Cl 2 (s) + 2e \u2212 \u2192 2Hg(l) + 2 Cl \u2212 (aq) Net reaction: 2Ag(s) + Hg 2 Cl 2 (s) \u2192 2Ag Cl(s) + 2Hg (l)\n8.47 Electrochemistry HINTS AND EXPLANATIONS
Answer: B, C
\nSolution: Informative
Answer: D
\nSolution: n n n n eq eq eq eq Cu Mg Na Al = \u00d7 = = \u00d7 = = \u00d7 = = 63 5 63 5 2 2 24 24 2 2 11 5 23 1 0 5 . . ; . . ; 9 9 27 3 1 \u00d7 =
Answer: A, C
\nSolution: Theoretical
Answer: B, C
\nSolution: \u2227 \u00b0 eq = 60 + 80 = 140 ohm \u20131 cm 2 eq \u20131 \u2227 \u00b0 m = 140 \u00d7 6 = 840 ohm \u20131 cm 2 eq \u20131
Answer: B, C, D
\nSolution: Salt bridge does not change standard potential of any electrode.
Answer: A, B, C
\nSolution: Moles of electron involved = 0 25 9 65 3600 96500 0 09 . . . \u00d7 \u00d7 = \u2234 Mass of Zn involved = 0 09 2 65 4 2 943 . . . \u00d7 = gm Mass of MnO 2 involved = 0.09 \u00d7 87 = 7.83 gm Mass of NH 4 + involved = 0.09 \u00d7 18 = 1.62 gm
Answer: A, C
\nSolution: Net cell reaction is Cd(s) + 2AgCl(s) \u001f 2Ag(s) + Cd 2+ (aq) + 2Cl \u2013 (aq) \u0394 \u00b0 = \u2212 \u00b0 = \u2212 \u00d7 \u00d7 = \u2212 \u00b0 G 50 2 96500 0 6 115800 C nFE J . \u0394 \u00b0 = \u2212 \u00b0 = \u2212 \u00d7 \u00d7 = \u2212 \u00b0 G 0 2 96500 0 7 135100 C nFE J . \u0394 \u00b0 = \u22c5 \u00b0 \u2212 \u00b0 \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f = \u00d7 \u00d7 \u2212 = \u2212 S T T nF E E J/K 2 1 2 1 2 96500 0 6 0 7 50 386 . . \u0394 H \u00b0 = \u0394 G \u00b0 + T \u22c5 \u0394 S \u00b0 = (\u2013135100) + 273 \u00d7 (\u2013386) = \u2013240478 J
Answer: B, C
\nSolution: Anode: 2H 2 O(l) O 2 (g) + 4H + (aq) + 4e \u2013 Cathode: 2H 2 O(l) + 2e \u2013 H 2 (g) +2OH \u2013 (aq) n eq H + produced = n eq OH \u2013 produced = Q F or, n n H OH + \u2212 \u00d7 = = \u00d7 \u00d7 = 1 1 25 965 60 96500 0 75 . . Anode: HPO H H PO 4 2 2 4 \u2212 + \u2212 + \u001e \u21c0 \u001e \u21bd \u001e \u001e 1.0 M 0.75 M 1.0 M Final 0.25 M 1.75 M \u2234 P P HPO H PO H K = + = + = \u2212 \u2212 a log [ ] [ ] . log . . . 4 2 0 2 4 2 15 0 25 0 75 1 30 Cathode: H PO OH HPO H O 2 4 4 2 2 \u2212 \u2212 \u2212 + + \u001e \u21c0 \u001e \u21bd \u001e \u001e 1.0 M 0.75 M 1.0 M Final 0.25 M 0 1.75 M \u2234 P P HPO H PO H K = + = + = \u2212 \u2212 a log [ ] [ ] . log . . . 4 2 0 2 4 2 15 1 75 0 25 3 0
Answer: A, B, D
\nSolution: V, Fe and Hg will be oxidised by NO 3 \u2212
Answer: A, B
\nSolution: Resistance and heat capacity depends on quantity.
Answer: A, B
\nSolution: Theoretical
Answer: A, B, C, D
\nSolution: Net cell reaction of discharge is Pb + PbO 2 + 2H 2 SO 4 2PbSO 4 + 2H 2 O x mole x mole Initial mass of H 2 SO 4 , w 1 = 1000 \u00d7 1.26 \u00d7 40 100 504 = gm Final mass of H 2 SO 4 , w 2 = (1260 \u2013 98 x + 18 x ) \u00d7 28 100 = (352.8 \u2013 22.4 x ) gm From reaction, 504 \u2013 98 x = 352.8 \u2013 22.4 x \u21d2 x = 2
Answer: A
\nSolution: E\u00b0 Cell = E E H H Zn Zn \u00b0 \u2212 \u00b0 + + / / 2 2 because E H H \u00b0 + / 2 was higher or, 0.76 = 1.00 \u2013 E Zn Zn \u00b0 + 2 / \u21d2 E Zn Zn \u00b0 + 2 / = 0.24 V
Answer: B
\nSolution: E\u00b0 Cell = E E Cu Cu H H \u00b0 \u2212 \u00b0 + + 2 2 / / because E Cu Cu \u00b0 + 2 / was higher or, 0.34 = E Cu Cu \u00b0 + 2 / \u2013 1.00 \u21d2 E Cu Cu \u00b0 + 2 / = 1.34 V
Answer: A
\nSolution: E\u00b0 Cell = E E Cu Cu Zn Zn \u00b0 \u2212 \u00b0 + + 2 2 / / = 1.34 \u2013 0.24 = 1.10 V\n8.48 Chapter 8 HINTS AND EXPLANATIONS Comprehension II
Answer: A
\nSolution: E E Ag K Ag Ag Ag Ag sp + + = \u00b0 \u2212 \u22c5 = \u2212 + / / . log [ ] . . .log 0 06 1 1 0 80 0 06 1 1 = + 0.314 V
Answer: B
\nSolution: E E K V I |Ag I|Ag Ag Ag sp \u00b0 = \u00b0 \u2212 \u22c5 = \u2212 \u2212 + / . log . 0 06 1 0 172
Answer: B
\nSolution: E E I I |Ag I|Ag I /Ag I Ag \u2212 \u2212 = \u00b0 \u2212 \u22c5 \u2212 / . log[ ] 0 06 1 = \u20130.172 \u2013 0.06 \u22c5 log 0.04 = \u20130.088 V Comprehension III
Answer: C
\nSolution: As(0.40) > (\u20130.87), reduction of Ni 2 O 3 (s) will occur.
Answer: B
\nSolution: E\u00b0 cell = (0.40) \u2013 (\u20130.87) = 1.27 V
Answer: C
\nSolution: Net cell reaction is independent from OH \u2013 (aq)
Answer: A
\nSolution: \u2013 \u0394 G \u00b0 = nFE\u00b0 cell = 2 \u00d7 96500 \u00d7 1.27 = 245110 J Comprehension IV
Answer: A
\nSolution: H 3 O + is only needed in balancing cathode reaction: NO H O e HNO H O 3 3 2 2 3 2 4 \u2212 + \u2212 + + \u23af \u2192 \u23af +
Answer: B
\nSolution: Moles of electron needed = 2 \u00d7 moles of HNO 2 formed
Answer: B
\nSolution: n eq HNO Q F 2 = \u21d2 0 1 2 10 96500 . \u00d7 = \u00d7 t \u21d2 t = 1930 sec Comprehension V
Answer: C
\nSolution: \u0394 G\u00b0 = \u2013 n FE\u00b0 \u21d2 \u2013237.39 \u00d7 10 3 = \u20132 \u00d7 96500 \u00d7 E\u00b0 cell \u2234 E \u00b0cell = 1.23 V
Answer: A
\nSolution: Moles of H 2 needed = 23 739 237 39 0 1 . . . = \u2234 Volume of H 2 needed = 0.1 \u00d7 22.7 = 2.27 L
Answer: A
\nSolution: E cell is independent from [OH \u2013 ]
Answer: C
\nSolution: \u0394 \u00b0 = \u0394 \u00b0 \u2212 \u0394 \u00b0 = \u2212 \u00d7 \u2212 \u2212 \u00d7 S H G T ( . ) ( . ) 285 8 10 237 39 10 298 3 3 = \u2013162.4 J/K \u20131
Answer: B
\nSolution: \u03b7 = \u2212\u0394 \u00b0 \u2212\u0394 \u00b0 = = ( ) ( ) . . . G H 237 39 285 8 0 8306 or 83.06% Comprehension VI
Answer: B
\nSolution: Refer theory given is passage.
Answer: D
\nSolution: E E E V O H O H Fe Fe \u00b0 = \u00b0 \u2212 \u00b0 = \u2212 \u2212 = + + 2 2 2 1 229 0 447 1 676 / , / . ( . ) .
Answer: A
\nSolution: E E Cu Cu Pb Pb \u00b0 > \u00b0 + + 2 2 / / and hence Cu will not oxidise easily.
Answer: B
\nSolution: n eq Fe Q F = \u21d2 w 56 2 0 5 1 0 3600 96500 \u00d7 = \u00d7 \u00d7 . . \u21d2 w = 0.522 gm
Answer: B
\nSolution: Theoretical Comprehension VII
Answer: C
\nSolution: n eq Ni Q F = \u21d2 w 58 7 2 15 3600 0 6 96500 . . \u00d7 = \u00d7 \u00d7 \u21d2 w = 9.85 gm
Answer: C
\nSolution: V = A \u00d7 t \u21d2 9 85 8 9 4 0 2 . . ( . ) = \u00d7 \u00d7 t \u21d2 t = 0.138 cm\n8.49 Electrochemistry HINTS AND EXPLANATIONS
Answer: D
\nSolution: n eq H Q F 2 = \u21d2 V H 2 22 4 2 15 3600 0 4 96500 . . \u00d7 = \u00d7 \u00d7 \u21d2 V L H 2 2 5 = .
Answer: B
\nSolution: Anode produced is only O 2 gas. n eq O Q F 2 = \u21d2 w 8 15 3600 96500 = \u00d7 \u21d2 w = 4.477 gm Comprehension VIII
Answer: A
\nSolution: E\u00b0 cell = 0.8 \u2013 0.05 = 0.75 V Now, E RT nF K cell \u00b0 = \u22c5 ln \u21d2 0 75 1 2 38 92 . . ln = \u00d7 \u22c5 K \u21d2 ln K = 58.38
Answer: D
\nSolution: The oxidation reaction of glucose contains H + and E E H = \u00b0 \u2212 \u22c5 + 0 0592 2 2 . log[ ] \u21d2 E \u2013 E\u00b0 = 0.0592 P H = 0.6512 V
Answer: B
\nSolution: Standard potential is independent from ammonia concentration. Comprehension IX
Answer: A
\nSolution: Left electron: H 2 (g) 2H + (aq) + 2e \u2013 Right electrode: 2AgCl(s) + 2e \u2013 2Ag(s) + 2Cl \u2013 (aq) \u2234 Net reaction: H 2 (g) + 2AgCl(s) 2Ag(s) + 2H + (aq) + 2Cl \u2013 (aq)
Answer: B
\nSolution: \u0394 \u00b0 = \u22c5 \u2212 \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f = \u00d7 \u00d7 \u2212 = \u2212 S nF E E T T J/K 2 1 2 1 2 96500 0 21 0 23 20 193 . . \u0394 G\u00b0 = \u2013nFE\u00b0 = \u20132 \u00d7 96500 \u00d7 0.23 = \u2013 44390 J at 15\u00b0C Now, \u0394 H \u00b0 = \u0394 G \u00b0 + T \u22c5 \u0394 S \u00b0 = (\u201344390) + 288 \u00d7 (\u2013193) = \u201399974 J = \u201349987 J/mole AgCl
Answer: C
\nSolution: \u0394 S\u00b0 = \u2013193 J/K = \u201396.5 J/K per mole AgCl
Answer: B
\nSolution: \u0394 G \u00b0 298 = \u0394 H \u00b0 \u2013 T \u22c5 \u0394 S \u00b0 = (\u201349987) \u2013 298 \u00d7 (\u201396.5) or, \u20131 \u00d7 96500 \u00d7 E\u00b0 = \u2013 21230 \u21d2 E\u00b0 = 0.22 V = E Cl AgCl/Ag \u00b0 \u2212 / or, E E K Cl AgCl/Ag Ag Ag sp \u00b0 = \u00b0 + \u2212 + / / . log 0 058 1 or, 0 22 0 80 0 058 1 . . . log = + K sp \u21d2 K sp = 1 \u00d7 10 \u201310 Hence, Solubility, S = K M sp = \u2212 10 5 Comprehension X
Answer: D
\nSolution: Add Ag(s) in both sides of given reaction to get cell reaction. Now, for E\u00b0 cell \u0394 G \u00b0 = \u2013nFE\u00b0 \u21d2 [\u2013109] \u2013 [77 + (\u2013129)] \u00d7 10 3 = \u20131 \u00d7 96500 \u00d7 E\u00b0 cell \u2234 E\u00b0 cell = 0.59 V
Answer: A
\nSolution: E K cell eq \u00b0 = 0 059 . log n \u21d2 0 59 0 059 1 1 . . log = \u22c5 K sp \u2234 K sp = 10 \u201310
Answer: B
\nSolution: Zn(s) + 2Ag + (aq) \u00a1 Zn 2+ (aq) + 3Ag(s); E\u00b0 = 0.80 \u2013 (\u20130.76) Now, E K eq \u00b0 = \u22c5 0 059 . log n \u21d2 1 56 0 059 2 2 2 . . log [ ] [ ] = \u22c5 + + Zn Ag \u2234 log [ ] [ ] . Zn Ag 2 2 52 88 + + =
Answer: D
\nSolution: Moles of Zn added = 6 539 10 65 39 10 2 3 . . \u00d7 = \u2212 \u2212 Moles of Ag + present = 10 1000 100 10 5 6 \u2212 \u2212 \u00d7 = (L.R.) \u2234 Moles of Ag precipitated = 10 \u20136\n8.50 Chapter 8 HINTS AND EXPLANATIONS Comprehension XI
Answer: D
\nSolution: For KCl: \u2227 = = \u22c5 \u2217 m C G G C \u03ba \u21d2 G* = \u2227 m \u22c5 C/G or, G* = \u2227 m \u22c5 C \u22c5 R = 200 \u00d7 (0.02 \u00d7 10 \u20133 ) \u00d7 100 = 0.4 cm \u20131
Answer: B
\nSolution: \u03ba water G G cm = \u22c5 = \u00d7 = \u00d7 \u03a9 \u2217 \u2212 \u2212 \u2212 1 10000 0 4 4 10 5 1 1 .
Answer: D
\nSolution: For NaCl: \u2227 = = \u22c5 \u2217 m C G G C \u03ba or, 125 1 8000 1 10000 0 4 585 58 5 = \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f . / . V \u21d2 V = 1.25 \u00d7 10 8 cm 3 = 1.25 \u00d7 10 5 L Comprehension XII
Answer: A
\nSolution: E E Cl Cl I \u00b0 > \u00b0 \u2212 \u2212 2 2 / /I
Answer: C
\nSolution: E E Mn Mn O H O, H \u00b0 > \u00b0 + + + 3 2 2 2 / / Comprehension XIII
Answer: A
\nSolution: Net reaction: M + (1M) M + (0.05 M); Higher Conc. Lower Conc. E cell > 0, \u0394 G cell < 0
Answer: A
\nSolution: 70 mV = E RT F \u00b0 \u2212 \u22c5 ln . 0 05 1 (1) E E RT F E RT F req = \u00b0 \u2212 \u22c5 = \u00b0 \u2212 \u22c5 ln . ln ( . ) 0 0025 1 0 05 1 2 (2) and E\u00b0 =
Answer: C
\nSolution: Hence, E req = 140 mV Comprehension XIV
Answer: B
\nSolution: \u0394 G cell = \u2013nFE cell = \u20132 \u00d7 96500 \u00d7 0.059 = \u201311387 J
Answer: A
\nSolution: E E M M cell cell Left Right = \u00b0 \u2212 \u22c5 + + 0 059 2 2 2 . log [ ] [ ] or, 0 059 0 0 059 2 4 0 001 1 3 . . log ( / ) . / = \u2212 \u22c5 K sp \u21d2 K sp = 4 \u00d7 10 \u201315 Comprehension XV
Answer: C
\nSolution: H + + Cl \u2013 + (NaOH) Na + + Cl \u2013 + H 2 O Conductance first decreases and H + ions are replaced by Na + ions. After equivalent point, conductance increase due to increase in number of ions.
Answer: B
\nSolution: CH 3 COOH + (NaOH added) CH 3 COO \u2013 + Na + + H 2 O As number of ions increases, conductance increases. slight decrease initially was due to some dissociated CH 3 COOH. After equivalence point, in place of CH 3 COO \u2013 ion, number of OH \u2013 ions increases and hence slope becomes greater.
Answer: B
\nSolution: H + + Cl \u2013 + (NH 4 OH added) NH Cl H O 4 2 + \u2212 + + As H + ions are replaced by NH 4 + ions, conductance decreases. After equivalent point, it become almost constant as the dissociation of added NH4OH will be suppressed in presence of NH 4 + ions.
Answer: A
\nSolution: HCl will neutralize fi rst followed by CH 3 COOH.
Answer: D
\nSolution: Ionic mobilities of Ag + and K + ions do not differ largely
Answer: A
\nSolution: CuCl 2 Cu + Cl 2
Answer: C
\nSolution: E E informative Zn Zn Ag Ag \u00b0 < \u00b0 + + 2 / / ( )
Answer: B
\nSolution: Reason is different charges on ions.
Answer: A
\nSolution: Theoretical
Answer: B
\nSolution: Negative reduction potential means greater tendency to get oxidised.
Answer: C
\nSolution: Number of ions increases considerably only for weak electrolytes.
Answer: A
\nSolution: Informative
Answer: D
\nSolution: Theoretical
Answer: C
\nSolution: It is due to very high over voltage potential of hydrogen at mercury cathode.
Answer: C
\nSolution: Theoretical
Answer: B
\nSolution: Theoretical
Answer: C
\nSolution: E E RT nF Q = \u2212 \u22c5 \u00b0 log
Answer: B
\nSolution: Cl \u2013 will combine with Ag + to precipitate AgCl.
Answer: B
\nSolution: At Cathode: 2H O(l) + 2e H (g) + 2OH aq 2 2 \u2212 \u2212 \u23af \u2192 \u23af ( ) At Anode: 2 2 2 3 2 6 2 CH COO aq C H CO e \u2212 \u2212 ( ) ( ) ( ) \u23af \u2192 \u23af + + g g
Answer: C
\nSolution: Theoretical
Answer: A \u2192 P, Q; B \u2192 P, Q; C \u2192 Q, R; D \u2192 P, S
\nSolution: Informative
Answer: A \u2192 R; B \u2192 S; C \u2192 Q; D \u2192 P
\nSolution: Theoretical
Answer: A \u2192 R; B \u2192 S; C \u2192 P; D \u2192 Q
\nSolution: E E E V Fe Fe Fe Fe Fe Fe 3 3 2 2 1 2 1 2 0 037 + + + + \u00b0 \u00b0 \u00b0 = \u00d7 + \u00d7 + = \u2212 | , | . E V H ( H O OH ) . . . 4 4 4 2 0 40 1 23 0 83 \u2192 + \u00b0 + \u2212 = \u2212 = \u2212 E E E Cu Cu Cu Cu (Cu Cu ) | Cu | . . . . 2 2 2 0 34 0 52 2 1 0 52 0 + + + + + + \u2192 \u00b0 \u00b0 \u00b0 = \u2212 = \u2212 \u2212 \u2212 = \u2212 7 70 V E V Cr Cr 3 2 3 0 74 2 0 91 3 2 0 4 + + \u00b0 = \u00d7 \u2212 \u2212 \u00d7 \u2212 \u2212 = \u2212 , ( . ) ( . ) .
Answer: A \u2192 R, S; B \u2192 P, R; C \u2192 Q, S
\nSolution: For concentration cell, both half cell must have same configuration. P. Ag Cl AgCl sponteneous, K K eq sp + \u2212 + \u23af \u2192 \u23af = >> 1 1 Q. Ag Br + Cl AgCl Br Cl Non-Spontene eq sp sp \u2212 \u2212 \u23af \u2192 \u23af + = << ; (Ag Br) (Ag ) , K K K 1 o ous R. Ag Ag + + \u23af \u2192 \u23af ( . M) ( . M) 1 0 0 1 Higher to lower concentration, spontaneous S. Cl Cl \u2212 \u2212 \u23af \u2192 \u23af \u2212 ( . M) ( . M) Non spontaneous 0 1 1 0
Answer: 4
\nSolution: E = E P H \u00b0 \u22c5 + \u2192 + + \u2212 \u2212 0 06 2 2 2 2 2 . log [H ] (assuming H e H ) n \u2212 = \u2212 \u22c5 \u21d2 = + + \u2212 0 18 0 0 06 2 1 10 2 3 . . log [H ] [H ] M C H NH H O C H NH H O M 6 5 3 2 6 5 2 3 + \u2212 + + + ( )M C x x \u001e \u21c0 \u001e \u21bd \u001e \u001e h x c = = = \u2212 10 0 04 4 3 1 40 . % or\n8.52 Chapter 8 HINTS AND EXPLANATIONS
Answer: 5
\nSolution: E E = \u2212 \u22c5 \u00b0 + 0 06 2 1 . log [Cu ] 0 31 0 34 0 06 2 1 0 1 2 2 . . . log [Cu ] [Cu ] . M = \u2212 \u22c5 \u21d2 = + + \u2234 [OH ] . \u2212 + \u2212 \u2212 = = = \u21d2 = K sp H Cu p 2 19 9 10 0 1 10 5
Answer: 5
\nSolution: E E E Fe Fe Fe Fe Fe Fe 3 2 3 2 3 2 3 2 3 0 04 2 0 44 + + + + \u00b0 \u00b0 \u00b0 = \u00d7 \u2212 \u00d7 \u2212 = \u00d7 \u2212 \u2212 \u00d7 \u2212 | | | ( . ) ( . ) 1 1 0 76 = . V Now, E E Fe Fe Fe Fe 3 2 3 2 0 06 1 2 3 + + + + = \u2212 \u00b0 + + | | . log [Fe ] [Fe ] or, 0 7 8 0 76 0 06 5 2 3 2 3 . . . log [Fe ] [Fe ] [Fe ] [Fe ] 1 = \u2212 \u21d2 = + + + +
Answer: 5
\nSolution: 2Fe 3+ + 2I \u2013 \u2192 2Fe 2+ + I 2 ; 0.5 M excess 100 % 0 1.0 M 0.5 M Equ. CM 1.0 M 0.5 M E cell V; \u00b0 = \u2212 = 0 77 0 53 0 24 . . . K eq = 10 8 Now, 10 0 5 1 0 5 10 8 2 2 2 5 = \u00d7 \u21d2 = \u00d7 \u2212 ( . ) ( . ) C C M
Answer: 4
\nSolution: Assuming the cell as concentration cell, net cell reaction is Ag C Ag C sp sp(AgI) + \u2212 + = \u239b \u239d \u239c \u239e \u23a0 \u239f \u23af \u2192 \u23af = C K K 1 2 (AgCl) [ l ] ( ) and E C C cell O = \u2212 0 06 1 2 1 . log or, 0.102 = \u2013 0.06 \u22c5 log . . [Cl ] [Cl ] 8 1 10 1 8 10 4 10 17 10 4 \u00d7 \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f \u21d2 = \u00d7 \u2212 \u2212 \u2212 \u2212 \u2212 M
Answer: 1
\nSolution: n n eq eq Pb Tl = + Q F \u21d2 \u00d7 \u00d7 + \u00d7 \u00d7 = \u00d7 5 0 70 100 208 2 5 0 30 100 204 1 1 1 96500 . . . t \u2234 t = 3597.4 sec ; 1 hr
Answer: 0
\nSolution: n Q F x x eq r I = \u21d2 \u00d7 = \u00d7 \u00d7 \u21d2 = 0 36 192 0 075 2 3600 96500 3 . . Now, x + 6(\u20131) = y \u21d2 y = \u20133
Answer: 9
\nSolution: 2 2 2 2 2 NaCl + 2H O NaOH H Cl electrolysis \u23af \u2192 \u23af\u23af\u23af\u23af + + 2 2 2 NaOH + Cl NaCl NaClO H O \u23af \u2192 \u23af + + n NaClO formed = n Cl 2 produced from electrolysis or, ( . ) . . . 10 10 1 0 7 45 100 74 5 2 2 5 96500 3 \u00d7 \u00d7 \u00d7 \u00d7 = \u00d7 t ( n -factor of Cl 2 in electrolysis) \u2234 t = 7.72 \u00d7 10 5 sec = 8.93 days \u2248 9 days
Answer: 3
\nSolution: Cathode: Cu 2+ + 2e \u2013 Cu Anode 2H 2 O O 2 + 4H + + 4e \u2013 Moles of e \u2013 used = Q F = \u00d7 \u00d7 \u00d7 \u2212 0 161 5 60 96500 5 10 4 . \u001a Eq. of Cu 2+ present = 500 0 1 1000 2 0 1 \u00d7 \u00d7 = . . ( ) excess \u2234 Moles of H + produced = Moles of e \u2013 = 5 \u00d7 10 \u20134 \u2234 [H + ] fi nal = 5 10 500 1000 10 4 3 \u00d7 \u00d7 = \u2212 \u2212 M \u21d2 P H = 3.0
Answer: 2
\nSolution: cathode: Cu 2+ + 2e \u2013 Cu 250 0 1 1000 \u00d7 . 5 1351 96500 \u00d7 = 0.025 mole = 0.07 mole As Cu + will not remain fi nally in solution, no complex formation.
Answer: 3
\nSolution: n eq metal = n eq Cl 2 \u21d2 52 8 9 08 22 7 2 . . . E = \u00d7 \u21d2 E = 66 At. wt. (approx) = 6 4 0 032 200 . . = \u2234 valency At wt Eq.wt = = . . 200 66 3 \u001a\n8.53 Electrochemistry HINTS AND EXPLANATIONS
Answer: 2
\nSolution: Initial mass of H 2 SO 4 , w 1 = 2000 1 1 16 100 352 \u00d7 \u00d7 = . gm Final mass of H 2 SO 4 , w 2 = 2000 1 42 40 100 1136 \u00d7 \u00d7= . gm Now, n eq H 2 SO 4 produced = Q F or, ( ) 1136 352 98 1 965 9 3600 96500 \u2212 \u00d7 = \u00d7 \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f i \u21d2 i = 2A
Answer: 9
\nSolution: Ionic mobility, \u03bc \u03bb = \u00b0 = m F Speed of ion Potential gradient or, 7 5 10 96500 1 93 0 12 3 . / . / . \u00d7 = \u00d7 \u2212 distance 20 3600 \u21d2 distance = 0.09 m = 9 cm
Answer: 8
\nSolution: n eq Ag oxidised = Q F \u21d2 w 108 1 9 65 1 3600 96500 \u00d7 = \u00d7 \u00d7 . \u2234 w = 38.88 gm \u2234 Mass of anode dissolved = 38 88 100 60 64 8 . . \u00d7 \u00d7 gm \u2234 Final mass of anode = 72.8 \u2013 64.8 = 8 gm
Answer: 9
\nSolution: Cathode: 2H 2 O(l) + 2e \u2013 H 2 (g) + 2OH \u2013 (aq) Anode: 2ce \u2013 (aq) Cl 2 (g) + 2e \u2013 n eq OH\u2013 produced = Q F \u21d2 n OH \u2212 \u00d7 = \u00d7 = \u2212 1 9 65 10 96500 10 3 . \u2234 [OH \u2013 ] fi nal = 10 100 10 3 5 \u2212 \u2212 = M \u21d2 P H = 9.0 Four digit integer type
Answer: 0287
\nSolution: \u039b\u00b0 = \u00b0 + \u00b0 \u2212 m m m (AgCl) (AgCl) (Cl ) \u03bb \u03bb = \u00d7 + \u00d7 = \u00d7 \u2212 \u2212 \u2212 \u2212 \u2212 6 19 10 7 81 10 14 00 10 3 3 3 1 2 1 . . . ohm m mol Now, \u039b m C S = \u21d2 \u00d7 = \u00d7 \u2212 \u2212 \u03ba 14 10 2 8 10 3 4 . \u21d2 S = 0.02 mol m \u20133 = 2 \u00d7 10 \u20135 M = 287 \u00d7 10 \u20135 g/L
Answer: 0142
\nSolution: [S ] K [H S] [ ] . ( ) 2 1 2 2 2 8 13 3 16 10 10 0 1 10 10 \u2212 + \u2212 \u2212 \u2212 \u2212 = \u22c5 \u22c5 = \u00d7 \u00d7 = a a K H M \u2234 [Ag ] [S ] + \u2212 \u2212 \u2212 \u2212 = = \u00d7 = \u00d7 K M sp 2 48 16 16 4 10 10 2 10 Now, E Ag /Ag + = \u2212 \u00d7 \u2212 0 80 0 06 1 1 2 10 16 . . log = \u20130.142 V \u2234 Required potential = 142 mV
Answer: 0815
\nSolution: 2Hg(l) + 2Fe 3+ \u001f Hg Fe 2 2 2 2 + + + excess 10 \u20133 M Equ. 10 \u20133 \u2013 x x 2 x = \u00d7 \u2212 10 100 10 3 M \u2234 x = 9 \u00d7 10 \u20134 Now, K eq = \u00d7 \u2212 = \u00d7 \u00d7 \u00d7 \u00d7 = \u00d7 \u2212 \u2212 \u2212 \u2212 \u2212 x x x 2 10 9 10 2 9 10 1 10 9 10 2 2 3 2 4 4 2 4 2 3 4 ( ) ( ) ( ) Now, E E E K cell Fe /Fe Hg /Hg eq 3+ 2 2+ \u00b0 = \u00b0 \u2212 \u00b0 = 0 06 2 . log or, 0 7724 0 06 2 9 10 2 3 4 . . log \u2212 \u00b0 = \u22c5 \u00d7 \u2212 E Hg /Hg 2 2+ \u2234 E V Hg /Hg 2 2+ \u00b0 = 0 815 .
Answer: 0571
\nSolution: The cell reaction is 6 14 6 2 7 3 3 Fe Cr O H Fe Cr H O 2+ 2 7 2 2 + + \u2192 + + \u2212 + + + E E n cell cell = \u00b0 \u2212 \u22c5 + + + + + 0 06 3 6 3 2 2 6 2 7 2 8 . log [Fe ] [Cr ] [Fe ] [Cr O ][H ] = \u2212 \u2212 \u00d7 \u00d7 \u00d7 ( . . ) . log ( . ) ( ) ( . ) ( ) 1 35 0 77 0 06 6 0 75 4 0 75 2 1 6 2 6 8 = 0.571 V
Answer: 0090
\nSolution: Cell reaction: H 2 (g) + 2Ag + \u2192 2H + + 2Ag(s) E E P cell cell H 2 = \u00b0 \u2212 \u22c5 + + 0 06 2 2 2 . log [H ] [Ag ]\n8.54 Chapter 8 HINTS AND EXPLANATIONS or, 0.50 = 0.80 \u2013 0.03 log 1 1 2 2 \u00d7 + [Ag ] \u21d2 [Ag + ] = 10 \u20135 M \u2234 Mass of Ag in alloy = \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f \u00d7 = \u00d7 \u2212 \u2212 250 10 1000 108 2 7 10 5 4 . gm \u2234 % of Pb in alloy 2 7 10 2 7 10 2 7 10 100 90 3 4 3 . . . % \u00d7 \u2212 \u00d7 \u00d7 \u00d7 = \u2212 \u2212 \u2212
Answer: 0100
\nSolution: E K cell eq \u00b0 = 0 06 . log n \u21d2 (0.2 \u2013 0.08) = 0 06 1 . log K eq \u21d2 K eq = 100
Answer: 2633
\nSolution: Left electrode: Mn(s) \u2192 Mn 2+ + 2e \u2013 Right electrode: 2H 2 O(l) \u2192 O 2 (g) + 4H + + 4e \u2013 \u2234 Net cell reaction: 2Mn (s) + 2H 2 O(l) \u2192 2Mn e+ + O 2 (g) + 4H + Now, E E Mn P cell cell 2+ O = \u00b0 \u2212 \u22c5 \u22c5 + 0 06 4 1 2 4 2 . log [ ] [H ] = \u2212 \u2212 \u2212 \u00d7 \u00d7 [ . ( . )] . log ( . ) ( . ) . 1 229 1 185 0 06 4 0 001 0 01 0 25 1 2 4 = 2.633 V
Answer: 0193
\nSolution: \u0394 S nF E T P = \u22c5 \u2202 \u2202 \u239b \u239d \u239c \u239e \u23a0 \u239f = \u00d7 \u00d7 = 2 96500 0 001 193 . J/k
Answer: 1300
\nSolution: \u0394 \u03a3 \u0394 \u03a3 \u0394 r f f G nFE G G \u00b0 = \u2212 \u00b0 = \u00b0 \u2212 \u00b0 Products Reactants or, \u2212 \u00d7 \u00d7 = \u00d7 \u00b0 \u23a1 \u23a3 \u23a2 \u23a4 \u23a6 \u23a5 \u2212 \u00d7 + \u00d7 + \u00d7 \u2212 + \u00d7 \u2212 12 9600 2 5 1000 4 4 0 3 0 6 280 4 4 . [ ( ) (OH) \u0394 f G Al ( ( . )] \u2212 156 25 \u2234 \u0394 f G \u00b0 = \u2212 \u2212 Al KJ/mol (OH) 4 1300
Answer: 0028
\nSolution: n eq MnO 2 = Q F \u21d2 8 7 87 1 3 99 10 96500 3 . . \u00d7 = \u00d7 \u00d7 \u2212 t \u2234 t = 2.418 \u00d7 10 6 sec \u2248 28 days
Answer: 1520
\nSolution: SnCl 2 Sn 2+ + 2Cl \u2013 Cathode: Sn 2+ + 2e \u2013 Sn Anode: 2Cl \u2013 Cl 2 + 2e\u2013 Cl 2 + SnCl 2 SnCl 4 Moles of SnCl 2 taken = 19 190 0 1 = . Moles of Sn produced = 1 19 119 0 01 . . = = moles of Cl 2 produced = moles of SnCl 4 formed and moles of SnCl 4 left = 0.1 \u2013 (0.01 + 0.01) = 0.08 \u2234 m m SnCl SnCl 2 4 0 08 190 0 01 261 1520 261 = \u00d7 \u00d7 = . .
Answer: 0772
\nSolution: n eq Au = Q F \u21d2 \u00d7 \u00d7 \u00d7 \u00d7 = \u00d7 \u2212 80 8 0 10 19 7 3 197 2 4 96 500 4 . . . t \u2234 t = 772 sec
Answer: 0055
\nSolution: AgBr(s) \u001f Ag + + Br \u2013 (10 \u20137 + x )M x M Now, (10 \u20137 + x ) x = 12 \u00d7 10 \u201314 \u21d2 x = 3 \u00d7 10 \u20137 Final solution: [Ag + ] = 4 \u00d7 10 \u20137 M, [Br \u2013 ] = 3 \u00d7 10 \u20137 M; [ ] NO M 3 7 10 \u2212 \u2212 = Now, \u03ba solution = \u03bb \u00b0 m (Ag + ) \u00d7 [Ag + ] + \u03bb \u00b0 m (Br \u2013 ) \u00d7 [Br \u2013 ] + \u03bb \u00b0m ( ) [ ] NO NO 3 3 \u2212 \u2212 \u00d7 = 6 \u00d7 10 \u20133 \u00d7 (4 \u00d7 10 \u20137 \u00d7 10 3 ) + 8 \u00d7 10 \u20133 \u00d7 (3 \u00d7 10 \u20137 \u00d7 10 3 ) + 7 \u00d7 10 \u20133 (10 \u20137 \u00d7 10 \u20133 )
Answer: 0728
\nSolution: C 60 + 60O 2 60 CO 2 Moles of O 2 needed = 60 60 96 60 12 8 60 \u00d7 = \u00d7 \u00d7 = n C n eq O 2 = n eq azobenzene \u21d2 8 \u00d7 4 = w 182 8 \u00d7 \u21d2 w = 728 gm + 8H + + 8e \u2013 + 4H2O N N NO2 2
Answer: 0108
\nSolution: \u039b\u00b0 eq[Ba (PO ) ] 3 4 2 = 160 + 140 \u2013 100 = 200 Ohm \u20131 cm 2 eq \u20131 Now, \u2227 \u00b0 eq = \u2227 eq = \u03ba C \u21d2 200 1 2 10 5 5 = \u00d7 \u2212 . \u2234 S = 6 \u00d7 10 \u20138 eq/cm 3 = 6 \u00d7 10 \u20135 N = 10 \u20135 M \u2234 K sp = 108 S 5 = 10 8 \u00d7 10 \u201325 M 5
Answer: B
\nSolution: P gas + 15.6 = 53.3 + 76.3 \u21d2 P gas = 114 cm Hg = 1.5 atm
Answer: D
\nSolution: Fractional increase = V V V 2 1 1 \u2212 = V V 2 1 1 \u2212 = P P 1 2 1 \u2212 = H H H + \u2212 7 7 1 = 1 7
Answer: D
\nSolution: P 2 P 1 44.5 46 P 0 P 0 45.25 45.25 P 0 \u00d7 45.25 = P 1 \u00d7 46 = P 2 \u00d7 44.5 P 1 + 5 sin30\u00b0 = P 2 P 0 45 25 46 \u00d7 . + 5 \u00d7 1 2 = P 0 45 25 44 5 \u00d7 . . \u21d2 P 0 = 75.4 cm Hg
Answer: C
\nSolution: P 1 V 1 = P 2 V 2 \u21d2 10 \u00d7 2A = (10 + h ) \u00d7 h A \u21d2 h = 1.71 m
Answer: C
\nSolution: Theory based
Answer: C
\nSolution: dv dt = V 0 273 = 0.08 \u21d2 V 0 = 21.84 L
Answer: A
\nSolution: V T 1 1 = V T 2 2 \u21d2 1 0 0 . x + = 0 6 100 . ( ) x + \u2212 \u21d2 x = 250 \u21d2 0 K = \u2013250\u00b0C
Answer: C
\nSolution: V T 1 1 = V T 2 2 \u21d2 V T = V V T T + + \u0394 \u0394 \u21d2 \u0394 \u0394 V V T . = 1 T \u21d2 y = 1 x
Answer: B
\nSolution: V T 1 1 = V T 2 2 \u21d2 V t 1 1 273 + = 1.1 V t 1 2 273 + \u2234 Percentage increase in temperature = t t t 2 1 1 \u2212 \u00d7 100 = ( ) 10 2730 1 + t %
Answer: D
\nSolution: Number of SO 2 molecules = N \u21d2 Number of atoms = 3N
Answer: C
\nSolution: n n N O 2 2 = V V N O 2 2 \u21d2 m m N O 2 2 28 32 = 1 7 8 \u21d2 m m N O 2 2 = 1 1
Answer: A
\nSolution: V n 1 1 = V n 2 2 \u21d2 4 3 10 2 8 3 \u03c0 ( ) = 4 3 2 1 3 \u03c0 d ( ) \u21d2 d = 5 cm
Answer: A
\nSolution: P = P CO 2 + P air = 0 5 0 0821 300 1 1 . . \u00d7 \u00d7 + = 13.315 atm
Answer: D
\nSolution: Weight of fi lled balloon, W = 20 g + 40 \u00d7 0.6 = 44 g Weight of displaced air, B = 40 \u00d7 1.3 = 52 g \u2234 Balloon will lift upward with pay load = 52 \u2013 44 = 8 g B W\n3.45 Gaseous State HINTS AND EXPLANATIONS
Answer: B
\nSolution: Water will behave like ideal gas on disappearance of intermolecular forces. V = nRT P = 4 5 10 18 3 . \u00d7 \u00d7 (22.4 \u00d7 10 \u20133 ) m 3 = 5.6 m 3
Answer: C
\nSolution: d = m v \u21d2 1.5 = n n n n co co co co \u00d7 + \u00d7 + \u00d7 \u00d7 28 44 0 0821 300 1 2 2 ( ) . \u21d2 n co = 7 055 8 945 . . n co 2 Alkali will absorb all CO 2 . Hence, final pressure is due to CO. P co = n n n co co co + 2 \u00d7 P total = 7 055 7 055 8 945 760 . . . + \u00d7 mm = 335.1 mm
Answer: C
\nSolution: V V water vapour water = 1 0 0821 373 1 18 0 96 \u00d7 \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f \u239b \u239d \u239c \u239e \u23a0 \u239f . . l ml = 1633.24
Answer: C
\nSolution: V O 2 = 3 2 32 0 0821 310 1 . . \u00d7 \u00d7 = 2.5451 L V CO 2 = 8 8 44 0 0821 310 1 . . \u00d7 \u00d7 = 5.0902 L
Answer: B
\nSolution: n CO 2 = 200 0 1 1000 \u00d7 . = 0.02 \u2234 V CO 2 = 0.02 \u00d7 22.4 = 0.448 L
Answer: A
\nSolution: d d d ( ) \u03c1 = M RT = 1.2 \u00d7 10 \u20135 Kg m \u20133 Pa \u20131 \u21d2 M 8 314 300 . \u00d7 = 1.2 \u00d7 10 \u20135 \u2234 M air = 0.03 Kg/mol = 30 gm/mol Now, 30 = n n n n N O N O 2 2 28 + 32 + \u00d7 \u00d7 2 2 \u21d2 n n N O 2 2 : = 1:1
Answer: D
\nSolution: m T P 1 1 1 = m T P 2 2 2 \u21d2 4 \u00d7 T P = m T P 2 2 \u00d7 2 \u21d2 m 2 = 16 gm Hence, (16 \u2013 4) = 12 gm gas should be added.
Answer: B
\nSolution: 74 5 50 . = 1.49 times
Answer: C
\nSolution: V 1 d 1 = V 2 d 2 \u21d2 1500 \u00d7 1.25 = 3.92 \u00d7 d 2 \u21d2 d 2 = 478.3 kg/mol
Answer: A
\nSolution: P V T 1 1 1 = P V T 2 2 2 \u21d2 P r T \u00d7 4 3 1 3 \u03c0 = P r T 4 4 3 2 3 \u00d7 \u03c0 2 \u21d2 r 2 = 2 r 1 \u2234 % Increase in radius = r r r 2 1 1 \u2212 \u00d7 100 = 100 %
Answer: B
\nSolution: Constant = P 2 V = nRT V V \uf8eb \uf8ed \uf8ec \uf8f6 \uf8f8 \uf8f7 2 \u21d2 T V 2 = Constant \u2234 On expansion, temperature will increase.
Answer: C
\nSolution: r R M = \u21d2 r r r n H N e 2 > > 2
Answer: B
\nSolution: P \u00d7 3 = 7 28 0 0821 300 \u00d7 \u00d7 . \u21d2 2.0525 atm
Answer: B
\nSolution: P 10 atm 2 atm 4 L 20 L 1 V 2 T 4 L 20 L 12 L V 1 2 40 R T 2 atm 10 atm 6 atm P 40 R For T max , V = 12 L and P = 6 atm and hence, T max = 6 12 1 0 08 \u00d7 \u00d7 . = 900 K.\n3.46 Chapter 3 HINTS AND EXPLANATIONS
Answer: A
\nSolution: 2Al + 2NaOH + 2H 2 O \u2192 2NaAlO 2 + 3H 2 2 mole 3 mole \u2234 0.15 27 mole 3 2 \u00d7 0.15 27 mole \u2234 V H 2 = 1 5 0 15 27 0 0831 300 0 831 . . . . \u00d7 \u00d7 \u00d7 = 0.25L
Answer: A
\nSolution: N 2 \u2192 2N Initial mole a = 1 4 28 . 0 Final mole a \u2013 0.4 a 2 \u00d7 0.4 a = 0.6 a = 0.8 a Final total moles = 0.6 a + 0.8 a = 1.4 \u00d7 1 4 28 . = 0.07 \u2234 P = 0 07 0 0821 1800 5 . . \u00d7 \u00d7 \u001e 2.07 atm
Answer: B
\nSolution: CH 4 (g) + 2O 2 (g) 127 \u00b0 \u23af \u2192 \u23af\u23af C CO 2 (g) + 2H 2 O (g) As there is no change in mole of gases, P T 1 1 = P T 2 2 \u21d2 1 300 = P 2 400 \u21d2 P 2 = 1.33 atm
Answer: C
\nSolution: P c = X c P total \u21d2 10 \u2013 (1 + 3) = n C 10 10 \u00d7 \u21d2 n c = 6 \u2234 Mass of C = 6 \u00d7 2 = 12 gm
Answer: A
\nSolution: x x M R 4 + 5 = 760 2.4 300 \u2212 \u00d7 \u00d7 and x R 4 = 19 2.4 15 \u00d7 \u00d7 \u2234 M = 96
Answer: B
\nSolution: C 2 H 6 + 7 2 O 2 2CO 2 + 3H 2 O( l ) 1 vol 7 2 vol 2 vol 0 vol \u2234 10 ml 35 ml 20 ml 0 Final volume should be 20 + (40 \u2013 35) = 25 ml but it is 26 ml. Hence, volume occupied by water vapour is (26 \u2013 25) = 1 ml. \u2234 Vapour pressure of water = 1 26 \u00d7 1 atm = 760 26 = 29.23 mm Hg
Answer: C
\nSolution: n H 2 O vapour needed = ( . ) . 26 463 24 1 760 0 0821 300 \u2212 \u00d7 \u00d7 \u00d7 = 1.32 \u00d7 10 \u20134
Answer: B
\nSolution: P = 1 2 0 0821 300 18 50 760 . . \u00d7 \u00d7 \u00d7 \u00d7 = 24.96 mm Hg
Answer: B
\nSolution: Mass of water lost per day = \u2206 P.V RT M \u00d7 = ( ) . 45 5 760 0 0821 310 18 \u2212 \u00d7 \u00d7 \u00d710000 = 372.23 gm
Answer: C
\nSolution: Vapour pressure is a function of temperature only
Answer: D
\nSolution: After achievement of equilibrium with its liquid form which will form on continuous injection of vapour, the pressure due to vapours become constant.
Answer: B
\nSolution: Rate of evaporation will remain constant throughout because neither surface area nor temperature are changing
Answer: A
\nSolution: r r x y = 1 5 and r r y z = 1 6 \u21d2 r r z x = 30 1
Answer: D
\nSolution: Smaller the rate of diff usion of HX, more closer to the HX end, NH 4 X will form.
Answer: C
\nSolution: r r N H 2 2 = M M H N 2 2 \u21d2 \u0394 \u0394 P P t / / 60 = 2 28 \u21d2 t = 16.04 min
Answer: B
\nSolution: M dry air > M moist air
Answer: A
\nSolution: r r CH HBr 4 = P P CH HBr 4 M M HBr CH 4 \u21d2 1 1 = n n CH HBr 4 81 16 \u21d2 n n CH HBr 4 = 0.4 \u2234 X CH 4 = n n n CH CH HBr 4 4 + = 0.31
Answer: C
\nSolution: As HCl will diff use slowly, white fumes will form closer to HCl end.
Answer: B
\nSolution: In gases, the intermolecular distance is much higher than the size of molecules.
Answer: A
\nSolution: u u av,2 av,1 = T T 2 1 = 375 250 = 1.22
Answer: C
\nSolution: u u rms, O rms, O 2 = 3R 2T 16 3RT 32 \u00d7 2 1 \u21d2 u rms, o = 2 V\n3.47 Gaseous State HINTS AND EXPLANATIONS
Answer: B
\nSolution: Average speed for a gas depends on temperature and it is independent from the presence of other gas.
Answer: A
\nSolution: Difference in any two kind of speed, \u2206 u K T = \u00d7 Now, d u dT ( ) \u0394 = K 2 T \u21d2 On increasing temperature, \u0394 u decreases.
Answer: D
\nSolution: u u av x av y , , = 2 1 = T T X Y \u21d2 T T X Y = 4 1 Now, P P X Y = nRT V nRT V X X Y Y = T T V V X Y Y X \u00d7 = 4 1 2 1 \u00d7 = 8 1
Answer: A
\nSolution: 1 \u00d7 V = 1 M R T A \u00d7 \u00d7 \u21d2 M M B A = 4 1 0.5 \u00d7 V = 2 M RT B \u00d7 \u2234 u u av A av B , , = M M B A = 2 1
Answer: B
\nSolution: u av, A = u av, B \u21d2 8 RT M A A \u03c0 = 3 RT M B \u21d2 M M B A = 3 8 \u03c0 Now, u av, A = u av, B \u21d2 8 RT M A A \u03c0 = 8 RT M B B \u03c0 \u21d2 T T A B = M M A B = 8 3 1 \u03c0 <
Answer: C
\nSolution: 1 4 . N *. u av = 1 4 6 10 22 4 10 8 8 314 273 28 10 23 3 3 \u00d7 \u00d7 \u00d7 \u00d7 \u00d7 \u00d7 \u00d7 \u00d7 \u2212 \u2212 . . \u03c0 = 3.05 \u00d7 10 27 m \u20132 s \u20131
Answer: D
\nSolution: u u O O rms rms , , 2 3 = 3 600 32 48 3 300 R R \u00d7 \u00d7 \u00d7 = 3 \u21d2 u rms,O 2 = 3 v m/s
Answer: B
\nSolution: Average translational K.E. per gm = 3 2 RT M
Answer: D
\nSolution: T M A A = T M B B \u21d2 u rms = 3 RT M = Same for both
Answer: C
\nSolution: 3 2 KT = qV \u21d2 3 2 8 314 6 022 10 23 \u00d7 \u00d7 \u00d7 . . T = 1.602 \u00d7 10 \u201319 \u00d7 3 \u21d2 T = 23207.2 K
Answer: B
\nSolution: u 2 rms \u2260 u 2 av
Answer: B
\nSolution: Z w = 1 4 . N *. u av = 1 4 8 \u00d7 \u00d7 P N RT RT M A . \u03c0 \u21d2 Z w \u221d 1 T
Answer: A
\nSolution: u u rms, CH rms, SO 4 2 = 3 16 64 3 300 R T R \u00d7 \u00d7 \u00d7 = 4 1 \u21d2 T = 1200 K \u2234 Average K.E. per mole = 3 2 RT = 3 2 2 1200 \u00d7 \u00d7 = 3600 cal
Answer: A
\nSolution: u av \u221d T \u21d2 u u 2 1 = 432 300 = 1.2
Answer: C
\nSolution: Mole of gas cannot change.
Answer: C
\nSolution: Collision number, Z 1 = 2 \u03c0\u03c3 2 . u av . N * = 2 \u03c0\u03c3 2 . 8 RT M PN RT A \u03c0 \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f \u2234 Z 1 \u221d 1 T Collision frequency, Z 11 = 1 2 \u03c0\u03c3 2 u av . N* 2 = 1 2 \u03c0\u03c3 2 . 8 2 RT M PN RT A \u03c0 \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f \u2234 Z 11 \u221d 1 3 2 T Mean free path, \u03bb = 1 2 2 \u03c0\u03c3 N * = RT PN A 2 2 \u03c0\u03c3 . \u21d2 \u03bb \u221d T
Answer: A
\nSolution: \u03bb = RT PN A 2 2 \u03c0\u03c3 \u00d7 = 8 314 300 2 1 5 10 4 1 10 1 013 10 6 022 10 10 2 14 5 23 . ( . ) ( . . ) ( . \u00d7 \u00d7 \u00d7 \u00d7 \u00d7 \u00d7 \u00d7 \u00d7 \u00d7 \u2212 \u2212 \u03c0 ) ) = 1.0 \u00d7 10 7 m
Answer: B
\nSolution: Theory based
Answer: C
\nSolution: Velocity is a vector quality.
Answer: A
\nSolution: dN N = 4 2 3 2 2 2 2 \u03c0 \u03c0 m KT u e du mu KT \u239b \u239d \u239c \u239e \u23a0 \u239f \u00d7 \u00d7 \u00d7 \u2212 = 2 1 3 2 \u03c0 KT E e dE E KT \u239b \u239d \u239c \u239e \u23a0 \u239f \u00d7 \u00d7 \u00d7 \u2212\n3.48 Chapter 3 HINTS AND EXPLANATIONS For most probable K.E., d dN N dE ( ) = 0 \u21d2 E = 1 2 KT
Answer: D
\nSolution: Deviation from ideal behavior is maximum at low temperature and high pressure.
Answer: A
\nSolution: Z > 1 for H 2 at 0\u00b0C at all pressure.
Answer: D
\nSolution: Z < 1 at low pressure and Z > 1 at high pressure.
Answer: C
\nSolution: Theory based
Answer: B
\nSolution: V 1 = V \u2013 nb \u21d2 b = V V n \u2212 1 \u21d2 4 6 3 \u00d7 \u00d7 \u03c0 d N A = V V n \u2212 1 \u2234 d = 3 2 1 1 3 ( ) V V nN A \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f \u03c0
Answer: B
\nSolution: P i = P an V + 2 2 \u21d2 P = P an V i + \u239b \u239d \u239c \u239c \u239e \u23a0 \u239f \u239f 2 2 Greater the value of \u2018 a \u2019, smaller will be \u2018 P \u2019.
Answer: D
\nSolution: Gaseous mixture is always homogeneous.
Answer: A
\nSolution: P 1 V 1 = P 2 V 2 \u21d2 0.5 \u00d7 2000 = 100 \u00d7 V 2 \u21d2 V 2 = 10 ml < 13 ml As the real volume is greater than ideal, the volume occupied by the molecule is significant.
Answer: A
\nSolution: When attractive forces are dominant, V real < V ideal .
Answer: A
\nSolution: B = b a RT \u2212 = 0.03 \u2013 1 344 0 0821 273 . . \u00d7 = \u2013 0.03 l/mol
Answer: A
\nSolution: Z = PV RT m = V V b m m \u2212 = 10 10 b b b \u2212 = 10 9
Answer: A
\nSolution: P an V V nb + \u239b \u239d \u239c \u239c \u239e \u23a0 \u239f \u239f \u2212 2 2 ( ) = nRT may be expressed as P a d M M d b + \u239b \u239d \u239c \u239c \u239e \u23a0 \u239f \u239f \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f . 2 2 = RT as d = m v = n m v \u00d7 Now, P + \u00d7 \u23a1 \u23a3 \u23a2 \u23a4 \u23a6 \u23a5 \u2212 \u23a1 \u23a3 \u23a2 \u23a4 \u23a6 \u23a5 3 6 2 2 44 44 2 2 0 05 2 2 . ( . ) ( ) . . = 0.0821 \u00d7 300 \u21d2 P = 1.226 atm
Answer: A
\nSolution: B = b a RT \u2212 = \u20131.0 L/mol Now, PV m = RT 1 + \u239b \u239d \u239c \u239e \u23a0 \u239f B V m and d = M V m Hence, PM d = RT 1 + \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f B d M or, 1 40 \u00d7 d = 0.08 \u00d7 262.5 1 1 0 40 + \u2212 \u00d7 \u23a1 \u23a3 \u23a2 \u23a4 \u23a6 \u23a5 ( . ) d \u2234 d = 2.005 g/L
Answer: A
\nSolution: When P \u2192 0, V \u2192 \u221e and hence e a/VRT \u2192 1 and ( V \u2013 b ) \u2192 V . Hence, P = RT V = 0 0821 300 410 5 . . \u00d7 = 0.06 atm
Answer: A
\nSolution: For a van der Waals gas, Z = V V b a V RT m m m \u2212 \u2212 or, 0.8 = 0 5 0 5 0 04 0 5 0 08 300 . . . . . \u2212 \u2212 \u00d7 \u00d7 a \u21d2 a = 3.44 atm L 2 /mol 2
Answer: B
\nSolution: At Boyle\u2019s temperature, dz dp = 0 \u21d2 T = 168 0 35 . = 480 K
Answer: B
\nSolution: Theory based. The initial slope of Z vs. P curve increases with increase in temperature, above Boyle\u2019s temperature, only upto 2 \u00d7 T B . Then, the slope starts decreasing.
Answer: A
\nSolution: For van der Waals gas, Z = V V b a V RT m m m \u2212 \u2212 \u00d7 At Boyle\u2019s temperature, Z = V V b a V R a Rb m m m \u2212 \u2212 \u00d7 \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f = 1 2 + \u2212 b V V b m m ( )
Answer: D
\nSolution: Ideal gas can never be liquified.
Answer: A
\nSolution: For ideal behavior, Boyle\u2019s temperature should be closer to 600 K.
Answer: C
\nSolution: Theory based
Answer: A
\nSolution: T c < T B
Answer: B
\nSolution: T P c c = 8 27 27 2 a Rb a b = 8 b R \u2234 T P T P c c c c \u239b \u239d \u239c \u239e \u23a0 \u239f \u239b \u239d \u239c \u239e \u23a0 \u239f CO CH 2 4 = b b CO CH 2 4 = 304 72 190 45 = 1 1\n3.49 Gaseous State HINTS AND EXPLANATIONS
Answer: D
\nSolution: At T > T c , the gas can never be liquified.
Answer: D
\nSolution: Theory based
Answer: A
\nSolution: P c V c = 3 8 RT c \u21d2 V c = 3 8 0 0821 128 41 05 \u00d7 \u00d7 . . = 0.096 L/mol
Answer: C, D
\nSolution: Boyle\u2019s law constant = PV = nRT
Answer: B, C
\nSolution: P V RT P V RT 0 0 0 0 0 0 + = PV R T PV RT 0 0 0 0 2 \u00d7 + \u21d2 P = 4 3 P 0 and n = 4 3 2 0 0 0 P V R T \u00d7 \u00d7 = 2 3 0 0 0 P V RT
Answer: A, B, C, D
\nSolution: P \u0192 = P i V V V n + \u239b \u239d \u239c \u239e \u23a0 \u239f \u0394 (a) P \u0192 = 24.2 \u00d7 10 10 1 1 + \u239b \u239d \u239c \u239e \u23a0 \u239f = 22 atm (b) P \u0192 = 24.2 \u00d7 10 10 1 2 + \u239b \u239d \u239c \u239e \u23a0 \u239f = 20 atm (c) P \u03b7 = P 10 10 1 + \u239b \u239d \u239c \u239e \u23a0 \u239f n \u21d2 n = ln ln . \u03b7 1 1 (d) P \u0192 = 24.2 \u00d7 10 10 1 + \u239b \u239d \u239c \u239e \u23a0 \u239f n atm
Answer: B, C
\nSolution: As the average molar mass increases, the molar mass of vapours must be greater than that of N 2 .
Answer: B, C
\nSolution: r H 2 > r D 2
Answer: A, B, C
\nSolution: (a) Number of molecules colliding at the wall per unit time per unit area, Z w = 1 4 . u av . N * N * is same for both but u av , He > u av , Ne (b) Average force per collision \u221d Change in momentum \u221d M
Answer: B, C
\nSolution: At valve \u2013 I: P 1 V 1 = P 2 V 2 \u21d2 1 \u00d7 60 A = P 2 \u00d7 45 A \u21d2 P 2 = 1.33 atm < 1.5 atm Hence, valve \u2013 I will not open. At valve \u2013 II: P 1 V 1 = P 2 V 2 \u21d2 1 \u00d7 60 A = P 2 \u00d7 30 A \u21d2 P 2 = 2 atm < 2.2 atm Hence, valve \u2013 II will not open. At valve \u2013 III: P 1 V 1 = P 2 V 2 \u21d2 1 \u00d7 60 A = P 2 \u00d7 20 A \u21d2 P 2 = 3 atm > 2.5 atm Hence, valve \u2013 III will open fi rst. As the piston will reach at valve \u2013 III, the gas will come out till the pressure of gas becomes 2.5 atm. Now, 2 5 20 821 1000 . \u00d7 \u00d7 = n \u00d7 0.0821 \u00d7 300 \u21d2 Moles of gas remained, n = 5 3 At valve \u2013 IV: P 1 V 1 = P 2 V 2 \u21d2 2.5 \u00d7 20 A = P 2 \u00d7 15 A \u21d2 P 2 = 3.33 atm < 4.4 atm Hence, valve - IV will not open. At valve \u2013 V: P 1 V 1 = P 2 V 2 \u21d2 2.5 \u00d7 20 A = P 2 \u00d7 10 A \u21d2 P 2 = 5 atm > 4.8 atm Hence, valve \u2013 V will open until the gas pressure becomes 4.8 atm.
Answer: C
\nSolution: u av \u221d T
Answer: A, B, C
\nSolution: u u rms, A rms, B = 3 300 3 400 R M M R A B \u00d7 \u00d7 : = 3 2 \u21d2 M A = M B
Answer: C, D
\nSolution: On increasing the temperature at constant volume, the average speed of molecules as well as number of molecular collisions at wall increases.
Answer: A, D
\nSolution: Theory based
Answer: A, B, C
\nSolution: Theory based
Answer: A, B
\nSolution: At very high pressure, P a V + \u239b \u239d \u239c \u239e \u23a0 \u239f 2 \u001e P \u21d2 Z = 1+ b P RT . and Z = PV RT \u21d2 P RT = Z V \u21d2 Z = V V b \u2212\n3.50 Chapter 3 HINTS AND EXPLANATIONS
Answer: A, B, C, D
\nSolution: T c = 273 + (\u2013177) = 96 K \u21d2 T B = 27 8 \u00d7 96 = 324K = 51\u00b0C (a) Z = PV RT m . = 0 821 9 6 0 0821 96 . . . \u00d7 \u00d7 = 1 But at T = T c , Z < 1 at low pressure (b) Z = PV RT m . = 0 821 40 0 0821 400 . . \u00d7 \u00d7 = 1 But at T > T B , Z > 1 at all pressure (c) Z = PV RT m . = 82 1 0 310 0 0821 324 . . . \u00d7 \u00d7 = 0.96 < 1 But at T = T B and P > 50 atm, Z > 1 (d) Z = PV RT m . = 0 821 32 4 10 0 0821 324 3 . . . \u00d7 \u00d7 \u00d7 \u2212 = 10 \u20133 < 1 But at T = T c and P < 50 atm, Z = 1
Answer: A, B, D
\nSolution: a. 9 8 27 36 8 R a Rb \u00d7 \u00d7 = a b. 3 \u00d7 a b 27 2 \u00d7 (3 b ) 2 = a c. 3 8 27 36 8 27 2 \u00d7 \u00d7 a b a Rb \u2260 a d. 27 64 8 27 27 2 2 2 \u00d7 \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f R a Rb a b = a
Answer: A, D
\nSolution: Theory based
Answer: A, B, C
\nSolution: Theory based
Answer: C
\nSolution: Real gas may behave ideally at Boyle\u2019s temperature. P = RT V m = R a Rb V m \u00d7 = a bVm .
Answer: A, C
\nSolution: Z = PV RT m = 1 + B \u2032 .P + C \u2032 .P 2 + \u2026. (1) Z = PV RT m = 1 + B Vm + C V m 2 + \u2026. (2) Or, P = RT V m (1 + B V m + C V m 2 + \u2026.) Substituting this value in Equation (1), we get: Z = 1 + B \u2032 . RT V B V C V m m m 1 2 + + + \u239b \u239d \u239c \u239c \u239e \u23a0 \u239f \u239f \u23a7 \u23a8 \u23aa \u23a9 \u23aa \u23ab \u23ac \u23aa \u23ad \u23aa ... + C \u2032 . RT V B V C V m m m 1 2 2 + + + \u239b \u239d \u239c \u239c \u239e \u23a0 \u239f \u239f \u23a7 \u23a8 \u23aa \u23a9 \u23aa \u23ab \u23ac \u23aa \u23ad \u23aa ... + \u2026 = 1 + \u2032 B RT V m + \u2032 + \u2032 B RT B C RT V m . ( ) 2 2 + \u2026. Comparing this with Equation (2), we get: B \u2032 RT = B and B \u2032 RT . B + C .( RT ) 2 = C
Answer: A, B, C, D
\nSolution: Theory based
Answer: C
\nSolution: Number of strokes = ( ) ( ) ( ) ( ) 8 bar cm 1 bar cm \u00d7 \u00d7 \u00d7 1000 25 4 3 3 = 80
Answer: B
\nSolution: F = P.A = 8 10 5 2 \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f N m \u00d7 (4 \u00d7 10 \u20134 m 2 ) = 320 N
Answer: C
\nSolution: m = F g = 320 10 = 32 kg\n3.51 Gaseous State HINTS AND EXPLANATIONS Comprehension \u2013 II
Answer: C
\nSolution: PV m = RT (Let 0K = \u2013 x \u00b0N) 28 = R (0 + x ) x = 233.33 40 = R (100 + x ) \u2234 0K = \u2013233.33\u00b0N
Answer: B
\nSolution: R = 28 x = 0.12 L \u2013 atm/K-mol
Answer: A
\nSolution: V = nRT P = 2 0 12 66 67 233 33 2 \u00d7 \u00d7 + . ( . . ) = 36 L Comprehension \u2013 III
Answer: C
\nSolution: A 20 A B D C 20 60 A B D C B 20 A B D C 20 60 P 1 1 atm 30 A B D C 20 50 P 2 1 atm For AB column: P 1 = 1 atm = 7 6 cm Hg V 1 = 20 A cm 3 P 2 = 1 atm \u2013 20 cm Hg = 76 \u2013 20 = 56 cm Hg V 2 = 30 A cm 3 Now, P 1 V 1 \u2260 P 2 V 2 Hence, only end A is not closed. C 20 A B D C 20 60 P 1 1 atm 30 A B D C 20 50 P 2 1 atm For CD column: P 1 = 1 atm = 76 cm Hg V 1 = 60 A cm 3 P 2 = 1 atm + 20 cm Hg = 76 + 20 = 96 cm Hg V 2 = 50 A cm 3 As P 1 V 1 \u2260 P 2 V 2 , only end D is not closed. Hence, both the ends are closed.
Answer: B
\nSolution: P 0 P 0 20 20 60 A B D C A B D C P 1 P 2 30 20 50 For AB column: P 0 \u00d7 20 = P 1 \u00d7 30 For CD column: P 0 \u00d7 60 = P 2 \u00d7 50 As P 1 + 20 cm Hg = P 2 or, 20 30 0 P + 20 cm Hg = 60 50 0 P \u21d2 P 0 = 37.5 cm Hg
Answer: A
\nSolution: D D C A B P 4 P 3 (80 \u2013 x ) 20 x P 4 + 20 cm Hg = P 3 or, 60 80 0 P x ( ) \u2212 + 20 = 20 0 P x \u21d2 x = 13.88 If both ends are open, then mercury will fall down.\n3.52 Chapter 3 HINTS AND EXPLANATIONS Comprehension \u2013 IV
Answer: B
\nSolution: Total moles of product gases = PV RT = 410 5 2 9 0 0821 2000 . . . \u00d7 \u00d7 = 7.25 \u2234 Moles of gases per 0.04 mole of nitroglycerine = 0.04 \u00d7 7.25 = 0.29
Answer: B
\nSolution: Moles of gases except A = 4 75 0 821 0 0821 250 . . . \u00d7 \u00d7 = 0.19 \u2018A\u2019 must be H 2 O because it solidifies at \u201323\u00b0C and its mole = 0.29 \u2013 0.19 = 0.10.
Answer: A
\nSolution: Moles of gases C and D = 2 1 0 821 0 0821 300 . . . \u00d7 \u00d7 = 0.07 \u2234 Mole of gas \u2018B\u2019, which is CO 2 = 0.19 \u2013 0.07 = 0.12
Answer: D
\nSolution: The gas remained, D must be N 2 and its mole = 1 8 0 821 0 0821 300 . . . \u00d7 \u00d7 = 0.06 and gas \u2018C\u2019 is O 2 and its mole = 0.07 \u2013 0.06 = 0.01. Comprehension \u2013 V
Answer: B
\nSolution: All H 2 O(g) will solidify in bulb \u2018B\u2019.
Answer: D
\nSolution: n H O 2 + n CO 2 + n N 2 = 570 1 642 760 0 0821 300 \u00d7 \u00d7 \u00d7 . . = 0.05 n CO 2 + n N 2 = 0 21 1 642 0 0821 300 0 21 1 642 0 0821 200 . . . . . . \u00d7 \u00d7 + \u00d7 \u00d7 = 0.035 \u2234 n H O 2 = 0.05 \u2013 0.035 = 0.015
Answer: B
\nSolution: H 2 O(g) will solidify in \u2018B\u2019 as well as \u2018C\u2019 but CO 2 (g) will solidify only in \u2018C\u2019.
Answer: A
\nSolution: n N 2 = 22 8 1 642 760 0 0821 1 300 1 200 1 80 . . . \u00d7 \u00d7 + + \u23a1 \u23a3 \u23a2 \u23a4 \u23a6 \u23a5 = 0.0125 \u2234 n CO 2 = 0.035 \u2013 0.0125 = 0.0225 Comprehension VI
Answer: C
\nSolution: Let x mole, NH 4 Cl was present initially. NH 4 Cl(s) \u2192 NH 3 (g) + HCl(g) x mole x mole Now, PV = nRT 114 \u00d7 V = 0.01 \u00d7 R \u00d7 300 and 908 \u00d7 V = (0.01 + 2 x ) \u00d7 R \u00d7 600 \u2234 x \u2248 0.015 \u2234 Mass of NH 4 Cl = x \u00d7 53.5 \u2248 0.8 gm
Answer: B
\nSolution: P NH 3 = 908 114 2 2 \u2212 \u00d7 340 mm Hg
Answer: B
\nSolution: V = 0 01 0 0821 300 760 114 . . \u00d7 \u00d7 \u00d7 = 1.642 L Comprehension \u2013 VII
Answer: A
\nSolution: Water will vaporize till P H O 2 = 0.04 atm Now, PV = nRT \u21d2 0.04 \u00d7 (40 \u00d7 10 3 ) = w 18 \u00d7 0.08 \u00d7 300 or w = 1200 = 1.2 kg \u2234 Percentage of water vaporized = 1 2 5 . \u00d7 100 = 24 %
Answer: A
\nSolution: V = nRT P = 5000 18 0 08 300 0 04 \u00d7 \u00d7 . . = 1.67 \u00d7 10 5 L Comprehension \u2013 VIII
Answer: B
\nSolution: \u2212 dP dt = K ( P \u2013 P 0 ) \u21d2 \u2212 \u2212 \u222b dP P P P P 0 1 2 = K dt t 0 \u222b \u21d2 ln P P P P 1 0 2 0 \u2212 \u2212 = Kt or, ln 20 1 1 2 \u2212 \u2212 P = 0.001 \u00d7 3600 = ln38 \u21d2 P 2 = 1.5 atm
Answer: C
\nSolution: Number of balloons = ( . ) 20 1 5 10 1 2 \u2212 \u00d7 \u00d7 = 92.5 \u2248 92
Answer: B
\nSolution: ln P P P P 1 0 2 0 \u2212 \u2212 = Kt \u21d2 ln 20 1 2 1 \u2212 \u2212 = 0.001 \u00d7 t \u21d2 t = 2900 sec\n3.53 Gaseous State HINTS AND EXPLANATIONS Comprehension \u2013 IX
Answer: A
\nSolution: \u2212 dP dt = K.P \u21d2 \u2212 \u222b dP P P 1 5 . atm = K dt t 0 \u222b \u21d2 P = (1.5 atm). e \u2013 Kt
Answer: A
\nSolution: A 38 cm 2 A x x 2 The pressure of gas in closed arm, P = 1 atm + 38 3 2 \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f x cm = 114 3 2 \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f x cm Hg Now, 114 3 2 \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f x = 114 \u00d7 e \u2013 kt \u21d2 x = 76 (1\u2013 e \u2013 kt ) cm Hg
Answer: A
\nSolution: x 2 = 38(1 \u2013 e \u2013 kt ) cm Hg Comprehension \u2013 X
Answer: C
\nSolution: u mp = 2 RT M \u21d2 T = M u R \u00d7 mp 2 2 = ( ) ( ) 32 10 400 2 8 3 2 \u00d7 \u00d7 \u00d7 \u2212 = 320K = 47\u00b0C
Answer: A
\nSolution: u rms \u2013 u mp = 400 m/s \u21d2 3 2 RT M RT M \u2212 = 400 m/s or, T = 400 3 2 2 \u2212 \u239b \u239d \u239c \u239c \u239e \u23a0 \u239f \u239f \u00d7 M R = ( ) 400 3 2 2 6 2 10 8 2 3 + \u2212 \u00d7 \u00d7 \u2212 = 400 K = 127\u00b0C
Answer: D
\nSolution: c 1 f ( c ) c 2 C 4 \u03c0 M RT 2 3 2 \u03c0 \u239b \u239d \u239c \u239e \u23a0 \u239f \u00d7 C 1 2 \u00d7 e MC RT \u2212 1 2 2 = 4 \u03c0 M RT 2 3 2 \u03c0 \u239b \u239d \u239c \u239e \u23a0 \u239f \u00d7 C 2 2 \u00d7 e MC RT \u2212 2 2 2 or, C C 1 2 2 2 = e M C C RT ( ) 1 2 2 2 2 \u2212 or, M C C RT ( ) 1 2 2 2 2 \u2212 = 2 ln C C 1 2 \u239b \u239d \u239c \u239e \u23a0 \u239f \u2234 T = M C C R C C ( ) .ln 1 2 2 2 1 2 4 \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f = ( )( ) ln 28 10 300 600 4 8 300 600 3 2 2 \u00d7 \u2212 \u00d7 \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f \u2212 = 337.5 K = 64.5\u00b0C
Answer: A
\nSolution: c = ? f ( c ) T n . T C 4 2 3 2 2 2 2 \u03c0 \u03c0 M RT c e MC RT \u239b \u239d \u239c \u239e \u23a0 \u239f \u00d7 \u00d7 \u2212 = 4 2 3 2 2 2 2 \u03c0 \u03c0 M RT n c e MC RT n . . \u239b \u239d \u239c \u239e \u23a0 \u239f \u00d7 \u00d7 \u2212 or, n 3/2 = e MC RT n 2 2 1 1 \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f or, 3 2 ln n = MC RT n n 2 2 1 \u00d7 \u2212 \u2234 C = 3 1 nRT n M n ln ( ) \u2212\n3.54 Chapter 3 HINTS AND EXPLANATIONS
Answer: A
\nSolution: f ( c ) T 1 T 2 > T 1 C
Answer: C
\nSolution: f ( c ) M 1 M 2 > M 1 C
Answer: B
\nSolution: ( ) ( ) dN dN 1 2 = 4 2 2 4 2 3 2 2 2 2 3 2 \u03c0 \u03c0 \u03c0 \u03c0 \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f \u00d7 \u00d7 \u00d7 \u00d7 \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f \u2212 M RT u e du N M RT M u RT ( ) ( ) mp mp 2 2 2 2 2 \u00d7 \u00d7 \u00d7 \u00d7 \u2212 \u00d7 u e du N M u RT mp mp = 4 \u00d7 e Mu RT mp \u2212 \u2212 2 2 4 1 ( ) =
Answer: A
\nSolution: e \u20133 Comprehension \u2013 XI
Answer: C
\nSolution: n A = m 2 , n B = m 16 , n C = m 32
Answer: D
\nSolution: Z W = 1 4 \u00d7 u av \u00d7 N * = 1 4 8 RT M \u03c0 \u00d7 N * \u21d2 Z W \u221d N M *
Answer: C
\nSolution: \u03bb = 1 2 2 \u03c0\u03c3 N * \u21d2 \u03bb A : \u03bb B : \u03bb C = 1 1 2 1 2 16 1 2 32 2 2 2 \u00d7 \u00d7 \u00d7 m m m : : = 1 : 2 : 4
Answer: A
\nSolution: E total = 3 2 nRT and n max for A
Answer: A
\nSolution: Z = 1 for all (Ideal behaviour)
Answer: D
\nSolution: Z 1 = 2 \u03c0 \u03c3 2 u av \u00d7 N * Z A : Z B : Z C = 1 2 \u00d7 1 2 2 \u00d7 m : 2 2 \u00d7 1 16 16 \u00d7 m : 2 2 \u00d7 1 32 32 \u00d7 m = 1 2 2 1 64 1 128 2 : :
Answer: B
\nSolution: u av \u221d 1 M Comprehension \u2013 XII For critical point, dP dV = 0 and d P dV 2 2 Now, dP dV = 0 \u21d2 \u2212 \u2212 + RT V b a T V ( ) . 2 3 2 = 0 \u21d2 RT V b a T V ( ) . \u2212 = 2 3 2 (1) and d P dV 2 2 = 0 \u21d2 2 6 3 4 RT V b a T V ( ) . \u2212 \u2212 = 0 \u21d2 2 3 3 4 RT V b a T V ( ) . \u2212 = (2) From (1) \u00f7 (2) : V \u2013 b = 2 3 V \u21d2 V C = 3 b Eq 1 : RT b b ( ) 3 2 \u2212 = 2 3 3 a T b .( ) \u21d2 T C = 8 27 a Rb\n3.55 Gaseous State HINTS AND EXPLANATIONS and P C = RT V b a T V \u2212 \u2212 . 2 = aR b 216 3
Answer: B
\nSolution: T C = 8 27 a Rb
Answer: C
\nSolution: P C = aR b 216 3
Answer: A
\nSolution: Avogadro\u2019s hypothesis is valid only for gases due to large intermolecular distance.
Answer: A
\nSolution: Charle\u2019s law
Answer: C
\nSolution: d = PM RT but M is independent from d , P or T .
Answer: D
\nSolution: H 2 and Cl 2 are reactive gases.
Answer: C
\nSolution: Escaping tendency increases only on increasing the energy of molecules.
Answer: C
\nSolution: Graham\u2019s law is valid for ideal as well as non-ideal gases.
Answer: B
\nSolution: Theory based
Answer: B
\nSolution: Theory based
Answer: A
\nSolution: Volume of ideal gas should be the total volume minus the volume occupied by gas molecules.
Answer: A
\nSolution: As the average K.E. is same, increase in mass decreases their speed.
Answer: A
\nSolution: Total K.E. = 3 2 nRT As the pressure exerted by the vapour is same in both but volume is in 1 : 2 ratio, the moles is also in 1 : 2 ratio.
Answer: D
\nSolution: f ( c ) T 1 T 2 > T 1 C
Answer: A
\nSolution: Concept based
Answer: D
\nSolution: Excluded volume is \u2018nb\u2019.
Answer: D
\nSolution: T C < T B and hence, attractive forces are dominant.
Answer: D
\nSolution: PV is constant at constant temperature.
Answer: A
\nSolution: Above Boyle\u2019s temperature, gases show positive deviation.
Answer: B
\nSolution: Theory based
Answer: B
\nSolution: Theory based
Answer: A
\nSolution: K.E. of molecules is the function of T.
Answer: A \u2192 P, R; B \u2192 Q ; C \u2192 S ; D \u2192 T
\nSolution: Boyle\u2019s law : PV = K \u21d2 dP dV T \u239b \u239d \u239c \u239e \u23a0 \u239f = \u2212 K V 2 = \u2212 P V And d PV dP T ( ) \u23a1 \u23a3 \u23a2 \u23a4 \u23a6 \u23a5 = 0 Charle\u2019s law V T = K \u21d2 dV dT P \u239b \u239d \u239c \u239e \u23a0 \u239f = K = V T Avogadro\u2019s law : V n = K = RT P \u21d2 dV dn P T \u239b \u239d \u239c \u239e \u23a0 \u239f , = RT P Graham\u2019s law r = \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f dP dt \u221d 1 d
Answer: A \u2192 Q, R; B \u2192 S; C \u2192 P; D \u2192 Q
\nSolution: Average translational K.E. per mole = 3 2 RT Average translational K.E. per gram = 3 2 RT M
Answer: A \u2192 R; B \u2192 P, S; C \u2192 Q
\nSolution: T C = 8 27 a Rb a b X \u239b \u239d \u239c \u239e \u23a0 \u239f = 120, a b Y \u239b \u239d \u239c \u239e \u23a0 \u239f = 333.33, a b Z \u239b \u239d \u239c \u239e \u23a0 \u239f = 171.4 V C = 3 b P C = a b 27 2 , a b X 2 \u239b \u239d \u239c \u239e \u23a0 \u239f = 4800, a b Y 2 \u239b \u239d \u239c \u239e \u23a0 \u239f = 11111.11, a b Z 2 \u239b \u239d \u239c \u239e \u23a0 \u239f = 4898
Answer: A \u2192 S; B \u2192 R; C \u2192 Q; D \u2192 P
\nSolution: A. P V = P nRT P \u239b \u239d \u239c \u239e \u23a0 \u239f = P nRT 2 P V P\n3.56 Chapter 3 HINTS AND EXPLANATIONS B. P V = nRT V V / = nRT V 2 P V V C. V P = nRT P 2 1 P 2 V P D. P V = P nRT 2 = 10 2 log P nRT P V log P
Answer: A \u2192 Q; B \u2192 S; C \u2192 P; D \u2192 R
\nSolution: Moles of water vapour formed, n = PV RT = 22 8 827 6 760 0 0821 300 . ( ) . \u00d7 \u2212 \u00d7 \u00d7 = 1
Answer: A \u2192 Q; B \u2192 R; C \u2192 P; D \u2192 S
\nSolution: Pressure correction = a n V . 2 2 = 4 5 10 2 2 \u00d7 = 1 atm Ideal volume = V \u2013 nb = 10 \u2013 5 \u00d7 0.05 = 9.75 L Volume occupied by molecules = nb 4 = 0 25 4 . = 0.0625 L Volume correction = nb = 0.25L
Answer: A \u2192 P, R, S; B \u2192 P, Q, S; C \u2192 P, Q, S; D \u2192 P, R, S
\nSolution: Theory based
Answer: A \u2192 P, S; B \u2192 Q, R; C \u2192 Q
\nSolution: (A) \u03bb = 1 2 2 \u03c0\u03c3 N * = RT PN A 2 2 \u03c0\u03c3 At constant volume, N * = constant \u21d2 \u03bb \u03bb 2 1 = 1 At constant pressure, \u03bb \u221d T \u21d2 \u03bb \u03bb 2 1 = 2 (B) Z 1 = 2 \u03c0\u03c3 2 \u00d7 u av \u00d7 N * = 2 \u03c0\u03c3 2 \u00d7 8 RT M PN RT A \u03c0 \u239b \u239d \u239c \u239e \u23a0 \u239f At constant volume : Z 1 \u221d T \u21d2 Z Z 1 2 1 1 , , = 2 At constant pressure : Z 1 \u221d 1 T \u21d2 Z Z 1 2 1 1 , , = 1 2 (C) Z 11 = 1 2 2 2 \u03c0\u03c3 . . * u N av = 1 2 8 2 2 \u03c0\u03c3 \u03c0 RT M PN RT A \u239b \u239d \u239c \u239e \u23a0 \u239f At constant volume : Z 11 \u221d T \u21d2 Z Z 11 2 11 1 , , = 2 At constant pressure : Z 11 \u221d 1 3 2 ( ) T \u21d2 Z Z 11 2 11 1 , , = 1 2 2
Answer: A \u2192 P, S; B \u2192 Q; C \u2192 R
\nSolution: A. P = 1 atm + 38 cm Hg = 1.5 atm Q. P = 1 atm + 57 cm Hg = 1.75 atm R. P = 38 cm Hg = 0.5 atm S. P = 1 atm + 1.9 m glycerine = 1+ 190 2 72 13 6 76 \u00d7 \u00d7 . . = 1.5 atm
Answer: A \u2192 Q; B \u2192 S; C \u2192 P; D \u2192 R; E \u2192 T
\nSolution: T C = 273 + (\u2013177) = 96 K = \u2013177\u00b0C T B = 27 8 96 \u00d7 = 324 K = 51\u00b0C A. V i = 0 2 0 08 96 20 10 3 . . \u00d7 \u00d7 \u00d7 = 76.8 ml But V real < V ideal in given condition \u21d2 V r < 76.8 ml B. Z = 1 \u21d2 V r = V i = 0 2 0 08 324 6 48 10 3 . . . \u00d7 \u00d7 \u00d7 = 800 ml C. Above Boyle\u2019s temperature, Z > 1 \u2234 V r > V i = 0 2 0 08 350 7 10 3 . . \u00d7 \u00d7 \u00d7 = 800 ml D. Below Boyle\u2019s temperature, Z < 1 \u2234 V r < V i = 0 2 0 08 300 6 10 3 . . \u00d7 \u00d7 \u00d7 = 800 ml E. T = T B but P > 50 atm \u21d2 Z > 1 \u2234 V r > V i = 0 2 0 08 324 64 8 10 3 . . . \u00d7 \u00d7 \u00d7 = 80 ml
Answer: 7
\nSolution: V T 1 1 = V T 2 2 \u21d2 45 300 = 42 2 T \u21d2 T 2 = 280 K = 7\u00b0C\n3.57 Gaseous State HINTS AND EXPLANATIONS
Answer: 5
\nSolution: 1 2 P 1 = 1 atm + hm water = (10 + h ) m water V 1 = 4 3 \u03c0 (1 mm) 3 A = \u03c0 r 2 = \u03c0 mm 2 \u2234 r = 1 mm P 2 = 1 atm = 10 m water V 2 = 2 \u03c0 mm 3 Now, P 1 V 1 = P 2 V 2 \u21d2 (10 + h ) \u00d7 4 3 \u03c0 = 10 \u00d7 2 \u03c0 \u21d2 h = 5 m Hence, water holding capacity of pool, V = 1 3 \u03c0 r 2 h = 1 3 \u03c0 (10 m) 2 \u00d7 5 m = 500 3 \u03c0 m 3
Answer: 5
\nSolution: Number of cosmic events = Number of Ar-atoms = 1 911 10 22 7 6 . . \u00d7 \u2212 l l \u00d7 6 \u00d7 10 23 = 5.05 \u00d7 10 16
Answer: 4
\nSolution: For lifting of balloon, B > W or, ( V \u00d7 \u03c1 outside air \u00d7 g ) > ( V \u00d7 \u03c1 inside air + m additional ) g or, V ( \u03c1 outside air \u2013 \u03c1 inside air ) > m additional or, 91 1 29 0 08314 290 1 29 0 08314 10 8 314 \u00d7 \u00d7 \u2212 \u00d7 \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f > . . . T \u2234 T > 293.22 Hence, diff erence in temperature, \u0394 T > 3.22 \u2248 4 K.
Answer: 3
\nSolution: n released gas = n taken \u2013 n remained or, m 22 4 1 25 . . \u00d7 = P p \u00d7 \u00d7 \u2212 \u2212 \u00d7 \u00d7 3 0 0821 273 0 8 3 0 0821 273 . ( . ) . \u21d2 m = 3 gm
Answer: 3
\nSolution: I T Vacuum VL 4 atm 300 K VL II I VL 600 K P atm VL 600 K ( P + 2) atm II Final moles of gases in vessel I and II = Initial mole in vessel II or, P V R P V R \u00d7 \u00d7 + + \u00d7 \u00d7 600 2 600 ( ) = 4 300 \u00d7 \u00d7 V R \u21d2 P = 3 atm
Answer: 4
\nSolution: Let the mixture contains a moles C x H 8 and b moles C x H 10 . Now, a \u00d7 (12 x + 8) + 6 \u00d7 (12 x + 10) = 28.4 (1) a + b = PV RT = 2 46 5 0 082 300 . . \u00d7 \u00d7 = 0.5 (2) and 28.4 \u00d7 84 5 100 . = a \u00d7 12 x + b \u00d7 12 x (3) On solving, x \u2248 4
Answer: 6
\nSolution: From PV = nRT , V = nR P T . As the pressure is constant, the change in slope is only due to change in moles. X n \u2192 nX Initial a mole o Final a \u2013 0.6 a 0.6 axn = 0.4 a Now, a a an 0 4 0 6 . . + = ( . . ) / ( . . ) / 50 2 49 9 20 49 1 47 9 20 \u2212 \u2212 \u21d2 n = 6
Answer: 4
\nSolution: 2H 2 + O 2 \u2192 2H 2 O 2a a 0 Final 2a \u2013 1.6a a \u2013 0.8a 1.6a = 0.4a = 0.2a Now, P nT = R V = Constant \u21d2 P n T 1 1 1 = P n T 2 2 2 \u21d2 4 5 3 330 . a \u00d7 = P a 2 2 2 400 . \u00d7 \u2234 P 2 = 4 atm
Answer: 8
\nSolution: n total = n n O N 2 2 + or, 1 1 30 . \u00d7 RT = Po RT RT 2 30 0 9 10 \u00d7 + \u00d7 . \u21d2 P O 2 = 0.8 atm
Answer: 4
\nSolution: P H O 2 = 40 100 \u00d7 (V.P.)\n3.58 Chapter 3 HINTS AND EXPLANATIONS First drop of liquid will form when P = V.P. Now, P 1 V 1 = P 2 V 2 \u21d2 40 100 \u00d7 (V.P.) \u00d7 10 = (V.P.) \u00d7 V 2 \u2234 V 2 = 4 ml
Answer: 9
\nSolution: H 2 O(l) \u2192 H 2 (g) + 1 2 O 2 (g) a mole 0 0 Final 0 a mole 0.5 a mole \u0394 P.V = \u0394 n .RT or, (1.86 \u2013 0.96) \u00d7 20 = (1.5a) \u00d7 0.08 \u00d7 300 \u21d2 a = 0.5 \u2234 Mass of water present initially = 0.5 \u00d7 18 = 9 gm
Answer: 4
\nSolution: Initial total moles = PV RT = 24 63 3 0 0821 300 . . \u00d7 \u00d7 = 3 \u2234 Initial mole of H 2 = 3 \u2013 1 = 2 Final mole ratio, n n H D 2 2 1 2 4 4 1 2 = = / / Now, n n n n M M f f i i n H D H D D H 2 2 2 2 2 2 = \u00d7 \u239b \u239d \u239c \u239c \u239e \u23a0 \u239f \u239f or, 1 2 2 1 4 2 = \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f n \u21d2 n = 4
Answer: 2
\nSolution: For critical point, dP dV m = 0 and d P dV m 2 2 0 = On solving, b =
Answer: 8
\nSolution: But b \u2260 0 from question. Hence, the gas does not have critical condition.
Answer: 0775
\nSolution: Boyle\u2019s temperature: T a Rb B = = \u00d7 = 4 105 0 0821 0 1 500 . . . K Hence, at 500 K, the gas will behave ideally. Not, d PM RT = = \u00d7 \u00d7 2 164 2 0 0821 500 . . = 8 g/L = 8 kg/m 3 Four-digit Integer Type
Answer: 3900
\nSolution: l mm 760 750 l \u2013 750 800 770 l \u2013 770 P 760 l \u2013 760 (760 \u2013 750) \u00d7 ( l \u2013 750) = (800 \u2013 770) \u00d7 ( l \u2013 770) = ( P \u2013 760) \u00d7 ( l \u2013 760) \u2234 P = 775 mm Hg
Answer: 0404
\nSolution: Mass of LNG = 10 m 3 \u00d7 416 kg/m 3 = 416 \u00d7 10 4 gm \u2234 Moles of CH 4 = 916 10 16 4 \u00d7 = 26 \u00d7 10 4 Now, V = nRT P = \u00d7 \u00d7 \u00d7 26 10 0 021 300 1 692 4 . . = 3.9 \u00d7 10 6 L = 3900 m 3
Answer: 8724
\nSolution: V n T V n T V n V n T 1 1 1 2 2 2 2 303 1 6 1 2 = \u21d2 \u00d7 = \u00d7 . . \u21d2 T 2 = 404 K
Answer: 0450
\nSolution: V initial = a = nRT P = \u00d7 \u00d7 \u00d7 64 0 08 300 64 3 . = 8 L V initial = b = nRT P = \u2212 \u00d7 \u00d7 \u00d7 ( ) . 64 8 0 08 300 64 3 = 7 L P = nRT V = \u00d7 \u00d7 \u00d7 = 64 0 08 300 64 7 24 7 . atm\n3.59 Gaseous State HINTS AND EXPLANATIONS
Answer: 2421
\nSolution: P 1 320 K P 0 P 2 T K P 0 P \u2032 2 P \u2032 1 P 2 + P 0 = P 1 P \u2019 2 + P 0 = P \u2019 1 P 0 = P 1 \u2013 P 2 = P \u2019 1 \u2013 P \u2019 2 or, n R V n R V n R T V n R T V \u00d7 \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f \u2212 \u00d7 \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f = \u00d7 \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f \u2212 \u00d7 \u00d7 \u239b \u239d \u239c 320 5 320 4 5 4 3 4 \u239e \u239e \u23a0 \u239f \u2234 T = 450 K
Answer: 1254
\nSolution: Mass of gas used = 28.8 \u2013 23.2 = 5.6 kg Volume of gas used up, V = nRT P = \u00d7 \u00d7 \u00d7 \u00d7 ( . ) . 5 6 10 0 08 300 56 1 3 = 2400 L Now, P M P M P 1 1 2 2 2 35 28 8 14 8 23 2 14 8 = \u21d2 \u2212 = \u2212 ( . . ) ( . . ) \u21d2 P 2 = 21 atm
Answer: 0065
\nSolution: n NO = \u00d7 \u00d7 1 6 0 75 0 08 3 00 . . . . = 0.05 n O 2 1 2 0 25 0 08 3 00 = \u00d7 \u00d7 . . . . = 0.0125 2NO + O 2 \u2192 2NO 2 \u2192 N 2 O 4 0.05 0.0125 0 0 Final \u2212 0 025 0 025 . . \u2212 0 0125 0 . 0 0 0 0125 0 0125 . . But at 200 K, N 2 O 4 is solid. Hence, the only gas is NO. Millimoles of NO remained = 0.025 \u00d7 1000 = 25 Now, P = 0 025 0 08 200 0 75 0 25 . . ( . . ) \u00d7 \u00d7 + = 0.4 atm
Answer: 0500
\nSolution: 76 70 6 76 l = ? 3 P 1 = (6 \u2013 1) = 5 cm Hg P 2 = ? V 1 = 6 A cm 3 V 2 = 3 A cm 3 P 2 = 5 6 3 \u00d7 A A = 10 cm Hg Hence, fi nal total pressure in table above mercury = 10 + 1 = 11 cm Hg \u2234 Barometer reading = 76 \u2013 11 = 65 cm
Answer: 0320
\nSolution: Mass of water vapour present initially, m V R 1 756 100 24 760 300 18 = \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f \u00d7 \u00d7 \u00d7 Mass of water vapour fi nally remained, m V R 2 8 4 760 280 18 = \u00d7 \u00d7 \u00d7 . \u2234 Fraction of water condensed = m m m 1 2 1 \u2212 = 0.5
Answer: 3610
\nSolution: Let the process time = t min Mass of water vapour in inlet air, m t 1 3 20 100 38 760 10 10 0 08 500 = \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f \u00d7 \u00d7 ( ) \u00d7 \u00d7 . 18 gm = 2.5 t gm Mass of water vapour in outlet air, m t 2 3 80 100 19 760 10 10 0 08 400 = \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f \u00d7 \u00d7 ( ) \u00d7 \u00d7 . 18 gm = 6.25 t gm From question, m 1 + 200 kg \u00d7 36 100 = m 2 \u21d2 t = 19200\n3.60 Chapter 3 HINTS AND EXPLANATIONS
Answer: 1836
\nSolution: At a depth of 10 m, P 1 = 1 atm + 10 m water = 1 + 1 013 1000 1000 1 013 10 6 . . \u00d7 \u00d7 \u00d7 = 2 atm V 1 = 24 ml n 1 = 2 24 10 0 08 300 3 \u00d7 \u00d7 \u00d7 \u2212 . = 2 \u00d7 10 \u2212 3 At surface, P 2 = 1 atm, V 2 = ?, n 2 = 2 \u00d7 10 \u2212 3 \u2212 0.05 \u00d7 10 \u2212 3 \u00d7 10 = 1.5 \u00d7 10 \u2212 3 Now, PV n P V n V V 1 1 1 2 2 2 3 2 3 2 2 24 2 10 1 1 5 10 = \u21d2 \u00d7 \u00d7 = \u00d7 \u00d7 \u21d2 \u2212 \u2212 . = 36 ml For volume remaining uncharged, P n P n 1 1 2 2 = or, 2 2 10 1 2 10 10 10 3 3 4 \u00d7 = \u00d7 \u2212 \u00d7 \u21d2 = \u2212 \u2212 \u2212 r r mol/min
Answer: 0013
\nSolution: N 2 O 4 \u2192 2NO 2 Initial mole 20 \u00d7 V RT 0 Final mole 20 10 \u00d7 \u2212 V RT V RT 20 V RT = 10 V RT NO 2 will eff use through SPM till its pressure becomes same in both chamber and hence, mole ratio of NO 2 in chamber-I and II should be 1 :
Answer: 0140
\nSolution: Final moles in chamber-I = 10 V RT of N 2 O 4 and 5 V RT of NO 2 Final moles in chamber-II except H 2 O vapour = 15 V RT of NO 2 \u2234 Pressure of gas in chamber-I = 15 1 2 V RT R T V \u00d7 \u00d7 . = 18 mm and pressure of gases in chamber-II = 15 1 2 3 V RT R T V \u00d7 \u00d7 . + 30 = 36 mm
Answer: 0015
\nSolution: 5 2 n \u2264 0.01 \u21d2 n \u2265 12.28
Answer: 0050
\nSolution: H2 H2 H2 H2 O2 N2 N2 N2 Inital 30 mole 5 mole 5 mole O2 = 5 mole N2 = 2.5 mole N2 = 2.5 mole Final H2 = 10 mole H2 = 10 mole H2 = 10 mole P 1 : P 2 : P 3 = 10 : 17.5 : 12.5 = 4 : 7 : 5
Answer: 9450
\nSolution: A1 = n 2 A1 = n 2 A3 = n 4 A3 = n 4 A3 = n 4 A2 = n 3 A2 = n 3 A3 = n 4 A2 = n 3 \u2234 P P n n N A A N 4 1 5 \u2212 = / / = 3 \u21d2 N = 15
Answer: 0100
\nSolution: r r M M t t M H air air H air 2 2 100 26 2 = \u21d2 = / / and r r M M t t M M gas air air H air gas 2 = \u21d2 = 100 130 / / \u2234 M gas = 50
Answer: 0250
\nSolution: 3 2 kT = mgh \u21d2 3 2 \u00d7 8 4 . N A \u00d7 300 = ( ) 40 10 3 \u00d7 \u2212 N A \u00d7 10 \u00d7 h \u2234 h = 9450 m
Answer: 0096
\nSolution: At Boyle\u2019s temperature, the second virial coefficient is zero. B = a + b . e c T / 2 = 0 or, e c T \u2212 / 2 = \u2212 a b \u21d2 e T \u2212 950 2 / = \u2212 \u2212 0 02 0 22 . . \u21d2 T = 100 k\n3.61 Gaseous State HINTS AND EXPLANATIONS
Answer: C
\nSolution: [ ] NH M 2 30 15 10 10 \u2212 \u2212 \u2212 = = \u2234 Number of NH 2 \u2212 ions per ml = \u00d7 \u00d7 \u00d7 = \u00d7 \u2212 10 1 1000 6 10 6 10 15 23 5 ( )
Answer: D
\nSolution: On increasing temperature, the dissociation of water will increase. It will result increase in [H + ] and as well as in [OH \u2013 ] and hence, decreases in P H and as well as P OH .
Answer: B
\nSolution: For maximum dissociation, [H + ] = [OH \u2013 ].
Answer: B
\nSolution: [H + ] = 10 \u20132 M \u21d2 n H + = \u00d7 = \u00d7 \u2212 \u2212 200 10 1000 2 10 2 3 [OH \u2013 ] = 10 \u20132 M \u21d2 n OH \u2212 = \u00d7 = \u00d7 \u2212 \u2212 300 10 1000 3 10 2 3 \u2234 Moles of excess OH \u2013 remained = 1 \u00d7 10 \u20133 [OH \u2013 ] = 1 10 500 1000 2 10 3 3 \u00d7 \u00d7 = \u00d7 \u2212 \u2212 M \u2234 P OH = \u2013 log (2 \u00d7 10 \u20133 ) = 2.7 \u21d2 P H = 11.3
Answer: A
\nSolution: [OD \u2013 ] excess = 80 0 1 20 0 2 100 0 04 \u00d7 \u2212 \u00d7 = . . . M \u2234 P OD = \u2013 log (0.04) = 1.4 Now, P Kw of D 2 O = P D + P OD = 13.6 + 1.4 = 15 \u2234 Kw = 1 \u00d7 10 \u201315
Answer: D
\nSolution: [OT - ] excess = 400 0 2 100 0 4 500 0 08 \u00d7 \u2212 \u00d7 = . . . M \u2234 P OT = \u2013 log(0.08) = 1.1 Now, PT = P Kw \u2013 P OT = 2 \u00d7 7.60 \u2013 1.1 = 14.1
Answer: A
\nSolution: K Kw Ka b = = \u00d7 = \u00d7 \u2212 \u2212 \u2212 10 2 10 5 10 14 10 5
Answer: B
\nSolution: NH + H O NH OH 3 2 \u001f \u21c0 \u001f \u21bd \u001f \u001f 4 + \u2212 + \u0394 H \u00b0 = (\u201352.21) + (54.70) = 2.49 kJ \u0394 S \u00b0 = 1.6 + (\u201376.3) = \u2013 74.7 J/K Now, \u0394 H \u00b0 = \u2013 RT. ln K eq or, 2490 \u2013 300 \u00d7 (\u201374.7) = \u20138.3 \u00d7 300 \u00d7 ln K eq \u2234 K eq = e \u201310
Answer: A
\nSolution: [ ] ] H [H HCOOH CH COOH 3 + + = or, 2 4 10 0 6 1 8 10 8 4 5 . . . \u00d7 \u00d7 = \u00d7 \u00d7 \u21d2 = \u2212 \u2212 C C M \u2234 Moles of CH 3 COOH added = 100 8 1000 0 8 \u00d7 = .
Answer: D
\nSolution: K a ( HA ) = K b ( A \u2013 ) = Kw = \u2212 10 7 Now, [ ] . . H M P H + \u2212 \u2212 = \u00d7 = \u21d2 = 10 0 1 10 4 0 7 4
Answer: A
\nSolution: NH +OH K M S K NH +H O 4 + f b 3 2 \u2212 \u2212 \u2212 = \u00d7 = 3 4 10 10 1 1 . ? Given : NH NH H K M 4 + 3 \u001f \u21c0 \u001f \u21bd \u001f \u001f + = \u00d7 + \u2212 ; . 1 10 5 6 10 and H O H OH K M 2 \u001f \u21c0 \u001f \u21bd \u001f \u001f + \u2212 \u2212 + = \u00d7 ; . 2 14 2 1 0 10 \u2234 NH OH NH H O K K K 4 + 3 2 eq + + = \u2212 \u001f \u21c0 \u001f \u21bd \u001f \u001f ; 1 2 Now, 3 4 10 5 6 10 10 6 07 10 10 10 14 5 1 . . . \u00d7 = \u00d7 \u21d2 = \u00d7 \u2212 \u2212 \u2212 K K S b b
Answer: B
\nSolution: CH COOH CH COO H 3 3 0 1 0 1 0 1 . . . \u2212 + + \u2248 + \u2212 + + x x y y x \u001f \u21c0 \u001f \u21bd \u001f \u001f Cl CHCOOH Cl CHCOO H 2 2 0 1 0 1 0 1 . . . \u2212 + + \u2248 + \u2212 + + y x y y y \u001f \u21c0 \u001f \u21bd \u001f \u001f 0 15 0 1 0 1 0 05 . ( . ) ( . ) . = \u00d7 + \u2212 \u21d2 = y y y y \u2234 [H + ] = 0.1 + x + y \u2248 0.1 + y = 0.15 M \u2234 P H = \u2013 log(0.15) = 0.82
Answer: A
\nSolution: [ ] . . / / . OH M \u2212 = \u00d7 = 0 4 100 4 25 17 250 1000 0 004 P OH = \u2013 log(0.004) = 2.4 \u2234 P H = 14 \u2013 2.4 = 11.6
Answer: B
\nSolution: [ ] . . OH K C M b \u2212 \u2212 \u2212 = \u00d7 = \u00d7 \u00d7 = \u00d7 1 6 10 0 0025 4 10 6 9 P P OH H = \u2212 \u00d7 = \u21d2 = \u2212 log . . 4 10 4 2 9 8 9 EXERCISE II (JEE ADVANCED)\n7.42 Chapter 7 HINTS AND EXPLANATIONS
Answer: A
\nSolution: [ ] / O HSaC M = \u00d7 = \u00d7 \u2212 \u2212 4 10 200 1000 2 10 4 3 and P H = 3.0 \u21d2 [H + ] = 10 \u20133 M Now, 2 10 10 2 10 4 10 12 3 3 12 \u00d7 = \u00d7 \u00d7 \u21d2 = \u00d7 \u2212 \u2212 \u2212 \u2212 \u2212 \u2212 [S ] [ ] aC SaC M
Answer: B
\nSolution: [ ] . . HA M O = \u00d7 = 20 0 5 50 0 2 [ ] . . HB M O = \u00d7 = 30 0 2 50 0 12 HA H A 0 2 . \u2212 + + \u2212 + x x y x \u001f \u21c0 \u001f \u21bd \u001f \u001f HB H 0 12 . \u2212 + + \u2212 + y x y y B \u001f \u21c0 \u001f \u21bd \u001f \u001f Now, 2 10 0 2 0 2 4 \u00d7 = + \u22c5 \u2212 \u2248 + \u22c5 \u2212 ( ) x . ( ) . x y x x y x \u2234 ( x + y ) \u22c5 x = 4 \u00d7 10 \u20135 (1) and, 5 10 0 12 0 12 5 \u00d7 = + \u22c5 \u2212 \u2248 + \u22c5 \u2212 ( ) ( . ) ( ) . x y y y x y y \u2234 ( x + y ) \u22c5 y = 6 \u00d7 10 \u20136 (2) From (1) and (2), [H + ] = x + y = 6.78 \u00d7 10 \u20133 M
Answer: C
\nSolution: 10 0 01 10 5 8 \u2212 \u2212 = \u00d7 \u21d2 = K K a a . Now, [ ] . . . OH P OH \u2212 \u2212 \u2212 = \u00d7 = \u21d2 = 10 0 1 10 4 5 8 4 5 \u2234 P H = 9.5
Answer: B
\nSolution: RNH +H O RNH + OH 2 2 0 01 3 10 4 . \u2212 + \u2212 + \u2212 x x x \u001f \u21c0 \u001f \u21bd \u001f \u001f 2 10 10 0 01 10 6 4 4 \u00d7 = + \u2212 \u21d2 = \u2212 \u2212 \u2212 x x x x ( ) . \u2234 [OH \u2013 ] = 2 \u00d7 10 \u20134 M
Answer: C
\nSolution: As on adding HCl, [H + ] is not changing and will remain unchanged.
Answer: A
\nSolution: Co +H O HCo H 2 2 Decrease Shiftleft \u2193 \u2190 \u2212 + + \u001f \u21c0 \u001f\u001f\u001f\u001f \u21bd \u001f \u001f\u001f\u001f\u001f 3 As [H + ] decreases, P H increases.
Answer: C
\nSolution: [ ] . / / . W H M O 2 4 0 16 32 500 1000 0 01 = = \u2234 \u221d = 4 10 0 01 0 02 2 6 \u00d7 = \u2212 . . % or
Answer: B
\nSolution: P H = \u2013 log (2 \u00d7 10 \u20136 ) = 5.70
Answer: B
\nSolution: [OH\u2013] = 6.67 \u00d7 10 \u20133 + 6 67 10 2 0 10 3 2 . \u00d7 + \u2248 \u2212 \u2212 M \u2234 P OH = \u2013 log (10 \u20132 ) = 2.0 \u21d2 P H = 12.0
Answer: D
\nSolution: en + H O enH OH M M M b 2 0 09 5 1 8 1 10 ( . ) ( ) ( ) ;K . \u2212 + \u2212 \u2212 + \u2212 + = \u00d7 x x y x y \u001f \u21c0 \u001f \u21bd \u001f \u001f enH + H O enH OH M M M b + \u2212 + \u2212 + \u2212 + = \u00d7 2 2 2 8 2 7 0 10 ( ) ( ) ;K . x y y x y \u001f \u21c0 \u001f \u21bd \u001f \u001f Now, 8 1 10 0 09 0 09 2 7 10 5 3 . ( )( ) ( . ) . . \u00d7 = \u2212 + \u2212 \u2248 \u22c5 \u21d2 = \u00d7 \u2212 \u2212 x y x y x K x x x and 7 0 10 7 0 10 8 8 . ( ) ( ) . \u00d7 = \u22c5 + \u2212 \u2248 \u22c5 \u21d2 = \u00d7 \u2212 \u2212 y x y x y y x x y \u2234 [en H + ] = ( x \u2013 y ) \u2248 x M = 2.7 \u00d7 10 \u20133 M [enH ]= = 7.0 10 M 2 2+ y \u00d7 \u2212 8 \u2234 [OH \u2013 ] = ( x + y ) = x = 2.7 \u00d7 10 \u20133 M and P OH = \u2013 log (2.7 \u00d7 10 \u20133 ) = 2.56 \u21d2 P H = 11.44
Answer: B
\nSolution: K K H H S a a 2 1 2 2 2 5 \u22c5 = + \u2212 [ ] [ ] [ ] or, ( . ) ( . ) ( . ) [ ] . [ ] . 1 4 10 1 0 10 0 1 5 0 2 5 2 8 10 7 14 2 2 2 20 \u00d7 \u00d7 \u00d7 = \u00d7 \u21d2 = \u00d7 \u2212 \u2212 \u2212 \u2212 \u2212 M
Answer: A
\nSolution: [ ] . H M + \u2212 \u2212 = \u00d7 \u00d7 = \u00d7 0 2 2 10 2 10 5 3 Now, ( ) ( ) ( ) ( ) [ ] [ ] 2 10 5 10 4 10 2 10 5 9 12 3 3 3 \u00d7 \u00d7 \u00d7 \u00d7 \u00d7 = \u00d7 \u00d7 \u2212 \u2212 \u2212 \u2212 + A H A 3 \u2234 = \u00d7 \u2212 \u2212 [ ] [ ] A H A 3 3 17 5 10
Answer: D
\nSolution: [ ] . H M + \u2212 \u2212 = \u00d7 = 0 1 10 10 5 3 Now, K a 3 3 2 3 2 13 3 10 10 10 10 = \u21d2 = = + \u2212 \u2212 \u2212 \u2212 \u2212 \u2212 \u2212 [ ][ ] [ ] [ ] [ ] H A HA A HA \u2234 P X = 10
Answer: C
\nSolution: OH ACOH ACO H O m mol Final 0 m mol 1 m mol m mol \u2212 \u2212 + + 2 3 0 2 2 \u001f \u21c0 \u001f \u21bd \u001f \u001f p H = + = 4 74 2 1 5 04 . l og . (Acidic) Addition of 1 ml ACOH will decrease P H by 0.3 unit.\n7.43 Ionic Equilibrium HINTS AND EXPLANATIONS
Answer: A
\nSolution: Millimoles of ACOH = 6 \u00d7 0.1 = 0.6 Millimoles of ACO \u2013 =
Answer: A
\nSolution: \u00d7 0.1 = 1.2 \u2234 = + = p H 4 75 1 2 0 6 5 05 . log . . .
Answer: B
\nSolution: 1st solution will finally have 1 mole of CH 3 COOH. \u2234 = P P H Ka 1 1 2 and for 2nd solution, P P H Ka 2 =
Answer: D
\nSolution: P P H H 2 1 0 6 = + . p M C P /M C K K a a + = + + log / log log . y x 3 98 \u2234 = y x 3 98 .
Answer: C
\nSolution: 4 0 5 0 0 5 0 05 1 1 . . log . . = + \u21d2 = C C M 6 0 5 0 0 5 5 0 2 2 . . log . . = + \u21d2 = C C M Now, fi nal P 0.05 + 5.0 0.5 + 0.5 H = + \u00d7 \u00d7 \u00d7 \u00d7 = 5 0 5 7 . log . V V V V
Answer: A
\nSolution: For maximum \u03b2 \u03b1\u03b2 \u03b1 , [ ] H P P H K a + = \u21d2 = 0
Answer: A
\nSolution: BOH H B C Final C-0.1 V 0.1 40 40 40 40 40 0 0 0 1 40 \u00d7 + + + \u00d7 + + + V M V V V M V \u001f \u21c0 \u001f \u21bd \u001f \u001f . + + + V H O 2 H + must be a limiting reagent because both P H are above > .0. Now, P =P +log 0.1V 40C 0.1V OH K b \u2212 14 =P +log 0.1 5 40C 0.1 5 K b \u2212 \u00d7 \u2212 \u00d7 10 (1) 14 =P +log 0.1 40C 0.1 K b \u2212 \u00d7 \u2212 \u00d7 9 20 20 (2) \u2234 K b = 2 \u00d7 10 \u20135
Answer: B
\nSolution: For maximum buffer capacity: [ [ [ ACOH] NaOH] NaOH]= M M = \u21d2 = 2 1 2 2 1 \u2234 Mass of NaOH added = \u00d7 \u00d7 = 500 1 1000 40 20 gm
Answer: B
\nSolution: n n HA OH M M= 80 = \u21d2 = \u00d7 \u21d2 \u2212 0 28 35 0 1 1000 . .
Answer: C
\nSolution: Fe H O Fe(OH H M 2 M 3 0 9 2 0 1 + + + + + . . ? ) x x \u001f \u21c0 \u001f \u21bd \u001f \u001f 9 10 0 1 0 9 0 081 1 08 3 \u00d7 = \u00d7 \u21d2 = \u21d2 = \u2212 + + . [ ] . [ ] . . x x H H P H
Answer: A
\nSolution: In fi nal solution: [HA] = [A \u2013 ]
Answer: D
\nSolution: [ ] NH K K K C = 8.33 10 M w a b 3 4 = \u22c5 \u00d7 \u00d7 \u2212
Answer: C
\nSolution: For equivalence point, 2 5 2 5 2 15 . \u00d7 = \u00d7 V Hcl \u2234 V HCl = 7.5 ml BOH + H B +H M Eqn. M + M 0 XM + M (0.1 )M 2 5 2 5 10 0 7 5 2 15 10 0 0 1 . . . \u00d7 \u00d7 \u2212 x x \u001f \u21c0 \u001f \u21bd \u001f \u001f 2 2 aq O; K = = \u2212 \u2212 10 10 10 12 14 2 100 0 1 2 7 10 2 = \u2212 \u22c5 \u21d2 = \u00d7 = \u2212 + . . ( ) x x x x M H
Answer: D
\nSolution: At 2nd equation point: P P +P H K K 2 3 = = + = 1 2 8 12 12 10 ( ) a a Now, K K K A H A 3 a a a 1 2 3 3 \u22c5 \u22c5 = + \u2212 [H ][ ] [ ] or 7 5 10 10 10 10 4 8 12 10 3 3 . ( ) [ ] [ ] \u00d7 \u00d7 \u00d7 = \u00d7 \u2212 \u2212 \u2212 \u2212 \u2212 A H A 3 \u2234 = = \u00d7 \u2212 \u2212 \u2212 [ ] [ ] . . H A A 3 3 6 7 10 7 5 1 33 10
Answer: B
\nSolution: CO H HCO K M 100% run 0 M 0 M 3 2 0 35 0 35 3 0 0 35 11 1 4 10 \u2212 + \u2212 \u2212 + = \u00d7 . . . ; \u001f \u21c0 \u001f \u21bd \u001f \u001f For HCO 3 \u2212 solution, [ ] , . H K K M + \u2212 = = \u00d7 a a 1 2 1 4 10 8 Now, K H O HCO a 2 3 2 3 = + \u2212 \u2212 [ ][C ] [ ] \u2234 = \u00d7 \u00d7 \u00d7 = \u2212 \u2212 \u2212 \u2212 [ ] . . CO M 3 2 11 8 3 4 10 0 35 1 4 10 10\n7.44 Chapter 7 HINTS AND EXPLANATIONS
Answer: D
\nSolution: [ ] . . . LaC M \u2212 = \u00d7 = 0 0 125 0 5 2 0 5 Now, P P C OH K a = \u2212 + 7 1 2 ( log ) 5 6 7 1 2 0 5 . ( log . ) = \u2212 + P K a \u2234 = \u21d2 = \u00d7 \u2212 P K K a 3 1 8 10 4 . a
Answer: A
\nSolution: As HA is stronger acid, it will react first. For first equivalent point, V V ml NaOH NaOH \u00d7 = \u00d7 \u21d2 = 0 2 50 0 05 12 5 . . . At fi rst equivalent point, [ ] . . . A M \u2212 = \u00d7 = 50 0 05 62 5 0 04 [ . . . HB]= M 50 0 08 62 5 0 064 \u00d7 = Now, A HB B HA ; K = K (HB) K (HA eq a a \u2212 \u2212 \u2212 \u2212 + + 0 04 0 04 0 064 0 064 . . . . x x x x \u001b \u001b \u001f \u21c0 \u001f \u21bd \u001f \u001f ) ) = = = \u00d7 \u2212 \u2212 \u2212 \u2212 10 10 10 4 10 8 2 3 8 4 4 5 . . . 4 10 0 04 0 064 3 2 10 5 4 \u00d7 = \u22c5 \u00d7 \u21d2 = \u00d7 \u2212 \u2212 x x x . . . Now, K (HA H A HA H a ) [ ][ ] [ ] . [ ] . . = \u21d2 \u00d7 = \u00d7 \u00d7 + \u2212 \u2212 + \u2212 1 6 10 0 04 3 2 10 4 4 \u2234 = \u00d7 \u21d2 = + \u2212 [ ] . . H P H 1 28 10 5 9 6
Answer: C
\nSolution: CrO +H O HCrO +OH ; K Kw K 4 2 2 4 h \u2212 \u2212 \u2212 \u2212 \u2212 = = \u00d7 0 005 8 2 2 10 . x x x a \u001f \u21c0 \u001f \u21bd \u001f \u001f 2 10 0 005 0 005 10 8 2 5 \u00d7 = \u22c5 \u2212 \u2248 \u21d2 = \u2212 \u2212 x x x x x . . \u2234 = = \u2212 h 10 0 005 0 002 5 . .
Answer: A
\nSolution: P and P K K a a 1 2 2 40 9 60 = = . . \u2234 Required pH = + = 1 2 2 40 9 60 6 00 ( . . ) .
Answer: C
\nSolution: HA H A Red Blue \u001f \u21c0 \u001f \u21bd \u001f \u001f + \u2212 + [ ] [ ] [ ] H K HA A + \u2212 = \u22c5 a \u2234 = \u2212 = \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f + + + [ ] [ ] [ ] H required H H K a 2 1 75 25 25 75 = 8 \u00d7 10 \u20135 M
Answer: A
\nSolution: 5 5 4 75 . . log [ ] [ = + \u2212 ACo ACOH] O O \u2234 = \u2212 [ ] [ . ACo ACOH] O O 5 62 1
Answer: C
\nSolution: n n CO H 2 and used = = = \u00d7 = + 224 22400 0 01 30 1 1000 0 03 . . CO +OH HCO 2 mole (L.R. 3 mole or less \u2212 \u2212 \u23af \u2192 \u23af 0 01 0 01 . ) . Hence, moles of H + used should be 0.01 or less and titration of HCO 3 \u2212 and H+ should not be detected by phenolphthalein. Hence, OH \u2013 must be in excess. CO + 2OH CO H O 2 mole mole 3 mole 2 0 01 0 02 2 0 01 . . . \u2212 \u2212 \u23af \u2192 \u23af + Thus, 0.01 mole of CO 3 2 \u2212 will require only 0.01 mole of H + in the presence of phenolphthalein. As the mole of H + used is 0.03, 0.02 mole OH \u2013 must be present in excess. Hence, total moles of OH \u2013 used = 0.02 + 0.02 = 0.04, \u2234 = = [ ] . . NaOH M used 0 04 1 0 04
Answer: A
\nSolution: PbSO g/100ml 4 = \u00d7 \u00d7 \u00d7 \u2212 \u2212 2 10 304 10 1 36 10 9 3 \u001b . ZaS g/100ml = \u00d7 = \u00d7 \u2212 \u2212 10 97 10 9 7 10 22 11 . AgBr g/100ml = \u00d7 \u00d7 = \u00d7 \u2212 \u2212 4 10 188 10 1 19 10 13 5 . CuCo g/100ml 3 8 3 10 123 10 1 23 10 = \u00d7 = \u00d7 \u2212 \u2212 .
Answer: C
\nSolution: Hg 4Cl HgC 2+ M 100% ram 0 Eqn. 1.6 M M 0.5M 0.5M 0 1 10 0 9 17 . . \u00d7 \u2212 \u2212 + \u001f \u21c0 \u001f \u21bd \u001f \u001f l l M 0.1M 4 2 0 0 1 \u2212 . \u2234 = \u00d7 \u00d7 = \u2212 K form 0 1 1 6 10 0 5 10 17 4 17 . . ( . )\n7.45 Ionic Equilibrium HINTS AND EXPLANATIONS
Answer: B
\nSolution: AgBr(s) S O Ag(S O Eqn/final 0 aM a 0 1 2 3 2 0 2 2 3 2 3 0 0 1 2 . . . ) + + \u2212 \u2212 \u2212 \u001f \u21c0 \u001f \u21bd \u001f \u001f B Br \u2212 0 0 1 . K eq = \u00d7 \u00d7 \u00d7 = \u00d7 \u2212 \u21d2 = \u2212 4 10 1 6 10 0 1 0 1 0 2 0 325 13 12 2 . . . ( . ) . a a
Answer: A
\nSolution: Tl S Tl S SM 2 2 2 2 (S) \u001f \u21c0 \u001f \u21bd \u001f \u001f + \u2212 + x S H O HS OH Kw K M SM SM h a 2 2 2 \u2212 \u2212 \u2212 + + = x \u001f \u21c0 \u001f \u21bd \u001f \u001f ; K 10 10 2 10 2 10 4 10 14 14 6 6 12 \u2212 \u2212 \u2212 \u2212 \u2212 = \u00d7 \u00d7 \u00d7 \u21d2 = \u00d7 x x \u2234 = \u00d7 \u00d7 \u00d7 \u00d7 = \u00d7 \u2212 \u2212 \u2212 K sp ( ) . 2 2 10 4 10 6 4 10 6 2 12 23
Answer: A
\nSolution: M SCN M Eqn. M M 3 2 10 2 10 5 10 1 51 10 1 51 10 1 3 3 4 3 3 + \u00d7 \u00d7 \u2212 = \u00d7 \u2212 \u00d7 \u00d7 \u2212 = \u2212 \u2212 \u2212 \u2212 \u2212 + x x . . . . . ) 0 10 2 0 1 5 10 5 3 \u00d7 + = \u00d7 \u2212 \u2212 M M M M(SCN \u001f \u21c0 \u001f \u21bd \u001f \u001f x \u2234 = \u00d7 \u00d7 \u00d7 \u00d7 = \u00d7 \u2212 \u2212 \u2212 K f 1 5 10 5 10 1 10 3 10 3 4 5 5 .
Answer: B
\nSolution: SrCO s) Sr CO M 3 2 2 10 3 2 2 10 4 4 ( \u001f \u21c0 \u001f \u21bd \u001f \u001f + \u00d7 \u2212 \u00d7 \u2212 \u2212 \u2212 + x CO H O HCO OH M M M 3 2 2 10 2 3 4 10 4 6 \u2212 \u00d7 \u2212 \u2212 \u2212 \u00d7 \u2212 \u2212 + + ( ) x x \u001f \u21c0 \u001f \u21bd \u001f \u001f Now, 10 5 10 4 10 2 10 0 01 51 14 11 6 4 \u2212 \u2212 \u2212 \u2212 \u00d7 = \u00d7 \u00d7 \u00d7 \u2212 \u21d2 = x x x ( ) . \u2234 = \u00d7 \u00d7 \u00d7 \u2212 = \u00d7 \u2212 \u2212 \u2212 K sp ( ) ( ) 2 10 2 10 4 51 10 4 4 8 x
Answer: C
\nSolution: MnS(S) Mn S SM S M \u001f \u21c0 \u001f \u21bd \u001f \u001f 2 2 + \u2212 \u2212 + ( ) x S H O HS OH S M M M 2 2 \u2212 \u2212 \u2212 \u2212 + + ( ) x x x \u001f \u21c0 \u001f \u21bd \u001f \u001f 10 10 14 14 10 \u2212 \u2212 \u2212 = \u22c5 \u2212 \u00d7 = \u22c5 \u2212 x x x x ( ) ( ) S and 2.5 10 S S \u2234 S = 6.3 \u00d7 10 \u20134 M
Answer: A
\nSolution: AgCl(s) Br AgBr(s) Cl aq M M 0 1 0 075 0 075 0 075 . . . . (aq) ( \u2212 \u2212 \u2212 + + x \u001f \u21c0 \u001f \u21bd \u001f \u001f ) ) K Br K (AgCl) K (AgBr) Br eq sp sp = = \u21d2 = \u00d7 \u00d7 \u2212 \u2212 \u2212 \u2212 \u2212 [Cl ] [ ] . [ ] 0 075 2 10 4 10 10 13 3 \u2234 = \u00d7 \u2212 \u2212 [ ] . Br M 1 5 10 4
Answer: C
\nSolution: [ ) ] . Ag(CN M 2 0 01 \u2212 = K Ag CN Ag CN Ag diss = \u21d2 \u00d7 = \u00d7 \u00d7 + \u2212 + \u2212 \u2212 + \u2212 [ ][ ] [ ( ) ] [ ] ( . ) . 2 2 20 7 2 1 10 2 5 10 0 01 \u2234 = \u00d7 + \u2212 [ ] . Ag M 1 6 10 9
Answer: B
\nSolution: [ ] . CO M 3 2 2 0 \u2212 = Now, K (CaCO K (CaF CO F F F sp sp 3 2 3 2 2 3 3 4 2 8 ) ) [ ] [ ] [ ] ( ) = \u21d2 = \u21d2 = \u2212 \u2212 \u2212 \u2212 x y y x
Answer: C
\nSolution: BaF s O BaC O s M M 2 2 4 2 0 1 2 4 2 2 ( ) C (aq) ( ) F (aq) ( . ) + \u21d2 + \u2212 \u2212 \u2212 x x K F C O K BrF K O eq 2 sp = = = = \u2212 \u2212 \u2212 \u2212 [ ] [ ] ( ) (BrC ) 2 4 2 2 4 6 10 4 10 10 10 sp \u2234 x \u2248 0.1 \u21d2 [F \u2013 ] = 0.2 M \u2234 [ ] ( . ) . Ba M 2 6 2 5 10 0 2 2 5 10 + \u2212 \u2212 = = \u00d7
Answer: C
\nSolution: S Zn(OH) Zn(OH) Zn Zn(OH) Zn(OH) = + + + + + + \u2212 \u2212 [ (aq)] [ ] [ ] [ ] [ ] 2 2 3 4 2 = + \u22c5 + \u22c5 + + \u2212 \u2212 \u2212 \u2212 K K K OH K K K OH K K OH K K K [OH 5 4 1 1 2 1 3 2 1 2 4 1 2 [ ] [ ] [ ] ] = + \u00d7 + \u00d7 \u00d7 + \u00d7 \u00d7 + \u00d7 \u2212 \u2212 \u2212 \u2212 \u2212 \u2212 \u2212 10 10 10 0 1 10 10 10 0 1 10 10 0 1 10 10 6 7 6 4 7 6 2 3 6 . ( . ) . 3 3 6 2 10 0 1 \u00d7 \u00d7 \u2212 ( . ) = 10 \u20136 +10 \u201312 + 10 \u201315 + 10 \u20134 + 10 \u20134 \u2248 2 \u00d7 10 \u20134 M
Answer: D
\nSolution: For molecular solubility of CaCl 2 , \u0394 H solution = 209.2 + (\u201333.5) = 175.7 KJ > 30 KJ For ionic solubility of CaCl 2 , \u0394 H solution = 209.2 + 1004.2 + 1715.4 \u2013 1598.3 \u2013 719.6 \u2013 711.2 = \u2013 100.3 KJ For molecular solubility of HgCl 2 , \u0394 H solution = 83.7 \u2013 66.9 = 16.8 KJ < 30 KJ For ionic solubility of HgCl 2 , \u0394 H solution = 83.7 + 460.2 + 2815.8 \u2013 1845.1 \u2013 719.6 \u2013 711.2 = 83.3 KJ > 30 KJ Hence, CaCl 2 is ionic and HgCl 2 is molecular solubility.
Answer: C
\nSolution: [ ] . C O M 2 4 2 5 6 0 001 5 250 2 6 10 \u2212 \u2212 = \u00d7 \u00d7 \u00d7 = \u00d7 \u2234 = \u00d7 = \u00d7 \u2212 \u2212 K sp ( ) . 6 10 3 6 10 5 2 9\n7.46 Chapter 7 HINTS AND EXPLANATIONS
Answer: A
\nSolution: Sr NO M Eqn. 0.001 M 0.05 M 2 0 001 0 001 75 100 3 0 05 0 05 + \u2212 = \u00d7 \u2212 \u2212 + . . . . x x \u001b \u001f \u21c0 \u001f \u001f \u21bd \u001f \u001f Sr(NO 3 0 ) + x \u2234 x = 0.00025 Now, K f = \u00d7 \u00d7 \u00d7 = \u2212 \u2212 2 5 10 7 5 10 0 05 20 3 4 4 . . .
Answer: A
\nSolution: ACOAg(s) H Cl ACOH AgCl(s) Mole M M 0 1 0 1 0 1 . . . + + + + \u2212 \u001f \u21c0 \u001f \u21bd \u001f \u001f K K K eq sp sp ACOH H Cl ACO ACO Ag Ag ACOAg] ( = \u00d7 \u00d7 = + \u2212 \u2212 \u2212 + + [ ] [ ][ ] [ ] [ ] [ ] [ ] [ A AgCl) \u00d7 K a = \u00d7 = \u21d2 \u2212 \u2212 \u2212 10 10 10 10 8 10 5 7 Almost complete reaction \u2234 = + \u2212 [ . , [H . ACOH] M ] =10 M \u001b 0 1 0 1 10 7 4 and [ ] [ ] [ ] . ACO Ka ACOH H M \u2212 + = \u00d7 = 0 01
Answer: B
\nSolution: A B (s) A B x y y x x y \u001f \u21c0 \u001f \u21bd \u001f \u001f + \u2212 + K x y s s K x y x y x y x y x y sp sp = \u22c5 \u22c5 \u21d2 = \u22c5 \u239b \u239d \u239c \u239c \u239e \u23a0 \u239f \u239f + + 1 As K sp << 1, greater the value of ( x + y ), greater is s.
Answer: A
\nSolution: Theory based
Answer: C
\nSolution: s = + = + + \u2212 \u2212 \u2212 [ ] [ [ ] [ ] Zn Zn(OH) K OH K OH sp f 2 4 2 2 For maximum or minimum S, d d OH S [ ] \u2212 = 0 or, \u2212 + = \u21d2 = \u239b \u239d \u239c \u239e \u23a0 \u239f = \u2212 \u2212 \u2212 \u2212 2 2 0 10 3 1 4 4 K OH K OH OH M sp f sp [ ] [ ] [ ] K K f \u2234 P H = 10 and S min = 2.4 \u00d7 10 \u20139 M
Answer: A
\nSolution: Al(OH) s) OH Al(OH) From question M 3 4 33 10 3 8 10 1 ( ; . ? + = \u00d7 \u2212 \u2212 \u2212 \u2212 \u001f \u21c0 \u001f \u21bd \u001f \u001f K 6 6 10 50 34 \u00d7 = \u2212 50 10 2 10 9 30 3 5 = \u21d2 = \u00d7 \u21d2 = \u2212 \u2212 \u2212 \u2212 [OH ] [ ] . OH M P H As the calculated [OH \u2013 ] is minimum OH \u2013 , P H is minimum. Al(OH) (s) Al OH K 3 M \u001f \u21c0 \u001f \u21bd \u001f \u001f 3 33 10 3 3 8 10 + \u2212 \u2212 \u2212 + = \u00d7 ; 8 10 10 2 10 4 30 33 3 3 10 \u00d7 = \u00d7 \u21d2 = \u00d7 \u21d2 = \u2212 \u2212 \u2212 \u2212 \u2212 [ ] [ ] . OH OH M P H As the calculated [OH \u2013 ] is maximum OH \u2013 , P H is maximum.
Answer: C
\nSolution: From the question, [ . Cu(CN) M 4 3 0 1 \u2212 = and [CN \u2013 ] = 0.2 M \u2234 = \u22c5 = \u00d7 \u00d7 = \u00d7 + \u2212 \u2212 \u2212 \u2212 [ ] [ ] [ ] . . ( . ) Cu Cu(CN) CN Instab K 4 3 4 15 4 6 4 10 0 1 0 2 4 10 1 13 M Now, [ ] [ ] . ( ) . S (Cu S) Cu M sp 2 2 2 27 13 2 2 2 56 10 4 10 1 6 10 \u2212 + \u2212 \u2212 \u2212 = = \u00d7 \u00d7 = \u00d7 K \u2234 = \u00d7 = \u00d7 \u00d7 \u00d7 = + \u2212 \u2212 \u2212 \u2212 [ ] [ ] [ ] . . . H H S S M 2 K a 2 21 2 10 1 6 10 0 1 1 6 10 10 and P H = 10.0
Answer: B
\nSolution: S M ppm = \u00d7 = \u00d7 = \u00d7 \u00d7 \u00d7 = \u00d7 \u2212 \u2212 \u2212 1 6 10 4 10 4 10 136 10 10 4 136 5 3 3 3 6 . For increase in concentration 4 times, volume should be 1 4 th . Hence, 75 % water should be evaporated.
Answer: C
\nSolution: During precipitation, the concentration of both Ba 2+ and SO 4 2 \u2212 ions will decrease.
Answer: B
\nSolution: K sp AgCl) ( = \u00d7 = \u2212 \u2212 \u2212 10 10 10 4 6 10 K sp (Ag CrO 2 4 4 2 4 12 10 8 10 8 10 ) ( ) = \u00d7 \u00d7 = \u00d7 \u2212 \u2212 \u2212 After precipitation of AgCl, fi nd the concentration of Cl \u2013 . [ ] . Cl M final \u2212 \u2212 \u2212 \u2212 = \u00d7 \u2212 \u00d7 \u00d7 \u00d7 1 0 10 8 10 2 10 6 7 7 Now, [ ] [ ] ( CrO ) ( Cl) CrO Cl K Ag K Ag final final sp sp 4 2 2 4 2 \u2212 \u2212 = [ ] ( ) ( ) [CrO ] . CrO final final 4 2 7 2 12 10 2 4 2 2 10 8 10 10 3 2 1 \u2212 \u2212 \u2212 \u2212 \u2212 \u00d7 = \u00d7 \u21d2 = \u00d7 0 0 5 \u2212 M\n7.47 Ionic Equilibrium HINTS AND EXPLANATIONS Hence, moles of Ag 2 CrO 4 precipitated = \u00d7 \u2212 \u00d7 = \u00d7 \u2212 \u2212 \u2212 8 10 3 2 10 7 68 10 4 5 4 . .
Answer: A
\nSolution: To prevent precipitation of AgCl, the concentration of Ag + needed in solution = K sp AgCl) Cl ( ( ) \u2212 = \u00d7 << \u2212 1 8 10 0 16 1 8 10 . . . M Hence, almost all Ag + ion must form complete with CN \u2013 ions. Ag CN Ag(CN) M CM M + \u00d7 \u2212 \u2212 \u2212 + = \u00d7 1 8 10 0 16 2 1 8 17 10 2 6 4 10 . . . ; . \u001f \u21c0 \u001f \u21bd \u001f \u001f K f
Answer: B
\nSolution: [CO ] ( ) [ ] [ ] 3 2 2 3 2 \u2212 + = \u22c5 K overall H CO H a To prevent precipitation of MCO 3 , or, [ ][ ] M CO K sp 2 3 2 + \u2212 \u2264 or K K sp , [M ] [H CO ] [H ] 2 2 3 2 + + \u22c5 \u22c5 \u2264 a \u2234 \u2265 \u22c5 + + [H ] [ ] [ ] M K H CO K a sp 2 2 3 For MgCO 3 : [H ] . . . + \u2212 \u2212 \u2212 \u2265 \u00d7 \u00d7 \u00d7 \u00d7 = \u00d7 0 1 5 10 0 05 9 10 2 5 10 17 8 6 M \u2234 P H \u2264 5.6 For SrCO 3 : [H ] . . + \u2212 \u2212 \u2212 \u2265 \u00d7 \u00d7 \u00d7 \u00d7 = \u00d7 0 1 5 10 0 05 9 10 5 10 3 17 10 5 M \u2234 P H \u2264 4.78 For precipitation of SrCO 3 without any precipitation of MgCO 3 , the P H range should be 4.78 to 5.6
Answer: A
\nSolution: Mn 2+ (aq) + H 2 S(aq) \u001c MnS(s) + 2H + (aq) To just start precipitation of MnS, Q < K eq or, [ ] [ ][ ] ) ( ) H Mn H S (H S MnS sp + + < 2 2 2 2 K K a or, [H ] . . . . [ ] + \u2212 \u2212 + \u2212 \u00d7 < \u00d7 \u00d7 \u21d2 < \u00d7 2 21 13 6 0 04 0 1 1 0 10 2 5 10 4 10 H M Now, in the given buffer, [ ] [ ] [ ] H CH COOH CH COO O O + \u2212 = \u22c5 K a 3 3 = \u00d7 \u00d7 = \u00d7 > \u00d7 \u2212 \u2212 \u2212 2 10 0 25 0 15 3 33 10 4 10 5 5 6 . . . M Hence, no precipitation. To start precipitation [H + ] should decrease and hence, CH 3 COONa should be added. Now, 4 10 2 10 0 25 6 5 \u00d7 = \u00d7 \u00d7 \u2212 \u2212 . [CH COONa] 3 O \u2234 [CH 3 COONa] O = 1.25 M
Answer: C
\nSolution: Mg(OH) S NH Mg NH OH 2 4 2 4 2 2 ( ) + + + + \u001f \u21c0 \u001f \u21bd \u001f \u001f To re-dissolve Mg(OH) 2 , Q \u2264 K eq or, [ ][ ] [NH ] Mg NH OH sp 2 4 2 4 2 2 + + \u2264 K K b or, 0 15 0 1 0 5 0 35 0 1 0 5 0 5 1 2 10 2 2 2 11 . . . . . . . . ( \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f \u239b \u239d \u239c \u239e \u23a0 \u239f \u2264 \u00d7 \u2212 n . . ) 0 10 5 2 \u00d7 \u2212 \u2234 n \u2265 0.035 Hence, minimum mass of (NH 4 ) 2 SO 4 needed = \u00d7 = 0 035 2 132 2 31 . . gm
Answer: B
\nSolution: Ag Cl M Eq 10 M + \u00d7 = \u00d7 \u00d7 \u2212 + \u2212 \u00d7 = \u00d7 \u2212 \u2212 + 500 0 01 1000 5 10 5 250 0 02 1000 5 3 3 . ( ) . x y 1 10 5 10 3 3 \u2212 \u2212 \u00d7 \u2212 M M AgCl(S) ( ) x \u001f \u21c0 \u001f \u21bd \u001f \u001f Ag Br AgBr( M 10 M M M + = \u00d7 \u00d7 \u2212 + \u2212 = \u00d7 \u00d7 \u2212 \u2212 \u2212 \u2212 \u2212 + 5 10 5 5 10 5 10 3 3 3 3 ( ) ( ) x y y \u001f \u21c0 \u001f \u21bd \u001f \u001f S S) As both reactions will tend towards completion, ( x + y ) = 5 \u00d7 10 \u20133 Now, [Ag ]( ) [ ] + \u2212 \u2212 + \u2212 \u00d7 \u2212 = \u21d2 \u22c5 = 5 10 10 10 3 10 10 x y Ag (1) and [Ag ]( ) + \u2212 \u2212 \u00d7 \u2212 = \u00d7 5 10 5 10 3 13 y (2) From (1) \u00f7 (2), y y y 5 10 200 1 201 3 \u00d7 \u2212 = \u21d2 = \u2212 \u2234 = \u00d7 \u2212 \u2248 \u00d7 \u2212 \u2212 \u2212 [ ] . Br M 5 10 2 5 10 3 5 y\n7.48 Chapter 7 HINTS AND EXPLANATIONS
Answer: D
" } } ] }, { "title": "Chem Sec 2", "originalName": "Section B - Multi Correct", "questions": [ { "question_id": "ionic-equilibrium-chem-sec-2-1-81", "marks": 4.0, "negMarks": 1.0, "partialMarks": 1, "subject": "chemistry", "chapter": "ionic-equilibrium", "chapterTitle": "Ionic Equilibrium", "type": "mcqm", "rawChapterType": "MSQ", "originalNumber": 81, "displayNumber": 1, "image": "./assets/chapter-packs/ionic-equilibrium/Chemistry Section 1__--__81__--__1.png", "solutionImage": null, "question": { "content": "Answer: D
\nSolution: (a) Complete neutralization \u21d2 P H = 7.0 (b) [ ] . . . . H M P final H + = \u00d7 \u2212 \u00d7 = \u21d2 = 55 0 1 45 0 1 100 0 01 2 0 (c) OH \u2013 is in excess. (d) [ ] . . H M P final H + = \u00d7 \u2212 \u00d7 = \u21d2 = 75 1 5 25 1 5 100 0 1 1 0
Answer: A, B, C
\nSolution: For basic solution: [ ] [ ] ] H OH and [H Kw + \u2212 + < < \u2234 P H > P OH or P P or P P H kw OH kw > < 2 2
Answer: C
\nSolution: HA H A C(1 M C M C M \u2212 + \u2212 + \u03b1 \u03b1 \u03b1 ) \u001f \u21c0 \u001f \u21bd \u001f \u001f K C C C C C K C a a = \u22c5 \u2212 = \u22c5 \u2212 \u2248 \u22c5 \u21d2 = \u03b1 \u03b1 \u03b1 \u03b1 \u03b1 \u03b1 \u03b1 ( ) 1 1 2 2 Now, K C C K K a a a = \u22c5 \u2212 = \u22c5 \u2212 \u21d2 = + + + + [ ] ( ) [ ] [ ] H H H \u03b1 \u03b1 \u03b1 \u03b1 \u03b1 1 1 = + = + + \u2212 1 1 1 1 10 [ ] ( ) H P P Ka H K a
Answer: A, B, C
\nSolution: Dilution results in increased degree of dissociation but decrease in concentrations of all active components.
Answer: A, B
\nSolution: Relation is valid only for conjugate pairs.
Answer: B
\nSolution: P H may decrease only on increasing [H + ].
Answer: A, B, C, D
\nSolution: CH COOH CH COO H M 0.1 M M M 0.1 M 3 3 0 1 0 1 ( . ) ( . ) \u2212 + \u2212 + + x x x \u001b \u001b \u001f \u21c0 \u001f \u21bd \u001f \u001f Now, 1 8 10 0 1 0 1 1 8 10 5 5 . . . . \u00d7 = \u00d7 \u21d2 = \u00d7 \u2212 \u2212 x x and \u03b1 = = \u00d7 \u2212 x 0 1 1 8 10 4 . . Now, [ ] [ ] [ ] H OH Kw H M T from water acid = = = \u2212 + \u2212 10 13
Answer: A, B, C, D
\nSolution: RNH g) H O(l) RNH aq OH aq bar M M 2 1 2 3 ( ( ) ( ) + + + \u2212 \u001f \u21c0 \u001f \u21bd \u001f \u001f x x 10 1 10 3 0 11 0 6 3 \u2212 \u2212 = \u22c5 \u21d2 = \u21d2 = \u21d2 = x x x P P OH H . .
Answer: D
\nSolution: NH OH(aq) NH OH 4 4 \u001f \u21c0 \u001f \u21bd \u001f \u001f + \u2212 + Addition of solid NH 4 OH will increase NH 4 OH(aq) concentration and hence, [OH \u2013 ] will increase.
Answer: B, C
\nSolution: (c) H O H O H O OH ve 2 2 3 + + \u0394 \u00b0 = + + \u2212 \u001f \u21c0 \u001f \u21bd \u001f \u001f ; H (d) HA OH A H O Final a a a a 2 2 0 2 2 + + \u2212 \u2212 \u001f \u21c0 \u001f \u21bd \u001f \u001f P P P H K K a a = + = log / / a a 2 2
Answer: D
\nSolution: [ ][ ] H C O + \u2212 >> 3 2
Answer: A, B, D
\nSolution: Theory based
Answer: A, C
\nSolution: NH Cl NaOH NH OH NaCl For buffer: ( 4 4 0 0 a a b b a b b \u2212 \u2248 \u21d2 > + + ) \u001f \u21c0 \u001f \u21bd \u001f \u001f CH COONa HCl CH COOH NaCl For buffer: ( 3 3 0 0 a a b b a b b \u2212 \u2248 \u21d2 > + + ) \u001f \u21c0 \u001f \u21bd \u001f \u001f
Answer: A, B, C, D
\nSolution: KCN is a salt of weak acid (HCN) and strong base (KOH).
Answer: B, C
\nSolution: BOH H B mole th run a mole Equivalent point a a b a 1 5 5 0 0 0 0 \u2212 + + + \u001b \u001b \u001f \u21c0 \u001f \u21bd \u001f \u001f 5 5 2 a mole H O + Now, for 1/5th reaction, P P B BOH] OH K b = + \u2212 log [ ] [ or, ( ) log / / 14 9 5 4 5 \u2212 = + P K b a a \u2234 = \u21d2 = \u00d7 \u2212 P K K b b 5 6 2 5 10 6 . . At equivalent point: P P C) H K b => \u2212 + 1 2 ( log or, 4 5 1 2 5 6 . ( . log => \u2212 + \u21d2 = C) C 0.25 M Now, n n HCl used B formed = + or, V 0.5 V V ml HCl \u00d7 = + \u00d7 \u21d2 = 1000 100 0 25 1000 100 ( ) .\n7.49 Ionic Equilibrium HINTS AND EXPLANATIONS Finally, n n BOH takes Hcl used for equivalent point = or gm , . . w w 45 100 0 5 1000 2 25 = \u00d7 \u21d2 = \u2234 Percentage purity of base = \u00d7 = 2 25 2 5 100 90 . . %
Answer: A, C, D
\nSolution: CO H O HCO OH K Kw K M M M h a 1 2 3 2 2 3 0 5 2 1 \u2212 \u2212 \u2212 \u2212 \u2212 + + + = = \u00d7 ( . ) ( ) ( ) ; x x y x y \u001f \u21c0 \u001f \u21bd \u001f \u001f 0 0 4 \u2212 HCO H O H CO OH K Kw K M M M 2 h a 3 2 3 9 2 1 2 5 10 \u2212 \u2212 \u2212 \u2212 + + + = = \u00d7 ( ) ( ) ; . x y y x y \u001f \u21c0 \u001f \u21bd \u001f \u001f Now, 2 10 0 5 0 5 10 4 2 \u00d7 = \u2212 \u22c5 + \u2212 \u2248 \u22c5 \u21d2 = \u2212 \u2212 ( ) ( ) ( . ) . x y x y x x x x and 2 5 10 2 5 10 9 9 . ( ) ( ) . \u00d7 = \u22c5 + \u2212 \u2248 \u22c5 \u21d2 = \u00d7 \u2212 \u2212 y x y x y y x x y Now, h x = = 0 5 0 02 . . P P OH H = \u2212 + \u2248 \u2212 = \u21d2 = \u2212 log( ) log( ) . . x y 10 2 0 12 0 2 and [H 2 CO 3 ] = y = 2.5 \u00d7 10 \u20139 M
Answer: A, B, D
\nSolution: N H CH COOH N H CH COO NH CH P P K a 1 K a2 + = + \u2212 = \u23af \u2192 \u23af\u23af\u23af\u23af \u23af \u2192 \u23af\u23af\u23af\u23af 3 2 2 22 3 2 9 78 3 2 . . C COO \u2212 P P P H K K = + = + = 1 2 1 2 2 22 9 78 6 0 1 2 ( ) ( . . ) . a a Now, K a 1 3 3 2 22 6 3 10 0 01 10 = \u21d2 = \u00d7 \u2295 \u2212 + \u2295 \u2212 \u2212 \u2295 [NH ][H ] [NH ] . [NH . CH COO CH COOH C 2 2 H H COOH 2 ] \u2234 = = \u00d7 \u2295 \u2212 \u2212 [NH ] . . 3 5 78 6 10 1 7 10 CH COOH M 2 % of glycine in cationic form = \u00d7 \u00d7 = \u2212 1 7 10 0 01 100 0 017 6 . . . %
Answer: C
\nSolution: At equivalent point, the solution should be acidic.
Answer: C
\nSolution: Sodium acetate solution is basic.
Answer: C
\nSolution: For precipitation of Fe(OH) 2 , [ ] . . . min max min OH P P OH H \u2212 \u2212 \u2212 = \u00d7 = \u00d7 \u21d2 = \u21d2 = 8 10 0 02 2 10 6 7 7 3 16 7 For precipitation of Fe(OH) 3 , [ ] . . . min / max min OH P P OH H \u2212 \u2212 \u2212 = \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f = \u00d7 \u21d2 = \u21d2 = 4 10 0 05 2 10 8 7 5 3 28 1 3 9
Answer: C
\nSolution: K x x x = \u00d7 = = \u2212 \u21d2 = 1.5 10 Dimer] [Monomer] 2 2 2 0 1 2 5 120 [ ( . ) \u2234 [ ] [ . Dimer Monomer] = \u2212 = x x 0 1 2 5 2
Answer: C
\nSolution: K x x x = \u00d7 = = \u2212 \u21d2 = \u00d7 \u2212 \u2212 3.6 Dimer Monomer 10 0 1 2 3 6 10 2 2 4 ( ) [ ] ( . ) . \u2234 (D ) [Monomer] . . imer = \u2212 \u21d2 = x x x 0 1 2 0 1 9 2500
Answer: B
\nSolution: [ ] . H M + \u2212 \u2212 \u2248 \u00d7 \u00d7 = \u00d7 0 1 2 10 2 10 5 3 (Dimerization is negative as Q. 2) \u2234 P H = 2.85\n7.50 Chapter 7 HINTS AND EXPLANATIONS Comprehension II
Answer: D
\nSolution: [ ] . CH COOH M O 3 3 3 8 0 7 10 10 10 7 10 = \u00d7 \u00d7 = \u00d7 \u2212 \u2212 The solution is so dilute that we may assume almost complete dissociation of acid. \u2234 \u2248 \u00d7 + \u2212 [ ] H M acid 7 10 8 Now, H O H OH M M 2 7 10 8 \u001f \u21c0 \u001f \u21bd \u001f \u001f + \u00d7 + \u2212 \u2212 + ( ) x x 10 7 10 7 10 14 8 8 \u2212 \u2212 \u2212 = \u00d7 + \u22c5 \u21d2 = \u00d7 ( ) x x x \u2234 = \u2212 \u00d7 + = \u2212 P H log( ) . 7 10 6 85 8 x
Answer: A
\nSolution: CH COOH CH COO H Eqn M 7 10 M M 3 3 8 7 10 8 14 10 8 . y x \u001f \u21c0 \u001f \u21bd \u001f \u001f \u2212 + \u00d7 \u00d7 \u2212 + = \u00d7 \u2212 \u2212 + Now, 2 0 10 7 10 14 10 5 8 8 . \u00d7 = \u00d7 \u00d7 \u00d7 \u2212 \u2212 \u2212 y \u2234 y = 4.9 \u00d7 10 \u201310 M Comprehension III
Answer: C
\nSolution: K a = \u00d7 = \u00d7 \u2212 \u2212 ( ) . . 8 10 0 2 3 2 10 3 2 4
Answer: C
\nSolution: K a = \u00d7 = \u00d7 \u00d7 \u2212 + 3 2 10 1 0 0 8 0 2 4 . [ ] ( . . ) . H \u2234 [H + ] = 8 \u00d7 10 \u20135 = \u21d2 P H = 4.1 Comprehension IV
Answer: B
\nSolution: p p NH NH NH H Ka O = + = + = + + ( ) log [ ] [ ] . log . . . 4 3 4 9 3 0 8 0 2 9 9
Answer: C
\nSolution: NH OH H NH H O M M M M Final M 4 0 8 0 3 0 4 0 2 0 5 2 0 5 . . . . . + + + \u2248 + \u001f \u21c0 \u001f \u21bd \u001f \u001f p p NH NH H K O O a = + = + = + + (NH ) log [ ] [ ] . log . . . 4 3 4 9 3 0 5 0 5 9 3
Answer: A
\nSolution: H + added in excess. Final [H + ] = 1.0 \u2013 0.8 = 0.2 M \u2234 P H = \u2013 log(0.2) = 0.7 Comprehension V
Answer: B
\nSolution: No hydrolysis \u21d2 p H = 7.0
Answer: A
\nSolution: Concentration of KAl(SO H O M 4 2 2 12 11 85 474 100 1000 0 25 ) . . / / . = = Al H O Al(OH H M M M 3 0 25 2 2 + \u2212 + + + + ( . ) ) x x x \u001f \u21c0 \u001f \u21bd \u001f \u001f 1 4 10 0 25 0 25 1 87 10 5 2 3 . ( . ) . . ] \u00d7 = \u22c5 \u2212 \u21d2 = \u00d7 \u2212 \u2212 + x x x x x \u001b M = [H
Answer: B
\nSolution: SO H O HSO OH M 4 2 0 5 2 4 \u2212 \u2212 \u2212 \u2212 + + ( . ) x x x \u001f \u21c0 \u001f \u21bd \u001f \u001f 10 1 25 10 0 5 0 5 6 32 10 19 2 2 7 \u2212 \u2212 \u2212 \u00d7 = \u22c5 \u2212 \u21d2 = \u00d7 . ( . ) . . x x x x x \u001b M \u2234 = \u00d7 = \u00d7 + \u2212 \u2212 \u2212 [ ] . . H M 10 6 32 10 1 58 10 14 7 8
Answer: B
\nSolution: 1 4 10 1 25 10 0 25 0 5 0 25 0 5 5 2 2 . . ( . )( . ) . . \u00d7 \u00d7 = \u22c5 \u2212 \u2212 \u00d7 \u2212 \u2212 x x x x x \u001b \u2234 x = 1.18 \u00d7 10 \u20132 Now, 1 4 10 1 18 10 0 25 5 2 . . [ ] . \u00d7 = \u00d7 \u00d7 \u2212 \u2212 + H \u2234 [H + ] = 2.97 \u00d7 10 \u20134 M Al SO H O Al(OH HSO M M M 3 0 25 0 25 4 2 0 5 2 2 0 4 + \u2212 \u2212 + \u2212 + + + . Eqn. ( . ) . ) x x \u001f \u21c0 \u001f \u21bd \u001f \u001f 0 0 x (0.5 \u2013 x )M\n7.51 Ionic Equilibrium HINTS AND EXPLANATIONS Comprehension VI
Answer: A
\nSolution: PuO H O PuO (OH H 2 2 0 01 0 01 2 2 1 6 10 4 + \u2212 \u2248 + + = \u00d7 + + \u2212 . . . ) x x x \u001f \u21c0 \u001f \u21bd \u001f \u001f \u2234 = \u22c5 \u2212 = \u00d7 \u2212 K x x x x h 0 01 0 1 2 56 10 2 6 . . . \u001b
Answer: C
\nSolution: K Kw K K a b b = \u21d2 = \u00d7 \u2212 3 9 10 9 . Comprehension VII
Answer: D
\nSolution: p P P H K K a 2 a = + = + = 1 2 1 2 8 13 1 0 5 3 ( ) ( ) .
Answer: A
\nSolution: For 2nd equivalent point, n n NaoH H PO = \u00d7 2 3 4 V V 40 ml \u00d7 = \u00d7 \u00d7 \u21d2 = 0 5 1000 2 100 0 1 1000 . . After adding HCl, HPO H H O millimole Final 5 millimole 0 4 2 10 5 2 4 0 5 \u2212 + \u2248 \u2212 + \u001f \u21c0 \u001f \u21bd \u001f \u001f \u2234 = + = P H 8 5 5 8 0 log .
Answer: A
\nSolution: S K OH M sp = = \u00d7 = \u00d7 \u2212 \u2212 \u2212 \u2212 [ ] . ( ) . 2 30 6 2 18 4 0 10 10 4 0 10 Comprehension VIII
Answer: A
\nSolution: P P [HCO H CO H K O O a = + = + = \u2212 log ] [ ] . log . 3 2 3 6 4 8 1 7 3
Answer: C
\nSolution: 7 4 6 4 1 10 3 2 3 2 3 3 . . log [ ] [ ] [ ] [ ] = + \u21d2 = \u2212 \u2212 HCO H CO H CO HCO O O O O
Answer: C
\nSolution: More CO 2 should dissolve in solution. Comprehension IX
Answer: C
\nSolution: pH P C)=7+ a 2 = + + + = 7 1 2 1 2 10 6 1 12 3 ( log ( . log ) . K
Answer: A
\nSolution: CO H HCO M Final M M M 3 2 50 1 75 25 75 25 1 75 3 0 25 75 0 \u2212 \u00d7 + \u00d7 \u2212 + \u2248 \u001f \u21c0 \u001f \u21bd \u001f \u001f pH CO HCO a 2 O O = + = + = \u2212 \u2212 P K log [ ] [ ] . log / / . 3 2 3 10 6 1 3 1 3 10 6
Answer: A
\nSolution: CO H HCO M Final M M 3 2 50 1 100 0 50 1 100 3 0 50 100 0 \u2212 \u00d7 \u2248 + \u00d7 \u2212 + \u2248 \u001f \u21c0 \u001f \u21bd \u001f \u001f \u2234 = + = + = pH a 1 a 1 2 1 2 5 4 10 6 8 0 2 ( ) ( . . ) . P P K K
Answer: D
\nSolution: CO H HCO M Final M M M 3 2 50 1 125 0 75 1 125 25 125 3 0 50 125 \u2212 \u00d7 + \u00d7 \u2212 + \u001f \u21c0 \u001f \u21bd \u001f \u001f HCO H H CO M Final M M M 3 50 125 25 125 25 125 0 2 3 0 25 125 \u2212 + = + \u001f \u21c0 \u001f \u21bd \u001f \u001f \u2234 = + = + = \u2212 pH [HCO H CO a O O P K log ] [ ] . log / / . 3 2 3 5 4 25 125 25 125 5 4
Answer: A
\nSolution: [H CO ] M 2 3 = \u00d7 = 50 10 150 1 3 \u2234 = + = + = pH C a 1 1 2 1 2 5 4 1 3 2 94 ( log ) ( . log ) . P K\n7.52 Chapter 7 HINTS AND EXPLANATIONS Comprehension X
Answer: A
\nSolution: N H CH COOH N H CH 3 3 K K 3 3 b + = \u00d7 = \u00d7 + \u2212 \u2212 \u23af \u2192 \u23af\u23af\u23af\u23af\u23af \u2190 \u23af \u23af\u23af\u23af\u23af\u23af \u2212 \u2212 \u2212 2 12 1 3 2 5 10 4 10 . a \u2212 \u2212 \u23af \u2192 \u23af\u23af\u23af\u23af\u23af \u2190 \u23af \u23af\u23af\u23af\u23af\u23af \u2212 \u2212 \u2212 = \u00d7 = \u00d7 \u2212 \u2212 \u2212 COO N H CH COO K K a b 2 10 1 5 1 6 10 6 25 10 2 2 . . Required K K b b = = \u00d7 \u00d7 \u2212 1 6 25 10 10 5 .
Answer: C
\nSolution: \u2234 = + = + = pH a 1 a 1 2 1 2 2 4 9 8 6 1 2 ( ) ( . . ) . P P K K Comprehension XI
Answer: D
\nSolution: Moles of Cu reacted = 6 35 10 63 5 10 3 4 . . \u00d7 = \u2212 \u2212
Answer: A
\nSolution: ln . K G RT K eq eq = \u0394 \u2212 = \u2212 \u00d7 \u2212 \u00d7 = \u21d2 >>> \u00b0 120 10 8 0 300 50 1 3 \u2234 = + [ ] Ag 2 \u00d7 Mole of Cu reacted = 2 \u00d7 10 \u20134 M
Answer: C
\nSolution: K sp Ag M = = \u00d7 = \u00d7 + \u2212 \u2212 \u2212 [ ][BRO ] ( ) 3 4 2 8 2 2 10 4 10 Comprehension XII
Answer: B
\nSolution: S K AgCN CN M sp = = \u00d7 = \u00d7 \u2212 \u2212 \u2212 [ ] [ ] . . . 1 0 10 0 02 5 0 10 16 15
Answer: D
\nSolution: AgCN(s) CN Ag(CN S M SM eq sp + = \u00d7 = \u2212 \u2212 \u2212 ( . ) ) 0 02 2 15 \u001f \u21c0 \u001f \u21bd \u001f \u001f K K K f 15 0 02 0 3 16 1 875 10 2 = \u2212 \u21d2 = = \u00d7 \u2212 s s s M . . .
Answer: C
\nSolution: s [Ag ] [Ag(Cu Cu Cu M sp sp = + + \u22c5 \u22c5 + \u2212 \u2212 \u2212 ) ]; [ ] [ ] 2 5 K K K f For minimum solubility: ds d[Cu ] \u2212 = 0 or, \u2212 + \u22c5 = \u21d2 = = \u00d7 \u2212 \u2212 \u2212 K K K f sp sp f Cu Cu K M [ ] [ ] . 2 9 0 1 2 58 10
Answer: D
\nSolution: For acidic solution, pH < 7.0 at 25\u00b0 C.
Answer: A
\nSolution: If there were no common ion effect, P H should lie in between 7.0 and 7.3
Answer: D
\nSolution: [ ] ] . H M [H M HCl HCOOH + \u2212 + \u2212 = < = \u00d7 10 3 16 10 4 2
Answer: D
\nSolution: Dilution results decrease in concentration of BOH (aq), B + (aq) as well as OH \u2013 (aq).
Answer: D
\nSolution: The pH of buffer remains constant on slight dilution but for acidic solution, the dilution results in the decrease in [H + ] and hence, increase in pH.
Answer: C
\nSolution: pH of buff er containing H A and A \u2013 may be less than, greater than or equal to 7.0.
Answer: C
\nSolution: K K a b ( ( ) CH COOH) NH OH 3 = 4 and hence, CH 3 COONH 4 solution is also neutral. But, it undergoes hydrolysis.
Answer: A
\nSolution: Theory based
Answer: C
\nSolution: Reaction occurs but at equivalent point, pH will be less than
Answer: D
\nSolution: 10. As dilution does not change the concentration of ions in saturated solution, the mole of ions will increase.\n7.53 Ionic Equilibrium HINTS AND EXPLANATIONS
Answer: A \u2192 P, Q; B \u2192 Q, R; C \u2192 R, S; D \u2192 T
\nSolution: True electrolytes produce ions in pure liquid form as well as in solution. Potential electrolytes are molecular in pure liquid state but it produces ions in solution.
Answer: A \u2192 Q, S; B \u2192 T; C \u2192 P; D \u2192 R
\nSolution: (P) [OH\u2013] to just start precipitation = K sp Mg [ ] 2 + = \u00d7 \u00d7 = \u2212 \u2212 \u2212 2 10 2 10 10 6 3 1 5 . \u2234 p OH = 1.5 \u21d2 P H = 12.5 (Q) [ ] [ ] . max / / OH Al M sp \u2212 + \u2212 \u2212 = \u239b \u239d \u239c \u239e \u23a0 \u239f = \u239b \u239d \u239c \u239e \u23a0 \u239f = K 3 1 3 28 1 3 9 10 0 1 10 \u2234 = \u21d2 = p p OH H min max . 9 5 0 (R) CH COOH CH COO 3 M = M 3 M = 0 1 0 1 10 11 0 1 11 0 1 10 11 1 1 . . . . \u2212 \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f \u2212 \u00d7 \u001f \u21c0 \u001f \u21bd \u001f \u001f 1 1 M H + + ? Now, K a = = \u00d7 \u21d2 = \u2212 + + \u2212 10 1 11 0 1 11 10 5 6 [H ] . [ ] H M \u2234 p H = 6.0 (S) [ ] . . H C M pH a + \u2212 \u2212 = \u22c5 = \u00d7 = \u21d2 = K 10 0 001 10 5 0 7 5 (T) A H O HA OH M M M \u2212 \u00d7 \u2212 \u2212 \u2212 \u2212 + + = \u00d7 ( ) ; 6 10 2 6 5 2 10 x x x a Kw K \u001f \u21c0 \u001f \u21bd \u001f \u001f 2 10 6 10 1 10 6 5 5 \u00d7 = \u22c5 \u00d7 \u2212 \u21d2 = \u00d7 \u2212 \u2212 \u2212 x x x x ( ) \u2234 P OH = 5 \u21d2 pH = 9
Answer: A \u2192 R; B \u2192 Q, T; C \u2192 P, S; D \u2192 P, Q, R
\nSolution: Theory based
Answer: A \u2192 R; B \u2192 Q; C \u2192 P; D \u2192 S; E \u2192 T
\nSolution: (A) pH [H A H A] a 2 O O = + = \u2212 P K 1 3 4 0 log ] [ . (B) pH [HA H A ] a O O = + = \u2212 \u2212 P K 2 2 2 8 0 log ] [ . (C) pH [A HA ] a O O = + = \u2212 \u2212 P K 3 3 2 12 0 log ] [ . (D) pH a a 2 = + = + = 1 2 1 2 4 8 6 0 1 ( ) ( ) . P P K K (E) pH K K a a = + = + = 1 2 1 2 8 12 1 0 0 2 3 ( ) ( ) . P P
Answer: A \u2192 Q, R; B \u2192 P; C \u2192 P, S
\nSolution: (A) K K a b ( ( [ ] . H O) H O)= Kw H O 2 2 2 14 16 10 1000 18 1 8 10 = = \u239b \u239d \u239c \u239e \u23a0 \u239f = \u00d7 \u2212 \u2212 (C) On increasing temperature, Kw increases and hence, P Kw decreases.
Answer: 8
\nSolution: [ ] log[ ] D K M P D W D + \u2212 \u2212 + = = = \u21d2 = \u2212 = 10 10 8 16 8
Answer: 4
\nSolution: [ . HCOOH] M C O = \u00d7 = = 1 15 10 46 25 3 HCOOH HCOOH HCOOH HCOO C C \u2212 \u2212 + \u2212 + + x x x x \u001f \u21c0 \u001f \u21bd \u001f \u001f 2 From given data: x = = \u2212 K M 10 3 \u2234 Percentage of HCOOH molecules converted into HCOO \u2013 = \u00d7 = \u00d7 = \u00d7 \u2212 \u2212 x C 100 10 25 100 4 10 3 3
Answer: 1
\nSolution: [ ] / / . NH M O 3 3 10 17 100 0 85 10 5 = \u00d7 =\n7.54 Chapter 7 HINTS AND EXPLANATIONS [ ] ) OH C Kw K (NH C M a \u2212 + \u2212 \u2212 \u2212 = \u22c5 = \u22c5 = \u00d7 \u00d7 = K b 4 14 10 2 10 5 10 5 10 \u2234 = = = + \u2212 \u2212 \u2212 \u2212 [ ] [ ] H O Kw OH M 3 14 2 12 10 10 10
Answer: 6
\nSolution: [ ] . H K C M a + \u2212 \u2212 = \u22c5 = \u00d7 \u00d7 = 4 10 0 0025 10 10 6 \u2234 = \u2212 = \u2212 P H log10 6 6
Answer: 2
\nSolution: [ . . H SO ] M 2 3 O = = 1 28 64 0 02 H SO H HSO 2 3 M M M ( . ) 0 02 3 \u2212 + \u2212 + x x x \u001f \u21c0 \u001f \u21bd \u001f \u001f Now, K x x x x x a M pH = = \u22c5 \u2212 \u21d2 = \u21d2 = \u2212 = \u2212 10 0 02 0 01 2 2 ( . ) . log
Answer: 3
\nSolution: pH HC H O [H C H O a 4 4 O O = + = + = \u2212 P K 1 6 2 4 4 6 3 3 18 8 18 8 30 150 3 log [ ] ] . log . / . /
Answer: 5
\nSolution: Buff er capacity, \u03b2 = \u2212 \u0394 = \u2212 \u2212 = + [ ] . / . ( . ) H added H P 0 05 0 2 0 05 5
Answer: 5
\nSolution: HA OH A millimole 0 millimole Equ.point a \u001b \u001b \u001f \u21c0 \u001f \u21bd \u001f \u001f + \u2212 \u2212 \u00d7 36 12 0 1 0 0 3 . . . .612 2 millimole H O + A H HA mmole Final mmole mmole \u2212 + + \u00d7 3 612 1 806 18 06 0 1 0 0 1 . . . . . \u001b \u001f \u21c0 \u001f \u21bd \u001f \u001f 8 806 mmole \u2234 = + = + = \u2212 pH A HA] a O O P K log [ ] [ log . . 5 1 806 1 806 5
Answer: 6
\nSolution: \u2234 = + = + = pH a a 2 1 2 1 2 2 28 9 72 6 1 ( ) ( . . ) P P K K
Answer: 4
\nSolution: For appearance of only ln + colour, log [ln ] [ln . . . log + = \u2212 = = OH] 4 6 3 4 2 0 6 4 \u2234 = + [ln ] [ln OH] 4 Four-digit Integer Type
Answer: 0160
\nSolution: C H NH H O C H NH OH M Eqn. M M 2 M M 6 5 2 0 2 0 2 0 2 6 5 3 0 10 8 . ( . ) . \u2212 + = + + \u2212 x x \u001b \u001f \u21c0 \u001f \u21bd \u001f \u001f \u2212 \u2212 + CM (C ) M CM x \u001b Now, K C b = \u21d2 \u00d7 = \u00d7 + \u2212 \u2212 \u2212 [C H NH ][OH ] C H NH . 6 5 3 6 5 2 10 8 4 10 10 0 2 \u2234 C = 8 \u00d7 10 \u20133 M Now, mass of NaOH added = \u00d7 \u00d7 \u00d7 = = \u2212 8 10 1000 500 40 0 16 160 3 . gm mg
Answer: 0369
\nSolution: K a = \u21d2 \u00d7 = \u00d7 \u00d7 + \u2212 \u2212 \u2212 \u2212 [ ][CH COO ] [CH COOH . [CH COO ] . H ] 3 3 5 4 3 1 8 10 4 10 0 2 \u2234 = \u00d7 \u2212 \u2212 [CH COO ] 3 3 9 10 M \u2234 Mass of CH 3 COONa added 9 10 500 1000 82 0 369 3 \u00d7 \u00d7 \u00d7 = = \u2212 . gm 369 mg
Answer: 1100
\nSolution: H SO H HSO 2 3 CM Final 0 CM CM \u001b \u001f \u21c0 \u001f \u21bd \u001f \u001f + \u2212 + 0 4 0 H A OH HA mmole Final 0 mmole mmole 2 20 0 09 30 0 06 0 1 8 + \u2212 \u00d7 \u00d7 + + . . . \u001b \u001b \u001f \u21c0 \u001f \u21bd \u001f \u001f H H O 2\n7.55 Ionic Equilibrium HINTS AND EXPLANATIONS HSO H SO Fianl C )M C M =0.01 M M 4 4 2 \u2212 \u2212 + + \u2212 + ( ( ) x x x \u001f \u21c0 \u001f \u21bd \u001f \u001f Now, 1 2 10 0 01 6 11 2 . . \u00d7 = \u00d7 \u2212 \u21d2 = \u2212 x x x C C and C + x = 0.01 \u21d2 C M = \u00d7 0 01 11 17 .
Answer: 0080
\nSolution: P P NaHCo [H Co H K O O a = + log [ ] ] 3 2 3 or, 7 4 6 1 10 2 80 . . log = + \u00d7 \u00d7 \u21d2 = V 5 V ml
Answer: 0060
\nSolution: [ ] . / / . HSO M O 4 1 8 120 100 1000 0 15 \u2212 = = HSO H SO M M M 4 0 15 4 2 \u2212 \u2212 + \u2212 + ( . ) x x x \u001f \u21c0 \u001f \u21bd \u001f \u001f 4 10 0 15 6 10 60 2 2 \u00d7 = \u22c5 \u2212 \u21d2 = \u00d7 = \u2212 \u2212 x x x x . M millimole/L
Answer: 5340
\nSolution: Al H O H O Al H O H O C M Now, 10 OH M M ( ) ( ) ( ) 2 6 3 2 2 5 3 1 5 2 + \u2212 = + = \u2212 + + + x x x \u001f \u21c0 \u001f \u21bd \u001f \u001f 0 0 3 \u2212 \u2248 M C 0.1M \u2234 Mass of Al(OH) 3 added = 400 \u00d7 0.1 \u00d7 133.5 = 5.34 gm = 5340 mg
Answer: 0050
\nSolution: 9 18 60 40 . log = + P K a (1) 9 00 100 . log = + \u2212 P x x K a (2) \u2234 x = 50
Answer: 0960
\nSolution: I I I M Eqn. (0.05 M M 2 12 7 254 0 05 0 1 0 1 3 0 . . ) . . = \u2212 \u2212 \u2212 \u2212 + x x x \u001f \u21c0 \u001f \u21bd \u001f \u001f = = 0 254 254 0 001 . . \u2234 x = 0.049 Now, K x x x c = \u2212 \u2212 = \u00d7 = ( . )( . ) . . . 0 05 0 1 0 049 0 001 0 051 960
Answer: 0740
\nSolution: Concentration of Ca(OH) 2 in its saturated solution = \u239b \u239d \u239c \u239e \u23a0 \u239f = \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f = \u2212 K sp M 4 3 2 10 4 0 02 1 3 5 1 3 / / . . Now, Ca OH Ca(OH) s M M 2 0 02 0 04 2 2 + \u2212 + . . ( ) \u001f \u21c0 \u001f \u21bd \u001f \u001f 0.01 M (0.02 + 0.8) M = 0.82 M Equ (0.01 \u2013 x )M = 0 (0.82 \u2013 2 x ) M = 0.8M \u2234 = \u00d7 = \u00d7 << + \u2212 \u2212 [ ] . ( . ) . Ca left 2 5 2 5 3 2 10 0 8 5 10 0 01 \u2234 Ca(OH) 2 precipitated = 0.01 mole = 0.01 \u00d7 74 = 0.74 gm = 740 mg
Answer: 5500
\nSolution: S M = = \u2212 0 0055 550 100 1000 10 4 . / / \u2234 K sp of Ca(pam) 2 = 4S 3 = 4 \u00d7 10 \u201312 M Now, Ca pam Ca(pam) M Final =0 M 0.1M 2 40 40 10 10 10 0 1 2 6 2 3 2 + \u00d7 = \u2212 \u2212 + / . ( \u001f \u21c0 \u001f \u21bd \u001f \u001f s s) \u2234 Ca(pam) 2 participated = 10 \u20133 \u00d7 10 = 0.01 mole = 0.01 \u00d7 550 = 5.50 gm = 550 mg On adding NaOH
Answer: A
\nSolution: P K X N H N 2 2 = \u22c5 ( ) Solution or 5 0 8 1 0 10 10 10 10 5 5 2 2 2 \u00d7 = \u00d7 \u00d7 + \u00d7 . ( . ) n n n N N N \u001f \u2234 n N 2 4 10 4 = \u00d7 \u2212
Answer: C
\nSolution: p K X K n n m P V H H = \u22c5 \u2248 \u22c5 \u21d2 \u22c5 gas liq gas liq \u03b1 \u2234 = \u22c5 \u22c5 \u21d2 = \u00d7 \u00d7 \u21d2 = m m P V P V m m m 2 1 2 2 1 1 2 2 5 2 1 1 10 m Now, P K n n K n P V RT H H = \u22c5 = \u22c5 \u22c5 gas liq liq \u2234 Volume of gas dissolved, V RT K n H = \u22c5 liq (Volume of gas dissolved is independent of pressure of gas) \u2234 = \u21d2 = \u21d2 = V V V V V V V 2 1 2 1 2 2 2 1 2 , , liquid liquid ml V ml\n10.31 Liquid Solution HINTS AND EXPLANATIONS
Answer: B
\nSolution: P X P Hg Hg = \u22c5 = \u00d7 \u00d7 + \u00d7 = \u00d7 \u2212 \u2212 \u2212 total t 0 8 10 200 0 8 10 200 50 4 28 720 1 6 10 3 3 3 . . . . o orr
Answer: C
\nSolution: Final vapour pressure and hence, the composition of both solutions must be same. As solution in beaker (A) has higher concentration, its vapour pressure is low. Hence, water from (B) will transfer in (A) as vapour. \u2234 + = \u2212 \u21d2 = 20 200 10 100 33 33 x x x .
Answer: B
\nSolution: Y P P X P P A A A A = = \u22c5 \u00b0 total total 1 1 1 1 X Y P Y P Y P Y P P P P P A A A A A A B A A B B A B = \u22c5 \u00b0 \u22c5 \u00b0 + \u2212 \u00b0 \u239b \u239d \u239c \u239e \u23a0 \u239f = \u22c5 \u00b0 \u00b0 + \u00b0 \u2212 \u00b0 \u00b0 ( )
Answer: C
\nSolution: n n n n P P P Q P Q P Q n \u239b \u239d \u239c \u239e \u23a0 \u239f = \u239b \u239d \u239c \u239e \u23a0 \u239f \u22c5 \u00b0 \u00b0 \u239b \u239d \u239c \u239e \u23a0 \u239f 2 nd condense initial Where n = number of condensation steps. Now, n n P Q \u239b \u239d \u239c \u239e \u23a0 \u239f = \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f = final 1 1 300 100 9 1 2 \u2234 = + = X P 9 9 1 0 90 .
Answer: B
\nSolution: The mole fraction of A in distillate, \u2032 = = \u22c5 \u00b0 = \u00d7 \u00d7 + \u00d7 = X Y X P P A A A A total 1 4 100 1 4 100 3 4 80 5 17 Now, V.P. of distillate, P X P X P A A B B = \u2032 \u22c5 \u00b0 + \u2032 \u22c5 \u00b0 = \u00d7 + \u00d7 = 5 17 100 12 17 80 85 88 . m m Kg
Answer: D
\nSolution: Let the final composition: liquid (10 mole): A = x mole, B = (10 \u2013 x ) mole Vapour (10 mole): A = (10 \u2013 x ) mole, B = x mole Now, Y Y X X P P x x x x A B A B A B = \u22c5 \u00b0 \u00b0 \u21d2 \u2212 = \u2212 \u22c5 10 10 200 100 \u2234 x = 4.14 Now, p X P X P x x A A B B = \u22c5 \u00b0 + \u22c5 \u00b0 = \u00d7 + \u2212 \u00d7 10 200 10 10 100 = 141.4 torr
Answer: A
\nSolution: 1 0 4 0 4 0 6 1 2 3 2 2 3 P Y P Y P P A A B B total total atm = \u00b0 + \u00b0 = + = \u21d2 = . . . .
Answer: A
\nSolution: If the solution were ideal, P total mm kg = \u00d7 + \u00d7 = 10 30 90 20 30 87 88 As the solution of phenol and aniline shows negative deviation, the V.P. must be less than 88 mm kg.
Answer: C
\nSolution: For ideal behavior, P total = 0.25 \u00d7 512 + 0.725 \u00d7 344 = 386 mm Hg < 600 mm Hg Hence, the solution shows positive deviation \u21d2 \u0394 H mix = positive
Answer: C
\nSolution: For increase in temperature, the solution shows negative deviation ( \u0394 H = negative).
Answer: C
\nSolution: n n P P Chlorobenzene water Chlorobenzene water = \u00b0 \u00b0 or, x x x / . ( ) / . . . 112 5 100 18 9 031 10 7 031 10 7 031 10 64 4 4 4 \u2212 = \u00d7 \u2212 \u00d7 \u00d7 \u21d2 =
Answer: D
\nSolution: As the available surface area for solvent molecules decreases, the rate of vaporization decreases.
Answer: B
\nSolution: P P P n n m m m m \u00b0 \u2212 = \u21d2 = \u21d2 = 1 2 1 2 2 1 5 95 0 3 57 10 / / . M M
Answer: B
\nSolution: P P X P \u00b0 \u2212 = \u22c5 \u00b0 1 10 = 0.2 \u00d7 P \u00b0 (1) 20 = X 1 \u00d7 P \u00b0 (2) From (1) and (2): X 1 = 0.4 \u21d2 X 2 = X solvent = 0.6
Answer: C
\nSolution: P P P n n \u00b0 \u2212 = \u21d2 \u2212 = \u21d2 = 1 2 3000 2985 2985 5 100 18 179 1 / / . M M
Answer: A
\nSolution: P P P n n \u00b0 \u2212 = \u21d2 \u2212 = \u21d2 = 1 2 0 85 0 845 0 845 0 5 39 78 169 . . . . / / M M\n10.32 Chapter 10 HINTS AND EXPLANATIONS
Answer: B
\nSolution: P X P = \u22c5 \u00b0 = + \u00d7 = 2 1000 18 1 1000 18 12 3 12 08 . . K Pa
Answer: A
\nSolution: P X P = \u22c5 \u00b0 2 2 8 90 18 30 90 18 . = + \u00d7 \u00b0 M P and 2 9 108 18 30 108 18 . = + \u22c5 \u00b0 M P \u2234 = M 23
Answer: A
\nSolution: P P X P \u00b0 \u2212 = \u22c5 \u00b0 = + \u00d7 = 1 1 1 1000 18 760 13 44 . m m kg
Answer: D
\nSolution: P P P n n \u00b0 \u2212 = \u21d2 \u2212 = \u21d2 = 1 2 89 78 89 89 2 100 78 178 . / / M M Now, the number of C-atoms in each molecule = 178 94 4 100 12 14 \u00d7 = . and the number of H-atoms in each molecule = 178 5 6 100 1 10 \u00d7 = . \u2234 Hydrocarbon is C 14 H 10 .
Answer: B
\nSolution: Loss is mass of solvent, w 1 a ( P \u00b0 \u2013 P ) and gain is mass of absorbent, w 2 a P \u00b0 \u2234 = \u00b0 \u2212 \u00b0 = \u21d2 = + \u21d2 = w w P P P X 1 2 1 0 05 2 05 40 40 100 18 288 . . M M M
Answer: C
\nSolution: C C M M M M M 1 2 10 20 6 67 30 1 3 = \u21d2 + = + \u21d2 = A B A B A B M .
Answer: A
\nSolution: As solution have same concentration, mixing will not change the total molar concentration.
Answer: A
\nSolution: Isopiestic refers to the same pressure. Blood is isotonic with 0.9 % ( w / v ) NaCl solution. \u2234 Osmolarity = \u00d7 \u2248 0 9 58 5 100 1000 2 0 31 . / . / . M
Answer: B
\nSolution: \u03c0 = = \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f \u00d7 \u00d7 = CRT atm 2 5 58 5 100 1000 2 0 0821 300 21 05 . / . / . .
Answer: C
\nSolution: Theory based
Answer: D
\nSolution: \u0394 = \u22c5 \u21d2 = \u00d7 \u21d2 = T K m b b 0 104 0 52 2 98 1000 104 . . / / M M
Answer: D
\nSolution: Theory based
Answer: A
\nSolution: Clausius\u2013Clapeyron equation: dT dP RT H P = \u0394 = \u00d7 \u00d7 = 2 2 3 2 350 4 9 10 50 . ( ) . K/atm
Answer: C
\nSolution: \u0394 = \u22c5 = \u00d7 = T K m b b 0 52 72 180 0 208 . . K \u2234 B.P. of solution = 373.208 K
Answer: B
\nSolution: \u0394 = \u22c5 \u21d2 \u2212 = \u00d7 \u00d7 T K m b b ( . . ) . . / 354 11 353 23 2 53 1 8 90 1000 M \u2234 M = 57.5
Answer: C
\nSolution: If molality of solution is \u2018 m \u2019, then P P P n n m m \u00b0 \u2212 = \u21d2 \u2212 = \u21d2 = 1 2 760 750 750 1000 18 20 27 / Now, \u0394 = \u22c5 = \u00d7 = T K m b b 0 52 20 27 0 385 . . K \u2234 B.P. of solution = 100.385\u00b0 C
Answer: A
\nSolution: K H b = \u0394 = \u00d7 = 0 002 0 002 320 80 2 56 2 2 . ( ) . ( ) . T Cal/gm K/m o vap Now, \u0394 = \u22c5 \u21d2 = \u00d7 \u21d2 = T K m x x b b 0 32 2 56 6 4 32 200 1000 8 . . . / / \u2234 Molecular formula of sulphur = S x = S 8
Answer: B
\nSolution: \u0394 = \u22c5 = \u00d7 + \u2248 \u00b0 T K X b x b , / . solute C 32 1 128 1 128 94 94 0 25 \u2234 B.P. of solution = 110.75 + 0.25 = 111\u00b0 C
Answer: B
\nSolution: \u0394 = \u22c5 \u21d2 = \u00d7 \u21d2 = T K m m m b b 1 04 0 52 2 . . Now, P P P n n P P \u00b0 \u2212 = \u21d2 \u00b0 \u2212 = \u21d2 \u00b0 = 1 2 750 750 2 1000 18 777 / torr
Answer: C
\nSolution: Theory based
Answer: B
\nSolution: \u0394 = \u22c5 \u21d2 = \u00d7 \u21d2 = T K m f f 0 93 1 86 7 93 1000 150 5 . . / / . M M\n10.33 Liquid Solution HINTS AND EXPLANATIONS
Answer: D
\nSolution: \u0394 = \u22c5 \u21d2 = \u00d7 \u21d2 = T K m f f 0 93 1 86 36 1 2 60 . . / . M M If the molecular formula is C x H 2 x O x , then 12 x + 2 x + 16 x = 60 \u21d2 x = 2 \u2234 Molecular formula = C 2 H 4 O 2
Answer: A
\nSolution: \u0394 = \u22c5 \u21d2 = \u00d7 \u21d2 = T K m m m f f 15 1 86 8 06 . .
Answer: C
\nSolution: \u0394 \u0394 = \u21d2 \u0394 = \u21d2 \u0394 = \u00b0 T T K K T T f b f b f f 0 78 1 86 0 52 2 79 . . . . C \u2234 F.P. of solution = \u20132.79\u00b0 C
Answer: B
\nSolution: \u0394 + \u0394 = + \u22c5 T T K K m f b f b ( ) or 4.76 = (1.86 + 0.52) \u00d7 w w / / . 342 100 1000 68 4 \u21d2 = gm
Answer: A
\nSolution: \u0394 = \u22c5 \u21d2 = \u00d7 \u21d2 = T K m w w f f 5 8 5 120 425 1000 30 . / / gm
Answer: C
\nSolution: \u0394 \u0394 = \u21d2 \u0394 = \u21d2 \u0394 = \u00b0 T T K K T T f b f b b b 0 7 5 17 5 0 2 . . . C \u2234 B.P. of solution = 90.2\u00b0C
Answer: A
\nSolution: If the molarity of solution is m , then P P P n n m m \u00b0 \u2212 = \u21d2 = \u21d2 = 1 2 2 100 1000 78 10 39 / Now, \u0394 = \u22c5 \u21d2 = \u00d7 \u21d2 = T K m K K f f f f 1 3 10 39 5 07 . . K/m
Answer: C
\nSolution: K K T H T H H H T T K K f b f b f b b f = \u00b0 \u0394 \u00b0 \u0394 \u21d2 \u0394 \u0394 = \u00b0 \u00b0 \u239b \u239d \u239c \u239e \u23a0 \u239f \u00d7 2 2 2 / / fus vap fus vap = \u239b \u239d \u239c \u239e \u23a0 \u239f \u00d7 = 280 350 2 5 5 6 2 7 2 . .
Answer: B
\nSolution: \u0394 = \u22c5 \u21d2 = \u00d7 \u21d2 = T K m K K f f f f 2 0 0 25 8 . . K/m
Answer: B
\nSolution: \u0394 = \u22c5 T K m f f For cane sugar solution: 273 15 272 85 5 342 95 1000 . . \u2212 ( ) = \u22c5 K f / / (1) For glucose solution: 273 15 5 180 95 1000 . / / \u2212 ( ) = \u22c5 T K f f (2) From (1) and (2), T f = 272.58 K
Answer: A
\nSolution: \u0394 = \u22c5 T K m f f For AB 2 solution: 2.55 = 5 1 1 2 20 1000 . / ( ) / \u00d7 + x y (1) For AB 4 solution: 1.7 = 5 1 1 4 20 1000 . / ( ) / \u00d7 + x y (2) \u2234 Atomic mass of A = x = 50 Atomic mass of B = y = 25
Answer: B
\nSolution: \u0394 = \u22c5 \u21d2 = \u00d7 \u21d2 = T K m m m f f 0 744 1 86 0 4 . . . Now, \u03c0 = CRT = 0.4 \u00d7 0.0821 \u00d7 300 = 9.852 atm
Answer: C
\nSolution: \u03d5 = \u0394 \u22c5 = \u00d7 = T K m f f 0 93 1 86 0 4 1 25 . . . .
Answer: A
\nSolution: \u0394 = \u22c5 \u21d2 = \u00d7 \u21d2 = T K m m m f f 1 0 1 80 1 1 8 . . . P X P = \u22c5 \u00b0 = + \u00d7 = 2 1000 18 1 1 8 1000 18 24 24 24 . . mm Hg
Answer: B
\nSolution: \u0394 = \u22c5 T K m f f \u2234 = \u00d7 = \u00d7 0 2 100 1000 1000 . / / K x K x y f f and 0.25 \u2234 Mass of ice separated out = 100 \u2013 y = 20 gm
Answer: B
\nSolution: 2 KI(aq) HgI K HgI aq particles added 2 4 particles ( ) ( ) ( ) ( ) 4 2 3 + \u23af \u2192 \u23af As the number of ions in solution decreases, osmotic pressure decreases.
Answer: A
\nSolution: Osmolarity of both solution should be equal \u2234 \u00d7 = \u00d7 + \u21d2 = 0 1 2 0 1 1 2 0 5 . . ( ) . \u03b1 \u03b1
Answer: A
\nSolution: Osmolarity = 0.2 \u00d7 3 = 0.6 M
Answer: C
\nSolution: \u0394 = \u22c5 \u21d2 = \u00d7 \u00d7 \u21d2 = T K m n n f f 3 72 1 86 1 0 2 . . . \u2234 From each particle, two ions should form.
Answer: C
\nSolution: \u0394 = \u22c5 \u21d2 = \u00d7 + \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f \u23a1 \u23a3 \u23a2 \u23a4 \u23a6 \u23a5 T K m K b b b 1 43 1 1 0 9 1 2 1 . . \u2234 K b = 2.6 K/m
Answer: B
\nSolution: \u0394 = \u22c5 + \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f \u23a1 \u23a3 \u23a2 \u23a4 \u23a6 \u23a5 T K m n f f 1 1 1 \u03b1 \u2234 = \u00d7 + \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f \u23a1 \u23a3 \u23a2 \u23a4 \u23a6 \u23a5 \u21d2 = 1 96 4 9 2 122 25 1000 1 1 2 1 0 78 . . / / . \u03b1 \u03b1\n10.34 Chapter 10 HINTS AND EXPLANATIONS
Answer: B
\nSolution: HA H A M M =0.01M M ( . ) 0 1 \u2212 + + x x x \u001f \u21c0 \u001f \u21bd \u001f \u001f \u2212 \u2234 Osmolarity = (0.1 \u2013 x ) + x + x = 0.11 M Now, \u03c0 = CRT = 0.11 RT
Answer: A
\nSolution: \u0394 = \u22c5 = \u00d7 \u00d7 + \u00d7 = \u00b0 T K m f f 1 86 0 1 2 0 025 2 0 465 . [ . . ] . C \u2234 F.P. of solution = \u2013 0.465\u00b0 C
Answer: A
\nSolution: The eff ective molality will be in the range of 0.2 to 0.3 and \u0394 T f = K b \u22c5 m will be in the range of 1.86 \u00d7 0.2 = 0.372\u00b0 C to 1.86 \u00d7 0.3 = 0.558\u00b0 C.
Answer: A
\nSolution: Let the mixture contain x mole KCl and y mole NaCl. Then x \u00d7 74.5 + y \u00d7 58.5 = 3.125 (1) and ( ) . . . x y T K f f + \u00d7 = \u0394 = = 2 0 186 1 86 0 1 (2) \u2234 x y = 1 3
Answer: B
\nSolution: 2 2 0 2 A A Initial conc. nM ( M M Equilibrium conc. n x x \u2212 ) \u001f \u21c0 \u001f \u21bd \u001f \u001f Now, \u0394 = \u22c5 = \u22c5 \u2212 + \u23a1 \u23a3 \u23a2 \u23a4 \u23a6 \u23a5 T K m K n x x b b b 2 \u2234 x n T K b b = \u2212 \u0394 2 2 Now, K A A x n x n T K T K n c b b b b = = \u239b \u239d \u239c \u239e \u23a0 \u239f \u2212 = \u2212 \u0394 \u0394 \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f [ ] [ ] ( ) 2 2 2 2 2 2 = \u22c5 \u2212 \u0394 \u0394 \u2212 \u22c5 K n K T T n K b b b b b ( ) ( ) 2 2
Answer: C
\nSolution: \u0394 \u0394 = \u22c5 \u22c5 = \u00d7 \u00d7 = T A T B K A m K B m m m f f f f ( ) ( ) , , . / . / 1 2 1 86 2 2 79 3 1 1
Answer: B
\nSolution: Colloidal solutions have low value of any colligative property than the true solution of same composition.
Answer: C
\nSolution: If the complex dissociates into n ions, then \u0394 = \u22c5 \u21d2 = \u00d7 \u00d7 \u21d2 \u2248 T K m n n f f 0 0054 1 86 0 001 3 . . [ . ]
Answer: A
\nSolution: XCl X Cl M M 3 3 3 3 (s) \u001f \u21c0 \u001f \u21bd \u001f \u001f + \u2212 + S S P P P n n \u00b0 \u2212 = \u21d2 \u2212 = \u21d2 = \u00d7 \u2212 1 2 2 17 25 17 20 17 20 4 1000 18 4 04 10 . . . / . S M S
Answer: C
\nSolution: (a) \u0394 = \u22c5 = \u00d7 = \u00b0 T K m f f 1 86 1 1 86 . . C \u2234 F.P. of solution = \u2013 1.86\u00b0 C (b) \u0394 = \u22c5 = \u00d7 = \u00b0 T K m b b 0 52 1 0 52 . . C \u2234 B.P. of solution should be 100.52\u00b0 C. As the solute dissociates completely above 100.26\u00b0 C, its actual \u0394 = \u00d7 = \u00b0 T b 0 52 2 1 04 . . C and hence, B.P. = 101.04\u00b0 C. (c) \u0394 = \u22c5 \u21d2 = \u00d7 T K m m f f 7 44 1 86 1 2 . . / solvent (as solute dimerizes) \u2234 m Solvent left = 0.125 kg \u2234 Percentage of water separated as ice = (1 \u2013 0.125) \u00d7 100 = 87.5 % (d) \u0394 = \u22c5 \u21d2 = \u00d7 T K m m b b 2 08 0 52 2 . . solvent (Complete dissociation) \u2234 m Solvent left = 0.5 kg Percentage of water evaporated = (1 \u2013 0.5) \u00d7 100 = 50 %
Answer: D
\nSolution: At constant temperature, the vapour pressure may be changed by changing the composition.
Answer: A
\nSolution: Theory based
Answer: A, C
\nSolution: Theory based
Answer: B, C, D
\nSolution: P AB = + = 75 22 2 48 5 . torr P BC = + = 22 10 2 16 torr P AC = + = 75 10 2 42 5 . torr P ABC = + + = 75 22 10 2 35 67 . t orr\n10.35 Liquid Solution HINTS AND EXPLANATIONS
Answer: A, B, C, D
\nSolution: Theory based
Answer: A, C, D
\nSolution: A, C, D shows negative deviation.
Answer: C
\nSolution: (a) Mass percent of A = 50 \u21d2 n A : n B = 1 : 2 \u21d2 Azeotrope Hence, vapour will have the same composition of liquid. (b) Mass percent of A > 50 \u21d2 n A : n B = 1 : 2 L L V 0.0 1.0 V \u00b0 A \u03c1 \u00b0 B \u03c1 mole-fraction of B \u03c1 2 3 In this case, the vapour must be more rich in A than liquid. (c) X B = = > 3 4 0 75 2 3 . \u21d2 Pure A cannot be obtained. (d) X B = = < 3 5 0 60 2 3 . \u21d2 Pure A cannot be obtained in traces.
Answer: B, C, D
\nSolution: (a) On changing the solvent, K f will change.
Answer: A, B
\nSolution: F. P. and V. P. will become lower for X .
Answer: B, C
\nSolution: Informative
Answer: A
\nSolution: Theory based
Answer: B, C, D
\nSolution: Theory based
Answer: A, B, C, D
\nSolution: (a) P P P n n n n \u00b0 \u2212 = \u21d2 \u2212 = \u21d2 = 1 2 2 2 760 740 740 1 37 \u2234 Moles of water separated as ice = 200 \u2013 37 = 163 (b) \u0394 = \u22c5 = \u00d7 \u00d7 = \u00d7 T K m f f 2 0 1 37 18 1000 2000 37 18 . ( ) / K \u2234 = \u2212 \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f T K 273 2000 37 18 (c) For original solution: \u0394 = \u22c5 = \u00d7 \u00d7 T K m f f 2 1 200 18 1000 ( ) / \u2234 F. P. = 0 10 18 \u2212 \u0394 = \u2212 \u00b0 T C f (d) For final solution: P P P X \u00b0 \u2212 \u00b0 = = + = 1 1 1 37 1 38
Answer: B, C, D
\nSolution: Theory based
Answer: C, D
\nSolution: 0.0 1.0 \u00b0 T A B T \u00b0 T B
Answer: B
\nSolution: Y X X P P X P P X A A A A B A A B A \u2212 = \u22c5 \u00b0 \u00b0 + \u22c5 \u00b0 \u2212 \u00b0 \u2212 ( ) = \u00b0 \u2212 \u00b0 \u2212 \u00b0 \u2212 \u00b0 \u00b0 + \u22c5 \u00b0 \u2212 \u00b0 = X P P X P P P X P P f X A A B A B A B A A B A ( ) ( ) ( ) ( ) 2 For maximum ( ), ( ) Y X d Y X dX A A A A A \u2212 \u2212 = 0 \u2234 X P P P P P A A B B A B = \u00b0 \u22c5 \u00b0 \u2212 \u00b0 \u00b0 \u2212 \u00b0
Answer: A
\nSolution: P P X P P P P B A A B A B total = \u00b0 + \u00b0 \u2212 \u00b0 = \u00b0 \u22c5 \u00b0 \u22c5 ( )\n10.36 Chapter 10 HINTS AND EXPLANATIONS Comprehension II
Answer: B
\nSolution: 1 2 5 0 4 3 5 0 6 0 5 0 3 P Y P Y P P A A B B total total bar = \u00b0 + \u00b0 = + \u21d2 = > / . / . . . As the applied pressure is less than equilibrium pressure, the system must be 100 % vapour.
Answer: A
\nSolution: First drop of liquid will form at 0.5 bar.
Answer: B
\nSolution: P 1 V 1 = P 2 V 2 \u21d2 0.3 \u00d7 10 = 0.5 \u00d7 V 2 \u21d2 V 2 = 6.0 dm 3
Answer: D
\nSolution: P X P X P X X A A B B A A total = \u22c5 \u00b0 + \u22c5 \u00b0 \u21d2 = \u00d7 + \u2212 \u00d7 0 5 0 4 1 0 6 . . ( ) . \u2234 X A = 0.5
Answer: B
\nSolution: Liquid composition: A \u2248 2 mole, B \u2248 3 mole \u2234 P total bar = \u00d7 + \u00d7 = 2 5 0 4 3 5 0 6 0 52 . . .
Answer: A
\nSolution: Y X P P A A A = \u22c5 \u00b0 = \u00d7 = total 2 5 0 4 0 52 4 13 . .
Answer: B
\nSolution: P X P X P X X A A B B B A total = \u22c5 \u00b0 + \u22c5 \u00b0 \u21d2 = \u00d7 + \u2212 \u00d7 0 51 0 4 1 0 6 . . ( ) . \u2234 X A = 0 45 . Now, Y X P P A A A = \u22c5 \u00b0 = \u00d7 = total 0 45 0 4 0 51 6 17 . . . Let moles of A and B in liquid form is x and y , respectively. X x x y Y x x y A A = + = = \u2212 \u2212 + \u2212 = 0 45 2 2 3 6 17 . ( ) ( ) and \u2234 n A (liquid) = x = 12 11 and n A (vapour) = 2 \u2013 x = 10 11
Answer: A
\nSolution: Final total moles of liquid = 5 20 100 1 \u00d7 = and total moles of vapour = 5 \u2013 1 =
Answer: B
\nSolution: Let the liquid contain x mole A . P x x x total = \u00d7 + \u2212 \u00d7 = \u2212 1 0 4 1 1 0 6 0 6 0 2 . . . . (1) and Y X P P x x x A A A = \u22c5 \u00b0 \u21d2 \u2212 = \u00d7 \u2212 total 2 4 1 0 4 0 6 0 2 . . . (2) From (2): x = 0.48 \u2234 From (1): P total = 0.504 bar Comprehension III
Answer: C
\nSolution: \u0394 H mix = 0
Answer: C
\nSolution: \u0394 G mix, m = RT [ X 1 \u22c5 ln X 1 + X 2 \u22c5 ln X 2 ] = \u00d7 \u22c5 + \u22c5 \u23a1 \u23a3 \u23a2 \u23a4 \u23a6 \u23a5 2 300 1 3 1 3 2 3 2 3 ln ln = \u2013380 cal/mol
Answer: D
\nSolution: \u0394 = \u2212 \u0394 = \u2212 \u2212 = S G T m m mix, mix, . / 380 300 3 8 3 cal K-mol and \u0394 S mix = 3 3 8 3 3 8 \u00d7 = . . / cal K Comprehension IV
Answer: D
\nSolution: P X P X P B B T T total mm Hg = \u22c5 \u00b0 + \u22c5 \u00b0 = \u00d7 + \u00d7 = 10 20 100 10 20 40 70
Answer: B
\nSolution: Y X P P A B B = \u22c5 \u00b0 = \u00d7 = = total 0 5 100 70 5 7 0 714 . .
Answer: B
\nSolution: The vapour will contain almost 10 moles of both 1 0 5 100 0 5 40 57 14 P Y P Y P P B B T T total total mm kg = \u00b0 + \u00b0 = + \u21d2 = . . .
Answer: C
\nSolution: X Y P P B B B = \u22c5 \u00b0 = \u00d7 = total 0 5 57 14 100 0 286 . . .
Answer: A
\nSolution: Final system contains 10 moles of liquid and 10 moles of vapour. Let the moles of benzene in liquid be x . P X P X P x x B B T T total = \u22c5 \u00b0 + \u22c5 \u00b0 = \u00d7 + \u2212 \u00d7 10 100 10 10 40 or, P total = 40 + 6 x (1) Y X P P x x x x B B B = \u22c5 \u00b0 \u21d2 \u2212 = \u00d7 + \u21d2 = total 10 10 10 100 40 6 3 87 . From Equation (1): P total = 63.25 mm kg\n10.37 Liquid Solution HINTS AND EXPLANATIONS Comprehension V A + B Residual solution A = x mole B = y mole Condensate ( n A + n B) moles = ( n A + n B) moles 1 4 A = ( n A \u2013 x ) mole B = ( n B \u2013 y ) mole = ( n A + n B) 3 4 From question: 700 = + \u00d7 \u00b0 + + \u00d7 \u00b0 n n n P n n n P A A B A B A B B (1) 600 = + \u00d7 \u00b0 + + \u00d7 \u00b0 x x y P y x y P A B (2) x y n n A B + = + 1 4 ( ) (3) x x y + = 0 3 . (4) n x n n A A B \u2212 + = 3 4 0 75 ( ) . (5)
Answer: D
\nSolution: n B : n A = 29.51
Answer: B
\nSolution: P A \u00b0 = 807 4 . mm
Answer: D
\nSolution: P B \u00b0 = 511 1 . mm Comprehension VI
Answer: B
\nSolution: P RT V A m \u00b0 = = \u00d7 \u00d7 \u00d7 = 0 08 300 100 1 25 3800 1000 760 60 . . torr
Answer: B
\nSolution: P RT V B m \u00b0 = = \u00d7 \u00d7 \u00d7 = 0 08 300 50 1 00 7600 1000 760 48 . . torr
Answer: B
\nSolution: 1 1 54 60 1 48 5 9 P Y P Y P Y Y Y A A B B A A A total = \u00b0 + \u00b0 \u21d2 = + \u2212 \u21d2 = Comprehension VII
Answer: C
\nSolution: \u0394 = \u22c5 = \u22c5 T K m K f f f 50 M \u2234 \u0394 \u0394 = = T A T B f f B B ( ) : ( ) M : : M 3 1
Answer: A
\nSolution: Average molar mass of solute in S 1 M(S M M M 1 2 3 2 3 11 5 ) = \u00d7 + \u00d7 + = A B A and average molar mass of solute in S 2 . M( S 2 ) = 3 2 2 3 9 5 M M A B A M + \u00d7 + = \u2234 \u0394 \u0394 = ( ) ( ) = T S T S M S M S f f ( ) : ( ) : : 1 2 2 1 9 11 Comprehension VIII
Answer: D
\nSolution: \u0394 = \u22c5 = \u00d7 \u00d7 \u00d7 = T K m f f 2 0 0 1 0 9 46 1000 4 8 . . . . K \u2234 Freezing point of solution = 155.7 \u2013 4.8 = 150.9 K
Answer: B
\nSolution: P X P = \u22c5 \u00b0 = \u00d7 = 2 0 9 40 36 . mm kg
Answer: A
\nSolution: \u0394 = \u22c5 = \u00d7 \u00d7 \u00d7 = T K m b b 0 52 0 1 0 9 18 1000 3 2 . . . . K \u2234 B.P. of solution = 373 + 3.2 = 376.2 K Comprehension IX
Answer: A
\nSolution: Increase in mass of absorber a P \u00b0 and decrease in mass of pure solvent a ( P \u00b0 \u2013 P ). \u2234 P P P X x x x \u00b0 \u2212 \u00b0 = = = + \u2212 0 02 0 24 180 180 100 18 1 . .\n10.38 Chapter 10 HINTS AND EXPLANATIONS \u2234 Mass percent of glucose, x = 1000 21 %
Answer: B
\nSolution: AlCl Al Cl 3 3 1 0 8 0 2 0 8 3 0 8 2 4 3 \u2212 = \u00d7 = + \u2212 + . . . . . \u001f \u21c0 \u001f \u21bd \u001f \u001f Total effective mole of solute = 0.2 + 0.8 + 2.4 = 3.4 Now, decrease in mass of solution a P and increase in mass of absorber a P \u00b0. \u2234 P P X m \u00b0 = = + = = \u0394 2 17 17 3 4 5 6 0 18 . . absorber \u2234 Increase in mass of absorber = 0.216 gm
Answer: A
\nSolution: Henry\u2019s law
Answer: A
\nSolution: Theory based
Answer: D
\nSolution: Both have same \u0394 T f
Answer: A
\nSolution: KCl will dissociate.
Answer: C
\nSolution: Theory based
Answer: A
\nSolution: Theory based
Answer: A
\nSolution: Theory based
Answer: A
\nSolution: Theory based
Answer: D
\nSolution: Relative lowering of V.P. is also independent of solvent.
Answer: D
\nSolution: Deviation may occur in non-ideal solution.
Answer: A \u2192 Q, B \u2192 P, R, C \u2192 P, R; D \u2192 P, S
\nSolution: Informative
Answer: A \u2192 P, B \u2192 Q, R, C \u2192 R, S; D \u2192 R, S
\nSolution: Informative
Answer: A \u2192 P, B \u2192 R, C \u2192 P; D \u2192 Q
\nSolution: (A) Some concentrations (B) Osmolarity : NaCl = 0.2 M, Na 2 SO 4 = 0.3 M (C) Osmolarity : NaCl = KCl = 0.2 M (D) Osmolarity : CuSO 4 = 0.2 M, Sucrose = 0.1 M
Answer: A \u2192 P; B \u2192 Q; C \u2192 R; D \u2192 S
\nSolution: Higher the B.P. of solvent, normally higher is its K b value.
Answer: A \u2192 Q; B \u2192 R; C \u2192 S; D \u2192 P
\nSolution: (A) 2 = 1 + a (2 \u2013 1) \u21d2 a = 1.00 (B) 2 = 1 + a (3 \u2013 1) \u21d2 a = 0.50 (C) 2 = 1 + a (5 \u2013 1) \u21d2 a = 0.25 (D) 2 = 1 + a (4 \u2013 1) \u21d2 a = 0.33
Answer: A \u2192 Q; B \u2192 P; C \u2192 R; D \u2192 S
\nSolution: (A) i = 1 (B) i = 1 + 1(2 \u2013 1) = 2 (C) i = 1 + 1 (3 \u2013 1) = 3 (D) i = 1 + 1(4 \u2013 1) = 4
Answer: A \u2192 P; B \u2192 Q, R, S; C \u2192 T; D \u2192 P, Q, R, S, T
\nSolution: (P) Eff ective conc. = 0.1 \u00d7 3 = 0.3 M = 0.3 m (Q) Eff ective conc. = 0.14 \u00d7 2 = 0.28 M = 0.28 m (R) Eff ective conc. = 0.1 [1+0.9(3 \u2013 1) = 0.28 M = 0.28 m (S) Eff ective conc. = 0.28 M = 0.28 m (T) HA H A M M M ( . ) 0 1 \u2212 + \u2212 + x x x \u001f \u21c0 \u001f \u21bd \u001f \u001f K x x x x a = = \u22c5 \u2212 \u21d2 = 0 81 0 1 0 09 . . . \u2234 Eff ective conc. = (0.1 + x ) = 0.19 M = 0.19 m
Answer: 2
\nSolution: V gas a n Solvent but independent of pressure. \u2234 \u239b \u239d \u239c \u239e \u23a0 \u239f = \u239b \u239d \u239c \u239e \u23a0 \u239f \u21d2 = \u21d2 = V V V V V V 2 1 2 1 2 2 4 0 5 1 2 gas Solvent ml ml .
Answer: 2
\nSolution: m solution = m water + m ethanol or, V \u00d7 0.9344 = 50 \u00d7 1.000 + 50 \u00d7 0.7939 \u2234 V \u2248 96 ml < 100 ml Solution is non-ideal with negative deviation.\n10.39 Liquid Solution HINTS AND EXPLANATIONS
Answer: 2
\nSolution: Ideal gas can never be liquefied.
Answer: 1
\nSolution: P = X 2 \u22c5 P \u00b0 = 0.8 \u00d7 233.5 =
Answer: 3
\nSolution: 8 torr = P exp \u2234 Solution is ideal.
Answer: 2
\nSolution: \u03c0 \u03c0 \u03c0 = + + = \u00d7 + \u00d7 + = 1 1 2 2 1 2 2 4 2 4 2 2 3 V V V V V V V V ( ) . . atm
Answer: 5
\nSolution: \u03c0 = CRT = \u03c1 g h or 0 2 100 1000 0 0821 300 1 013 1000 0 2463 1 013 10 6 . / / . . . . M \u00d7 \u00d7 = \u00d7 \u00d7 \u00d7 \u2234 M = 2 \u00d7 10 5
Answer: 5
\nSolution: \u03c0 = CRT = \u03c1 g h or n 1 0 08 298 1 013 1000 7 45 1 013 10 6 \u00d7 \u00d7 = \u00d7 \u00d7 \u00d7 . . . . \u2234 n = \u00d7 \u2212 25 10 80 3 \u2234 Millimoles in 320 gm = 25 80 320 20 5 \u00d7 =
Answer: 4
\nSolution: V V C C T T 2 1 1 2 1 1 2 2 500 283 105 3 298 5 = = = \u2248 \u03c0 \u03c0 / / / . /
Answer: 2
\nSolution: \u0394 = \u22c5 \u21d2 = \u00d7 \u00d7 \u21d2 \u2248 T K m n n f f 0 29 1 86 1 04 267 100 1000 4 . . . / /
Answer: 0054
\nSolution: \u0394 = \u22c5 T K m f f For KCN solution: 0.80 = K f \u00d7 0.2 \u00d7 2 (1) Hg(CN) mCN Hg(CN mole Final 0 mole (0.2 m mole 2 0 1 0 2 0 1 . . . ) ) + \u2212 \u2212 \u001f \u21c0 \u001f \u21bd \u001f \u001f m m m + \u2212 2 0 0 1 . mole Final eff ective molality = (0.2 \u2013 0.1 m) + 0.1 + 0.2 = 0.5 \u2013 0.1 m Now, 0.60 = K f \u00d7 (0.5 \u2013 0.1 m) (2) From (1) and (2): m = 2 Four Digit Integer Type
Answer: 0380
\nSolution: P X P = \u22c5 \u00b0 2 20 180 18 6 180 18 = + \u00d7 \u00b0 / M P (1) and 20.02 = 11 6 11 M + \u00d7 \u00b0 P (2) \u2234 M = 54
Answer: 0060
\nSolution: Mole fraction of solvent is same in both. \u2234 90 18 10 90 18 95 18 5 180 95 18 380 M M + = + \u21d2 = X
Answer: 0795
\nSolution: P P P X \u00b0 \u2212 \u00b0 = = \u2212 = 1 2400 2300 2400 1 24 1 mole solution Urea mole gm Water mole = = \u00d7 = = = 1 24 1 24 60 2 5 23 24 23 2 . 4 4 18 17 25 \u00d7 = \u23a7 \u23a8 \u23aa \u23aa \u23a9 \u23aa \u23aa . gm \u2234 Volume of 1 mole solution = + = 2 5 17 25 1 185 50 3 . . . ml Now, \u03c0 = = \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f \u00d7 \u00d7 = CRT 1 24 50 3 1000 0 08 300 60 / / . atm
Answer: 0125
\nSolution: \u0394 = \u22c5 T K m f f or, 30 1 86 62 795 1000 795 = \u00d7 \u21d2 = . / / w w gm
Answer: 0058
\nSolution: \u0394 = \u22c5 T K m f f or, 6 1 86 50 62 1000 250 = \u00d7 \u21d2 = . / / w w water (final) gm \u2234 Mass of water separated as ice = 375 \u2013 250 = 125 gm
Answer: 0744
\nSolution: \u0394 = \u22c5 T K m f f Naphthalene solution: 13 5 38 4 128 185 1000 . . / / = \u00d7 K f (1) Unknown substance solution: 9 0 11 6 185 1000 . . / / = \u00d7 K f M (2) \u2234 M = 58
Answer: 0160
\nSolution: \u0394 = \u22c5 = \u00d7 + \u239b \u239d \u239c \u239e \u23a0 \u239f \u00d7 = \u00b0 T K m f f 1 86 3 6 180 3 6 60 200 1000 0 744 . . . . C \u2234 F. P. of solution = \u2013 0.744\u00b0C\n10.40 Chapter 10 HINTS AND EXPLANATIONS
Answer: 0096
\nSolution: \u0394 = \u22c5 T K m f f ( . . ) . / M / 26 84 25 64 8 2 4 10 100 10 1000 3 3 \u2212 = \u00d7 \u00d7 \u00d7 \u21d2 = \u2212 \u2212 M 160
Answer: 0020
\nSolution: \u0394 = \u22c5 \u21d2 = \u00d7 \u00d7 + \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f \u23a1 \u23a3 \u23a2 \u23a4 \u23a6 \u23a5 T K m f f 1 60 4 88 2 122 26 1000 1 1 2 1 . . / / \u03b1 \u2234 a = 0.96 or 96 %
Answer: 0098
\nSolution: \u03c0 = \u21d2 = + \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f \u00d7 \u00d7 CRT x 4 92 200 0 05 2 0 08 300 . . . \u2234 x = 20
Answer: 0050
\nSolution: P P X P n n n P n n P \u00b0 \u2212 = \u22c5 \u00b0 = + \u00b0 \u2248 \u22c5 \u00b0 1 1 1 2 1 2 , Urea solution: 0 03 0 1 1000 18 . . = \u239b \u239d \u239c \u239e \u23a0 \u239f \u00d7 \u00b0 P KCl solution: 0 0594 0 1 1 2 1 1000 18 . . [ ( )] = + \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f \u00d7 \u00b0 \u03b1 P \u2234 a = 0.98 or 98 %
Answer: 0075
\nSolution: Loss in weight of solution a P Loss in weight of water a ( P \u00b0 \u2013 P ) Now, P P P n n \u00b0 \u2212 = \u21d2 = + \u2212 1 2 0 01 0 98 1 25 90 1 3 1 49 18 . . . [ ( )] \u03b1 \u2234 a = 0.50 or 50 %
Answer: 0377
\nSolution: \u0394 = \u22c5 \u21d2 = \u00d7 \u00d7 + \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f \u23a1 \u23a3 \u23a2 \u23a4 \u23a6 \u23a5 T K m f f 7 14 75 2 94 1 1 2 1 . \u03b1 \u2234 a = 0.75 or 75 %
Answer: 0060
\nSolution: 100 gm solution (Say) Water = 100 \u2013 (12 + 9.5) = 78.5 gm MgCl2 = 9.5 gm = = 0.1 mole 9.5 9.5 MgSO4 = 12 gm = = 0.1 mole 12 120 Eff ective moles of solute = 0 \u22c5 1 [1 + 0.8(2 \u2013 1)] + 0.1 [1 + 0.6 (3\u2013 1)] = 0.4 Now, \u0394 = \u22c5 = \u00d7 = T K m b b 0 785 0 4 78 5 1000 4 . . . / K \u2234 B. P. of solution = 373 + 4 = 377 K
Answer: D
\nSolution: Nuclear forces are same in between any two nucleon and it is attractive at 1 fm but repulsive forces are also there between protons
Answer: C
\nSolution: Informative (B.E./nucleon is maximum for Fe)
Answer: B
\nSolution: For lighter nuclei, n p > 1 may make the nucleus unstable
Answer: C
\nSolution: Theory based
Answer: C
\nSolution: Informative
Answer: D
\nSolution: Number of n and p , both is even in 30 Zn 64 .
Answer: D
\nSolution: r A N \u221d 1 3 / \u21d2 r r 1 2 1 2 = \u00d7 \u21d2 ( ) ( ) / / A 1 1 3 1 3 1 2 56 = \u00d7 \u21d2 A 1 = 7
Answer: B
\nSolution: Informative
Answer: A
\nSolution: For 1 H 1 , n p = = 0 1 0
Answer: B
\nSolution: For 1 H 3 , 0 693 32 365 24 730 90 0 693 64 90 . . \u00d7 \u00d7 \u00d7 \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f = \u00d7 \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f N w N A A (Some isotopes having n p ratio greater that, 1 H 3 are also know, like 2 He 8 )
Answer: C
\nSolution: Informative
Answer: A
\nSolution: Informative Radioactivity
Answer: C
\nSolution: Experimental reason behind considering \u03b1 -particle as He- nucleus.
Answer: A
\nSolution: Experimental fact
Answer: C
\nSolution: Experimental reason behind considering b-emission as nuclear charge.
Answer: D
\nSolution: Isotope formation
Answer: C
\nSolution: Reason of g -emission
Answer: C
\nSolution: For an increase in mass, large amount of energy is needed and hence, it is non-spontaneous.
Answer: C
\nSolution: b N a N N c \u2212 \u00d7 = \u2212 \u00d7 + \u00d7 = \u03b1 \u03b1 \u03b2 \u03b1 4 2 1 and \u2234 N b N c a b d \u03b1 \u03b2 \u03b1 = \u2212 = \u2212 + \u00d7 \u2212 4 2 4 and ( )
Answer: C
\nSolution: n p n p F F \u239b \u239d \u239c \u239e \u23a0 \u239f < \u239b \u239d \u239c \u239e \u23a0 \u239f 18 19 , Hence, F 18 should undergo a -decay on b + - decay on k -capture. Normally, a -decay and k -capture is not found in lighter nuclei.
Answer: C
\nSolution: 11 23 10 23 Na Ne \u2192 + F
Answer: A
\nSolution: n p n p \u239b \u239d \u239c \u239e \u23a0 \u239f > \u239b \u239d \u239c \u239e \u23a0 \u239f Na Na 24 23 Hence, Na 24 should undergo \u03b2 -decay.
Answer: A
\nSolution: c N 14 14 \u2192 \u2212 \u03b2
Answer: D
\nSolution: Informative
Answer: A
\nSolution: \u0394 m m m u = \u2212= \u2212 = Au Hg 198 198 197 968 197 966 0 002 . . . \u2234 Q \u2013 value = 0.002 \u00d7 931.5 = 1.8630 MeV But Hg 198 is having energy 1.063 MeV greater than Hg 198 and hence, maximum K.E. of emitted b \u2013particle = 1.863 \u2013 1.063 = 0.8MeV. HINTS AND EXPLANATIONS EXERCISE (JEE ADVANCED)\n14.16 Chapter 14 HINTS AND EXPLANATIONS Rate Law
Answer: C
\nSolution: r \u221d N
Answer: B
\nSolution: r \u221d N \u2032
Answer: D
\nSolution: Rate is independent from all external factors.
Answer: D
\nSolution: r N = = \u00d7 \u00d7 \u00d7 \u00d7 \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f = \u00d7 \u03bb 0 693 28 3 15 10 1 90 6 10 5 24 10 7 23 12 . . . dpspg
Answer: B
\nSolution: r N A 1 0 693 10 10 = \u00d7 \u00d7 . ( ); r N A 2 0 693 5 1 = \u00d7 \u00d7 . ( ) r N A 3 0 693 2 5 = \u00d7 \u00d7 . ( ); r N A 4 0 693 1 2 = \u00d7 \u00d7 . ( )
Answer: C
\nSolution: t \u00bd is independent from amount.
Answer: B
\nSolution: N N n = \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f 0 1 2
Answer: D
\nSolution: w w o n = \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f 1 2 \u21d2 3 1 2 12 3 g w o = \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f \u21d2 w o = 48gm
Answer: D
\nSolution: Moles of He formed = \u00d7 \u00d7 4 5 10 6 10 23 23 . = 0.75 = Moles of decayed \u2234 t t = \u00d7 = 2 2 0 1 2 / hrs
Answer: D
\nSolution: Number of atoms present at time T 1 , N R T 1 1 0 693 = . / Number of atoms present at time T 2 , N R T 2 2 0 693 = . / \u2234 Number of atoms decayed = \u2212 ( ) . R R T 1 2 0 693
Answer: B
\nSolution: Rate should decrease 1 64 1 2 6 = times and hence, t t = \u00d7 = 6 12 1 2 / hrs
Answer: D
\nSolution: P w w w Q w w : : 10 20 40 20 20 1 1 2 0 day 0 day 0 day \u23af \u2192 \u23af\u23af \u23af \u2192 \u23af\u23af \u23af \u2192 \u23af\u23af (As fi nal mass ratio is 1 : 4) Hence, Q is non-radioactive.
Answer: A
\nSolution: t \u00bd = 30 min Now, r = \u03bb N \u21d2 28 0 7 30 1200 = \u00d7 \u21d2 = . N N
Answer: C
\nSolution: r r o = \u00d7 \u00d7 = = 3 10 3 10 1 8 1 2 8 8 3 \u21d2 t t = \u00d7 = \u00d7 = 3 3 12 26 36 78 1 2 / . . yrs
Answer: C
\nSolution: w w o n = \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f 1 2 \u21d2 10 1 2 3 6 mg = \u239b \u239d \u239c \u239e \u23a0 \u239f w o / \u21d2 w o = 14 14 . mg
Answer: A
\nSolution: r N = = \u00d7 \u00d7 \u00d7 \u00d7 \u00d7 \u00d7 \u00d7 \u00d7 \u00d7 \u03bb 0 693 1 3 10 365 24 3600 75 10 0 35 100 0 012 100 40 6 9 3 . . . . .0 022 10 017 64 23 \u00d7 \u239b \u239d \u239c \u239c \u239c \u239c \u239e \u23a0 \u239f \u239f \u239f \u239f = . dps
Answer: C
\nSolution: t t r r t o = \u22c5 \u21d2 > \u22c5 1 2 1 2 2 2 2 5 / / log log log log . \u2234 t 1/2 = 5.25 days
Answer: A
\nSolution: Let the sample contains x gm Pu 239 . Now, r N N Pu Pu = + ( ) ( ) \u03bb \u03bb 239 240 or 6 10 0 693 2 4 10 365 24 3600 239 6 022 10 0 693 9 4 23 \u00d7 = \u00d7 \u00d7 \u00d7 \u00d7 \u00d7 \u00d7 \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f + . . . . x 7 7 17 10 365 24 3600 1 240 6 022 10 3 23 . . \u00d7 \u00d7 \u00d7 \u00d7 \u00d7 \u2212 \u00d7 \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f x \u21d2 x = 0.3112 Hence, mass percent of Pu 239 = \u00d7 = x 1 100 31 12 . %\n14.17 Nuclear Chemistry HINTS AND EXPLANATIONS
Answer: A
\nSolution: U Pb 238 206 \u23af \u2192 \u23af Initial a 0 Present a \u2013 x x From question, ( ) . a x x \u2212 \u00d7 \u00d7 = 238 206 1 0 1 \u21d2 x a = 238 2298 Now, age of ore, t t a a x = \u22c5 \u2212 1 2 2 / log log = \u00d7 \u22c5 \u2212 = \u00d7 4 5 10 0 3 1 1 238 2298 7 2 10 9 8 . . log . years
Answer: C
\nSolution: Th Pb He 232 208 4 6 \u23af \u2192 \u23af + Initial a mole 0 Present ( a \u2013 x ) mole 6 x mole = \u00d7 = \u00d7 \u2212 \u2212 4 64 10 232 2 10 7 9 . = \u00d7 \u21d2 \u00d7 \u2212 \u2212 6 72 10 22400 5 10 5 10 . \u2234 a = 2.5 \u00d7 10 \u20139 Now, age of sample, t t a a x = \u22c5 \u2212 1 2 2 / log log = \u00d7 \u00d7 \u00d7 \u2212 \u00d7 = \u00d7 \u2212 \u2212 \u2212 1 38 10 0 3 2 5 10 2 5 10 2 5 10 4 6 10 10 9 9 9 9 . . log . . . . years Parallel and Sequential Decay
Answer: B
\nSolution: 224 is an integer multiple of 4 and hence, Ra 224 belongs to 4n series, which is thorium series.
Answer: C
\nSolution: Informative
Answer: B
\nSolution: Informative
Answer: D
\nSolution: Informative
Answer: B
\nSolution: r r Th Ra = \u21d2 N t N t Th Th Th Ra ( ) ( ) / / 1 2 1 2 = \u21d2 N N Th Ra = 80000 1600
Answer: D
\nSolution: (a) \u03bb AC Yr 227 0 693 22 3 15 10 2 1 = = \u00d7 \u2212 \u2212 . . (b) l for the formation of Th 229 = \u00d7 \u00d7 = \u00d7 \u2212 \u2212 2 100 3 15 10 6 3 10 2 4 . . Yr (c) l for the formation of Fr 223 = \u00d7 \u00d7 = \u00d7 \u2212 \u2212 \u2212 98 100 3 15 10 3 087 10 2 2 1 . . Yr (d) N N m m Th Fr Th Fe 227 223 227 223 2 98 2 227 98 223 1 49 = \u21d2 = \u00d7 \u00d7 \u2260
Answer: C
\nSolution: The net rate of formation of radioisotope, + = \u2212 dn dt R N \u03bb . After very long time, steady state will be achieved, at which + = dn dt
Answer: A
\nSolution: Hence, N R = \u03bb .
Answer: C
\nSolution: Pb Bi hr hr 212 8 212 1 1 2 1 2 t t / / = = \u23af \u2192 \u23af\u23af\u23af \u23af \u2192 \u23af\u23af\u23af Time for maximum nuclei and hence, maximum activity of Bi212, t max ln = \u2212 \u22c5 1 2 1 2 1 \u03bb \u03bb \u03bb \u03bb = = \u22c5 = = 1 2 1 2 8 1 1 1 8 3 429 205 7 ln ln ln . . min hr
Answer: B
\nSolution: \u03bb \u03bb \u03bb \u03b1 \u03b2 overall = + \u21d2 = + 0 693 0 693 20 0 693 60 1 2 . . . / t \u2234 t 1/2 = 15 min \u2234 For 87.5 % decay, t t = \u00d7 = 3 4 5 1 2 / min
Answer: D
\nSolution: Average energy released = \u00d7 + \u00d7 + = 0 05 40 0 15 80 0 05 0 15 70 . . . . MeV\n14.18 Chapter 14 HINTS AND EXPLANATIONS Nuclear Reactions
Answer: D
\nSolution: 92 235 0 1 54 139 38 94 0 1 3 U n Xe Sr n + \u23af \u2192 \u23af + +
Answer: A
\nSolution: 25 55 0 1 25 56 Mn n Mn + \u23af \u2192 \u23af + \u03b3
Answer: C
\nSolution: 4 9 1 1 5 10 Be H B (proton) + \u23af \u2192 \u23af + \u03b3
Answer: C
\nSolution: 13 27 2 4 15 30 0 1 Al He P n particle + \u23af \u2192 \u23af + \u2212 ( ) \u03b1
Answer: A
\nSolution: Informative
Answer: A
\nSolution: Informative
Answer: D
\nSolution: Theory based
Answer: B
\nSolution: Theory based
Answer: A
\nSolution: Informative
Answer: A, C, D
\nSolution: Theory based
Answer: A, D
\nSolution: Q -value is distributed between \u03b2 -particle and anti- neutrino.
Answer: C
\nSolution: 92 235 90 23 88 227 89 227 93 235 U Th Ra Ac Np \u2212 \u2212 \u2212 \u2212 \u23af \u2192 \u23af \u23af \u2192 \u23af \u23af \u2192 \u23af \u23af \u2192 \u23af \u03b1 \u03b1 \u03b2 \u03b2 \u03b1 9 91 231 Pa \u2212 89 AC 235 is not possible.
Answer: A, B, C
\nSolution: Activity is independent from all external factors.
Answer: A, B, C, D
\nSolution: Half-life of a radio isotope is its characteristic property, independent from all factors.
Answer: A
\nSolution: Actinum series: 92 235 82 207 U Pb \u2192
Answer: A
\nSolution: Informative
Answer: B, C, D
\nSolution: 96 242 2 4 97 293 0 1 2 Cm He Bk Incorrect + \u23af \u2192 \u23af + n ( ) 5 10 2 4 7 13 0 1 7 19 0 1 6 14 1 1 1 B He N n Correct N n C H (Correct) + \u23af \u2192 \u23af + + \u23af \u2192 \u23af + ( ) 9 9 28 1 2 15 29 0 1 Si H P n (Correct) + \u23af \u2192 \u23af +
Answer: C
\nSolution: Neutron is projectile and proton is emitted particles.
Answer: D
\nSolution: Informative
Answer: A
\nSolution: 13 27 2 4 15 30 0 1 Al He P n + \u23af \u2192 \u23af + 6 96 C He N P Si e Au He BK 12 1 1 7 13 15 30 14 30 +1 0 241 2 4 97 24 + \u23af \u2192 \u23af + \u23af \u2192 \u23af + + \u23af \u2192 \u23af \u03b3 4 4 1 1 H +
Answer: A, B
\nSolution: 4 9 4 8 0 1 Be Be n + \u23af \u2192 \u23af + \u03b3 4 9 1 1 4 8 1 2 Be H Be H + \u23af \u2192 \u23af +
Answer: B
\nSolution: \u0394 m u = \u00d7 + \u00d7 \u2212 = ( . . ) . 8 1 0072 8 1 0086 16 0 1264 \u2234 B.E. per nucleon = \u00d7 = 0 1264 931 5 16 7 36 . . . MeV
Answer: A
\nSolution: 8 16 2 4 4 O He \u23af \u2192 \u23af \u0394 m u = \u2212 \u00d7 = \u2212 15 9944 4 4 0026 0 016 . . . \u2234 Energy required in separation = \u00d7 = 0.016 MeV 931 5 14 904 . .
Answer: C
\nSolution: 10 20 6 12 2 4 2 Ne C He \u23af \u2192 \u23af + Energy required = \u00d7 \u2212 \u00d7 + \u00d7 \u00d7 (20 8 03 12 7 68 2 4 7 07 . ) ( . . ) = 11.88 MeV\n14.19 Nuclear Chemistry HINTS AND EXPLANATIONS Comprehension II
Answer: C
\nSolution: SC 50 50 \u23af \u2192 \u23af + + \u03c4 \u03b2 \u03bd i Q \u2212 = \u2212 \u00d7 = value (49.9516 49.94479 MeV ) . . 931 5 6 34 \u2234 K.E. of MeV \u03bd = \u2212 = 6 34 0 80 5 54 . . .
Answer: D
\nSolution: \u03bb = = \u00d7 \u00d7 \u00d7 \u2212 \u00d7 \u00d7 \u00d7 = \u00d7 \u2212 \u2212 LC E \u0394 6 626 10 3 10 4 795 4 611 10 1 6 10 6 75 1 39 8 6 19 . ( . . ) . . 0 0 12 \u2212
Answer: B
\nSolution: Th Ra Ra 228 224 224 \u2212 \u2212 \u23af \u2192 \u23af \u23af \u2192 \u23af \u03b1 \u03b3 Q-value = 228 028726 224 020196 4 0026 4 0026 931 6 . . . . . ( ) ( ) \u23a1 \u23a3 \u23a4 \u23a6 \u2212 + + \u00d7 \u2212 2 217 10 3 5 307 \u00d7 \u2212 = . MeV \u2234 \u00d7 = K.E. of -particle = MeV \u03b1 224 228 5 307 5 214 . . Comprehension III
Answer: A
\nSolution: Overall rate is the rate of slowest step and hence, T required = 270 days
Answer: A
\nSolution: At transient equilibrium, N N A B B A A = \u2212 \u22c5 \u03bb \u03bb \u03bb or, N N N N A B Th Ra Ra Th Th = = \u2212 = \u2212 \u00d7 \u00d7 = \u03bb \u03bb \u03bb ln . ln . ln . 2 3 64 2 1 913 365 2 1 913 365 190 0 8 1 .
Answer: A
\nSolution: At secular equilibrium, N N B A B A = \u22c5 \u03bb \u03bb or, N N N N A B Ra Rn Rn Ra = = = \u00d7 \u00d7 = \u03bb \u03bb ln ln . . 2 55 2 3 65 24 3600 5733 8 Comprehension IV
Answer: B
\nSolution: U Pb 238 206 \u23af \u2192 \u23af Initial a mole 0 Present ( a \u2013 x ) mole x mole = 59 5 238 . = \u00d7 12 875 206 80 100 . \u2234 a = 0.30 Now, t a a x = \u22c5 \u2212 = \u00d7 \u22c5 \u2212 1 1 1 52 10 0 3 0 25 10 \u03bb ln . ln . . = 1.33 \u00d7 10 9 Yrs
Answer: C
\nSolution: K Ar 40 40 \u23af \u2192 \u23af Initial a mole 0 Present ( a \u2013 x ) mole x mole = 1 = 10.3 \u2234 a = 11.3 Now, t t a a x = \u22c5 \u2212 = \u00d7 \u22c5 1 2 9 2 1 25 10 0 3 11 3 1 / log log . . log . = 4.375 \u00d7 10 9 years
Answer: B
\nSolution: U Pb 238 206 \u23af \u2192 \u23af K Ar 40 40 \u23af \u2192 \u23af Initial mole 0 b mole 0 Present ( a \u2013 x ) mole x mole ( b \u2013 y ) mole y mole = \u00d7 \u2212 0 86 10 238 3 . = \u00d7 \u2212 0 15 10 206 3 . = \u00d7 \u2212 10 10 40 3 = \u00d7 \u2212 1 6 10 40 3 . t t a a x t b b y U K = \u22c5 \u2212 = \u22c5 \u2212 ( ) log log ( ) log log / / 1 2 1 2 238 40 2 2 \u2234 w = 1.7 mg\n14.20 Chapter 14 HINTS AND EXPLANATIONS Comprehension V
Answer: A
\nSolution: Given in paragraph
Answer: A
\nSolution: For radioactive tracing, time should be comparable to t 1/2 .
Answer: C
\nSolution: T c c T c c 1 1 2 2 1 1 = \u22c5 = \u22c5 \u03bb \u03bb ln ln and As c c T T T T c c 1 2 1 2 1 2 1 2 1 > > \u2212 = \u22c5 , ln and \u03bb Comprehension VI
Answer: C
\nSolution: Let the volume of blood be V ml. t T N N N av 1 0 0 1 2 = = \u22c5 \u2212 ln \u21d2 5 = \u22c5 15 2 1260 15 60 log log / V \u2234 V = 4000
Answer: D
\nSolution: r 0 1260 60 4000 18 9 = \u00d7 = . dpm per ml Now, r r r r 0 5 5 10 = \u21d2 18 9 15 15 10 . = r \u21d2 r 10 = 11.9 dpm per ml Comprehension VII
Answer: A
\nSolution: Isotopes are B and E, C and F, D and G.
Answer: A
\nSolution: Mass number of H = 230 \u2013 4 \u00d7 4 = 214
Answer: A
\nSolution: Z A \u2013 4 \u00d7 2 + 3 \u00d7 1 = 88 \u21d2 Z A = 93 Comprehension VIII
Answer: C
\nSolution: Net rate of formation + = \u2212 dn dt N \u03b1 \u03bb or, dN N dt N N t \u03b1 \u03bb \u2212 = \u222b \u222b 0 0 \u21d2 N N e t = \u2212 \u2212 \u23a1 \u23a3 \u23a4 \u23a6 \u2212 1 0 \u03bb \u03b1 \u03b1 \u03bb \u03bb ( ).
Answer: B
\nSolution: t t N = = = 1 2 0 2 2 / ln \u03bb \u03b1 \u03bb and \u2234 N = 1.5 N 0
Answer: C
\nSolution: t N N \u2192 \u221e = = , then \u03b1 \u03bb 2 0 Comprehension IX
Answer: A
\nSolution: \u03bb \u03bb \u03bb = + 1 2 \u21d2 ln ln ln / 2 2 24 2 8 1 2 t = + \u21d2 t 1/2 = 6 hours.
Answer: A
\nSolution: Activity of excreted material in 48 hours N N T av 1 0 0 116 16 = = . and sec. But as T c is simultaneously decaying with t 1/2 = 8 hrs. Final activity after 48 hrs = = 24 2 0 375 6 . . \u03bc ci
Answer: C
\nSolution: n p ratio does not increases continuously.
Answer: C
\nSolution: Binding energy increases but the binding energy per nucleon first increases and then decreases.
Answer: A
\nSolution: Stable\n14.21 Nuclear Chemistry HINTS AND EXPLANATIONS
Answer: A
\nSolution: Informative
Answer: A
\nSolution: Theory based
Answer: D
\nSolution: All heavy nuclei should not produce 82 Pb 206 .
Answer: A
\nSolution: Theory based
Answer: C
\nSolution: \u03b2 -decay occurs to decrease n p ratio.
Answer: D
\nSolution: Same mass of U 238 and U 238 F 6 have diff erent numbers of U 238 nuclei.
Answer: A
\nSolution: t T av 1 2 0 693 1 / . / / = \u03bb \u03bb = 0.693 = Same for all
Answer: C
\nSolution: Mesons have mass 200 to 300 times mass of electrons.
Answer: A
\nSolution: Theory based
Answer: B
\nSolution: 13 Ae 30 have high n p ratio than its stable nucleus 13 Ae 27 .
Answer: A
\nSolution: Theory based
Answer: D
\nSolution: Theory based
Answer: A \u2192 R; B \u2192 Q; C \u2192 P; D \u2192 S
\nSolution: Informative
Answer: A \u2192 P; B \u2192 Q; C \u2192 R; D \u2192 P; E \u2192 P
\nSolution: Theory based
Answer: A \u2192 Q, S; B \u2192 P, T; C \u2192 R, T; D \u2192 T; E \u2192 T
\nSolution: Theory based
Answer: A \u2192 S; B \u2192 Q; C \u2192 R; D \u2192 P
\nSolution: 53 I 127 is stable and hence, I 333 is beta emitter and I 121 is positron emitter.
Answer: A \u2192 S; B \u2192 R; C \u2192 Q; D \u2192 P
\nSolution: (a) 92 U 235 82 Pb 207 N N \u03b1 \u03b2 = \u2212 = = \u2212 \u2212 \u00d7 = 235 207 4 7 82 92 2 7 4 ( ) (b) 92 U 238 82 Pb 206 N N \u03b1 \u03b2 = \u2212 = = \u2212 \u2212 \u00d7 = 238 206 4 8 82 92 2 8 6 ( ) (c) 94 Pu 241 83 Bi 209 N N \u03b1 \u03b2 = \u2212 = = \u2212 \u2212 \u00d7 = 241 209 4 8 83 94 2 8 5 ( ) (d) 90 Th 232 82 Pb 208 N N \u03b1 \u03b2 = \u2212 = = \u2212 \u2212 \u00d7 = 232 208 4 6 82 90 2 6 4 ( )
Answer: 2
\nSolution: Number of half-lifes 28 1 2 81 10 . . = \u2234 Mass of Sr 90 remained = \u00d7 = \u00d7 \u2212 2 048 10 2 2 10 10 6 . gm gm
Answer: 3
\nSolution: r N = = \u00d7 \u00d7 \u00d7 \u00d7 \u00d7 \u00d7 \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f = \u00d7 \u2212 \u03bb 0 7 14 24 3600 86 4 10 164 0 164 100 6 10 3 10 3 23 . . . 1 11 dps
Answer: 9
\nSolution: Z m Z m A B \u23af \u2192 \u23af + \u2212 \u2212 6 12 2 4 3 He t = 0 1 mole 0 t = 20 days 1 3 4 \u2212 3 3 4 \u00d7 mole = 1 4 mole \u2234 V = \u00d7 9 4 22.4 = 9 \u00d7 5.6 L at 0\u00b0C and 1 atom
Answer: 1
\nSolution: 1 2 0 335 2 mV eV = .\n14.22 Chapter 14 HINTS AND EXPLANATIONS \u2234 V = \u00d7 \u00d7 \u00d7 \u00d7 = \u2212 \u2212 2 0 335 1 6 10 1 675 10 8000 19 27 . . . m/s Hence, time for travelling 80 km, t d v = = \u00d7 = 80 10 8000 10 3 sec Now, t t N N = \u22c5 1 2 0 2 / ln ln or, 10 700 2 100 100 = \u22c5 \u2212 ln x \u21d2 x = 0.99 \u2248 1
Answer: 6
\nSolution: t t N N t N N U = \u22c5 \u239b \u239d \u239c \u239e \u23a0 \u239f = \u22c5 \u239b \u239d \u239c \u239e \u23a0 \u239f 1 2 0 1 2 0 2 2 238 235 / / log log log log U or, 4 5 10 2 140 7 2 10 2 9 0 8 0 . log log . log log \u00d7 \u22c5 = \u00d7 N x N x \u2234 log . N x 0 2 5 = \u2234 Age of earth, t N x = \u00d7 \u22c5 = \u00d7 7 2 10 2 6 10 8 0 9 . log log years Four-digit Integer Type
Answer: 0535
\nSolution: t t r r = \u22c5 1 2 0 2 / log log or, 6 93 6 93 2 5 10 0 15 . . log log = \u22c5 \u00d7 r \u21d2 r 0 = 5.35 \u00d7 10 15 dpm Now, r N 0 0 = \u03bb . or, 5.35 \u00d7 = \u00d7 \u00d7 \u00d7 \u00d7 10 0 693 69 3 60 6 10 15 23 . . ( ) n \u2234 n = 5.35 \u00d7 20 \u20135
Answer: 0060
\nSolution: Initial number of H 3 atoms = 0 93 10 18 6 10 2 8 10 3 23 18 . \u00d7 \u00d7 \u00d7 \u00d7 \u00d7 \u00d7 \u2212 \u2212 = 4.8 \u00d7 10 2 Number of half lives = = 36 9 12 3 3 . . \u2234 Final number of H 3 atoms
Answer: 0016
\nSolution: t T N N N av 1 0 0 1 2 = = \u22c5 \u2212 ln and 3 6 8 3 1 0 0 1 t T N N N av = = \u22c5 \u2212 ln \u2234 N N T av 1 0 0 116 16 = = . and sec
Answer: 0050
\nSolution: Let the mass of water present in body = w gm. Now, 9 \u00d7 10 9 = 2.25 \u00d7 10 5 \u00d7 w \u21d2 w = 4 \u00d7 10 4 gm = 40 kg \u2234 Mass per cent of water in body = \u00d7 = 40 80 100 50%
Answer: 0040
\nSolution: \u03bb \u03bb \u03b2 \u2212 = \u00d7 32 100 overall \u2234 t t 1 2 1 2 100 32 100 32 12 8 40 / / . ( ) = \u00d7 ( ) = \u00d7 = \u2212 \u03b2 overall hr
Answer: 0449
\nSolution: \u03bb = + 1 1620 1 405 \u21d2 \u03bb = \u2212 1 324 1 Yr \u2234 t t required = \u00d7 = \u00d7 \u00d7 2 2 0 693 324 1 2 / . = 449.064 years
Answer: 0167
\nSolution: Sr Y 90 90 \u23af \u2192 \u23af \u23af \u2192 \u23af other format For radioactive equilibrium, ( .N) ( .N) Sr Y 90 90 \u03bb \u03bb = or, 0 693 32 365 24 730 90 0 693 64 90 . . \u00d7 \u00d7 \u00d7 \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f = \u00d7 \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f N w N A A \u2234 w = 0.1667 gm
Answer: 0216
\nSolution: Energy released = 2 \u00d7 120 \u00d7 8.1 \u2013 240 \u00d7 7.2 = 216 MeV\n14.23 Nuclear Chemistry HINTS AND EXPLANATIONS
Answer: 0960
\nSolution: \u0394 m = 2 \u00d7 2.0021 \u2013 4.0026 = 0.0016 amu Now, let n moles of H 2 be required. n 2 6 10 0 0016 1 5 10 25 100 200 10 3600 24 23 10 6 \u00d7 \u00d7 \u00d7 \u00d7 \u00d7 \u00d7 = \u00d7 \u00d7 \u00d7 \u2212 . . \u2234 n = 960
Answer: 0002
\nSolution: After 2 8 2 2 = half-life, detectable activity = 100 2 % But actual detected activity is 10 % . Hence, mass of iodine migrated in thyroid gland = \u00d7 = 10 2 10 100 2 2 mg
Answer: B
\nSolution: a b c \u2260 \u2260 = = = \u00b0 , \u03b1 \u03b2 \u03b3 90 \u21d2 Orthorhombic
Answer: B
\nSolution: Volumeof metal taken cm 3 = = = m d 100 6 25 16 . Volumeof each unit cell cm cm 3 = \u00d7 ( ) = \u00d7 \u2212 \u2212 4 10 64 10 8 3 24 \u2234 Number of unit cells = \u00d7 = \u00d7 \u2212 16 64 10 2 5 10 24 23 .
Answer: A
\nSolution: d Z M N V = \u22c5 \u22c5 A \u21d2 45 16 27 6 10 4 10 23 8 3 = \u00d7 \u00d7 ( ) \u00d7 \u00d7 ( ) \u2212 Z \u21d2 Z = 4 Hence, Al crystal is FCC. For FCC: 2 4 a r = \u21d2 r = \u00d7 = 2 4 0 4 1 414 . . \u00c5
Answer: B
\nSolution: d Z M N V FCC A 3 gm/cm = \u22c5 \u22c5 = \u00d7 \u00d7 ( ) \u00d7 \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f = \u2212 4 56 6 10 4 125 10 2 8 45 23 10 3 . d Z M N V BCC A 3 gm/cm = \u22c5 \u22c5 = \u00d7 \u00d7 ( ) \u00d7 \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f = \u2212 2 56 6 10 4 50 2 10 3 42 87 23 10 3 . As the density of iron is increased, there is contraction.\n9.30 Chapter 9 HINTS AND EXPLANATIONS
Answer: C
\nSolution: Packing fraction = \u00d7 ( ) \u00d7 \u00d7 \u00d7 ( ) \u239b \u239d \u239c \u239e \u23a0 \u239f = \u2212 6 10 4 3 144 10 108 10 6 0 7 23 10 3 \u03c0 . . 3 36 Hence, the crystal should be FCC.
Answer: B
\nSolution: d Z M N V = \u22c5 \u22c5 A \u21d2 10 5 4 198 5 10 8 3 . = \u00d7 \u00d7 \u00d7 ( ) \u2212 N A \u21d2 N A = \u00d7 6 034 10 23 .
Answer: D
\nSolution: Void space per unit cell = 0.26 \u00d7 V unit cell = \u00d7 ( ) = 0 26 4 16 64 3 3 . . \u00c5 \u00c5
Answer: C
\nSolution: d Z M N V = \u22c5 \u22c5 A \u21d2 12 5 4 6 10 4 100 2 10 2 23 10 3 . gm cm gm cm 3 3 = \u00d7 \u00d7 ( ) \u00d7 \u00d7 \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f \u2212 M \u2234 M = 120 \u21d2 Metal is Sn.
Answer: A
\nSolution: P.F. / / particle unit cell particle unit cell particle par d V V m V V m = = t ticle or, 0 1 4 3 1 0 10 6 10 8 3 23 . . / = \u00d7 \u00d7 ( ) \u00d7 \u00d7 \u2212 Z Z M \u03c0 \u21d2 M = 8 \u03c0
Answer: B
\nSolution: Fraction of edge covered by atoms = = = 2 2 4 2 0 707 r a r r / . Hence, fraction of edge not covered = 0.293
Answer: C
\nSolution: d Z M N V = \u22c5 \u22c5 A \u21d2 8 3 4 6 10 5 10 23 8 3 = \u00d7 \u00d7 ( ) \u00d7 \u00d7 ( ) \u2212 M \u21d2 M = 50 gm/mol At 0\u00b0C, the substance will exist as gas and its density = = 50 22 4 2 23 gm L g/L . .
Answer: B
\nSolution: M HCl MCl 1 2 H 2 + \u2192 + 1 1 2 mole mole = A gm = \u00d7 \u00b0 1 2 22 7 0 1 . L at C and bar \u2234 (7.68 \u00d7 4.5) gm 11 35 7 68 4 5 4 54 . . . . A \u00d7 \u00d7 ( ) = \u2234 A = 86.4 Now, d Z M N V = \u22c5 \u22c5 A \u21d2 4 5 86 4 6 10 400 10 23 10 3 . . = \u00d7 \u00d7 ( ) \u00d7 \u00d7 ( ) \u2212 Z \u2234 Z = 2 \u21d2 Unit cell is BCC.
Answer: C
\nSolution: Percentage of occupied space in water = \u00d7 = x x 0 99 0 96 33 32 . . \u2234 Percentage of empty space in water is 100 33 32 \u2212 x .
Answer: B
\nSolution: (Volume of crystal containing one mole metal) \u00d7 70 100 = \u00d7 ( ) \u00d7 \u00d7 \u00d7 ( ) \u2212 6 10 4 3 0 2 10 23 7 3 \u03c0 . cm \u2234 V crystal cm = 64 7 3 \u03c0 \u2234 Density gm/cm = \u239b \u239d \u239c \u239e \u23a0 \u239f = 32 64 7 3 5 3 \u03c0 \u03c0 .
Answer: D
\nSolution: There is no octahedral voids in BCC. However, all the face centres are distorted octahedral voids, which are not considered because they are not regular voids.
Answer: C
\nSolution: None of the tetrahedral as well as octahedral voids will be in contact with other tetrahedral and octahedral voids, respectively.\n9.31 Solid State HINTS AND EXPLANATIONS
Answer: C
\nSolution: Packing is FCC for which 2 4 a r = . \u2234 r = \u00d7 = 2 10 4 2 5 2 \u00c5 \u00c5 . Now, density of metal atom = m V atom atom = \u00d7 ( ) \u00d7 \u00d7 \u00d7 ( ) = \u2212 60 22 6 022 10 4 3 2 5 2 10 0 54 23 8 3 . . . . \u03c0 gm/cm 3
Answer: B
\nSolution: P.F. particle unit cell = = \u00d7 \u00d7 \u00d7 ( ) \u23a7 \u23a8 \u23aa \u23a9 \u23aa \u23ab \u23ac \u23aa \u23ad \u23aa \u00d7 = V V r r r 3 4 3 6 3 4 2 2 3 3 3 2 \u03c0 \u03c0
Answer: B
\nSolution: 1 2 4 3 3rd layer 2nd layer 1st layer 1 4 3 2
Answer: C
\nSolution: A B A 0.V 0.V 0 h 3 h 4 h 2 h 4
Answer: D
\nSolution: Number of NaCl formula units in 1 gm = \u00d7 \u00d7 ( ) 1 58 5 6 10 23 . Each unit cell contains 4 NaCl formula units and hence, volume of crystal = \u00d7 \u00d7 \u00d7 \u00d7 = \u2212 6 10 58 5 4 4 7 10 0 12 23 23 . . . ml
Answer: C
\nSolution: d Z M N V = \u22c5 \u22c5 = \u00d7 \u00d7 ( ) \u00d7 \u00d7 ( ) = \u2212 A 3 gm/cm 4 58 5 6 10 600 10 1 8 23 10 3 . .
Answer: A
\nSolution: The structure is simple cubic for both metals.
Answer: A
\nSolution: \u2018B\u2019 should occupy all the tetrahedral voids and hence, its C.N. =
Answer: B
\nSolution: 25. P.E. A ( ) = \u00d7 + \u00d7 ( ) \u239b \u239d \u239c \u239e \u23a0 \u239f = 4 4 3 8 4 3 0 225 4 2 0 76 3 3 3 \u03c0 \u03c0 r r r . . P.E. B ( ) = \u00d7 + \u00d7 ( ) \u239b \u239d \u239c \u239e \u23a0 \u239f = 4 4 3 4 4 3 0 414 4 2 0 79 3 3 3 \u03c0 \u03c0 r r r . . P.E. C ( ) = \u00d7 + \u00d7 ( ) ( ) = 1 4 3 1 4 3 0 732 2 0 72 3 3 3 \u03c0 \u03c0 r r r . .
Answer: B
\nSolution: Z Z x Z O X Y 2 2+ 3+ \u2212 = = \u00d7 = \u00d7 = 4 8 100 4 50 100 2 ; ; For neutrality of crystal, 4 2 8 100 2 2 3 0 \u00d7 \u2212 ( ) + \u00d7 + ( ) + \u00d7 + ( ) = x \u21d2 x = 12.5
Answer: D
\nSolution: r r x Rb Cl pm + \u2212 + = = 328 5 . r r y r r z r r w K Cl Na Br K Br pm pm pm + \u2212 + \u2212 + \u2212 + = = + = = + = = 313 9 298 1 329 3 . . . \u2234 r r x w y Rb Br pm + \u2212 + = + \u2212 = 343 9 .\n9.32 Chapter 9 HINTS AND EXPLANATIONS
Answer: A
\nSolution: r r + \u2212 = 0 414 . \u21d2 r \u2212 = = 200 0 414 483 1 . . pm
Answer: B
\nSolution: Z Z Z A B C = = \u00d7 = = \u00d7 = 4 2 1 2 1 4 1 4 1 ; ; \u2234 Formula = A 4 BC
Answer: A
\nSolution: Z Z Z O Metal I M Metal II N = = \u00d7 = = \u00d7 = ( ) ( ) 4 1 8 8 1 1 2 4 2 ; \u2234 Formula = MN 2 O 4 For neutrality, M Zn and N Al 3+ = = + 2
Answer: A
\nSolution: Z M 3+ If M are at corners = \u00d7 = ( ) + 8 1 8 1 3 Z X If F are at face centres \u2212 = \u00d7 = ( ) \u2212 6 1 2 3
Answer: A
\nSolution: Octahedral voids in FCC = 4, but only one is occupied by \u2018 x \u2019 and one by \u2018 y \u2019.
Answer: D
\nSolution: (II) r r + \u2212 = = \u2212 ( ) \u21d2 = 20 95 0 21 0 155 0 225 3 . . . C.N.
Answer: B
\nSolution: 0.V. T.V. T.V. Fraction of body diagonal covered = + \u00d7 + \u00d7 \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f = 2 2 0 414 4 0 225 3 4 2 0 76 r r r r . . . \u2234 Fraction, not covered = 1 \u2013 0.76 = 0.24
Answer: A
\nSolution: For electrical neutrality, A = bivalent and B = trivalent. Now, one octahedral void is occupied by \u2018A\u2019 and one by \u2018B\u2019.
Answer: A
\nSolution: Frenkel defect does not change the density.
Answer: B
\nSolution: ln f f H R T T 2 1 1 2 1 1 = \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f \u0394 or, ln / / 1 2 10 1 10 1 1100 1 1200 9 10 \u00d7 ( ) ( ) = \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f \u0394 H R \u21d2 \u0394 H = 176 8 . KJ/mol
Answer: D
\nSolution: Informative
Answer: D
\nSolution: Informative
Answer: A
\nSolution: Theory based
Answer: A, C
\nSolution: Amorphous
Answer: C, D
\nSolution: Informative
Answer: C, D
\nSolution: Informative
Answer: A, D
\nSolution: A B A 0.V. T.V. T.V. T.V. T.V. 0.V 0 h 7 h 8 6 h 8 5 h 8 4 h 8 3 h 8 2 h 8 h 8
Answer: A, B, D
\nSolution: Shortest distance between two T.V. is a 2 .
Answer: A, B, C, D
\nSolution: Both have same packing efficiency and C.N.
Answer: A, B, D
\nSolution: There are the voids per sphere in hexagonal close packing.
Answer: C, D
\nSolution: For octahedral void, the orientation of both tetrahedral voids should be opposite to each other.
Answer: A, C, D
\nSolution: a r r = + ( ) + \u2212 2 Na Cl
Answer: B
\nSolution: Informative
Answer: B, C, D
\nSolution: Informative
Answer: A, B, C, D
\nSolution: Informative\n9.33 Solid State HINTS AND EXPLANATIONS
Answer: A, B, C, D
\nSolution: (a) P P c K b H = + + ( ) 7 1 2 log \u21d2 5 0 7 1 2 4 4 . . log = \u2212 + ( ) c \u2234 c M M = = \u21d2 = 0 4 80 2 100 . / (b) h K K c h b = \u00d7 = \u00d7 \u00d7 = \u00d7 \u2212 \u2212 \u2212 10 4 10 0 4 2 5 10 14 5 5 . . (c) 3 2 2 160 186 4 3 400 a r r a x y = + ( ) \u21d2 = \u00d7 + ( ) = + \u2212 . pm (d) d M N V = \u22c5 = \u00d7 \u00d7 \u00d7 \u00d7 ( ) = \u2212 Z gm/cm A 3 1 100 6 10 400 10 2 6 23 10 3 .
Answer: A, B
\nSolution: (a) Z Z Z C A B 3 2 4 4 1 2 8 4 \u2212 + + = = = \u00d7 = , , Crystal is electrically neutral and hence, it is possible. (b) Z Z Z B A C 2 3 6 6 1 2 12 6 + + \u2212 = = = \u00d7 = , , Crystal is electrically neutral and hence, it is possible. (c) Z Z Z A B C + + \u2212 = \u00d7 = = \u00d7 = = 4 1 8 1 2 4 1 8 1 2 1 2 3 , , Crystal is negatively changed and hence, it is not possible. (d) Z Z Z B C A 2 3 4 8 4 + \u2212 + = = = , , Crystal is negatively changed and hence, it is not possible.
Answer: A, B, D
\nSolution: Informative
Answer: B, C
\nSolution: CaF Al O 2 2 3 8 4 6 4 : , : ( ) ( )
Answer: B, C, D
\nSolution: Informative
Answer: B, C
\nSolution: Informative
Answer: C
\nSolution: Metal defi cient defect can occur with extra anion present in the interstitial voids, but it is very rare.
Answer: B, C
\nSolution: Informative
Answer: A, B, C
\nSolution: \u0394 H = + ve and hence, the surroundings must lose heat.
Answer: A, C
\nSolution: O CCP 2 \u2212 = Fe O.V. 2 1 4 1 + = \u00d7 = Fe in T.V. and 1 in O.V. 3 1 + =
Answer: B, C
\nSolution: Informative
Answer: C, D
\nSolution: Informative
Answer: C, D
\nSolution: Informative
Answer: B
\nSolution: 3 4 2 3 5 0 2 4 33 a r r = \u21d2 = \u00d7 = . . \u00c5
Answer: A
\nSolution: Next nearest neighbours are at \u2018 a \u2019 distance.
Answer: A
\nSolution: C.N. = 8
Answer: B
\nSolution: Number of next nearest neighbours = 6
Answer: C
\nSolution: d Z M N V = \u22c5 \u22c5 = \u00d7 \u00d7 ( ) \u00d7 \u00d7 ( ) = \u2212 A 3 gm/cm 2 39 6 10 5 10 1 04 23 8 3 .
Answer: A
\nSolution: Fractional space occupied by atoms = = V V V V atoms liquid atoms/mol liquid/mol = \u00d7 ( ) \u00d7 \u00d7 \u00d7 \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f ( ) = \u2212 6 10 4 3 3 5 10 4 39 0 9 0 5883 23 8 3 \u03c0 / . . \u2234 Percentage of empty space = (1 \u2013 0.5883) \u00d7 100 = 41.17 %\n9.34 Chapter 9 HINTS AND EXPLANATIONS Comprehension II
Answer: A
\nSolution: 2 4 2 0 5 2 4 0 125 a r r = \u21d2 = \u00d7 = . . nm Now, size of octahedral void = 0.414 \u00d7 0.0125 = 0.052 nm
Answer: C
\nSolution: Size of tetrahedral void = 0.225 \u00d7 0.0125 = 0.028 nm Comprehension III
Answer: B
\nSolution: Volume Mass Density cm 3 = = \u00d7 \u00d7 ( ) \u00d7 = \u00d7 \u2212 6 24 6 10 1 92 1 25 10 23 22 . .
Answer: C
\nSolution: Volume occupied by particles = \u00d7 \u03c0 3 2 V unit all or, 6 4 3 3 2 1 25 10 1 5625 10 3 22 8 \u00d7 = \u00d7 \u00d7 \u21d2 = \u00d7 \u2212 \u2212 \u03c0 \u03c0 r r . . cm
Answer: C
\nSolution: Height of unit cell = \u22c5 = 4 2 3 5 r \u00c5
Answer: B
\nSolution: Number of nearest neighbours = 12 Comprehension IV
Answer: B
\nSolution: Number of T.V. per particle = 2 Number of O.V. per particle = 1
Answer: A
\nSolution: T.V. are smaller than O.V. Comprehension V
Answer: D
\nSolution: a a r r r r r r r r r r KCl NaCl K Cl Na Cl K Na Cl Na Cl N = + ( ) + ( ) = + + + \u2212 + \u2212 + + \u2212 + \u2212 2 2 1 a a + = + + = 1 0 7 1 0 5 1 1 0 5 1 143 . . . .
Answer: A
\nSolution: d d a a NaCl KCl NaCl KCl = = \u00d7 ( ) = 58 5 74 5 58 5 74 5 1 143 1 172 3 3 3 . . . . . . Comprehension VI
Answer: B
\nSolution: Centre is octahedral void.
Answer: B
\nSolution: Number of O.V. = 4, but only one is occupied.\n9.35 Solid State HINTS AND EXPLANATIONS Comprehension VII
Answer: C
\nSolution: C.N. of Zn 2+ = 4
Answer: A
\nSolution: C.N. of S 2 \u2212 = 4
Answer: B
\nSolution: Zn 2+ is tetrahedrally linked with 4 S 2 \u2212 ions.
Answer: C
\nSolution: r r + \u2212 > 0 225 . Comprehension VIII
Answer: C
\nSolution: Z Mn = \u00d7 = 8 1 8 1 Z F = \u00d7 = 12 1 4 3 \u2234 Formula = MnF 3
Answer: A
\nSolution: C.N. of Mn = 6
Answer: B
\nSolution: a r r = + ( ) = + ( ) = + \u2212 2 2 0 65 1 35 4 00 3 Mn F . . . \u00c5
Answer: B
\nSolution: d = \u00d7 + \u00d7 ( ) \u00d7 ( ) \u00d7 \u00d7 ( ) = \u2212 1 55 3 19 6 10 4 10 2 92 23 8 3 . gm/cm 3 Comprehension IX
Answer: B
\nSolution: V m d formula unit 3 cm = = + ( ) \u00d7 \u00d7 = \u00d7 \u2212 132 5 35 5 6 10 3 5 8 10 23 23 . . .
Answer: A
\nSolution: d Z M N V A = . . \u21d2 3 5 1 168 6 10 23 3 . = \u00d7 \u00d7 ( ) \u00d7 a \u21d2 a = \u00d7 \u2212 4 3 10 8 . cm Hence, nearest Cs \u2013 Cs distance = a = 4.3 \u00c5
Answer: B
\nSolution: Nearest Cs Cl distance \u2212 = = 3 2 3 72 a . \u00c5 Comprehension X
Answer: B
\nSolution: N N e e O E RT = = = \u00d7 \u2212 \u2212 \u00d7 \u00d7 \u00d7 \u2212 2 46 10 2 2 1000 5 3 1 0 10 .
Answer: D
\nSolution: In NaCl, all O.V. are occupied. Hence, the available voids are only tetrahedral. \u2234 N i = 2 \u00d7 N o Now, N N N e o o E RT = \u00d7 \u22c5 \u2212 2 2 / \u2234 N N e e o E RT = \u00d7 = \u00d7 = \u00d7 \u2212 \u2212 \u00d7 \u00d7 \u00d7 \u2212 2 2 1 41 10 2 73 6 1000 2 2 1000 8 / . . Comprehension XI
Answer: B
\nSolution: d Z M N V = \u22c5 \u22c5 A = \u00d7 \u00d7 ( ) \u00d7 \u22c5 \u00d7 ( ) = = \u2212 4 6 023 6 023 10 2 10 1 200 5 23 1 3 7 3 . . / Y Y gm/cm kg/m 3 3
Answer: A
\nSolution: d observed >> d theoretical Such large diff erence is possible due to impurity defect.\n9.36 Chapter 9 HINTS AND EXPLANATIONS Comprehension XII
Answer: A
\nSolution: For diamond crystal, 3 8 a r = and Z = 8 Now, d Z M N V = \u22c5 \u22c5 A \u2234 3 6 8 12 6 10 8 3 23 3 . = \u00d7 \u00d7 ( ) \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f r \u2234 r = 0.76 \u00d7 20 \u20138 cm
Answer: B
\nSolution: For SiC, 3 4 a r r c si = + ( ) and Z = 4 Now, d Z M N V r r c si = \u22c5 \u22c5 \u21d2 = \u00d7 \u00d7 ( ) \u00d7 + ( ) \u239b \u239d \u239c \u239e \u23a0 \u239f A 3 2 4 40 6 10 4 3 23 3 . d Z M N V r r c si = \u22c5 \u22c5 \u21d2 = \u00d7 \u00d7 ( ) \u00d7 + ( ) \u239b \u239d \u239c \u239e \u23a0 \u239f A 3 2 4 40 6 10 4 3 23 3 . \u2234 r si = \u00d7 \u2212 1 12 10 8 . cm
Answer: C
\nSolution: Interchange between C and si atoms will neither change \u2018 Z \u2019 nor \u2018 a \u2019.
Answer: D
\nSolution: For the same volume, m m dia sic 3 6 3 2 . . = or, n n n n c sic sic c \u00d7 = \u00d7 \u21d2 = 12 3 6 40 3 2 1 3 75 . . .
Answer: D
\nSolution: Packing efficiency of SiC is greater due to unequal size of particles.
Answer: C
\nSolution: Voids are named according to the orientation of spheres constituting the voids.
Answer: D
\nSolution: Relative increase in packing efficiency is high when larger voids are occupied.
Answer: A
\nSolution: BCC : 2 3 2 r a = FCC : 2 2 2 r a =
Answer: A
\nSolution: Informative
Answer: A
\nSolution: Informative
Answer: A
\nSolution: Informative
Answer: D
\nSolution: Density depends on mass and size of particles but not the packing efficiency.
Answer: A
\nSolution: Theoretical
Answer: C
\nSolution: Packing efficiency of diamond is only 0.34.
Answer: A
\nSolution: Informative
Answer: A
\nSolution: Theoretical
Answer: B
\nSolution: Informative
Answer: A
\nSolution: Informative
Answer: B
\nSolution: Informative
Answer: A
\nSolution: \u0394 \u0394 H S = + = + ve ve , Hence for \u2013ve \u0394 G , the temperature should be high.
Answer: A
\nSolution: Informative
Answer: B
\nSolution: Theoretical
Answer: A
\nSolution: Theoretical
Answer: B
\nSolution: Informative
Answer: B
\nSolution: Informative
Answer: A \u2192 Q, W; B \u2192 P, U; C \u2192 R, V
\nSolution: Theoretical (Cubic crystals)
Answer: A \u2192 Q, W; B \u2192 P, X; C \u2192 R, Z; D \u2192 S, Y
\nSolution: Informative (Ionic solids)
Answer: A \u2192 P; B \u2192 Q, R; C \u2192 Q, S, T; D \u2192 Q, R
\nSolution: Theoretical (Classification of solids)
Answer: A \u2192 Q; B \u2192 P; C \u2192 S; D \u2192 R
\nSolution: Theoretical (Classification of solids)\n9.37 Solid State HINTS AND EXPLANATIONS
Answer: A \u2192 S; B \u2192 R; C \u2192 P; D \u2192 Q
\nSolution: Theoretical (Classification of solids)
Answer: A \u2192 P; B \u2192 Q, S; C \u2192 R
\nSolution: CaCl : 3 2 a r r = + ( ) + \u2212 CaF and 2 3 4 2 4 : . a r r a r = + ( ) = + \u2212 + Diamond : 3a r = 8 Nacl: + + \u2013 \u2013 = \u221a 2 a 2 a \u221a 2
Answer: A \u2192 R, S; B \u2192 P, Q, R, S; C \u2192 Q
\nSolution: Theoretical (Ionic solids)
Answer: A \u2192 P, S; B \u2192 P, Q; C \u2192 Q; D \u2192 Q, R
\nSolution: Informative (Basic crystal system)
Answer: A \u2192 P, Q, R, S; B \u2192 Q, S; C \u2192 P, R, T; D \u2192 T
\nSolution: Informative (Basic crystal system)
Answer: A \u2192 P, Q, R; B \u2192 T; C \u2192 S; D \u2192 P
\nSolution: CaF and 2 2 2 2 2 : d a d a F F Ca Ca \u2212 \u2212 + \u2212 \u2212 \u2212 = = NaCl Na Na : d a + \u2212 \u2212 = CsCl Cs Cs : d a + + \u2212 = d a T.V. from corner = 3 4 Na O and 2 Na Na O O 2 : d a d d a + + \u2212 \u2212 \u2212 = \u2212 = 2 2 2
Answer: A \u2192 P, T; B \u2192 R; C \u2192 Q; D \u2192 S
\nSolution: SC Nearest Next nearest : , = = a a 2 BCC : = = 3 2 a a FCC : = = 2 2 a a
Answer: 3
\nSolution: Volume cm cm , . . l m d l 3 3 58 5 2 167 27 3 = = = \u21d2 =
Answer: 4
\nSolution: d Z M N V Z Z = \u22c5 \u22c5 \u21d2 = \u00d7 \u00d7 \u00d7 \u00d7 ( ) \u21d2 = \u2212 A 4 72 6 10 0 493 10 4 23 7 3 .
Answer: 4
\nSolution: d d l s CH CH A Z M N V 4 4 ( ) ( ) = = \u22c5 \u22c5 or, 0 5 16 6 10 0 6 10 4 23 7 3 . . = \u00d7 \u00d7 \u00d7 \u00d7 ( ) \u21d2 = \u2212 Z Z
Answer: 4
\nSolution: d Z M N V = \u22c5 \u22c5 A \u21d2 0 92 18 6 10 2 3 4 4 53 10 7 41 10 23 8 2 8 . . . = \u00d7 \u00d7 \u00d7 \u00d7 \u00d7 \u00d7 ( ) \u00d7 \u00d7 \u23a7 \u23a8 \u23aa \u23a9 \u23aa \u23ab \u23ac \u23aa \u23ad \u23aa \u2212 \u2212 Z \u2234 Z = 4
Answer: 2
\nSolution: d Z M N V = \u22c5 \u22c5 A \u21d2 2 4 12 6 10 6 3 4 10 2 3 10 23 8 2 8 . = \u00d7 \u00d7 \u00d7 \u00d7 \u22c5 \u00d7 ( ) \u00d7 \u00d7 \u23a7 \u23a8 \u23aa \u23a9 \u23aa \u23ab \u23ac \u23aa \u23ad \u23aa \u2212 \u2212 Z x \u2234 x 2 200 108 =
Answer: 1
\nSolution: 2 4 a r = \u21d2 2 2 2 1 r a = = nm
Answer: 4
\nSolution: + + \u2013 \u2013 r 2 2 186 2 214 2 400 r = + \u239b \u239d \u239c \u239e \u23a0 \u239f = pm pm\n9.38 Chapter 9 HINTS AND EXPLANATIONS
Answer: 5
\nSolution: d Z M N V = \u22c5 \u22c5 A = \u00d7 \u00d7 \u00d7 \u00d7 ( ) = \u2212 4 58 5 6 10 500 10 3 12 23 10 3 . . gm/cm 3 \u2234 Percentage vacancy = \u2212 \u00d7 = 3 12 2 964 3 12 100 5 . . . %
Answer: 5
\nSolution: (ii), (iv), (v), (vi) , (vii) are true statements.
Answer: 2
\nSolution: d theo 3 gm/cm = \u00d7 \u00d7 \u00d7 \u00d7 ( ) = \u2212 \u2212 4 31 25 1 67 10 500 10 1 67 24 10 3 . . . Now, m theo = 1670 gm per litre m actual = 1607.5 gm per litre \u2234 Moles of metal missing per litre = \u2212 = 1670 1607 5 31 25 2 . .
Answer: 6
\nSolution: Z Z ZnS ZnS but due to defect, = = 4
Answer: 8
\nSolution: 12. Octahedron has eight triangular faces. Hence, truncated octahedron will have eight hexagonal faces.
Answer: 2
\nSolution: BCC: Fraction of edge covered by atom = 2 r a . \u2234 Fraction of edge uncovered = \u2212 = \u2212 \u00d7 = 1 2 1 2 3 4 0 134 r a . From question: 0.134 a = 67 pm \u21d2 a = 500 pm Now, d Z M N V = \u22c5 \u22c5 = \u00d7 \u00d7 \u00d7 \u00d7 = \u2212 A 3 gm/cm 2 75 6 10 500 10 2 23 10 3 ( ) ( )
Answer: 2
\nSolution: Let the ideal crystal was having 100 Si-atom, After doping, x Si-atom are missing and y B-atom are doped. Now, ( ) 100 30 11 100 30 88 100 \u2212 \u00d7 + \u00d7 = \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f \u00d7 x N y N N A A A or, 30 x \u2013 11 y = 360 (1) and ( ) ( . ) . 100 30 11 1 0 001 0 001 3000 30 11 999 \u2212 \u00d7 \u00d7 = \u2212 \u21d2 \u2212 = x N y N x y A A (2) From (1) and (2), y x \u00d7 = 100 2%
Answer: 2
\nSolution: In truncated octahedron, all corner of octahedron become square faces and hence, its number =
Answer: 0143
\nSolution: And, the number of c-atoms per unit cell in diamond =
Answer: 0066
\nSolution: Four-digit Integer Type
Answer: 0045
\nSolution: m d N r \u00d7 = \u00d7 0 74 4 3 3 . A \u03c0 or, 197 19 7 0 74 6 10 4 3 23 3 . . \u00d7 = \u00d7 \u00d7 \u00d7 \u03c0 r \u21d2 r = \u00d7 = \u2212 1 43 10 143 8 . cm pm
Answer: 0072
\nSolution: P.E. of diamond and here silicon is 0.34.
Answer: 0012
\nSolution: d Z M N V = \u22c5 \u22c5 A \u21d2 1 5 10 6 6 10 12 5 8 0 3 0 10 3 23 9 3 . . . . \u00d7 = \u00d7 \u00d7 ( ) \u00d7 \u00d7 \u00d7 \u00d7 ( ) \u23a7 \u23a8 \u23aa \u23a9 \u23aa \u23ab \u23ac \u23aa \u23ad \u23aa \u2212 M \u21d2 M = 45 Kg/mol
Answer: 0155
\nSolution: Let the percentage of fayalite be x . V V V olivine fayalite fosterite = + or, 100 3 9 4 2 100 3 3 . . . = + \u2212 x x \u21d2 x = 71.79
Answer: 0018
\nSolution: d Z M N V = \u22c5 \u22c5 A \u21d2 2 14 4 426 18 6 10 1 26 10 23 7 3 . . = \u00d7 + ( ) \u00d7 ( ) \u00d7 \u00d7 ( ) \u2212 x \u21d2 x = 12\n9.39 Solid State HINTS AND EXPLANATIONS
Answer: 0182
\nSolution: For diamond: 3 4 a d = \u00d7 \u2212 C C Now, d Z M N V = \u22c5 \u22c5 A \u21d2 2 3 8 12 6 10 4 3 23 3 = \u00d7 \u00d7 ( ) \u00d7 \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f \u2212 d C C \u2234 d C C cm pm \u2212 \u2212 = \u00d7 = 1 55 10 155 8 .
Answer: 6041
\nSolution: The maximum backing efficiency of identical spheres in 2D is 0.90. Hence, 40 0 90 10 2 2 2 ( ) \u00d7 = \u00d7 \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f . N \u03c0 \u21d2 N = 18
Answer: 0067
\nSolution: 4 2 520 . r Cl \u2212 = \u00d7 \u21d2 r Cl pm \u2212 = 182
Answer: 0059
\nSolution: Z Ti = \u00d7 = 1 2 1 1 2 Z O = 1 (assume) \u2234 Formula = Ti O T O i 1 2 1 2 / \u2245 Now, Ti = + \u00d7 = 48 48 32 100 60% and oxidation state of Ti = +4
Answer: 0125
\nSolution: For one litre crystal, m m m m initial Bremoved Cadded final \u2212 + = or, 4800 30 1 15 4795 \u2212 \u00d7 + \u00d7 = x \u21d2 x = 2 3 \u2234 Percentage of C-atoms which replaced B-atoms = \u00d7 = x 1 100 67%
Answer: 0073
\nSolution: Percentage of body diagonal covered = + + ( ) \u23a1 \u23a3 \u23a4 \u23a6 \u00d7 \u2212 + \u2212 2 4 3 2 100 r r r r B C A B % = + \u00d7 + ( ) \u00d7 = \u2212 \u2212 \u2212 \u2212 2 4 0 225 2 3 0 414 100 59 2 r r r r B B B B . . % . %
Answer: 4000
\nSolution: Out of 8 Fe 2+ in original crystal, 1 is missing. \u2234 Percentage of cation vacancy = \u00d7 = 1 8 100 12 5 . %
Answer: 1000
\nSolution: Percentage occupied space = \u00d7 = \u00d7 \u00d7 \u00d7 ( ) \u239b \u239d \u239c \u239e \u23a0 \u239f \u00d7 \u2212 V V particles solid 100 6 10 4 3 1 54 10 40 4 1 23 8 3 % . / \u03c0 \u03c0 0 00 27 % % = \u2234 Empty space = 73 %
Answer: D
\nSolution: Greater the specific surface area of adsorbent, greater will be the extent of adsorption.
Answer: B
\nSolution: Adsorption decreases the surface energy.
Answer: A
\nSolution: x m K P x m K n P n = \u21d2 = + \u22c5 . log log log 1 1 From question, log K = 0.3010 = log 2 \u21d2 K = 2 And 1 45 1 1 n n = \u00b0 = \u21d2 = tan \u2234 x m P = \u00d7 = \u00d7 = 2 2 0 2 0 4 . .
Answer: B
\nSolution: K A e E RT a = \u2212 . / \u2234 ln K K E R T T a 2 1 1 2 1 1 = \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f or, ln 10 cal/mol = \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f \u21d2 = E R E a a 1 600 1 1000 6900
Answer: A
\nSolution: For a particular combination of adsorbent, adsorbate and temperature, only one value of \u2018 n \u2019 is permissible.
Answer: B
\nSolution: x m K P n = \u22c5 1 \u21d2 0 2 4 1 . ( ) = \u00d7 K n (1) 0 5 25 1 . ( ) = \u00d7 K n (2) 0 8 64 1 . ( ) = \u00d7 K n (3) From (1), (2) and (3), K n = = 1 10 2 and \u2234 x m K n \u239b \u239d \u239c \u239e \u23a0 \u239f = \u00d7 = required ( ) . 36 0 6 1 Hence, moles of N 2 adsorbed per gm of iron = = 0 6 28 3 140 .
Answer: D
\nSolution: r r 1 2 < \u21d2 A is the catalyst. r r r r r 3 1 1 4 5 < \u21d2 = = \u21d2 B is the catalyst. C and D are not catalysts. . r r r r r r C 1 7 2 1 2 6 < < \u21d2 < < \u21d2 D is catalytic poison. is catalytic prom motor.
Answer: D
\nSolution: A catalyst always involve in the reaction.
Answer: C
\nSolution: Enzymes are specific.
Answer: A
\nSolution: A catalyst does not initiate the reaction.
Answer: A
\nSolution: Theory based
Answer: C
\nSolution: Catalyst does not alter the equilibrium position.
Answer: C
\nSolution: Homogeneous catalysis, because the physical states of both reactant and catalyst is aqueous (liquid).
Answer: D
\nSolution: Activation energy is decreased.
Answer: A
\nSolution: Catalyst lowers the activation energy.
Answer: D
\nSolution: Catalysis occurs through chemisorption.
Answer: C
\nSolution: Viscosity is higher and surface tension is smaller than water.
Answer: A
\nSolution: K , as it reacts vigorously in water.
Answer: A
\nSolution: SnCl 4 formed by reaction will adsorb some common Cl \u2013 ions.
Answer: A
\nSolution: Informative
Answer: A
\nSolution: True solution or suspension does not show Tyndall eff ect.
Answer: D
\nSolution: Blood, clay and smoke are negative sol. In strong acidic solution, gelatin adsorbs some H + ions and become positive.
Answer: B
\nSolution: Theory
Answer: B
\nSolution: Informative
Answer: B
\nSolution: Informative EXERCISE II (JEE ADVANCED)\n12.24 Chapter 12 HINTS AND EXPLANATIONS
Answer: A
\nSolution: Volume of metal used = \u00d7 = \u2212 \u2212 1 9 10 19 10 4 5 . cm 3 \u2234 N \u00d7 \u00d7 \u00d7 = \u2212 \u2212 4 3 10 10 10 7 3 5 3 ( ) cm cm \u21d2 N = \u00d7 2 39 10 12 . Hence, number of particles per cm 3 = \u00d7 = \u00d7 2 39 10 1000 2 39 10 12 9 . .
Answer: A
\nSolution: Larger the carbon chain, normally smaller is CMC.
Answer: C
\nSolution: Sulphide sol have negative charge on colloidal particles
Answer: B
\nSolution: Adsorption is physisorption.
Answer: B
\nSolution: Theory based
Answer: A, B, C
\nSolution: Entropy decreased in adsorption.
Answer: A, B, C, D
\nSolution: Theory based
Answer: A, C
\nSolution: Theory based
Answer: C
\nSolution: \u0394 H can never be equal to \u0394 S.
Answer: A, D
\nSolution: (a) Chemisorption does not change into physisorption at higher pressure. (b) CO or CO 2 gases leave the surfaces.
Answer: A, B, D
\nSolution: Theory based
Answer: A, B, C, D
\nSolution: Informative
Answer: A, B, C
\nSolution: In Lindlar\u2019s catalyst, catalytic poison is used
Answer: A, C
\nSolution: Theory based
Answer: B, D
\nSolution: All catalytic reaction is multistep reaction
Answer: A, B, C
\nSolution: Informative
Answer: A, C, D
\nSolution: (a) K K e A e A e e E RT E RT E E RT a a a a cat uncat = = = \u2032 \u2032 \u2032 \u2212 20 . . / / ( )/ \u2234 20 2 2 1000 300 = \u2212 = \u2212 \u00d7 \u2032 E E RT E a a a Kcal \u21d2 E a = 14 Kcal/mol (b), (c) Reaction: 2H 2 O 2 (aq) \u2192 2H 2 O(l)+O 2 (g) is fi rst order. (d) Rate of uncatalysed reaction increases to greater extent on increasing temperature because its activation energy is high.
Answer: C, D
\nSolution: Sol particles are restricted to move. Solvent particles move in opposite direction to the expected movement of sol particles. Fe(OH) 3 sol is positively charged.
Answer: A, B, C, D
\nSolution: Informative
Answer: B, C
\nSolution: Electrophoresis and electro-osmosis are the experimental methods to determine charge on colloidal particles.
Answer: A, C
\nSolution: Informative
Answer: D
\nSolution: Below CMC, the solution is true solution.
Answer: A, B, C
\nSolution: PO SO Cl 4 3 4 2 \u2212 \u2212 \u2212 > > \u21d2 Sol particles are positively charged.
Answer: A, B, C
\nSolution: RCOONa RCOO Na \u001c \u2212 + + As true solution, one mole of RCOONa will become two moles in solution. But, as micelle formation starts, the total number of particles start decreasing due to association.
Answer: C, D
\nSolution: Due to excess Ag+, sol particles will be positively charged.
Answer: C
\nSolution: (a) It is due to sharp decrease in number of ions. (b) Tyndall effect is better shown by lyophobic colloid. (c) Colloidal solutions have lower value of colligative properties. (d) Larger the carbon chain, normally lower CMC value.
Answer: A, B, D
\nSolution: (a) Basic dye is positively charged and hence, Fe(CN) HPO 6 4 3 2 \u2212 \u2212 > (c) Slope should not change in Freundlich\u2019s isotherm.
Answer: A, C
\nSolution: Tyndall effect is shown by colloids.
Answer: A, D
\nSolution: Theory based
Answer: C
\nSolution: Charge : Mg 2+ (2 unit) > Cl \u2013 (1 unit) Hence, better coagulation for negatively charged gold sol.\n12.25 Surface Chemistry HINTS AND EXPLANATIONS
Answer: D
\nSolution: Polarizability is maximum in Xe.
Answer: C
\nSolution: CO is polar and hence, more preferential adsorption.
Answer: B
\nSolution: Adsorption decreases on increasing temperature. Comprehension II
Answer: A
\nSolution: In case of concentrated KCl, KCl adsorbs on blood charcoal surface, but in case of dilute KCl, blood charcoal dissolves in KCl solution.
Answer: C
\nSolution: Greater critical temperature, greater the extent of adsorption.
Answer: A
\nSolution: Adsorption is always exothermic. Comprehension III
Answer: D
\nSolution: Initial surface area, A 1 2 2 6 2 24 = \u00d7 = ( ) cm cm Final volume of each cube = cm 3 8 10 12 \u2234 Final side length of each cube = \u00d7 ( ) 8 10 12 1 3 cm 3 / = \u00d7 \u2212 2 10 4 cm Hence, final surface area of each cube, A 2 = 6 \u00d7 (2 \u00d7 10 \u22124 cm ) 2 = 24 \u00d7 10 \u22128 cm 2 \u2234 Final total surface area Initial surface area = \u00d7 \u00d7 \u2212 24 10 10 2 8 12 4 4 10 4 =
Answer: C
\nSolution: Number of H 2 molecules = \u00d7 \u00d7 \u00d7 \u00d7 = \u00d7 2 0 112 0 0821 546 6 10 3 10 23 21 . . \u2234 Specific surface area = \u00d7 \u00d7 \u00d7 \u2212 3 10 0 4 10 5 21 7 2 . ( ) cm gm = \u00d7 2 4 10 6 . cm /gm 2 Comprehension IV
Answer: A
\nSolution: log log log x m K n P \u239b \u239d \u239c \u239e \u23a0 \u239f = + \u22c5 1 Slope = = \u21d2 = 1 0 25 4 n n . and for x -intercept, log x m \u239b \u239d \u239c \u239e \u23a0 \u239f = 0 \u21d2 log log K n P = \u2212 \u22c5 1 = \u2212 \u00d7 \u2212 = 1 4 4 1 0 ( ) . \u2234 K = 10
Answer: A
\nSolution: x m = \u00d7 = 10 16 20 1 4 ( ) / \u21d2 x = \u00d7 = 20 10 200 gm
Answer: B
\nSolution: 810 1 0 10 1 4 . ( ) / = \u00d7 P \u21d2 P = 3 atm\n12.26 Chapter 12 HINTS AND EXPLANATIONS Comprehension V
Answer: B
\nSolution: Positively charged due to adsorption of Ag + ions.
Answer: B
\nSolution: Theory based
Answer: C
\nSolution: Informative Comprehension VI
Answer: D
\nSolution: Positively charged due to adsorption of Fe 3+ ions.
Answer: A
\nSolution: AgI Ag NO fixed la , , - + \u2193 \u2193 3 y yer diffused layer
Answer: A
\nSolution: S 2\u2013 ions get adsorbed. Comprehension VII
Answer: B
\nSolution: Gold number = 0.025 \u00d7 1000 = 25
Answer: A
\nSolution: Theory based
Answer: C
\nSolution: Theory based
Answer: A
\nSolution: Colour become less intense due to adsorption.
Answer: B
\nSolution: Surface particles have higher energy due to unbalanced forces.
Answer: D
\nSolution: Word 'always' is not suitable because chemisorption increased with increase in temperature.
Answer: A
\nSolution: Theory based
Answer: A
\nSolution: Theory based
Answer: A
\nSolution: Theory based
Answer: B
\nSolution: Theory based
Answer: A
\nSolution: Theory based
Answer: A
\nSolution: Theory based
Answer: B
\nSolution: Theory based
Answer: D
\nSolution: Cellulose nitrate sol is lyophilic.
Answer: B
\nSolution: Scattering is not related to speed of particles.
Answer: B
\nSolution: Theory based
Answer: D
\nSolution: Theory based
Answer: A
\nSolution: Colloidal particles are negatively charged due to adsorption of I - ions and hence, it moves towards anode.
Answer: A
\nSolution: Theory based
Answer: C
\nSolution: Informative
Answer: C
\nSolution: Peptization occurs due to adsorption of common ion.
Answer: A
\nSolution: Natural colloids are normally lyophilic.
Answer: B
\nSolution: Theory based
Answer: A \u2192 Q, U; B \u2192 P, W; C \u2192 P, V; D \u2192 R, X
\nSolution: Informative
Answer: A \u2192 R; B \u2192 S; C \u2192 P; D \u2192 Q
\nSolution: Informative
Answer: A \u2192 Q; B \u2192 P; C \u2192 Q, R; D \u2192 P, S
\nSolution: Informative
Answer: A \u2192 Q; B \u2192 S; C \u2192 P; D \u2192 R
\nSolution: Informative\n12.27 Surface Chemistry HINTS AND EXPLANATIONS
Answer: 4
\nSolution: 6 48 0 01 10 162 10 3 3 . ( . ) = \u00d7 \u00d7 \u00d7 \u2212 n \u21d2 n = 4
Answer: 3
\nSolution: log log log x m K n P = + \u22c5 1 From the given graph, log K = 1.0 \u21d2 K = 10 and 1 0 25 n = . \u21d2 n = 4 Now, x m K P n = . 1 \u21d2 x 1 0 10 8 1 10 3 1 4 . ( . ) / = \u00d7 \u00d7 \u2212 \u21d2 x = 3 gm
Answer: 4
\nSolution: t k A e s e av E RT a = = = \u00d7 \u00d7 = \u2212 \u2212 \u2212 \u00d7 \u00d7 1 1 1 1 25 10 4 8 1 16 10 2 400 3 . ( . ) sec / /
Answer: 6
\nSolution: Soap solution of sodium palmitate, gold sol, silicic acid sol, acidic dye, metal sulphide sol, sol of AgCl by excess KCl in AgNO 3 .
Answer: 6
\nSolution: ln P P R T T ads 1 2 1 2 1 1 = \u2212 \uf8eb \uf8ed \uf8ec \uf8ec \uf8ec \uf8ec \uf8f6 \uf8f8 \uf8f7 \uf8f7 \uf8f7 \uf8f7 \u0394 H or, ln H 1 6 32 1 200 1 250 . = \u2212 \uf8eb \uf8ed \uf8ec \uf8ec \uf8ec \uf8f6 \uf8f8 \uf8f7 \uf8f7 \uf8f7 \uf8f7 \u0394 ads R \u21d2 \u2206 H ads = 6000 cal/mol
Answer: 5
\nSolution: Number of CH 3 COOH molecules adsorbed = \u00d7 \u2212 \u00d7 \u00d7 = \u00d7 100 0 5 0 49 1000 6 10 6 10 23 20 ( . . ) \u2234 Surface area of each molecule = \u00d7 \u00d7 = \u00d7 \u2212 3 10 6 10 5 10 2 20 19 m 2
Answer: 2
\nSolution: Number of N 2 molecules = \u00d7 \u00d7 \u00d7 \u00d7 \u00d7 = \u00d7 \u2212 0 001 2 46 10 0 082 300 6 023 10 6 023 10 3 23 16 . . . . . \u2234 Number of active sites per molecule = \u00d7 \u00d7 \u00d7 \u00d7 = 1000 6 023 10 20 100 6 023 10 2 14 16 . .
Answer: 9
\nSolution: Number of N 2 molecules absorbed = \u00d7 \u00d7 \u00d7 = \u00d7 \u2212 2 24 10 22 4 6 10 6 10 3 23 19 . . \u2234 Specific surface area = \u00d7 \u00d7 \u00d7 = \u2212 6 10 0 15 10 9 19 9 2 . ( )
Answer: 2
\nSolution: Colloid is a heterogeneous system \u21d2 min = 2 phases
Answer: 9
\nSolution: ln 20 1 600 1 1000 = \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f E R a \u21d2 E a = 9000 cal/mol. Four-digit Integer Type
Answer: 0600
\nSolution: Mass of NaCl used = \u00d7 \u00d7 = \u00d7 ( ) 585 1 2 1 100 5 85 1 2 . / . . gm Moles of NaCl used = \u00d7 = 5 85 1 2 58 5 0 12 . . . . \u2234 Coagulation value = \u00d7 = 0 12 10 200 1000 600 3 . / millimole/litre
Answer: 0020
\nSolution: Number of palmitic acid molecules needed = \u00d7 = \u00d7 \u2212 480 0 2 10 2 4 10 7 2 17 cm cm 2 . ( ) . Moles of palamitic acid molecules = \u00d7 \u00d7 = \u00d7 \u2212 2 4 10 6 10 4 10 17 23 7 . \u2234 Volume of solution needed = \u00d7 \u00d7 = \u00d7 = \u2212 \u2212 1 5 12 256 4 10 2 10 20 7 5 3 dm dm mm 3 3 . /\n12.28 Chapter 12 HINTS AND EXPLANATIONS
Answer: 0015
\nSolution: The radius of hydrogen molecule = \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f m d 3 4 1 3 \u03c0 / = \u00d7 \u00d7 \u00d7 \u239b \u239d \u239c \u239c \u239c \u239e \u23a0 \u239f \u239f \u239f = \u00d7 \u2212 2 6 10 0 16 3 4 2 5 10 23 1 3 8 . . / \u03c0 \u03c0 cm Number of hydrogen molecules at the surface per gm Cu = \u00d7 \u00d7 = \u00d7 224 22400 6 10 25 2 4 10 23 20 / . \u03c0 \u03c0 \u2234 Specific surface area of Cu = cm /gm m /gm 2 2 2 4 10 2 5 10 150000 15 20 2 8 . ( . ) \u00d7 \u00d7 \u00d7 \u00d7 = = \u2212 \u03c0 \u03c0
Answer: 0014
\nSolution: t k av = 1 Now, ln ln k k t t E R T T a 2 1 1 2 1 2 1 1 = = = \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f or, ln . . 0 36 0 72 1 2500 1 2000 = \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f E R a \u21d2 E a = 14000 cal/mol
Answer: 0010
\nSolution: t t K K A e A e Fe 1 2 1 2 20 10 2 600 8 3 / / / . . ( ) ( ) = = \u2212 \u00d7 \u00d7 \u2212 \u00d7 Fe charcoal charcoal 1 10 2 600 10 3 1 / \u00d7 = e
Answer: 0025
\nSolution: Number of adsorbate molecules = \u00d7 \u00d7 \u00d7 \u00d7 = \u00d7 \u2212 0 10 10 0 25 10 6 10 2 4 10 3 3 23 7 . . . \u2234 Eff ective surface area = \u00d7 = \u00d7 \u2212 0 06 2 4 10 25 10 17 20 2 . . m
Answer: 0091
\nSolution: Moles of gas adsorbed per gm of charcoal = \u2212 \u00d7 \u00d7 \u00d7 \u00d7 ( ) . 700 400 1 52 760 300 6 R Volume of gas adsorbed per gm of charcoal (at 0\u00b0C and 1 atm) = \u00d7 \u00d7 \u00d7 \u00d7 \u00d7 \u00d7 300 1 52 760 300 6 273 1 . R R = 0.091 litre
Answer: 0720
\nSolution: Specific surface area of silica gel = \u00d7 \u00d7 \u00d7 \u00d7 = \u2212 168 22400 6 10 0 16 10 720 23 9 2 . ( ) m /gm 2
Answer: 0400
\nSolution: ln P P R T T 1 2 1 1 1 1 = \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f \u0394 \u03b8 H or, ln . . . . 0 4 59 2 16 628 10 8 314 1 200 1 3 = \u2212 \u00d7 \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f T \u21d2 T = 400 K
Answer: 4000
\nSolution: r K K E S K K K S K K E S K K S = + + + \u2212 \u2212 1 2 0 1 2 1 1 2 0 1 1 [ ][ ] [ ] [ ][ ] [ ] \u001b For r max , K S K 1 1 [ ] \u001a \u2212 \u2234 r max = = = K K E S K S K E 1 2 0 1 2 0 0 02 [ ][ ] [ ] [ ] . M From question, K E K K E S K K S 2 0 1 2 0 1 1 2 [ ] [ ][ ] [ ] = + \u2212 \u21d2 K K S K S \u2212 + = 1 1 1 2 [ ] [ ] \u2234 K K S 1 1 3 3 6 3 1 1 250 250 10 4000 \u2212 \u2212 = = = \u00d7 = [ ] mg dm dm kg dm kg
Answer: B
\nSolution: N O NO Brown 2 4 2 2 \u001f On heating, colour deepens means reaction is endothermic \u0394 H ve = + ( ) .
Answer: B
\nSolution: As the process is endothermic but no heat is absorbed from surrounding, the temperature of the system will decrease, As initial temperature is used, the calculated mole will be lower.
Answer: B
\nSolution: \u0394 \u0394 \u0394 \u0394 \u0394 H H H H 2 1 2 1 2 1 3 2 0 \u2212 \u2212 = ( ) = ( ) \u2212 + ( ) = \u21d2 = T T C x x x P
Answer: D
\nSolution: I 2 (s) \u2192 I 2 (g); \u0394 \u0394 H cal/gm at K H at K 1 1 2 2 24 473 523 = = = = T T ? \u0394 \u0394 \u0394 H H H 2 1 2 1 2 2 2 24 523 473 0 055 0 031 \u2212 \u2212 = ( ) \u2212 ( ) \u21d2 \u2212 \u2212 = \u2212 T T C I g C I s P P , , . . \u2234 \u0394 H c al/gm 2 25 2 = .
Answer: B
\nSolution: For direct measurement, reaction must occur directly in the conditions to measure heat.
Answer: A
\nSolution: Greater the mass per cent of hydrogen, greater is the calorific value.
Answer: D
\nSolution: For the reaction: \u0394 \u0394 \u0394 U H RT kJ = \u2212 \u22c5 = \u2212 ( ) \u2212 \u2212 ( ) \u00d7 \u00d7 = \u2212 n g 72 3 1 8 314 1000 298 69 8 . . . As HCl is limiting reagent, for the given amount, \u0394 U kJ = \u00d7 \u2212 ( ) = \u2212 2 69 8 139 6 . .\n5.36 Chapter 5 HINTS AND EXPLANATIONS
Answer: C
\nSolution: HAuBr 4 + 4HCl \u2192 HAuCl 4 + 4HBr; \u0394 H = (\u201328) \u2013 (\u201336.8) = 8.8 kcal \u2234 Percentage reaction = \u00d7 = 0 44 8 8 100 5 . . %
Answer: A
\nSolution: (a) C + O CO; 1 2 9 0 4 5 2 \u2192 . . \u0394 H = \u201375 kcal Heat evolved = 75 \u00d7 9 = 675 kcal (b) C + O CO 2 2 2 2 \u2192 ; \u0394 H = \u201395 kcal Heat evolved = 95 \u00d7 2 = 190 kcal (c) 4C O CO 3CO + \u2192 + 3 5 2 2 . Heat evolved = 75 \u00d7 1 + 95 \u00d7 3 = 360 kcal (d) C + O CO 2 2 2 5 2 5 \u2192 . . \u0394 H = \u201395 kcal Heat evolved = 95 \u00d7 2.5 = 237.5 kcal
Answer: A
\nSolution: CH3 NO2 NO2 O2N l O g CO g H O l N g ( ) + ( ) \u2192 ( ) + ( ) + ( ) 21 4 7 5 2 3 2 2 2 2 2 \u0394 H kJ/mol k = \u00d7 \u2212 ( ) + \u00d7 \u2212 ( ) \u2212 ( ) = \u2212 = \u2212 \u00d7 7 395 5 2 285 65 3542 5 3542 5 227 1 816 . . . J J/mol kJ/mol MJ/L = \u2212 = \u2212 28 34 28 34 . .
Answer: D
\nSolution: For H A 2 , enthalpy of ionization = \u00d7 \u2212 ( ) = 2 13 5 13 1 . kcal/mol For B OH 2 ( ) , enthalpy of ionization = \u00d7 \u2212 ( ) = 2 13 5 10 7 . kcal/mol \u2234 Required \u0394 = \u00d7 \u2212 \u2212 = \u2212 H kcal 2 13 5 1 7 19 .
Answer: D
\nSolution: If BaSO 4 were water soluble, then \u0394 = \u00d7 \u2212 ( ) = \u2212 H kJ expected 2 57 114
Answer: B
\nSolution: Required \u0394 = \u2212 \u2212 \u00d7 ( ) = \u2212 H cal 13700 400 0 9 13340 .
Answer: A
\nSolution: Aerobic oxidation results release of energy and hence, it is biologically benefical by (2880 + 2530 = 5410 kJ/mol)
Answer: D
\nSolution: (a) Si H g H g SiH g H kcal 2 6 2 4 ( ) + ( ) \u2192 ( ) \u0394 = \u2212 2 11 7 ; . (b) SiH g SiH g H g H kcal 4 2 2 ( ) \u2192 ( ) + ( ) = + ; . \u0394 239 7 (c) 2Si s H g Si H g H kcal 2 ( ) + ( ) \u2192 ( ) = + 3 80 3 2 6 ; . \u0394 Required thermochemical equation is Si s H g SiH g 2 2 ( ) + ( ) \u2192 ( ) From (b) a) (c) H kcal/mol + + = + 1 2 1 2 274 ( : \u0394
Answer: B
\nSolution: Required thermochemical equation is Dy s Cl g DyCl s 2 3 ( ) + ( ) \u2192 ( ) 3 2 From (ii) + 3 \u00d7 (iii) \u2013 (i), we get: \u0394 H kJ/mol = \u2212 ( ) + \u2212 ( ) \u2212 \u2212 ( ) = \u2212 699 43 3 158 31 180 06 994 3 . . . .
Answer: A
\nSolution: \u0394 \u2212 ( ) + \u2212 ( ) + \u2212 ( ) + \u2212 ( ) \u23a1 \u23a3 \u23a4 \u23a6 \u2212 \u2212 ( ) + \u2212 H= 188 84 22 05 2 22 1 2 17 63 70 97 8 5 . . . . . . 6 6 0 2 68 32 74 18 ( ) + + \u2212 ( ) \u23a1 \u23a3 \u23a4 \u23a6 = \u2212 . . kcal
Answer: C
\nSolution: H aq OH aq H O l 2 + \u2212 ( ) + ( ) \u2192 ( ) \u0394 = \u0394 \u2212 \u0394 + \u0394 \u23a1 \u23a3 \u23a2 \u23a4 \u23a6 \u23a5 ( ) ( ) ( ) + \u2212 H H H H f H O l f H aq f OH aq 2 or, \u2212 = \u2212 ( ) \u2212 + \u0394 \u23a1 \u23a3 \u23a2 \u23a4 \u23a6 \u23a5 \u2212 ( ) 57 32 285 84 0 . . f OH aq H \u2234 \u0394 = \u2212 \u2212 ( ) f OH aq H kJ/mol 228 52 .
Answer: C
\nSolution: CH CH COOH l O g CO g H O l 3 2 2 2 2 7 2 3 3 ( ) + ( ) \u2192 ( ) + ( ) \u0394 = \u00d7 \u0394 + \u00d7 \u0394 \u23a1 \u23a3 \u23a4 \u23a6 \u2212 \u0394 ( ) ( ) ( ) C CH CH COOH l f CO g f H O l f CH CH COOH H H H H 3 2 2 2 3 2 3 3 l l f O g H ( ) ( ) + \u00d7 \u0394 \u23a1 \u23a3 \u23a2 \u23a4 \u23a6 \u23a5 7 2 2 or, 3 3 94 3 68 3 2 3 2 \u00d7 \u0394 = \u2212 ( ) + \u2212 ( ) \u23a1 \u23a3 \u23a4 \u23a6 \u2212 \u0394 \u23a1 \u23a3 \u23a4 ( ) ( ) f CH CH COOH f CH CH COOH H H +O l l \u23a6 \u23a6 \u2234 \u0394 = \u2212 ( ) f CH CH COOH H kcal/mol 3 2 121 5 l .\n5.37 Thermochemistry HINTS AND EXPLANATIONS
Answer: B
\nSolution: phOH(solution II) \u2192 phOH(solution I) \u0394 = \u2212 \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f \u2212 \u2212 \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f = \u2212 H kcal/mol 0 02 0 47 94 0 03 1 410 94 2 . . . .
Answer: A
\nSolution: \u0394 = \u2212 ( ) \u2212 \u2212 ( ) \u23a1 \u23a3 \u23a4 \u23a6 \u2212 \u2212 [ ] = + H kcal/mol required 14 7 2 7 13 75 1 75 . . . .
Answer: A
\nSolution: 2 1 2 197 2 65 2 3 FeO O Fe O H 2 + \u2192 = \u2212 ( ) \u2212 \u2212 ( ) ; \u0394 = \u2212 67 kcal Initial Final 2 2 2 a a a x a x \u2212 + 2 2 1 2 3 5 a x a x x a \u2212 + = \u21d2 = and heat released = 67 x kcal \u2234 Heat released per mole of initial mixture = = 67 3 13 4 x a . kcal
Answer: B
\nSolution: 3C H OH 25H O 3C H OH 25H O 2 5 2 2 5 2 + \u2192 ( ) ; \u0394 \u0394 H kcal H cal theo exp = \u2212 ( ) + \u2212 ( ) = \u2212 = \u2212 ( ) = \u2212 1120 2 1760 4640 3 1650 4950 As experimentally, more heat is released means the mixing is exothermic by (4950 \u2013 4640) = 310 cal.
Answer: D
\nSolution: Heat absorbed in solubility = Heat released from solution = \u0394 = + ( ) \u00d7 \u00d7 = m.s. T J 200 25 4 2 3 2835 . \u2234 \u0394 H J = + \u00d7 = + 2835 7 45 74 5 28350 . .
Answer: A
\nSolution: Heat released by reaction = Heat gained by ice = = \u00d7 = m.L cal 0 2 80 16 . \u0394 \u2212 = \u2212 \u00d7 \u2212 H= cal 16 10 16 10 3 3
Answer: B
\nSolution: let C 2 H 6 = x L, then CH 4 = (4 \u2013 x )L Volume of CO 2 produced, 2 x + (4 \u2013 x ) = 6 \u21d2 x = 2 \u2234 Total heat evolved = \u2212 \u00d7 + \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f \u00d7 \u00d7 = 1 2 1573 1 2 890 1 0 0821 300 50 . kJ
Answer: A
\nSolution: C O CO H cal + \u2192 \u0394 = \u2212 \u00d7 = \u2212 2 2 2400 0 3 12 96000 ; . C O CO H cal + \u2192 \u0394 = \u2212 \u00d7 = \u2212 1 2 1400 0 6 12 28000 2 ; . Now, CO O CO H cal + \u2192 \u0394 = \u2212 \u2212 ( ) \u2212 \u2212 ( ) = \u2212 1 2 96000 28000 68000 2 2 ; \u2234 Heat produced cal = \u00d7 = 68000 28 0 7 1700 .
Answer: B
\nSolution: Heat liberated from propane = Heat absorbed by water or, n n \u00d7 \u00d7 \u00d7 = \u00d7 \u00d7 \u00d7 \u21d2 = 500 10 40 100 160 10 1 50 40 3 3
Answer: B
\nSolution: q = v . i . t = 15 \u00d7 0.125 \u00d7 (14 \u00d7 60) J = 1575 J \u2234 \u0394 = \u00d7 = H J/mol 1575 0 1 1 15750 .
Answer: A
\nSolution: H g O g H O g H 240 kJ 2 2 2 1 1 2 ( ) + ( ) \u2192 ( ) \u0394 = \u2212 ; 2H g O g H O H 672 kJ ( ) + ( ) \u2192 ( ) \u0394 = \u2212 + ( ) = \u2212 1 2 240 432 2 2 2 g ; \u2234 \u0394 \u0394 = \u2212 \u2212 = H H 2 1 672 240 2 8 .
Answer: A
\nSolution: \u0394 = = \u00d7 \u00d7 \u00d7 ( ) = \u00d7 \u2212 m E C kg 2 3 8 2 12 103 10 4 2 3 10 4 8 10 . .
Answer: B
\nSolution: \u2212 = \u2212 ( ) + + ( ) \u23a1 \u23a3 \u23a4 \u23a6 \u2212 \u2212 ( ) + \u2212 ( ) + + \u2212 ( ) \u23a1 \u23a3 \u23a4 \u23a6 12 75 3 9 11 8 5 17 5 . . . x \u2234 X = \u2212 22 1 . kcal/mol
Answer: A
\nSolution: Dulong and petit\u2019s law : Atomic mass \u00d7 Specific heat \u2245 6.4 for greater temperature rise, heat lost should be high.\n5.38 Chapter 5 HINTS AND EXPLANATIONS
Answer: B
\nSolution: g O g CO g H O l ( ) + ( ) \u2192 ( ) + ( ) 9 2 3 3 2 2 2 \u0394 = \u00d7 \u0394 + \u00d7 \u0394 \u23a1 \u23a3 \u23a4 \u23a6 \u2212 \u0394 ( ) ( ) C cyclopropane f CO g f H O l f cyclopropan H H H H 3 3 2 2 e e f O H g + \u00d7 \u0394 ( ) \u23a1 \u23a3 \u23a2 \u23a4 \u23a6 \u23a5 9 2 2 = \u00d7 \u2212 ( ) + \u00d7 \u2212 ( ) \u23a1 \u23a3 \u23a4 \u23a6 \u2212 ( ) + ( ) + \u00d7 \u23a1 \u23a3 \u23a2 \u23a4 \u23a6 \u23a5 3 394 3 286 33 20 9 2 0 { } = \u2212 2093 kJ/mol
Answer: B
\nSolution: w = \u2212 = \u2212 = \u2212 \u00d7 \u00d7 = \u2212 P.V n RT cal H 2 1 5 2 298 894 .
Answer: C
\nSolution: w = \u2212 = \u2212 \u00d7 \u00d7 = \u2212 nRT cal 1 2 353 706 \u2234 \u0394 = + = + \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f = U kcal/mol q w 7 4 706 1000 6 694 . .
Answer: A
\nSolution: C H g +5O g 3CO g +4H O l 3 8 2 2 2 ( ) ( ) \u2192 ( ) ( ) \u0394 = \u00d7 \u0394 + \u00d7 \u0394 \u23a1 \u23a3 \u23a4 \u23a6 \u2212 \u0394 + \u00d7 \u0394 ( ) ( ) ( ) ( ) C C H f CO g f H O l f C H g f O g H H H H H 3 3 2 2 3 3 2 3 4 5 g ( ( ) \u23a1 \u23a3 \u23a4 \u23a6 = \u2212 ( ) + \u2212 ( ) \u23a1 \u23a3 \u23a4 \u23a6 \u2212 \u2212 ( ) = \u2212 3 393 5 4 285 8 103 8 2219 9 . . . . kJ \u0394 \u0394 \u0394 \u0394 \u0394 r required C C H g C CH g C C H g C H g H H H H H 2 6 4 3 8 2 = \u2212 + \u23a1 \u23a3 \u23a4 \u23a6 + + \u23a1 ( ) ( ) ( ) ( ) \u23a3 \u23a3 \u23a4 \u23a6 = \u2212 \u2212 ( ) + \u2212 ( ) \u23a1 \u23a3 \u23a4 \u23a6 + \u2212 ( ) + \u2212 ( ) \u23a1 \u23a3 \u23a4 \u23a6 = \u2212 1560 0 890 0 2219 9 285 8 55 7 . . . . . kJ J
Answer: B
\nSolution: The required thermochemical equation is K s Cl g KCl s ; H 2 ( ) + ( ) \u2192 ( ) = 1 2 \u0394 ? From (iv) + (iii) \u2013 (v) + (i) \u2013 (ii): we get, \u0394 H Kcal = \u2212 ( ) + \u2212 ( ) \u2212 ( ) + \u2212 ( ) \u2212 \u2212 ( ) = \u2212 116 5 39 3 4 4 13 7 68 4 105 5 . . . . . .
Answer: A
\nSolution: From 1 3 2 2 3 \u00d7 + \u00d7 + \u00d7 ( ) ( ) ( ) : i ii iii we get, \u0394 = \u2212 ( ) + ( ) + \u2212 ( ) = \u2212 H kJ Required 1 3 46 4 2 9 0 2 3 41 24 8 . . .
Answer: C
\nSolution: For the reaction, 1 2 1 2 2 2 H g I s HI g ( ) + ( ) \u2192 ( ) ; \u0394 = \u2212 ( ) \u2212 \u2212 ( ) + \u2212 ( ) \u2212 \u2212 ( ) + \u2212 ( H Required 1 2 44 20 1 2 52 42 17 31 19 21 13 74 . . . . . ) ) \u2212 \u2212 ( ) = 13 67 5 94 . . kcal
Answer: A
\nSolution: The required thermochemical equation is I s O g I O s 5 2 2 2 5 2 ( ) + ( ) \u2192 ( ) From 2 \u00d7 (ii) + 6 \u00d7 (v) + 5 \u00d7 (vii) \u2013 (i) \u2013 6 \u00d7 (iii) \u2013 6 \u00d7 (iv) \u2013 (vi) \u2013 10 \u00d7 (viii) \u2013 10 \u00d7 (ix), we get, \u0394 = \u2212 ( ) + \u2212 ( ) + \u2212 ( ) \u2212 ( ) \u2212 \u2212 ( ) \u2212 \u2212 ( ) \u2212 \u2212 H Required 2 322 6 100 5 255 4 0 6 44 6 57 22 . 4 4 10 92 10 75 169 ( ) \u2212 \u2212 ( ) \u2212 \u2212 ( ) = \u2212 kJ
Answer: B
\nSolution: Given thermochemical equations are (i) H g H g H kJ 2 ( ) \u2192 ( ) = 2 218 ; \u0394 (ii) Cl g 2Cl g H kJ 2 ( ) \u2192 ( ) = ; \u0394 124 (iii) 1 2 3 2 46 N g H g NH g H kJ 2 2 3 ( ) + ( ) \u2192 ( ) = \u2212 ; \u0394 (iv) 1 2 2 1 2 314 N g H g Cl g NH Cl s H kJ 2 2 2 4 ( ) + ( ) + ( ) \u2192 ( ) = \u2212 ; \u0394 (v) H g H g e H kJ ( ) \u2192 ( ) + = + \u2212 ; \u0394 1310 (vi) Cl g e Cl g H kJ ( ) + \u2192 ( ) = \u2212 \u2212 \u2212 ; \u0394 348 (vii) NH Cl s NH g Cl g H kJ 4 ( ) \u2192 ( ) + ( ) = + \u2212 4 683 ; \u0394 Required thermochemical equations are NH g H g NH g H 3 ( ) + ( ) \u2192 ( ) = + + 4 ; ? \u0394 From ( ) ( ) ( ) ( ) ( ) ( ) ( ) vii i v iii ii i v vi + \u2212 \u2212 \u2212 \u2212 \u2212 1 2 1 2 \u0394 = ( ) + \u2212 ( ) \u2212 \u2212 ( ) \u2212 \u00d7 ( ) \u2212 \u00d7 ( ) \u2212 ( ) \u2212 \u2212 H required 683 314 46 1 2 124 1 2 218 1310 348 8 718 ( ) = \u2212 kJ/mol\n5.39 Thermochemistry HINTS AND EXPLANATIONS
Answer: A
\nSolution: In such polymerization, one sigma bond is formed on cleavage of one pi bond. \u0394 H B.E. B.E. required C C bond C C bond = ( ) \u2212 ( ) = \u2212 ( ) \u2212 ( ) = \u2212 \u2212 \u2212 \u03c0 \u03c3 590 331 331 7 72 kJ/mole
Answer: D
\nSolution: Required thermochemical equation is C S 2H g O g CH OH l 2 2 3 ( ) + ( ) + ( ) \u2192 ( ) 1 2 \u0394 H kJ = + \u00d7 + [ ] \u2212 \u00d7 + + [ ] \u2212 = \u2212 715 4 218 249 3 415 356 463 38 266
Answer: B
\nSolution: 3 3 C s H g C H g H 53kJ 2 3 6 exp ( ) + ( ) \u2192 ( ) = ; \u0394 \u0394 H 3 715 6 218 3 356 6 408 63kJ theo = \u00d7 + \u00d7 [ ] \u2212 \u00d7 + \u00d7 [ ] = \u2212 \u2234 Strain energy H H kJ theo = \u2212 = \u0394 \u0394 exp 116
Answer: B
\nSolution: 2C g 6H g C H g 2 6 ( ) + ( ) \u2192 ( ) \u0394 H B.E B.E kJ C C C C = \u2212 = + [ ] \u2212 + \u00d7 [ ] \u21d2 = \u2212 \u2212 2839 0 0 6 412 367 and, 2C g 4H g C H g 2 4 ( ) + ( ) \u2192 ( ) \u0394 H B.E B.E. kJ C C C C = \u2212 = + [ ] \u2212 + \u00d7 [ ] \u21d2 = = = 2275 0 0 4 412 627 Now, 6C g 6H g C H g 6 6 ( ) + ( ) \u2192 ( ) \u0394 H R.E. = \u2212 = + [ ] \u2212 \u00d7 + \u00d7 + \u00d7 [ ] \u2212 5506 0 0 3 367 3 627 6 412 \u2234 R.E. kJ/mol = 52
Answer: A
\nSolution: C H S C H g S g C H S S C H g 2 5 2 5 2 5 2 5 \u2212 \u2212 ( ) + ( )\u2192 \u2212 \u2212 \u2212 ( ) \u0394 H kJ = \u2212 ( ) \u2212 \u2212 ( ) + \u23a1 \u23a3 \u23a4 \u23a6 = \u2212 202 143 222 276 B.E. kJ/mol s s \u2212 = 276
Answer: B
\nSolution: CH g CH g H g 4 3 ( ) \u2192 ( ) + ( ) 103 103 2 18 33 5 = + \u23a1 \u23a3 \u23a2 \u23a4 \u23a6 \u23a5 \u2212 \u2212 [ ] \u21d2 = ( ) ( ) \u0394 \u0394 f CH g f CH g H H kcal/mol 3 3 .
Answer: D
\nSolution: \u0394 H required = \u00d7 + \u00d7 + + \u00d7 [ ] \u2212 \u00d7 + + + + + 6 414 2 348 580 2 610 3 414 348 580 354 462 11 18 2 580 140 2 462 + \u00d7 + + \u00d7 \u23a1 \u23a3 \u23a2 \u23a4 \u23a6 \u23a5 = \u2212348 \u039a J
Answer: D
\nSolution: \u0394 H kJ = \u2212 = \u2212 50 70 20
Answer: A
\nSolution: \u0394 H B.E. B.E. bond in C C bond inC C = ( ) \u2212 ( ) = \u2212 ( ) \u2212 ( ) = \u2212 = \u2212 \u03c0 \u03c3 835 610 348 123 k kJ
Answer: B
\nSolution: n HCHO g HCHO 2 ( ) \u2192 ( ) \u0394 H n n = \u2212 = \u00d7 \u2212 ( ) \u2212 \u2212 ( ) \u21d2 = 72 134 732 6 \u2234 Molecular formula HCHO C H O 6 6 12 6 = ( ) =
Answer: B
\nSolution: There is 2.5 B-B bond per B atom. Hence, \u0394 H = \u20132.5 \u00d7 300 = \u2013750 kJ/mole of Boron.
Answer: C
\nSolution: Let the enthalpy of combustion of gauche form be \u2013 x kcal/mol. Now, 690 0 7 2 0 2 0 06 3 0 04 5 5 = \u00d7 \u2212 ( ) + \u00d7 + \u00d7 + ( ) + \u00d7 + ( ) . . . . . x x x x \u2234 x = 691
Answer: A
\nSolution: M s X g MX s H 1.5 H eg g ( ) + ( ) \u2192 ( ) \u0394 = \u00d7 \u0394 ( ) 2 2 ; x From Born\u2013Hafer Cycle, we get: \u0394 = \u0394 + \u0394 + \u0394 + \u0394 + \u00d7 \u0394 + \u0394 ( ) ( ) ( ) ( ) ( ) H H H H H H sub M s i M g i M g Bond X g eg X g lat 1 2 2 2 l lice MX s H 2 ( ) or, 1 5 96 1 2 2 8 0 8 1 2 . . . . . \u00d7 \u2212 ( ) = \u0394 + \u00d7 \u0394 + \u00d7 \u0394 + \u00d7 \u00d7 \u0394 ( ) ( ) ( ) sub M s sub M s sub M s s H H H u ub M s sub M s H H ( ) ( ) ( ) + \u2212 ( ) + \u00d7 \u2212 \u00d7 \u00d7 \u0394 \u23a1 \u23a3 \u23a4 \u23a6 2 96 5 0 8 1 2 . . \u0394 = ( ) sub M s H kcal/mol 41 38 .\n5.40 Chapter 5 HINTS AND EXPLANATIONS
Answer: B, C, D
\nSolution: Combustion is exothermic. Decomposition or elimination are endothermic. Graphite is more stable form.
Answer: A, B, C
\nSolution: Conversion of liquid into gas is endothermic.
Answer: A, D
\nSolution: \u0394 \u00b0 = f H 0 for elements in their reference state.
Answer: A
\nSolution: Endothermic compounds have +ve \u0394 \u00b0 f H .
Answer: B
\nSolution: For \u0394 = \u0394 \u0394 = H E, n g 0
Answer: A
\nSolution: One mole of the substance should burn completely.
Answer: A, B
\nSolution: \u0394 = + ( ) f NO g H ve
Answer: A, B, D
\nSolution: Heat released in reaction = Heat gained by calorimeter system = \u00d7 = 1 5 1 4 2 1 . . . kJ n H SO eq 2 4 100 0 5 1000 0 05 ( ) = \u00d7 = . . n NH OH Limiting reagent eq 4 200 0 2 1000 0 04 ( ) = \u00d7 = . . ( ) \u0394 neut NH OH H By strong acid kJ/eq kJ/mole 4 ( ) . . . . = \u2212 = \u2212 = \u2212 2 1 0 04 52 5 52 5 \u0394 = \u2212 ( ) \u2212 \u2212 ( ) = diss NH OH H kJ/mol 4 52 5 57 4 5 . . \u0394 = \u2212 ( ) \u2212 = diss CH COOH H kJ/mol 3 57 48 1 4 5 4 4 . . .
Answer: A, D
\nSolution: (a) \u0394 = + \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f \u2212 \u2212 = r H kJ 436 1 2 495 242 925 5 ( ) . (b) \u0394 = \u00d7 + \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f \u2212 \u2212 = r H kJ 1 2 436 1 2 495 42 423 5 ( ) . (c) \u0394 = \u00d7 = f H(g) H kJ mol 1 2 436 218 / (d) \u0394 = f OH(g) H kJ/mol 42
Answer: B, D
\nSolution: Resonance occurs in 1, 3-Butadiene and N 2 O.
Answer: A, B, C
\nSolution: Resonance occurs in product but not in reactant.
Answer: A, B
\nSolution: (a) \u0394 = \u00d7 \u2212 ( ) \u23a1 \u23a3 \u23a4 \u23a6 \u2212 \u2212 ( ) + \u2212 ( ) \u23a1 \u23a3 \u23a4 \u23a6 = \u2212 r H kJ 2 1263 2238 285 3 (b) \u2212 = \u0394 \u2212 \u00d7 \u0394 3KJ H H C -maltose C glucose \u03b1 2
Answer: A, B, C
\nSolution: (a) n C s ( ) = \u00d7 = 1 2 1000 12 100 . \u2234 Maximum obtainable heat = 100 \u00d7 94 = 9400 cal (b) Heat released = \u00d7 + \u00d7 = 100 68 100 68 13600 cal (c) Heat released = \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f = 13600 100 1200 30 5440 cal
Answer: A, D
\nSolution: C H COOH s O g CO g H O l 2 6 5 2 2 15 2 7 3 ( ) + ( ) \u2192 ( ) + ( ) \u0394 = \u2212 ( ) + \u2212 ( ) \u23a1 \u23a3 \u23a4 \u23a6 \u2212 \u2212 [ ] = \u2212 H kJ/mol 7 393 3 286 408 3201 and \u0394 = \u0394 \u2212 \u0394 = \u2212 ( ) \u2212 \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f \u00d7 \u00d7 = \u2212 U H n RT kJ/ g . . . 3201 7 15 2 8 314 1000 300 3199 75 m mol
Answer: C
\nSolution: \u0394 H kcal = = \u00d7 \u2212 ( ) = \u2212 q 3 35 105 \u0394 \u0394 \u0394 U H n RT kcal g = \u2212 = \u2212 ( ) \u2212 \u2212 ( ) \u00d7 \u00d7 \u23a1 \u23a3 \u23a2 \u23a4 \u23a6 \u23a5 \u00d7 = \u2212 . . 35 2 3 2 1000 300 3 103 2 and w u q = \u2212 = \u2212 ( ) \u2212 ( ) = 103 2 105 1 8 . . kcal
Answer: C
\nSolution: From \u2212 \u00d7 \u2212 \u00d7 + \u00d7 \u2212 \u00d7 3 4 1 4 1 4 9 4 a b c d , \u0394 = \u2212 \u00d7 \u2212 \u2212 \u00d7 + \u00d7 \u2212 = \u2212 H kcal required 3 4 76 1 4 240 1 4 36 9 4 68 147 ( ) ( ) ( ) ( )
Answer: C
\nSolution: From 3 4 1 4 1 4 1 4 \u00d7 + \u00d7 \u2212 \u00d7 + \u00d7 a bc d , \u0394 = \u00d7 \u2212 + \u00d7 \u2212 \u00d7 + = H kcal/mol required 3 4 76 1 4 240 1 4 36 1 4 68 11 ( ) ( ) ( ) ( )
Answer: C
\nSolution: Given data based\n5.41 Thermochemistry HINTS AND EXPLANATIONS Comprehension II
Answer: C
\nSolution: \u0394 = \u2212 \u00d7 + \u22c5 = \u2212 \u22c5 H kJ 75 5 5 1 8 55 15
Answer: D
\nSolution: \u0394 = \u2212 \u00d7 + \u22c5 = \u2212 \u22c5 H kJ 75 10 10 1 8 63 56
Answer: C
\nSolution: \u0394 = \u2212 + \u22c5 = \u2212 + \u22c5 \u221e = \u2212 H n kJ 75 1 1 8 75 1 1 8 75
Answer: B
\nSolution: \u0394 = \u2212 \u2212 \u2212 = \u2212 H kJ ( . ) ( . ) . 63 56 55 15 8 41
Answer: B
\nSolution: \u0394 = \u2212 \u2212 \u2212 = \u2212 H kJ ( ) ( . ) . 75 63 56 11 44 Comprehension III
Answer: D
\nSolution: \u0394 = \u2212 \u22c5 = \u2212 \u22c5 H kJ/mol H 1 2 483 636 2 241 818 \u0394 = \u2212 \u22c5 = \u2212 \u22c5 H kJ/mol H 2 2 868 2 3 289 4 \u0394 = \u2212 H kJ/mol H 3 2 347 33 .
Answer: B
\nSolution: \u0394 = \u2212 \u22c5 = \u2212 \u22c5 H kJ/gm O 1 2 483 636 32 15 11 \u0394 = \u2212 \u22c5 = \u2212 \u22c5 H kJ/gm O 2 3 868 2 48 18 09 \u0394 = \u2212 \u22c5 = \u2212 \u22c5 H kJ/gm H O 3 2 2 347 33 34 10 22
Answer: D
\nSolution: \u0394 = \u2212 \u22c5 = \u2212 \u22c5 H kJ/gm reactant 1 483 636 36 13 43 \u0394 = \u2212 \u22c5 = \u2212 \u22c5 H kJ/gm reactant 2 868 2 54 16 08 \u0394 = \u2212 \u22c5 = \u2212 \u22c5 H kJ/gm reactant 3 347 33 36 9 65 Comprehension IV
Answer: D
\nSolution: Heat released = 0.25 \u00d7 320 = 80 cal \u2234 Molar enthalpy of solution = \u2212 \u22c5 \u00d7 = \u2212 80 0 98 98 8000 cal
Answer: C
\nSolution: Heat released = 0.80 \u00d7 320 = 256 cal \u2234 \u0394 \u2212 \u22c5 \u00d7 = \u2212 r H = cal 256 0 49 98 51200 Comprehension V
Answer: D
\nSolution: 2 3 2 1 2 2 2 3 C(S)+ H g N g CH CN(g); ( ) ( ) + \u2192 \u0394 H B.E. B.E. C C C N = = \u00d7 + \u00d7 + \u00d7 \u2212 \u00d7 + + \u2192 \u2212 \u2261 88 2 719 3 2 435 1 2 948 3 414 1 [ ] [ ] 3 2 3 8 C(s) 4H g C H g + \u2192 ( ) ( ) \u0394 = \u2212 = \u00d7 + \u00d7 \u2212 \u00d7 + \u00d7 \u2192 \u2212 H B.E. C C 85 3 719 4 435 2 8 414 2 [ ] [ ] From (1) and (2), we get: B.E. kJ/mol and B.E. kJ/mol C C C N \u2212 \u2261 = = 335 899 5 . Now, CH CN(g) 2H g CH CH NH (g), 3 2 3 2 2 + \u2192 ( ) \u0394 = \u00d7 + + + \u00d7 \u2212 \u00d7 + + + \u00d7 H [ . ] [ ] 3 414 335 899 5 2 435 5 414 335 378 2 426 = \u2212 288 5 . kJ/mol\n5.42 Chapter 5 HINTS AND EXPLANATIONS Comprehension VI
Answer: B
\nSolution: C < C p,m,N (g) p,m,H O(g) 2 2
Answer: A
\nSolution: H g O g H O g 2 2 2 1 2 ( ) ( ) ( ); + \u2192 \u0394 = \u2212 \u22c5 \u0394 = \u2212 \u22c5 H kcal U kcal 55 85 56 0 Let x mole H 2 be burnt. x 2 mol O (g) 2 is needed and hence, x 2 4 2 \u00d7 = \u00d7 mol N 2 is also present. Now, Heat released from reaction = Heat gained by H O(g) 2 and N g 2 ( ) 56.0 \u00d7 10 3 = x \u00d7 6.2 \u00d7 ( T 2 \u2013 300) + 2 x \u00d7 4.9 \u00d7 ( T 2 \u2013 300) \u2234 T 2 = 3800 K
Answer: C
\nSolution: p n T p n T x x x p x x 1 1 1 2 2 2 2 1 2 2 300 2 3800 = \u21d2 + + \u239b \u239d \u239c \u239e \u23a0 \u239f \u00d7 = + \u00d7 ( ) \u2234 p 2 = 10.86 atm
Answer: B
\nSolution: q = 0, w = 0 \u21d2 \u2206 E = 0 Comprehension VII
Answer: B
\nSolution: C H l O g CO g H O(g) 18 8 2 2 2 25 2 8 9 ( ) ( ) ( ) + \u2192 + \u2206 c H = [8 \u00d7 (\u221294) + 9 \u00d7 (\u221258)] \u2212 [\u221274] = \u22121200 kcal/mol
Answer: C
\nSolution: \u2206 H required = [8 \u00d7 (\u221226.5) + 9 \u00d7 (\u221258)] \u2212 [\u221274] = \u2212660 kcal/mol
Answer: D
\nSolution: Let x mole of C 8 H 18 be converted into CO 2 . As temperature is increased, some heat is absorbed by product gases. Now, [ ( ) ] [ ( ) x x x x \u00d7 + \u22c5 \u2212 \u00d7 \u2212 \u00d7 \u00d7 \u00d7 + \u22c5 \u2212 \u00d7 \u22c5 \u00d7 + \u22c5 \u00d7 1200 0 1 660 1 1000 8 8 500 8 0 1 7 0 500 0 9 6 6 0 500 87 3 \u22c5 \u00d7 = \u22c5 ] \u2234 x = 0.05 Moles of CO 2 formed = 0.05 \u00d7 8 = 0.4
Answer: C
\nSolution: Moles of H 2 O formed = 9 x + (0.1\u2212 x ) \u00d7 9 = 0.9
Answer: D
\nSolution: w p v v p n RT p n RT p R n T n T = \u2212 \u2212 = \u2212 \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f = \u2212 \u2212 ( ) ( ) 2 1 2 2 1 1 2 2 1 1 = \u2212 \u00d7 \u22c5 \u00d7 + \u22c5 \u2212 \u22c5 \u00d7 + \u22c5 \u00d7 \u2212 \u22c5 \u00d7 + \u22c5 \u2212 \u22c5 \u00d7 2 0 05 8 0 1 0 05 8 0 9 800 0 05 25 2 0 1 0 05 17 2 [{ ( ) } ( ) { { } \u00d7 \u2212 300 2090 ] = cal Comprehension VIII
Answer: A
\nSolution: CO 2H CH OH rd mol + \u23af \u2192 \u23af\u23af 2 2 3 3 1000 = \u00d7 = 3 2 1000 1500 mol In reformer, CO and H 2 is forming in 1 : 3 ratio.
Answer: C
\nSolution: CO = 1500 \u22121000 = 500 mole H 2 = 4500 \u2212 2000 = 2500 mole
Answer: B
\nSolution: Heat produced in 1 min = 1000 \u00d7 100 R \u00d7 60 = 1.2 \u00d7 10 7 cal\n5.43 Thermochemistry HINTS AND EXPLANATIONS Comprehension IX
Answer: C
\nSolution: Let x mole C be converted into CO. Hence, x \u00d7 26 + (1 \u2013 x ) \u00d7 94 = 53.2 \u21d2 x = 0.6 Hence, moles of C formed = 0.6
Answer: A
\nSolution: O consumed +(1 ) ]32 gm 2 2 22 4 = \u2212 \u23a1 \u23a3 \u23a2 \u23a4 \u23a6 \u23a5 = \u22c5 x x Comprehension X Given thermochemical equations are (a) H 2 S (g) \u2192 H (g) + H S (g); \u2206 H = 376.0 kcal (b) H 2 (g) + S(s) \u2192 H 2 S (g); \u2206 H = \u221220.0 kcal (c) S (s) \u2192 S (g); \u2206 H = 277.0 kcal (d) H 2 (g) \u2192 2H (g); \u2206 H = 436.0 kcal
Answer: A
\nSolution: 1 2 2 H S(s) HS(g) ( ) g + \u2192 From a b d we get: + \u2212 \u00d7 1 2 , \u0394 = + \u2212 \u2212 \u00d7 = H kJ mol required 376 20 1 2 436 138 ( ) /
Answer: A
\nSolution: HS(g) \u2192 H(g) + S(g) From (d) \u2212 (a) \u2212 (b) + (c) \u2206 H required = 436 \u2212 376 \u2212 (\u221220) + 277 = 357 kJ/mol Comprehension XI
Answer: B
\nSolution: \u0394 \u00b0 = \u0394 \u00b0 + \u0394 \u22c5 = + \u00d7 \u00d7 = H E n RT kcal g 2 1 2 2 1000 298 3 292 . . Now, \u0394 \u00b0 = \u0394 \u00b0 \u2212 \u22c5 \u0394 \u00b0 = \u2212 \u00d7 = \u2212 G H T S kcal 3 292 298 1000 20 2 668 . .
Answer: A
\nSolution: Spontaneous as \u2206 G\u00b0 = \u2212ve
Answer: A
" } }, { "question_id": "thermochemistry-chem-sec-3-34-82", "marks": 4.0, "negMarks": 1.0, "partialMarks": 1, "subject": "chemistry", "chapter": "thermochemistry", "chapterTitle": "Thermochemistry", "type": "mcq", "rawChapterType": "MCQ", "originalNumber": 82, "displayNumber": 34, "image": "./assets/chapter-packs/thermochemistry/Chemistry Section 1__--__82__--__1.png", "solutionImage": null, "question": { "content": "Answer: C
" } } ] }, { "title": "Chem Sec 4", "originalName": "Section D - Assertion Reason", "questions": [ { "question_id": "thermochemistry-chem-sec-4-1-83", "marks": 4.0, "negMarks": 1.0, "partialMarks": 1, "subject": "chemistry", "chapter": "thermochemistry", "chapterTitle": "Thermochemistry", "type": "mcq", "rawChapterType": "MCQ", "originalNumber": 83, "displayNumber": 1, "image": "./assets/chapter-packs/thermochemistry/Chemistry Section 1__--__83__--__1.png", "solutionImage": null, "question": { "content": "Answer: A
\nSolution: Theory based
Answer: A
\nSolution: Theory based
Answer: D
\nSolution: H 2 SO 4 is dibasic but HCl is monobasic.
Answer: D
\nSolution: Information based
Answer: D
\nSolution: Heat liberated will be four times but as quantity is also four times, the change in the temperature will be same.
Answer: C
\nSolution: Solubility is exothermic but all gases are not highly soluble in all liquid.
Answer: C
\nSolution: If | \u2206 Hydration H| < | \u2206 lattice H|, the salt dissolves partially and the extent depends on the difference in two values.
Answer: A
\nSolution: H HDiamond + O2 Hgraphite + O2 HCO2
Answer: A
\nSolution: H H1 \u2013 Extent + O2 H2 \u2013 Extent + O2 HCO2 + H2O
Answer: C
\nSolution: Theory based\n5.44 Chapter 5 HINTS AND EXPLANATIONS
Answer: A \u2192 R; B \u2192 P; C \u2192 Q; D \u2192 P, R, S
\nSolution: C \u2192 CO, \u2206 H \u2260 \u2206 H combustion
Answer: A \u2192 Q; B \u2192 S; C \u2192 P; D \u2192 R
\nSolution: (A) \u2206 n g = 0 (B) \u2206 n g = \u22121 (C) \u2206 n g = 1 (D) \u2206 n g = \u2212 2
Answer: A \u2192 P, B \u2192 Q; C \u2192 R; D \u2192 S
\nSolution: (A) \u2206 H = (\u221257.3) + 15 = \u221242.3 kJ (B) \u2206 H = \u221242.3 \u2212 70.7 + 20 = \u221293.0 kJ (C) \u2206 H = \u221270.7 + 15 = \u221255.7 kJ (D) \u2206 H = 0
Answer: A \u2192 Q; B \u2192 R; C \u2192 P; D \u2192 S
\nSolution: (A) Mg (s) + Cl2 (g) \u2013(1110 + 790) (2360 \u2013 1110) Mg 2+ (aq) + 2Cl \u2013 (aq) Mg 2+ (g) + 2Cl \u2013 (aq) + 1110 Mg 2f (aq); +Cl2 (g) or = 1110 \u2013 (1110 + 790) + (2360 \u2013 1110) = 460 kJ/mol (B) 1 2 1 2 1110 2360 1 2 1 2 652 2 2 Cl g Cl ag H Mg g Mg g 2 ( ) ( ); ( ) ( ) ( ) \u2192 \u0394 = \u2212 + + = \u2212 \u2212 + + K KJ/mol (C) Mg 2+ (g) +2Cl \u2212 (aq) \u2192 Mg 2+ (aq) +2Cl \u2212 (aq); \u2206 H = \u2212790 \u2212 1110 = \u22121900 kJ (D) Mg 2+ (g) + 2Cl \u2212 (g) \u2192 MgCl 2 (s); \u2206 H = \u2212 640 \u2212 1870 = \u2212 2510 kJ
Answer: A \u2192 P, Q; B \u2192 P, R; C \u2192 S
\nSolution: Defi nition based
Answer: A \u2192 Q; B \u2192 R; C \u2192 P; D \u2192 S
\nSolution: 3O 2 (g) \u2192 2O 2 (g)
Answer: A \u2192 S; B \u2192 Q; C \u2192 P; D \u2192 R
\nSolution: Theory based
Answer: A \u2192 P; B \u2192 Q, T; C \u2192 Q, S; D \u2192 R, S
\nSolution: \u2206 n g = 0 \u21d2 \u2206 H = \u2206 U \u2206 n g = + ve \u21d2 \u2206 H > \u2206 U \u21d2 If \u2206 H = \u2212ve, then | \u2206 H| < | \u2206 U| \u21d2 If \u2206 H = +ve, then | \u2206 H| > | \u2206 U| \u2206 n g = \u2212 ve \u21d2 \u2206 H < \u2206 U \u21d2 If \u2206 H = \u2212ve, then | \u2206 H| > | \u2206 U| \u21d2 If \u2206 H = +ve, then | \u2206 H| < | \u2206 U|
Answer: A \u2192 P; B \u2192 P, Q; C \u2192 R, S; D \u2192 R
\nSolution: Defi nition based (10) (A) \u0394 = \u0394 + \u2212 \u00d7 \u2212 = + \u2212 \u00d7 \u00d7 H H c c T T p,A(g p,A(l 400 300 2 1 25 20 40 1 1000 40 [ ] [ ] ( ) ( ) ) 0 0 300 23 \u2212 = + ) kJ/mol (B) \u0394 = \u0394 + \u2212 \u00d7 \u2212 = + \u2212 \u00d7 \u00d7 H H c c T T p,A (g p,A (l 300 400 2 1 3 3 50 30 50 1 1000 [ ] [ ] ( ) ( ) ) 3 300 400 52 \u2212 = + ) kJ/mol (C) \u2206 H 300 = 3 \u00d7 25 \u2212 100 \u2212 52 = \u2212 77 kJ/mol (D) \u0394 = \u0394 + \u2212 \u00d7 \u00d7 \u2212 = \u2212 + \u2212 \u00d7 \u00d7 H H c c T T p,A (l p,A (l 400 300 2 1 3 3 3 77 50 3 40 [ ] [ ] ( ) ( ) ) ) 1 1 1000 400 300 84 \u00d7 \u2212 = \u2212 ( ) kcal/mol
Answer: A \u2192 Q; B \u2192 R; C \u2192 S; D \u2192 P
" } } ] }, { "title": "Chem Sec 6", "originalName": "Section F - Integer", "questions": [ { "question_id": "thermochemistry-chem-sec-6-1-103", "marks": 4.0, "negMarks": 1.0, "partialMarks": null, "subject": "chemistry", "chapter": "thermochemistry", "chapterTitle": "Thermochemistry", "type": "sa", "rawChapterType": "NAT", "originalNumber": 103, "displayNumber": 1, "image": "./assets/chapter-packs/thermochemistry/Chemistry Section 1__--__103__--__1.png", "solutionImage": null, "question": { "content": "Answer: 3
\nSolution: For HCl : 13.7 \u00d7 0.05 = c \u00d7 411 (1) For HCOOH : q \u00d7 0.05 = c \u00d7 321 (2) From (2) \u00f7 (1) \u21d2 q = 10.7 kcal \u2234 Enthalpy of ionisation of HCOOH = 13.7 \u2212 10.7 = 3.0 kcal/mol
Answer: 1
\nSolution: (C 6 H 10 O 5 ) x + 6 x O 2 (g) \u2192 6 x CO 2 (g) + 5 x H 2 O(l) \u2206 H = \u2212 4.6 \u00d7 162 x = [6x( \u221294.2) + 5 x (\u221268.4)] \u2212 \u0394 f C H O H 6 10 5 ( ) x + \u23a1 \u23a3 \u23a4 \u23a6 0 \u2234 \u0394 f C H O H 6 10 5 ( ) x = \u2212162 kcal/mol = \u22121kcal/gm
Answer: 5
\nSolution: 1 800 3120 800 3120 \u00d7 = \u00d7 \u21d2 = a a L/hr Butane C H O CO H O 2 4 10 2 2 13 2 4 5 + \u2192 + \u2234 Rate of Oxygen Supply L/hr = \u00d7 \u00d7 = 800 3120 13 2 3 5
Answer: 2
\nSolution: \u0394 = \u00d7 \u2212 = H kcal/mol required 100 75 13 7 12 2 2 ( . . )\n5.45 Thermochemistry HINTS AND EXPLANATIONS
Answer: 5
\nSolution: Total moles of gases = \u00d7 \u00d7 = 1 192 1 642 0 0821 298 0 08 . . . . Now, n n CH CH 4 4 210 10 1260 0 667 0 004 3 \u00d7 \u00d7 = \u00d7 \u21d2 = . . \u2234 Volume per cent of CH 4 = \u22c5 \u22c5 \u00d7 = 0 004 0 08 100 5%
Answer: 8
\nSolution: Heat released by 6 3 64000 \u22c5 mole haemoglobin = 25 4.2 J \u00d7 \u00d7 = 0 03 3 15 . . \u2234 Heat released per mole haemoglobin = \u22c5 \u00d7 \u22c5 = 3 15 64000 6 3 32000 J \u2234 Heat released per mole O 2 = = 32000 4 8000 J
Answer: 6
\nSolution: Heat released = \u00d7 \u00d7 \u00d7 \u2212 = 300 1 0 1 0 26 25 300 . . ( ) cal Now, n HA = \u00d7 \u22c5 = \u22c5 200 0 4 1000 0 08 n NaOH = \u00d7 \u22c5 = \u22c5 100 0 5 1000 0 05 Hence, NaOH is a limiting reagent. \u2234 \u0394 = \u2212 \u22c5 \u00d7 = \u2212 neut H cal/mol 300 0 05 1 6000
Answer: 5
\nSolution: There is 3 H-bond per NH 3 molecule because for each bond two NH 3 molecules are required. \u2234 Strength of H-bond = \u2212 = 30 4 15 4 3 5 0 . . . kcal/mol
Answer: 6
\nSolution: C(s)+ O g CO(g); H kcal/mol 2 1 1 2 7 5 3 12 30 ( ) \u2192 \u0394 = \u2212 \u22c5 \u00d7 = \u2212 C(s)+O g CO (g); H kcal/mol 2 2 ( ) \u2192 \u0394 = \u2212 \u00d7 = \u2212 2 32 4 12 96 Now, CO (g) CO(g)+ O g H H H kcal/mol 2 1 2 2 1 2 66 \u2192 \u0394 = \u0394 \u2212 \u0394 = + ( ); For 4 gm CO H kcal 2 \u22c5 \u0394 = \u00d7 = 0 66 44 4 6 ,
Answer: 3
\nSolution: ( ) 1 6900 3 4 \u2212 \u00d7 + \u00d7 = \u21d2 = a a a 2900 3900 \u2234 n n eq(HA) eq(HB) : ( ) : : : = \u2212 = = 1 1 4 3 4 1 3 a a Four-digit Integer Type
Answer: 0043
\nSolution: C 2 H 6 + H 2 \u2192 2 CH 4 ; \u2206 H = \u221265.2 kJ C 3 H 8 + 2H 2 \u2192 3 CH 4 ; \u2206 H = \u221287.4 kJ Hence, for CH 4 (g) + C 3 H 8 (g) \u2192 2 C 2 H 6 (g); \u2206 H = (\u221287.4) \u22122 \u00d7 (\u221265.2) = + 43 kJ
Answer: 0400
\nSolution: Moles of O 2 consumed = \u00d7 \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f \u00d7 \u00d7 \u23a7 \u23a8 \u23a9 \u23ab \u23ac \u23ad \u00d7 \u22c5 \u00d7 = 164 2 1000 20 10 100 20 60 1 0 0821 310 24 31 . C H O O CO HO H = kJ 6 2 2 2 12 6 6 6 6 3100 + \u2192 + \u0394 \u2212 ; \u2234 Heat produced in body per hr = \u00d7 = 3100 6 24 31 400 kJ
Answer: 0030
\nSolution: Number of glycogen units oxidized per day = \u00d7 \u00d7 \u00d7 \u00d7 = 150 60 60 24 432 10 30 3
Answer: 0085
\nSolution: Moles of C = = 15 12 1 25 . Moles of O 2 = \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f \u00d7 \u22c5 \u00d7 = 20 19 100 8 21 0 0821 380 1 . 1 25 0 5 2 2 . . C + O CO +0.75 CO \u2192 \u2234 Heat produced = \u00d7 + \u00d7 = 0 5 26 0 75 96 85 . . kcal\n5.46 Chapter 5 HINTS AND EXPLANATIONS
Answer: 2400
\nSolution: C 2 H 5 OH(l) + O 2 (g) \u2192 CH 3 COOH(g) + H 2 O(l) \u2206 H = [(\u2212118)+(\u221268)] \u2212 [(\u221266)+0] = \u2212120 kcal Hence, rate of heat removal = \u00d7 \u00d7 \u00d7 = 120 46 2 3 10 40 100 2400 3 . kcal/mol
Answer: 0216
\nSolution: C 6 H 12 O 6 (s) + 6 O 2 (g) \u2192 6 CO 2 (g) + 6 H 2 O(l) \u2206 H = [6 \u00d7 (\u2212395) + 6 \u00d7 (\u2212285)] \u2212 [(\u22121280) + 0] = \u22122800 kJ Moles of CO 2 released per astronaut = \u00d7 = 6 2800 2100 4 5 . \u2234 Mass of LiOH required = 4.5 \u00d7 2 \u00d7 24 = 216 gm
Answer: 0500
\nSolution: 16 1 322 100 10 500 3 3 . . \u00d7 \u00d7 = \u00d7 \u21d2 = = V 10000 V 0 5m L
Answer: 0120
\nSolution: Total heat absorbed = \u00d7 + \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f \u00d7 = 8 9 45 1 9 72 2 5 120 . KJ
Answer: 0075
\nSolution: For banana: q = c \u00d7 3.0 (1) For benzoic acid: 800 122 0 305 0 \u00d7 = \u00d7 . . c 4 (2) From (1) and (2), q = 1.5 kacl for 2.5 gm banana \u2234 Heat obtained per banana = \u22c5 \u22c5 \u00d7 = 1 5 2 5 125 75 kcal
Answer: 0241
\nSolution: H O H O(l); H kJ 2 2 2 298 1 2 286 ( ) ( ) g g + \u2192 \u0394 = \u2212 H O(l) H O(g); H kJ 2 2 \u2192 \u0394 = \u22c5 398 40 8 \u0394 = \u0394 + \u0394 \u22c5 \u0394 = \u2212 \u00d7 \u2212 = H H C T 40.8+ kJ 398 p 298 33 4 75 4 1000 298 398 45 . . ( ) \u2234 H O(g) H O H 2 2 \u2192 + \u0394 = \u2212 \u2212 2 298 1 2 45 286 ( ) ( ); [ ] g g = 241 kJ
Answer: 0700
\nSolution: \u0394 \u2212 \u0394 = \u0394 \u22c5 \u222b H H C dT p T T 1 2 1 2 or, 0 T dT T T \u2212 \u2212 = \u00d7 = \u00d7 \u2212 \u2212 \u2212 \u222b ( ) ( ) ( ) 4000 2 10 2 10 2 300 2 300 2 2 2 \u2234 T K = 700
Answer: 0120
\nSolution: 3C(s) + 3H 2 (g) \u2192 C 3 H 6 (g) \u2206 H theo = (3 \u00d7 715 + 6 \u00d7 218) \u2212 (3 \u00d7 356 + 6 \u00d7 408) = \u221263 kJ \u2206 H exp = [3 \u00d7 (\u2212393) +3 \u00d7 (\u2212285)] \u2212 [3 \u00d7 (\u2212697)] = 57 kJ \u2234 Strain energy = 57 \u2212 (\u221263) = 120 kJ/mol
Answer: 0060
\nSolution: KF.CH COOH s K g F CH COOH g H 734 kJ 3 3 ( ) \u2192 ( ) + ( ) \u0394 = + \u2212 . ; CH COOH l CH COOH g H 20 kJ 3 3 ( ) \u2192 ( ) \u0394 = ; KF s K aq F aq H kJ ( ) \u2192 ( ) + ( ) \u0394 = + \u2212 ; 35 K g K aq H kJ + + ( ) \u2192 ( ) \u0394 = \u2212 ; 325 F g F aq H kJ \u2212 \u2212 ( ) \u2192 ( ) \u0394 = \u2212 ; 389 KF s CH COOH l KF CH COOH s H kJ ( ) + ( ) \u2192 ( ) \u0394 = \u2212 3 1 3 25 ; Required: F g CH COOH g F CH COOH g \u2212 \u2212 ( ) + ( ) \u2192 ( ) 3 3 ; \u0394 = \u2212 ( ) + \u2212 \u2212 + \u2212 ( ) + \u2212 ( ) = \u2212 H kJ/mol 389 734 20 35 325 25 60
Answer: 0455
\nSolution: B s H BH g ( ) + ( ) \u2192 ( ) 3 2 2 3 g \u0394 = = + \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f \u2212 \u00d7 ( ) \u2212 H B E. B H 100 565 3 2 436 3 . \u2234 B E. kJ/mol B H . \u2212 = 373 2B(s) + 3H g B H g 2 ( ) ( ) \u2192 2 6 \u0394 = = \u00d7 + \u00d7 [ ] \u2212 \u00d7 + \u00d7 [ ] \u2212 H B E. 3c 2e 36 2 565 3 436 4 373 2 . \u2234 B E. kJ/mol 3c 2e . \u2212 = 455\n5.47 Thermochemistry HINTS AND EXPLANATIONS
Answer: 0292
\nSolution: XeF Xe F F F H kcal 4 + \u2192 + + + \u0394 = \u00d7 ( ) + + \u2212 ( ) + \u2212 ( ) = \u2212 2 4 34 279 85 38 292
Answer: 0021
\nSolution: (a) KF CH COOH s K ACOH F ACOH CH COOH l kJ . . 3 3 3 ( ) \u2192 ( ) + ( ) + ( ) = + \u2212 (b) KF s K ACOH F ACOH H 35 kJ ( ) \u2192 ( ) + ( ) \u0394 = + \u2212 ; (c) F CH COOH g F g CH COOH g H 46 kJ . ; \u2212 \u2212 ( ) \u2192 ( ) + ( ) \u0394 = 3 3 (d) KF CH COOH s K g F CH COOH g H 734 kJ . . ; 3 3 ( ) \u2192 ( ) + ( ) \u0394 = + \u2212 (e) KF s K g F g H 797 kJ ( ) \u2192 ( ) + ( ) \u0394 = + \u2212 ; Required: CH COOH l CH COOH g 3 3 ( ) \u2192 ( ) From (c) (a)+(d) e b \u2212 \u2212 + ( ) ( ), we get: \u0394 = \u2212 \u2212 ( ) + \u2212 + = H 46 kJ/mol 3 734 797 35 21
Answer: 0085
\nSolution: (l) + 3H2 (g) (g) N NH \u2206 H = (\u221250) \u2212 [ \u2206 f H py(l) + 0] = (40 + 125) + [2 \u00d7 {(\u2212156) \u2212 (\u221237)} + {(\u221218) \u2212 44}] \u2234 \u2206 f H py(l) = 85 kJ/mol
Answer: 0272
\nSolution: 1 2 5 2 847 2 2 5 I F g IF g H kJ s ( ) + ( ) \u2192 ( ) \u0394 = \u2212 ; \u2212 = \u00d7 + ( ) + \u00d7 \u2212 \u00d7 \u2212 847 B E. I F 1 2 62 149 5 2 155 5 . B E. kJ/mol I F . \u2212 = 268 1 2 3 2 470 2 2 3 I s F g IF g H kJ ( ) + ( ) \u2192 ( ) \u0394 = \u2212 ; \u2212 = + ( ) + \u00d7 \u2212 \u00d7 + \u23a1 \u23a3 \u23a4 \u23a6 \u2212 ( ) 470 1 2 62 149 3 2 155 2 268 B E. I F eq . B E. kJ/mol I F eq . \u2212 ( ) = 272
Answer: 0142
\nSolution: H g O g H O l H kJ 2 2 2 1 2 286 ( ) + ( ) \u2192 ( ) \u0394 = \u2212 ; \u2212 + \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f \u2212 \u00d7 \u2212 \u2212 \u2212 286 = B E. B E. H H O H . . 1 2 498 2 44 (1) H g O g H O l H kJ 2 2 2 2 188 ( ) + ( ) \u2192 ( ) \u0394 = \u2212 ; \u2212 + ( ) \u2212 \u00d7 + \u2212 \u2212 \u2212 \u2212 188 = B E. B E. B E. H H O H O O . ( . . ) 498 2 53 (2) From (1) (2), we get: B E. kJ/mol O O \u2212 = \u2212 . 142
Answer: 0120
\nSolution: (a) Ag + (aq) +Br - (aq) \u2192 AgBr(s); \u2206 H = \u221284.54 kJ (b) Ag(s) \u2192 Ag + (aq); \u2206 H = \u22128 x kJ (c) 1 2 9 Br l Br aq H kJ 2 ( ) \u2192 ( ) = \u2212 ; \u0394 x (d) Ag s Br l Ag Br s H kJ 2 ( ) + ( ) \u2192 ( ) = \u2212 1 2 99 54 ; . \u0394 As (a) + (b) + (c) = (d), we get: (\u221284.54) + (\u22128 x ) + 9 x = \u221299.54 \u21d2 x = \u221215 \u2234 \u2206 f H Ag + (aq) = \u22128 x = 120 kJ/mol
Answer: D
\nSolution: U n f R T = \u00d7 \u00d7 2 For larger U , n , f , T , should be higher.
Answer: B
\nSolution: w P dv a V b dv a V V b V V V V V V = \u2212 \u22c5 = \u2212 + \u239b \u239d \u239c \u239e \u23a0 \u239f = \u2212 \u2212 \u2212 \u222b \u222b 1 2 1 2 2 1 2 1 ln ( )
Answer: A
\nSolution: q = m.s. \u0394 T \u21d2 10 \u00d7 10 6 = 80 \u00d7 (4.2 \u00d7 10 3 ) \u00d7 \u0394 T \u21d2 \u0394 T = 29.76 K = 29.76\u00b0 C
Answer: C
\nSolution: Heat lost by water = Heat gained by ice or, 500 75 6 18 20 9 6000 18 \u00d7 \u00d7 = \u00d7 \u00d7 . ( ) N \u21d2 N = 14
Answer: A
\nSolution: Let T 1 > T 2 . Now, heat lost by gas (1) = Heat gained by gas (2) or, n 1 \u22c5 C m \u22c5 ( T 1 \u2013 T f ) = n 2 \u22c5 C m \u22c5 ( T f \u2013 T 2 ) or, PV RT T T P V RT T T f f 1 1 1 1 2 2 2 2 \u22c5 \u2212 = \u2212 ( ) ( ) \u21d2 T T T PV P V PV T P V T f = + + 1 2 1 1 2 2 1 1 2 2 2 1 ( )
Answer: A
\nSolution: w = \u2013 P ext ( V 2 \u2013 V 1 ) = \u2212 \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f P nRT P nRT P 2 2 1 = \u2212 \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f = \u2212 \u2212 \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f nRT P P nRT P P 1 1 1 2 1 1 1 = \u2212 \u00d7 nRT P 1 1\n4.37 Thermodynamics HINTS AND EXPLANATIONS Now, w total = w 1 + w 2 + \u2026 + w f = \u2212 \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f + \u2212 \u00d7 \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f + + \u2212 \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f nRT P nRT P nRT 1 1 1 1 2 \u001d = \u2212 + \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f = = \u2212 \u2211 nRT P i i i P 1 1 1 1
Answer: B
\nSolution: w P dV K V dV K V V K V V V V V 1 1 2 1 0 0 2 1 0 2 2 1 2 1 4 0 0 1 2 = \u2212 \u22c5 = \u2212 \u22c5 = + \u2212 \u23a1 \u23a3 \u23a2 \u23a4 \u23a6 \u23a5 = \u2212 \u222b \u222b = \u2212 = \u2212 = \u2212 2 4 2 0 5 0 0 2 0 0 0 0 0 P V V P V P V . w P dV K V dV K V V P V V V V V 2 1 2 0 0 2 0 0 2 0 7 0 0 1 2 = \u2212 \u22c5 = \u2212 \u22c5 = \u2212 = \u2212 \u00d7 \u222b \u222b ln .
Answer: A
\nSolution: As \u0394 T = 0, \u0394 U = 0
Answer: D
\nSolution: Area = P 2 \u00d7 \u0394 V \u21d2 P 2 \u00d7 4 = 49.26 L -atom Now, correct work, w = \u2212 \u22c5 = \u2212 \u22c5 nRT V V P V ln ln 2 1 2 2 4 2 = \u201349.26 \u00d7 0.693 = \u2013 34.137 L -atom
Answer: A
\nSolution: PV x = Constant \u21d2 = \u21d2 = \u239b \u239d \u239c \u239e \u23a0 \u239f \u21d2 = \u21d2 = PV P V P P V V x x x x x 1 1 2 2 1 2 2 1 8 4 3 2 Now, C C R x R R R m v m = + \u2212 = + \u2212 = \u2212 1 1 3 2 1 3 2 2
Answer: C
\nSolution: Dulong and Petit\u2019s law is applicable only for solid element. (Molar heat capacity \u2248 6.4 cal/K-mol = 26.8 J/K-mol).
Answer: C
\nSolution: PV P V 1 1 2 2 \u03b3 \u03b3 = \u21d2 P d P d 1 1 2 2 \u03b3 \u03b3 = \u21d2 P P d d 2 1 2 1 7 5 32 128 = \u239b \u239d \u239c \u239e \u23a0 \u239f = = \u03b3 ( ) /
Answer: C
\nSolution: U U T T rms rms , , 2 1 1 2 1 4 2 1 = = \u21d2 Now, T.V r \u2013 1 = Constant \u21d2 T T V V V V r 2 1 1 2 1 1 2 7 5 1 1 4 = \u239b \u239d \u239c \u239e \u23a0 \u239f \u21d2 = \u239b \u239d \u239c \u239e \u23a0 \u239f \u2212 \u2212 / \u2234 V 2 = 32 V 1
Answer: B
\nSolution: w w A B = \u00d7 2 But \u0394 U A = \u0394 U B , Hence q A > q B or, ( C A \u22c5 \u0394 T ) > ( C B \u22c5 \u0394 T ) \u21d2 C A > C B
Answer: A
\nSolution: q = n \u22c5 C m \u22c5 \u0394 T = 1 \u00d7 (0.22 \u00d7 32) \u00d7 (273 \u00d7 1.1 \u2013 273) \u00d7 4.2 J
Answer: C
\nSolution: q q n C T n C T V P V m P m = \u22c5 \u22c5 \u0394 \u22c5 \u22c5 \u0394 = , , 1 \u03b3 \u21d2 q V = \u239b \u239d \u239c \u239e \u23a0 \u239f \u00d7 = 1 74 3 66 793 660 . J ( C P m , . = \u00d7 = 743 5 2 74 3 \u21d2 C V m , = 74.3 \u2013 8.3 = 66.0)
Answer: D
\nSolution: Isothermal: P.V = P V n P n P i i \u00d7 \u21d2 = \u22c5 Adiabatic: P.V r = P V n P n P a a \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f \u21d2 = \u22c5 \u03b3 \u03b3 \u2234 P P n P n P n i a = \u22c5 \u22c5 = \u2212 \u03b3 \u03b3 1
Answer: C
\nSolution: \u0394 U = n C T T n R P V nR P V nR PV V m \u22c5 \u22c5 \u2212 = \u22c5 \u2212 \u22c5 \u22c5 \u2212 \u22c5 \u239b \u239d \u239c \u239e \u23a0 \u239f = \u2212 , ( ) 2 1 1 2 1 \u03b3 \u03b3
Answer: A
\nSolution: K.E. = \u0394 U \u21d2 1 2 40 1000 100 8 314 1 5 1 2 \u00d7 \u00d7 \u00d7 = \u00d7 \u2212 \u00d7 \u0394 ( ) ( ) . ( . ) n n T \u2234 \u0394 T = 12.03 K
Answer: B
\nSolution: For minimum pressure, compression should be irreversible. \u0394 = \u21d2 \u22c5 \u2212 \u22c5 \u2212 = \u2212 \u2212 = \u2212 \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f U w n R T T P V V P nRT P nRT P ext \u03b3 1 2 1 2 1 2 2 2 1 1 ( ) ( )\n4.38 Chapter 4 HINTS AND EXPLANATIONS or, T T T T P P P 2 1 2 2 2 1 2 1 700 400 1 4 1 700 400 100 \u2212 \u2212 = \u2212 \u2212 \u22c5 \u239b \u239d \u239c \u239e \u23a0 \u239f \u21d2 \u2212 \u2212 = \u2212 \u2212 \u00d7 \u239b \u239d \u239c \u239e \u03b3 . \u23a0 \u23a0 \u239f \u2234 P 2 = 362.5 kPa
Answer: C
\nSolution: q n C T n C T V m Ne V m SO = \u22c5 \u22c5 \u0394 + \u22c5 \u22c5 \u0394 ( ) ( ) , , 3 or, 12 \u00d7 10 3 = 2 \u00d7 3 \u00d7 ( T f \u2013 300) + 3 \u00d7 6 \u00d7 ( T f \u2013 400) \u21d2 T f = 875 K Now, P nRT V final atm = = \u00d7 \u00d7 = 5 0 08 875 10 35 .
Answer: A
\nSolution: Free expansion is isothermal.
Answer: B
\nSolution: C R R N R N R V m , ( ) ( ) = \u00d7 + \u00d7 + \u2212 \u00d7 = \u2212 3 1 2 3 1 2 3 6 3 3 \u2234 \u03b3 = = + = + \u2212 C C R C N P V V m 1 1 1 3 3 ,
Answer: C
\nSolution: C C P dV dT C P C RT V C R T V V m V m V m V m V m = + \u22c5 = + = + \u22c5 = + + , , , , ( ) \u03b1 \u03b1 \u03b1 \u03b1 0 = + C RT V P m , 0 \u03b1 Now, q C dT C dV C RT V dV m T T m V V P m V V = \u22c5 = \u22c5 \u22c5 = \u22c5 + \u239b \u239d \u239c \u239e \u23a0 \u239f \u222b \u222b \u222b 1 2 1 2 1 2 0 ( ) , \u03b1 \u03b1 = \u22c5 \u2212 + \u22c5 \u03b1 C V V RT V V P m , ( ) ln 2 1 0 2 1
Answer: D
\nSolution: w = \u201325 J = \u2013 nR \u22c5 \u0394 T \u0394 = \u22c5 \u22c5 \u0394 = \u22c5 \u239b \u239d \u239c \u239e \u23a0 \u239f \u22c5 \u0394 = U n C T n R T J V m , 6 2 75 \u2234 q = \u0394 U \u2013 w = 100 J
Answer: B
\nSolution: C C R x R R R m V m = + \u2212 = + \u2212 = , 1 3 2 1 5 2 5 6 \u2234 q n C T R J m = \u22c5 \u22c5 \u0394 = \u00d7 \u00d7 = \u22c5 1 5 6 26 180 14
Answer: C
\nSolution: PV K dP dV x P V x x x = \u21d2 = \u2212 \u22c5 \u21d2 \u2212 = \u2212 \u00d7 \u21d2 = 1 4 2 1 2 \u2234 C C R x R R R m V m = + \u2212 = + \u2212 = , . 1 3 2 1 1 2 3 5
Answer: C
\nSolution: V 0 2 V 0 V P 2 1 3
Answer: B
\nSolution: V 1 V 2 V Isobaric Isothermal Adiabatic P \u0394 E adiabatic = Negative \u0394 E isothermal = 0 \u0394 E isobaric = Positive
Answer: B
\nSolution: Boyle temperature, T B = 20 + 273 = 293 K Inversion temperature, T i = 2 \u00d7 T B = 586 K = 313\u00b0 C > 50\u00b0 C
Answer: C
\nSolution: V 0 Mono Initial Di V 0 V 1 Mono Di V 0 3 4 Monoatomic : P V P V 1 0 5 3 2 1 5 2 \u22c5 = \u22c5 / / Diatomic : P V P V 1 0 7 5 2 0 7 5 3 4 \u22c5 = \u22c5 \u239b \u239d \u239c \u239e \u23a0 \u239f / / \u2234 V V 1 0 21 25 3 4 = \u239b \u239d \u239c \u239e \u23a0 \u239f\n4.39 Thermodynamics HINTS AND EXPLANATIONS
Answer: B
\nSolution: V 0 2 V V P P 2 V = Constant Isothermal reversible Adiabatic irreversible Adiabatic reversible
Answer: A
\nSolution: For greater heat exchange, heat capacity should be high. C C R x m V m = + \u2212 , 1 for PV x = Constant
Answer: A
\nSolution: q ABC = 600 + 200 = 800 J w AB = 0 and w N m m BC = \u2212 \u00d7 \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f \u00d7 \u2212 \u00d7 \u2212 \u2212 8 10 5 10 2 10 4 2 3 3 2 ( ) = \u2013240 J \u2234 \u0394 U AC = \u0394 U ABC = q ABC + w ABC = 800 + (\u2013240) = 560 J
Answer: C
\nSolution: 3 60\u00b0 30\u00b0 A B C V 0 V 0 6 V 0 V 9 4 P 0 P 0 AB P V C : = + 3 1 P V C 0 0 1 3 = + (1) and 3 3 0 1 P V C B = + (2) BC P V C : = \u2212 + 1 3 2 P V C 0 0 2 1 3 6 = \u2212 \u22c5 + (3) 3 1 3 0 2 P V C B = \u2212 \u22c5 + (4) From equation (1), (2), (3) and (4), V V B = 9 4 0 Now, T T P V P V B A = \u22c5 \u22c5 = 3 9 4 27 4 0 0 0 0
Answer: A
\nSolution: F = P 0 = P P 0 P i P f dw = F \u22c5 dx = ( P 0 \u2013 P ) A \u22c5 dx = ( P 0 \u2013 P ) \u22c5 dV \u2234 w P nRT V dV P V V RT V V V V = \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f = \u2212 \u2212 \u22c5 \u22c5 \u22c5 \u222b 0 0 \u03b7 \u03b7 \u03b7 ( ) ln P 0 V ( \u03b7 -1) \u2212 RT.ln \u03b7 = RT [ \u03b7 -1-ln \u03b7 ]
Answer: B
\nSolution: T 0 ( V 0) P 0, P 0, P 1, P 2, T 0 \u22c5 ( V ) \u03b7 ( V 0) ( V ) Work performed on the piston = \u2212 \u22c5 \u23a1 \u23a3 \u23a2 \u23a4 \u23a6 \u23a5 \u22c5 + \u222b \u222b \u22c5 P P dV dV V V V V 1 2 0 0 \u03b7 and ( V + h \u22c5 V ) = 2 V 0 = \u22c5 + P V 0 0 2 1 4 ln ( ) \u03b7 \u03b7
Answer: C
\nSolution: Isothermal : P A \u22c5 V = P \u22c5 (2 V ) \u21d2 P A = 2 P Adiabatic : P B \u22c5 V 1.5 = P \u22c5 (2 V ) 1.5 \u21d2 P A = 2 2 P Isobaric : P C = P \u2234 P A : P B : P C = 2 : 2 2 : 1
Answer: B
\nSolution: A = ( V 0, T 0) , T 0 ( V 0, T 0) 27 8 = 27 8 P 0 , P 0 P 0 P 0 P f P f Chamber B P V P V B : 0 0 0 27 8 \u22c5 = \u22c5 \u03b3 \u03b3\n4.40 Chapter 4 HINTS AND EXPLANATIONS \u2234 V V T T B B = \u21d2 = 4 9 3 2 0 0 and V V V V T T A A = \u2212 = \u21d2 = 2 4 9 14 9 21 4 0 0 0 0 Now, q U U n C T T n C T T A A B V m A V m B = \u0394 + \u0394 = \u22c5 \u22c5 \u2212 + \u22c5 \u22c5 \u2212 , , ( ) ( ) 0 0 = \u22c5 \u22c5 \u22c5 \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f + \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f \u23a1 \u23a3 \u23a2 \u23a4 \u23a6 \u23a5 = \u22c5 P V R T R T T T T P V 0 0 0 0 0 0 0 0 0 2 21 4 3 2 19 2
Answer: A
\nSolution: h 1st step Final position x = ? mg A mg A P 0 P 1 After 1st step, the process is irreversible adiabatic. Hence, \u0394 U = w n C T T P V V V m \u22c5 \u22c5 \u2212 = \u2212 \u2212 , ( ) ( ) 2 1 2 1 ext or, n R P V nR PV nR P V V A H \u22c5 \u22c5 \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f = \u2212 \u2212 + \u22c5 3 2 1 2 1 1 1 2 1 [ ( )] \u2234 V 2 = V 1 + 0.4 H.A \u21d2 x = 0.4 H (The final pressure of gas after 2nd step will remain same as initial, beginning of processes.)
Answer: A
\nSolution: Smaller the heat capacity larger is \u0394 T.
Answer: B
\nSolution: \u0394 = \u22c5 \u22c5 \u2212 = \u00d7 \u00d7 \u2212 = H n C T T P m , ( ) ( ) 2 1 1 40 500 300 8000 J \u2234 \u0394 U = \u0394 H \u2013 P \u22c5 \u0394 V = 8000 \u2013 2(40 \u2013 30) \u00d7 100 = 6000 J
Answer: C
\nSolution: q = 0 \u21d2 \u0394 U = w = \u2013 P ext \u22c5 ( V 2 \u2013 V 1 ) = \u20134 \u00d7 (30 \u2013 40) = 40 l -bar Now, \u0394 H = \u0394 U + \u0394 ( PV ) = 40 + (4 \u00d7 30 \u2013 2 \u00d7 40) = 80 L -atom = 8000 J
Answer: C
\nSolution: \u0394 U = 0 \u0394 H = \u0394 U + \u0394 ( PV ) = 0 + B ( P 2 \u2013 P 1 ) = B RT V B RT V B \u22c5 \u2212 \u2212 \u2212 2 1 = \u00d7 \u00d7 \u2212 \u2212 \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f = \u2212 2 8 314 400 1 22 2 1 12 2 332 56 . . J
Answer: C
\nSolution: \u0394 = \u22c5 \u22c5 \u2212 U n C T T V m 1 2 1 , ( ) \u0394 H 1 = \u0394 U 1 + V \u22c5 \u0394 P = 1 \u00d7 C V,m \u00d7 ( T 2 \u2013 T 1 ) + V 1 ( P 2 \u2013 P 1 ) Now, \u0394 U 2 = w 2 = \u2013 P ext ( V 2 \u2013 V 1 ) = \u2013 P 3 ( V 2 \u2013 V 1 ) \u0394 H 2 = \u0394 U 2 + \u0394 ( PV ) = \u2013 P 3 ( V 2 \u2013 V 1 ) + ( P 3 V 2 \u2013 P 2 V 1 ) \u2234 \u0394 H total = \u0394 H 1 + \u0394 H 2 = C V ( T 2 \u2013 T 1 ) + V 1 ( P 3 \u2013 P 1 )
Answer: A
\nSolution: \u03b7 = \u2212 1 T T C H 1 6 1 = \u2212 T T C H and 1 3 1 65 390 = \u2212 \u2212 \u21d2 = T T T C H H K = 117\u00b0 C
Answer: C
\nSolution: q q T T q C H rej abs rej cal = \u21d2 = \u00d7 = 390 600 120 78
Answer: B
\nSolution: 1 1 \u2212 \u2212 \u0394 \u239b \u239d \u239c \u239e \u23a0 \u239f > \u2212 + \u0394 \u239b \u239d \u239c \u239e \u23a0 \u239f T T T T T T C H C H
Answer: A
\nSolution: T V T V H C \u22c5 = \u22c5 \u2212 \u2212 2 1 3 1 \u03b3 \u03b3 \u21d2 T T V V C M = \u239b \u239d \u239c \u239e \u23a0 \u239f = \u239b \u239d \u239c \u239e \u23a0 \u239f = \u2212 \u2212 2 3 1 1 4 1 1 2 75 1 1 5 \u03b3 . . . \u2234 \u03b7 = \u2212 = 1 1 3 T T C H
Answer: C
\nSolution: 2 1 3 4 V T 2 T 3 T 4 T 1 P T V T V 2 1 1 3 2 1 \u22c5 = \u22c5 \u2212 \u2212 \u03b3 \u03b3 and T V T V 1 1 1 4 2 1 \u22c5 = \u22c5 \u2212 \u2212 \u03b3 \u03b3 \u2234 T T T T T T T T T T 2 1 3 4 2 1 1 3 4 4 = \u21d2 \u2212 = \u2212 @TheBookCorner\n4.41 Thermodynamics HINTS AND EXPLANATIONS \u03b7 = \u2212 = \u2212 \u22c5 \u2212 \u22c5 \u2212 = \u2212 = \u2212 \u239b \u239d \u239c 1 1 1 1 3 4 2 1 4 1 1 2 q q n C T T n C T T T T V V V m V m rej abs , , ( ) ( ) \u239e \u239e \u23a0 \u239f \u2212 \u03b3 1 = \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f = \u2212 \u2212 1 1 10 0 6 7 5 1 .
Answer: C
\nSolution: \u0394 S unit = \u0394 S Source + \u0394 S Heat engine + \u0394 S Sink = \u2212 \u00d7 + + \u00d7 = + 40 10 500 0 30 10 300 20 3 3 J/K
Answer: B
\nSolution: \u0394 = \u22c5 \u22c5 S n C T T P m , ln 2 1 = \u00d7 \u00d7 = \u2212 3 2 32 14 900 1000 0 14 . ln . c al/K
Answer: A
\nSolution: \u0394 = \u22c5 S nR V V ln 2 1 = \u00d7 \u00d7 = 2 8 314 2 34 58 3 3 . ln ( ) . a a J/K
Answer: B
\nSolution: \u0394 = \u22c5 \u22c5 S n C T T P m , ln 2 1 = \u00d7 \u00d7 = 1 5 2 1000 250 7 0 R ln . cal/K
Answer: C
\nSolution: \u0394 = \u22c5 \u22c5 S n C T T V m , ln 2 1 S R K 500 46 2 1 3 2 500 250 \u2212 = \u00d7 \u00d7 . ln \u2234 S 500 K = 48.3 Ccal/K-mol
Answer: A
\nSolution: \u0394 = \u22c5 S nR V V ln 2 1 \u21d2 \u2212 = \u00d7 \u00d7 \u00d7 \u21d2 = \u2212 5 0 10 15 10 300 15 5 4 5 3 2 2 . ( ) ln . V V L
Answer: B
\nSolution: \u0394 = \u2212 \u00d7 = \u2212 S Surr J/K 1 5 10 300 5 3 . Now, \u0394 S unit = \u0394 S Sys + \u0394 S Surr = 5.51 + (\u20135) = + 0.51 J/K Hence, the process is irreversible.
Answer: C
\nSolution: \u0394 = \u22c5 \u22c5 + \u22c5 S n C T T nR P P P m , ln ln 2 1 1 2 or, 0 5 2 1200 300 1 32 2 2 = \u00d7 \u00d7 + \u00d7 \u21d2 = n R nR P P ln ln bar
Answer: C
\nSolution: \u0394 S = \u0394 S adiabatic + \u0394 S isobaric = 0 2 1 + \u22c5 \u22c5 n C T T P m , ln = \u00d7 \u00d7 = \u2212 1 6 4 5 2 1 3 2 2 . ln . R cal/K
Answer: A
\nSolution: \u0394 = \u22c5 \u22c5 + \u22c5 S n C T T nR V V V m , ln ln 2 1 2 1 = \u22c5 \u22c5 + \u22c5 \u22c5 n C P P n C V V V m P m , , ln ln 2 1 2 1 = \u00d7 \u2212 \u00d7 + \u00d7 \u2212 \u00d7 2 1 5 1 1 4 2 1 5 1 5 1 2 R R . ln . . ln = \u201311.64 J/K
Answer: C
\nSolution: S S n C T T nR V V V m 2 1 2 1 2 1 \u2212 = \u22c5 \u22c5 + \u22c5 , ln ln = \u00d7 \u00d7 + \u00d7 \u00d7 1 2 3 2 1 2 1 2 2 . l n . ln R R = \u20130.84 cal/K
Answer: A
\nSolution: \u0394 = \u22c5 \u22c5 + \u22c5 S n C T T nR V V V m , ln ln 2 1 2 1 = \u00d7 \u2212 \u00d7 + \u00d7 \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f \u2212 1 1 1 2 1 1 2 1 1 R T T R T T n \u03b3 ln ln / (as T.V n -1 = Constant) R T T n \u22c5 \u2212 \u2212 \u2212 \u23a1 \u23a3 \u23a2 \u23a4 \u23a6 \u23a5 ln 2 1 1 1 1 1 \u03b3 = \u2212 \u2212 \u2212 \u22c5 ( ) ( )( ) ln n R n \u03b3 \u03b3 \u03c4 1 1
Answer: A
\nSolution: \u0394 = \u22c5 \u22c5 + \u22c5 \u22c5 S n C P P n C V V V m P m , , ln ln 2 1 2 1 = \u00d7 \u00d7 + \u00d7 \u00d7 2 3 2 2 2 5 2 2 R R ln ln = + 11.2 cal/K
Answer: B
\nSolution: dS n C dT P dV T V m = \u22c5 \u22c5 + \u22c5 , For maximum entropy, dS dV = 0\n4.42 Chapter 4 HINTS AND EXPLANATIONS or, n C dT dV P V m \u22c5 + = , 0 (1) Now, P RT V P V dT dV R P V = = \u2212 \u21d2 = \u2212 0 0 1 2 \u03b1 \u03b1 ( ) (2) From (1) and (2), 1 1 1 2 0 0 0 \u00d7 \u2212 \u00d7 \u2212 + \u2212 = R R P V P V \u03b3 \u03b1 \u03b1 ( ) ( ) \u2234 V P = \u22c5 + \u03b3 \u03b1 \u03b3 0 1 ( )
Answer: C
\nSolution: dS C dT T P dV T a dT C T dT V m V m = \u22c5 + \u22c5 = \u22c5 + \u22c5 \u22c5 , , 1 or, R V dV a dT R V V a T T V V T T 0 0 0 0 \u222b \u222b \u22c5 = \u22c5 \u21d2 \u22c5 = \u2212 ln ( ) \u2234 T = T 0 + R a V V \u22c5 ln 0
Answer: A
\nSolution: dS C dT T T dT T S aT S T T 0 0 3 3 0 3 \u222b \u222b \u222b = \u22c5 = \u22c5 \u21d2 =
Answer: B
\nSolution: \u0394 = \u0394 + \u0394 = \u2212 + = + S S S A B 12000 600 12000 400 10 J/K
Answer: C
\nSolution: Heat lost by alloy = Heat gained by water or 4 \u00d7 4 \u00d7 (800 \u2013 T ) = 4 \u00d7 1.0 \u00d7 ( T \u2013 300) \u21d2 T = 700 K (As date is not given for vaporization of water) Now, \u0394 S mix = \u0394 S alloy + \u0394 S water = 4 \u00d7 4 \u00d7 ln 700 800 + 4 \u00d7 1 \u00d7 ln 700 300 = 1.0 K cal/K
Answer: A
\nSolution: Final temperature of both blocks = T T 1 2 2 + \u2234 \u0394 S = \u0394 S 1 + \u0394 S 2 = C T T T C T T T \u22c5 + + \u22c5 + ln ( ) / ln ( ) / 1 2 1 1 2 2 2 2 = \u22c5 + C T T T T ln ( ) 1 2 2 1 2 4
Answer: A
\nSolution: \u0394 S = \u2013 R [ n 1 \u22c5 ln x 1 + n 2 \u22c5 ln x 2 ] = \u2013 R [0.8 \u00d7 ln 0.8 + 0.2 \u00d7 ln 0.2] = + 0.96 Cal/K.
Answer: D
\nSolution: Larger molar mass, greater is the molar entropy.
Answer: C
\nSolution: Greater the number of atoms, greater is the molar entropy.
Answer: C
\nSolution: nC( s ) + (n + 1) H 2 ( g ) \u2192 C n H 2n + 2 ( g ) with increase in n , the decrease in entropy increases.
Answer: C
\nSolution: H 2 O ( l , 1 atm, 100\u00b0C) ( ) 1 \u23af \u2192 \u23af H 2 O ( g , 1 atm, 100\u00b0C) ( ) 2 \u23af \u2192 \u23af H 2 O ( g , 5 atm, 100\u00b0C) \u0394 G 1 = 0 and \u0394 G 2 = nRT ln P P 2 1 = 5 \u00d7 2 \u00d7 373 \u00d7 ln 5 1 = 3730 ln 5 Cal
Answer: B
\nSolution: q = \u0394 U \u2013 w = 0 \u2013 \u2212 \u22c5 \u239b \u239d \u239c \u239e \u23a0 \u239f nRT P P ln 1 2 = nRT P P \u22c5 ln 1 2 = \u2013 \u0394 G \u2234 \u0394 G = \u2013 q = \u2013(\u20131200) = +1200 cal
Answer: C
\nSolution: \u0394 G \u00b0 = \u2013 R T \u22c5 ln K eq \u21d2 \u20131743 = \u2013 8.3 \u00d7 300 \u00d7 ln K eq \u2234 K eq = 2 Now, (a) K eq = \u00d7 3 3 6 (b) K eq = \u00d7 6 3 3 2 (c) K eq = \u00d7 6 3 3 2 (d) K eq = \u00d7 3 3 6 2
Answer: B
\nSolution: H O C Pa H O C Pa H O 2 2 2 ( , , . ) ( , , . ) ( l s G G g \u2212 \u00b0 \u23af \u2192 \u23af \u2212 \u00b0 \u2193 \u0394 = \u2191 \u0394 = 10 0 28 10 0 26 0 0 1 3 , , , . ) ( , , . ) \u2212 \u00b0 \u23af \u2192 \u23af\u23af \u2212 \u00b0 \u0394 10 0 28 10 0 26 2 C Pa H O C Pa 2 G g \u0394 G 2 = nRT ln P P 2 1 = 1 \u00d7 R \u00d7 263 \u00d7 ln 0 26 0 28 . . \u2234 \u0394 G = \u0394 G 1 + \u0394 G 2 + \u0394 G 3 = 263 R ln 13 14
Answer: D
\nSolution: Free expansion is isothermal \u0394 G = n RT ln P P 2 1 = nRT ln V V 2 1 = 10 5 \u00d7 (1.2 \u00d7 10 \u20133 ) \u00d7 ln 1 2 2 4 . . = \u201384 J\n4.43 Thermodynamics HINTS AND EXPLANATIONS
Answer: D
\nSolution: \u0394 G 1 \u00b0 = \u2013 RT \u22c5 ln K 1 and \u0394 G 2 \u00b0 = \u2013 RT \u22c5 ln K 2 Now, \u0394 G 2 \u00b0 \u2013 \u0394 G 1 \u00b0 = \u2013 RT [ln K 2 \u2013 ln K 1 ] = \u2013 RT [ln e 4 ] = \u20132 \u00d7 300 \u00d7 4 = \u20132400 cal
Answer: A
\nSolution: At 0.04 atom, the system is in equilibrium.
Answer: B, C, D
\nSolution: Theory based
Answer: A, B, C
\nSolution: Theory based
Answer: A, B, C
\nSolution: For isolated system, there should not be any mass and energy transfer with surroundings.
Answer: A, B, D
\nSolution: 2 V V V Given 0.5 P Isothermal P P 0.5 P T P P
Answer: A, B
\nSolution: The internal energy of real gas may change on changing the volume of gas. Change in physical state also changes the physical state.
Answer: A, B, C
\nSolution: Theory based
Answer: C
\nSolution: Theory based
Answer: A, B, C, D
\nSolution: Option (c) should be changed with (c) adiabatic free expansion of any gas is also isothermal.
Answer: A, C
\nSolution: w rev \u2013 w irr = (\u2013 P \u22c5 dV ) \u2013 (\u2013 P ext \u22c5 dV ) = ( P ext \u2013 P ) \u22c5 dV = negative, always and q rev \u2013 q irr = positive, always
Answer: A, C
\nSolution: q = 0 w = \u0394 U = n C T T R V m \u22c5 \u22c5 \u2212 = \u00d7 \u00d7 \u2212 = \u2212 , ( ) ( ) cal 2 1 4 3 2 290 320 360 \u0394 H = \u03b3 \u22c5 \u0394 U = 5 3 360 600 \u00d7 \u2212 = \u2212 ( ) cal
Answer: A, B, C
\nSolution: Theory based
Answer: A, B, C
\nSolution: Process BC : P T P T P P B B C C B B = \u21d2 = \u21d2 = 500 1 250 2 bar and \u0394 = \u22c5 \u2212 = \u00d7 \u00d7 \u2212 = \u2212 U n C T T BC V m C B 1 2 1 5 250 500 750 ( ) . R ( ) R Process CD : \u0394 = \u21d2 \u22c5 \u2212 = \u2212 \u2212 U w n C T T P V V V m D C D C 1 ( ) ( ) ext or, n R T T P nRT P nRT P T D C D D D C C D \u00d7 \u00d7 \u2212 = \u2212 \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f \u21d2 = 1 5 450 . ( ) K and \u0394 = \u22c5 \u22c5 \u2212 = \u00d7 \u00d7 \u2212 = H n C T T R CD P m D C 1 2 2 5 450 250 1000 ( ) . ( ) R
Answer: C, D
\nSolution: PV K P K V dP dV K V P V \u03b3 \u03b3 \u03b3 \u03b3 \u03b3 = \u21d2 = \u22c5 \u21d2 = \u22c5 \u2212 = \u2212 \u22c5 \u2212 \u2212 \u2212 1 1 1 1 ( ). The gas having higher \u03b3 will have higher magnitude of slope of P vs. V curve. Now, n R dT P dV nRT V dV \u22c5 \u2212 \u22c5 = \u2212 \u22c5 = \u2212 \u22c5 \u03b3 1 or, dV dT V T = \u2212 \u2212 \u22c5 1 1 \u03b3\n4.44 Chapter 4 HINTS AND EXPLANATIONS Gas having higher \u03b3 will have lower magnitude of slope of V vs. T curve. Now, n C dT P dV nRdT V dP V m \u22c5 \u22c5 = \u2212 \u22c5 = \u2212 \u2212 \u22c5 , [ ] or, n C dT V dP n R dT nRT P dP P m \u22c5 \u22c5 = \u2212 \u22c5 \u21d2 \u22c5 \u2212 \u22c5 = \u22c5 , \u03b3 \u03b3 1 \u2234 dP dT P T = \u2212 \u22c5 \u03b3 \u03b3 1 Gas having higher \u03b3 will have lower magnitude of slope of P vs. T curve.
Answer: C
\nSolution: q = 0
Answer: A
\nSolution: dT = 0
Answer: A, B, C
\nSolution: \u03b3 V V 2 V CH4 SO2 O2 Ne P Here, \u03b3 decreases on increasing degree of freedoms. As fi nal pressure is minimum for Ne, its fi nal temperature is minimum (decrease in temperature is maximum). Now, for overall process, \u0394 T = 0 \u21d2 \u0394 U total = 0 or, \u0394 U I + \u0394 U II = 0 \u21d2 (0 + w I ) + ( q II + 0) = 0 \u2234 q II = \u2013 w I = maximum for CH 4
Answer: A, B
\nSolution: Theory based
Answer: B, D
\nSolution: C C R P m V m , , \u2212 = \u21d2 S S R M P V \u2212 = = 0 04545 . \u2234 M = 44 gm/mol
Answer: C, D
\nSolution: q = 0 \u21d2 \u0394 U = w = \u2013 P 0 (4 V 0 \u2013 V 0 ) = \u20133 P 0 V 0 Now, \u0394 H = \u0394 U + \u0394 ( PV ) = (\u20133 P 0 V 0 ) + ( P 0 \u22c5 4 V 0 \u2013 2 P 0 \u22c5 V 0 ) = \u2013 P 0 \u22c5 V 0
Answer: B, C, D
\nSolution: Theory based
Answer: A, B, C, D
\nSolution: Theory based
Answer: A, B, C, D
\nSolution: V i i a a P
Answer: A, D
\nSolution: In reversible cycle, heat rejected is minimum. For reversible cycle, q T T q C H rej abs J = \u00d7 = \u00d7 = 400 500 100 80
Answer: A, D
\nSolution: dS q T = rev and \u001e \u222b dS = 0
Answer: A, B, C
\nSolution: In rusting, moles of gas decreases.
Answer: B, C
\nSolution: Theory based.
Answer: B, C, D
\nSolution: \u0394 U = 0 \u21d2 q = \u2013 w
Answer: B, C, D
\nSolution: For a process to be spontaneous at low temperature and non-spontaneous at high temperature, \u0394 H = negative and \u0394 S = negative.
Answer: A, B, D
\nSolution: ( ) , \u0394 = \u00d7 = S J K K Vap atm mol 350 1 3 35 10 350 100 On increasing pressure at constant temperature entropy decreases. ( ) , \u0394 = G K Vap atm 350 1 0 On increasing pressure at constant temperature energy increases.
Answer: B, D
\nSolution: Theory based.
Answer: C
\nSolution: \u0394 = \u22c5 \u22c5 \u2212 = \u00d7 \u00d7 \u2212 \u2212 = \u2212 U n C T T R V m 1 2 1 4 5 2 50 0 1000 ( ) ( ) cal
Answer: A
\nSolution: \u0394 H = \u03b3 \u22c5 \u0394 U = 7 5 1000 1400 \u00d7 \u2212 = \u2212 ( ) cal
Answer: D
\nSolution: w = 0 ( V = Constant)\n4.45 Thermodynamics HINTS AND EXPLANATIONS Comprehension II
Answer: B
\nSolution: q U n C T n C T C C V m V m m V m = \u2212\u0394 \u21d2 \u22c5 \u22c5 \u0394 = \u2212 \u22c5 \u22c5 \u0394 \u21d2 = \u2212 , , ,
Answer: A
\nSolution: C C R x m V m = + \u2212 , 1 \u21d2 \u2212 = + \u2212 \u21d2 \u2212 \u22c5 \u2212 = \u2212 C C R x R R x V m V m , , 1 2 1 1 \u03b3 \u2234 x = + \u03b3 1 2 Now, T \u22c5 V x \u20131 = Constant \u21d2 T \u22c5 V ( \u03b3 \u20131)/2 = Constant
Answer: B
\nSolution: T T V V T T 2 1 1 2 1 2 2 5 3 1 2 2 300 1 8 150 = \u239b \u239d \u239c \u239e \u23a0 \u239f \u21d2 = \u239b \u239d \u239c \u239e \u23a0 \u239f \u21d2 = \u2212 \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f \u03b3 / / K \u2234 w nR T T x = \u2212 \u2212 \u2212 = \u2212 \u00d7 \u2212 \u2212 + \u239b \u239d \u239c \u239e \u23a0 \u239f = \u2212 ( ) ( ) / 2 1 1 1 2 150 300 1 5 3 1 2 900 cal Comprehension III
Answer: B
\nSolution: w = \u2013 nR \u22c5 \u0394 T = \u20131 \u00d7 8.314 \u00d7 72 = \u2013598.6 J = \u2013 0.6 kJ
Answer: C
\nSolution: \u0394 U = q + w = 1.6 + (\u20130.6) = 1.0 kJ
Answer: A
\nSolution: \u03b3 = \u0394 \u0394 = = H U 1 6 1 0 1 6 . . . Comprehension IV
Answer: D
\nSolution: V 2 = 4 V 0 \u21d2 P 2 = 4 P 0 As PV T P V T 1 1 1 2 2 2 = \u21d2 P V T P P T 0 0 0 0 0 2 4 4 = \u22c5 \u21d2 T T 2 0 16 = Now, \u0394 = \u22c5 \u22c5 \u0394 = \u00d7 \u2212 \u00d7 \u2212 = \u2212 = \u2212 U n C T n R T T P V V V m , ( . ) \u03b3 \u03b3 \u03b1 \u03b3 1 16 15 1 15 1 0 0 0 0 0 2
Answer: A
\nSolution: w nR T T x nR T V = \u2212 \u2212 = \u00d7 \u2212 \u2212 = ( ) ( ) 2 1 0 0 2 1 15 1 1 15 2 \u03b1
Answer: B
\nSolution: C C R x R R R m V m = + \u2212 = \u2212 + \u2212 \u2212 = + \u2212 1 1 1 1 1 1 2 1 \u03b3 \u03b3 \u03b3 ( ) ( ) ( ) Comprehension V
Answer: B
\nSolution: P = a \u22c5 T a = a \u22c5 PV nR \u239b \u239d \u239c \u239e \u23a0 \u239f \u03b1 \u21d2 P V \u22c5 = \u2212 \u03b1 \u03b1 1 Constant \u2234 w nR T x R T R T = \u22c5 \u0394 \u2212 = \u00d7 \u22c5 \u0394 \u2212 \u2212 = \u2212 \u22c5 \u0394 1 1 1 1 1 \u03b1 \u03b1 \u03b1 ( )
Answer: A
\nSolution: C C R x R R R m V m = + \u2212 = \u2212 + \u2212 \u2212 = \u2212 + \u2212 \u23a1 \u23a3 \u23a2 \u23a4 \u23a6 \u23a5 , ( ) . 1 1 1 1 1 1 1 \u03b3 \u03b1 \u03b1 \u03b3 \u03b1
Answer: C
\nSolution: 1 1 1 0 \u03b3 \u03b1 \u2212 + \u2212 < ( ) \u21d2 \u03b1 \u03b3 \u03b3 > \u2212 1\n4.46 Chapter 4 HINTS AND EXPLANATIONS Comprehension VI
Answer: C
\nSolution: U V n C T V m = \u22c5 = \u22c5 \u22c5 \u03b1 \u03b1 , \u21d2 T V \u22c5 = \u2212 \u03b1 Constant As T V x \u22c5 = \u2212 1 Constant \u21d2 x \u2013 1 = \u2013 a Now, w n R T x U = \u22c5 \u22c5 \u0394 \u2212 = \u2212 \u22c5 \u0394 \u2212 1 1 ( ) \u03b3 \u03b1
Answer: C
\nSolution: q U w U U U = \u0394 \u2212 = \u0394 + \u2212 \u22c5 \u0394 = \u0394 + \u2212 \u23a1 \u23a3 \u23a2 \u23a4 \u23a6 \u23a5 ( ) \u03b3 \u03b1 \u03b3 \u03b1 1 1 1
Answer: B
\nSolution: C R R x R R m = \u2212 + \u2212 = \u2212 + \u03b3 \u03b3 \u03b1 1 1 1 Comprehension VII
Answer: B
\nSolution: PV P V 1 1 2 2 \u03b3 \u03b3 = \u21d2 P 2 7 5 1 320 10 128 = \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f = / atm
Answer: A
\nSolution: PV T P V T 1 1 1 2 2 2 = \u21d2 1 320 300 128 10 2 \u00d7 = \u00d7 T \u21d2 T 2 1200 = K
Answer: C
\nSolution: w n C T V m = \u22c5 \u22c5 \u0394 = \u00d7 \u00d7 \u00d7 \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f \u00d7 \u2212 = , . . . ( ) . 1 0 32 0 082 300 5 2 8 314 1200 300 243 3 3 J Comprehension VIII
Answer: A
\nSolution: PV P V P 1 1 2 2 2 5 3 1 8 32 \u03b3 \u03b3 = \u21d2 = \u00d7 = ( ) / atm
Answer: C
\nSolution: For A : P 1 = 1 atm, P 2 = 32 atm V 1 = VL , V 2 = V + 7 8 V = 15 8 V L T 1 = 27.3 K; T 2 = ? Now, PV T P V T 1 1 1 2 2 2 = \u21d2 T 2 = 1638 K
Answer: A
\nSolution: \u0394 = \u22c5 \u22c5 \u0394 = \u00d7 \u00d7 \u00d7 \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f \u00d7 \u2212 H n C T A P m , . . . ( . ) 1 22 4 0 082 27 3 5 2 2 1638 27 3 = 80535 cal Comprehension IX
Answer: A
\nSolution: \u0394 U = 0 \u0394 H = \u0394 U + V \u22c5 \u0394 P = 0.9 L \u00d7 1 532 760 \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f atm = 0.27 L -atm = 27 J
Answer: C
\nSolution: \u0394 = \u2192 U 1 2 0 \u0394 = \u0394 \u2192 U U 2 3 for temperature increase + \u0394 U for vaporization of water. = m . s . \u22c5 \u0394 T + ( q + w ) = \u00d7 \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f + \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f + \u2212 \u00d7 \u00d7 \u239b \u239d \u239c \u239e \u23a0 900 4 2 1000 20 450 18 40 450 18 8 1000 373 . \u239f \u239f = 1001 kJ
Answer: C
\nSolution: \u0394 = \u0394 + = \u00d7 + \u00d7 \u00d7 \u239b \u239d \u239c \u239c \u239c \u239e \u23a0 \u239f \u239f \u239f \u2192 \u2192 \u2192 H H q 1 3 1 2 2 3 27 450 18 40 4 2 1000 20 J+ kJ 900 kJ . = 1075.573 kJ
Answer: C
\nSolution: w 1 2 0 \u2192 = w 2 3 450 18 8 1000 373 74 6 \u2192 = \u2212 \u00d7 \u00d7 = \u2212 . kJ\n4.47 Thermodynamics HINTS AND EXPLANATIONS Comprehension X
Answer: A
\nSolution: T A > T B < T C < T D = T A From question : T T A B = 4 and T A = 800 K \u21d2 T B = 200K Also, V A > V B > V C = V D From equation : V V A C = 8 2 and V V T T A B A B = = 4 For process BC : T V T V B B C C \u22c5 = \u22c5 \u2212 \u2212 \u03b3 \u03b3 1 1 \u21d2 T T V V C B B C r = \u22c5 \u239b \u239d \u239c \u239e \u23a0 \u239f \u2212 1 = \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f = \u2212 200 8 2 4 400 5 3 1 K
Answer: C
\nSolution: \u0394 U BC = C T T R V m C B , ( ) ( ) . \u22c5 \u2212 = \u00d7 \u00d7 \u2212 = 1 3 2 400 200 2 4942 kJ
Answer: B
\nSolution: q = 0 because \u0394 U = 0 and w = 0
Answer: C
\nSolution: Enthalpy of ideal gas is independent from pressure.
Answer: D
\nSolution: For non-ideal gas, U = f ( T , V )
Answer: D
\nSolution: In adiabatic free expansion, \u0394 T = 0
Answer: A
\nSolution: V 1 V 2 V Isothermal Adiabatic P P 1 Pi Pa
Answer: A
\nSolution: V 1 V 2 V Isothermal Adiabatic P Pa Pi P 2
Answer: A
\nSolution: Magnitude of work in adiabatic process depends on change in temperature.
Answer: A
\nSolution: Magnitude of work in adiabatic process depends on change in temperature.
Answer: A
\nSolution: q n C T P P m = \u22c5 \u22c5 \u0394 ,
Answer: C
\nSolution: Endothermic reactions may also be spontaneous.
Answer: A
\nSolution: At high temperature, process may become entropy driven.
Answer: A
\nSolution: At low temperature, process may become enthalpy driven.
Answer: C
\nSolution: If \u0394 G = negative and \u0394 S = negative, \u0394 H must be negative and \u0394 G = \u0394 H \u2013 T \u22c5 \u0394 S.
Answer: D
\nSolution: Theory based
Answer: D
\nSolution: \u0394 S sys = 0 but \u0394 S univ = +ve\n4.48 Chapter 4 HINTS AND EXPLANATIONS
Answer: A \u2192 P, R, S, T; B \u2192 P, Q, R, S; C \u2192 Q, R, S, T; D \u2192 P, Q, R, T
\nSolution: Theory based
Answer: A \u2192 Q; B \u2192 Q, S; C \u2192 Q, S; D \u2192 P
\nSolution: Theory based
Answer: A \u2192 P, R, S; B \u2192 Q, R, S; C \u2192 Q, R, S; D \u2192 R, S
\nSolution: Theory based
Answer: A \u2192 P, Q, R, S; B \u2192 R, S; C \u2192 Q
\nSolution: For ideal gas, H = f ( T ) but in general, H = f ( T , P )
Answer: A \u2192 P, S; B \u2192 Q, R, S; C \u2192 R
\nSolution: U f T V dU U T dT U V dV V T = \u21d2 = \u2202 \u2202 \u239b \u239d \u239c \u239e \u23a0 \u239f \u22c5 + \u2202 \u2202 \u239b \u239d \u239c \u239e \u23a0 \u239f \u22c5 ( , ) \u2202 \u2202 \u239b \u239d \u239c \u239e \u23a0 \u239f = \u22c5 \u2202 \u2202 \u239b \u239d \u239c \u239e \u23a0 \u239f = \u22c5 \u2202 \u2202 \u239b \u239d \u239c \u239e \u23a0 \u239f \u2212 U T n C U V T P T P V V m T V , and
Answer: A \u2192 P, R; B \u2192 Q, S; C \u2192 R; D \u2192 Q, S
\nSolution: N 2 ( g ) + O 2 ( g ) \u2192 2NO ( g ) ; \u0394 H = Positive, \u0394 S \u2248 0 2 KI ( aq ) + HgI 2 ( aq ) \u2192 K 2 [HgI 4 ]( aq ) ; \u0394 H = Negative, \u0394 S \u2248 Negative PCl 5 ( g ) \u2192 PCl 3 ( g ) + Cl 2 ; \u0394 H = Positive, \u0394 S \u2248 Positive NH 3 ( g ) + HCl ( g ) \u2192 NH 9 Cl ( s ); \u0394 H = Negative, \u0394 S = Negative
Answer: A \u2192 Q; B \u2192 P, S; C \u2192 R, S
\nSolution: (A) \u0394 H = 0, \u0394 U = 0, \u0394 S total = 0, \u0394 S Sys = Positive (B) q = 0, \u0394 S Sys = 0, \u0394 S total = 0, \u0394 S surr = 0 (C) q = 0, \u0394 S surr = 0, \u0394 S total = Positive.
Answer: A \u2192 P, S; B \u2192 P, R, S; C \u2192 Q; D \u2192 R, S
\nSolution: (A) \u0394 H = Negative, \u0394 V = \u00b1 V e (B) \u0394 H = \u00b1 V e, \u0394 S = \u2212 V e, \u0394 G = \u0394 H \u2013 T , \u0394 S = + V e, if \u0394 H = + V e = \u00b1 V e, if \u0394 H = \u2212 V e (C) \u0394 H = Ea f \u2013 Ea b = 10 kJ / mol = + V e \u0394 S = + V e \u2234 \u0394 G = \u0394 H \u2013 T \u0394 S = \u2212 V e, at high temperature (D) \u0394 H = + V e, \u0394 S \u2248 0 \u21d2 \u0394 G \u2248 \u0394 H
Answer: A \u2192 P, S, R; B \u2192 P, R; C \u2192 P; D \u2192 Q, R, S
\nSolution: (A) Solid \u001f Liquid ; \u0394 G = 0, \u0394 S = Positive, \u0394 V \u2248 0 \u21d2 \u0394 H \u2248 \u0394 U (B) Liquid \u001f Vapour : \u0394 G = 0, \u0394 S = Positive, (C) Triple point is equilibrium condition. (D) Melting at boiling point is spontaneous.
Answer: A \u2192 S; B \u2192 P; C \u2192 Q; D \u2192 R
\nSolution: dG = V \u22c5 dP \u2013 S \u22c5 dT \u21d2 ( dG ) T = V \u22c5 dP and ( dG ) P = \u2013 S \u22c5 dT
Answer: 6
\nSolution: w = \u2013 P ext ( V 2 \u2013 V 1 ) = \u2013 P ext [ V 0 (1 + \u03b3 \u22c5 t 2 ) \u2013 V 0 (1 + \u03b3 \u22c5 t 1 )] = \u2013 P ext \u22c5 V 0 \u22c5 \u03b3 \u22c5 ( t 2 \u2013 t 1 ) \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f \u00d7 \u00d7 \u00d7 \u00b0 \u00d7 \u00b0 \u2212 10 18 10 0 0002 10 5 2 6 3 N m ( ) . m C C = 0.0036 J
Answer: 6
\nSolution: q = \u0394 U \u2013 w = 0 \u2013 \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f nRT V V ln 2 1 or, 420 = 1 \u00d7 2 \u00d7 300 \u00d7 ln V 2 1 0 082 300 8 21 \u00d7 \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f . . \u21d2 V 2 = 6 L
Answer: 1
\nSolution: P = K \u22c5 l For initial condition, 1 bar = K \u00d7 1 m \u21d2 K = 1 bar/m And, V r l = = 4 3 6 3 3 \u03c0 \u03c0 \u21d2 dV l dl = \u22c5 \u03c0 2 2 Now, w P dV K l l dl K l l l l V V = \u2212 \u22c5 = \u2212 \u22c5 \u22c5 \u239b \u239d \u239c \u239e \u23a0 \u239f = \u2212 \u22c5 \u2212 \u222b \u222b ( ) \u03c0 \u03c0 2 2 4 2 2 4 1 4 1 2 1 2 = \u2212 \u00d7 \u00d7 \u2212 \u2212 \u00d7 10 2 4 1 4 1 10 5 2 4 4 4 7 N/m m J \u03c0 ( ) m \u001c
Answer: 6
\nSolution: \u2013 w = mgh \u21d2 P ext ( V 2 \u2013 V 1 ) = mgh = \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f \u00d7 \u2212 \u00d7 = \u00d7 \u00d7 \u21d2 = \u2212 4 10 8 2 10 40 10 6 5 2 3 3 N m m ( ) m h h
Answer: 6
\nSolution: q U n C T n C T Q Q Q m V m \u0394 = \u22c5 \u22c5 \u0394 \u22c5 \u22c5 \u0394 = \u2212 , 2 \u21d2 C R m 3 2 2 = \u21d2 C m = 6 cal/K mole\n4.49 Thermodynamics HINTS AND EXPLANATIONS
Answer: 7
\nSolution: q w n C T nR T P m = \u22c5 \u22c5 \u0394 \u2212 \u22c5 \u0394 , \u21d2 q R R \u2212 = \u2212 2 7 2 \u21d2 q = 7 J
Answer: 3
\nSolution: q = q 1 + q 2 = \u0394 U 1 + \u0394 H 2 = n C n C V m P m \u22c5 \u22c5 \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f + \u22c5 \u22c5 \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f , , 300 2 300 300 300 2 = \u22c5 \u22c5 = \u00d7 \u00d7 = n R 300 2 10 2 300 2 3000 cal
Answer: 5
\nSolution: State I State II Isothermal Isochoric \u23af \u2192 \u23af\u23af\u23af\u23af \u23af \u2192 \u23af\u23af\u23af T 1 = 273 K T 2 = 273 K T 3 = 5 \u00d7 273 K V 1 = V V 2 = 5 V V 3 = 5 V P 1 = P 0 P P 2 5 = P 3 = P q total = nRT V V n C T T V m \u22c5 + \u22c5 \u22c5 \u2212 ln ( ) 2 1 3 2 1 or, 80 \u00d7 10 3 = 3 \u00d7 8.314 \u00d7 273 \u00d7 ln5 + 3 \u00d7 C V m , \u00d7 4 \u00d7 273 \u2234 C V m , \u2248 21 J/K-mol = 5 cal/K-mol
Answer: 3
\nSolution: PT = Constant \u21d2 P \u22c5 V 1/2 = Constant Now, C C R x f f m V m = + \u2212 \u21d2 = \u00d7 + \u2212 \u21d2 \u2248 1 1 29 8 314 2 8 314 1 1 2 3 . .
Answer: 9
\nSolution: u u T T 2 1 2 2 2 300 1200 = = \u21d2 = K \u21d2 T 2 = 1200 K Now, q n C T T V V m = \u22c5 \u22c5 \u2212 = \u00d7 \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f \u00d7 \u2212 = , ( ) ( ) 2 1 56 28 5 2 2 1200 300 9000 cal
Answer: 6
\nSolution: Z u N RT M PN RT w aV A = \u22c5 \u22c5 = \u22c5 \u22c5 = 1 4 1 4 8 * \u03c0 Constant or, P T = \u21d2 \u22c5 = \u2212 Constant P V Constant 1 \u2234 C C R x R R R m V m = + \u2212 = + \u2212 \u2212 = = 1 1 5 2 1 1 3 6 ( ) cal/K mol
Answer: 5
\nSolution: \u0394 = \u22c5 \u0394 = \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f \u2212 \u00d7 = \u2212 G V P 13 0 78 10 3001 1 10 5000 6 3 5 . ( ) m N m J 2
Answer: 8
\nSolution: \u0394 G 1 = \u0394 H \u2013 T 1 \u22c5 \u0394 S and \u0394 G 2 = \u0394 H \u2013 T 2 \u22c5 \u0394 S \u2234 (\u2013 \u0394 G 2 ) \u2013 (\u2013 \u0394 G 1 ) = ( T 2 \u2013 T 1 ) \u22c5 \u0394 S = ( T 2 \u2013 T 1 ) \u00d7 \u0394 \u2212 \u0394 H G T 1 1 = (302 \u2013 298) \u00d7 ( ) ( ) \u2212 \u2212 \u2212 = 5737 6333 298 8 kJ
Answer: 3
\nSolution: Graphite \u001f Diamond; \u0394 G \u00b0 = 5.0 kJ P = 1 bar \u0394 G = 0 P = ? Now, \u0394 G 2 \u2013 \u0394 G 1 = ( V P \u2013 V G ) ( P 2 \u2013 P 1 ) or, 0 5000 12 3 6 12 2 4 10 10 3 10 6 2 5 9 \u2212 = \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f \u00d7 \u23a1 \u23a3 \u23a2 \u23a4 \u23a6 \u23a5 \u00d7 \u2212 \u21d2 \u00d7 \u2212 . . ( ) P N m 2
Answer: 5
\nSolution: \u0394 G \u00b0 = \u2013 RT \u22c5 ln K eq = \u20132 \u00d7 300 \u00d7 ln ( e \u201310 ) = + 6000 cal Now, \u0394 \u00b0 = \u0394 \u00b0 \u2212 \u0394 \u00b0 = \u2212 = S H G T 7500 6000 300 5 cal/k-mol\n4.50 Chapter 4 HINTS AND EXPLANATIONS Four-digit Integer Type
Answer: 5713
\nSolution: w nRT V nb V nb an V V = \u2212 \u2212 \u2212 + \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f \u23a1 \u23a3 \u23a2 \u23a4 \u23a6 \u23a5 ln 2 1 2 2 1 1 1 = \u2212 \u00d7 \u00d7 \u00d7 \u2212 \u00d7 \u2212 \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f + \u00d7 \u00d7 \u00d7 \u2212 1 8 314 300 20 1 0 03 2 1 0 03 1 42 10 1 1 20 1 12 2 . ln . . . 2 2 10 10 \u239b \u239d \u239c \u239e \u23a0 \u239f \u00d7 \u23a1 \u23a3 \u23a2 \u23a2 \u23a2 \u23a2 \u23a4 \u23a6 \u23a5 \u23a5 \u23a5 \u23a5 \u2212 = \u20135713.16 J
Answer: 0044
\nSolution: \u0394 H = \u0394 U + \u0394 ( PV ) = 30 + (4 \u00d7 5 \u2013 2 \u00d7 3) = 44 L -atm
Answer: 0209
\nSolution: Calcite \u2192 Aragonite \u0394 H = \u0394 U + P \u22c5 \u0394 V \u0394 V = 210 J + 2 7 10 100 3 100 2 7 10 5 6 3 . . \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f \u00d7 \u2212 N m m 2 = 209 J
Answer: 8400
\nSolution: \u0394 U = ( q + w ) path I = ( q + w ) path II or, 10 \u00d7 10 3 \u00d7 4.2 J + 0 = (11 \u00d7 10 3 \u00d7 4.2 J) + (\u20130.5 w max ) \u2234 w max = 8400 J
Answer: 0025
\nSolution: H = U + PV = 2.5 PV \u2234 \u0394 H = 2.5 \u00d7 10 200 100 10 25 5 3 3 N m m kJ 2 \u00d7 \u2212 \u00d7 = \u2212 ( ) 0 10 25 3 m kJ \u00d7 \u2212 \u00d7 = \u2212 ( )
Answer: 0020
\nSolution: q = \u0394 U \u2013 w = 1.5 nR \u22c5 \u0394 T + P ext \u22c5 A \u22c5 \u0394 l 42 = 1.5 \u00d7 1 \u00d7 8.314 \u00d7 2 + 100 \u00d7 10 3 \u00d7 8.5 \u00d7 \u00d7 10 -4 \u00d7 \u0394 l \u2234 \u0394 l \u2248 0.2 m
Answer: 6000
\nSolution: \u0394 H \u2013 \u0394 U = P ( V D \u2013 V G ) \u21d2 \u2212 = \u2212 \u23a1 \u23a3 \u23a2 \u23a4 \u23a6 \u23a5 \u00d7 \u2212 1000 12 3 6 12 2 4 10 6 P . . \u2234 P = 6000 \u00d7 10 5 Pa
Answer: 1567
\nSolution: Solid (70 C) Liquid (70 C) Liquid (450 C) Vapour (450 \u00b0 \u23af \u2192 \u23af \u00b0 \u23af \u2192 \u23af \u00b0 \u23af \u2192 \u23af 1 2 3 \u00b0 \u00b0 C) q = q 1 + q 2 + q 3 = 30 \u00d7 10 + 10 \u00d7 0.215 \u00d7 380 + 10 \u00d7 45 = 1567 cal
Answer: 0033
\nSolution: q = \u00d7 \u00d7 = 12 0 5 1805 6 1805 . J w = \u2013 P \u22c5 ( V g \u2013 V l ) = \u2013 P \u22c5 V g = \u2013 nRT = \u2212 \u00d7 \u00d7 = \u2212 0 9 18 8 314 373 155 . . J \u2234 \u0394 U = q + w = 1805 + (\u2013155) = 1650 J (for 0.9 g ) = \u00d7 \u00d7 = \u2212 1650 0 9 18 10 33 3 . kJ
Answer: 1990
\nSolution: q = \u0394 U \u2013 w = 0 \u21d2 \u0394 U = w = \u2013 P ext ( V 2 \u2013 V 1 ) = \u2013 100(\u20131) = 100 bar-ml Now, \u0394 H = \u0394 U + \u0394 PV = 100 + (100 \u00d7 99 \u2013 1 \u00d7 100) = 9900 bar-ml = 990 J
Answer: 1500
\nSolution: q = \u0394 U \u2013 w = 0 \u21d2 \u0394 U = w = \u00d7 \u00d7 1 2 10 1000 bar-ml = 500 J V 990 1000 1001 bar 1 bar (ml) 2 1 P
Answer: 0332
\nSolution: For adiabatic process: T T V V 2 1 1 2 1 1 4 1 300 2 400 = \u22c5 \u239b \u239d \u239c \u239e \u23a0 \u239f = \u00d7 = \u2212 \u22c5 \u2212 \u03b3 ( ) K Now, w = w 1 + w 2 = \u2212 \u22c5 \u239b \u239d \u239c \u239e \u23a0 \u239f + \u22c5 \u2212 nRT V V nC T T V m ln [ ( ) , 2 1 2 1 = [\u20131 \u00d7 8.3 \u00d7 300 \u00d7 ln 2] + 1 8 3 1 4 1 400 300 \u00d7 \u2212 \u00d7 \u2212 \u23a1 \u23a3 \u23a2 \u23a4 \u23a6 \u23a5 . . ( ) = 332 J
Answer: 0500
\nSolution: q n C T V m = \u22c5 \u22c5 \u0394 , or, 50 \u00d7 166 \u00d7 t = 1 8 21 10 2 9 10 0 0821 290 5 8 3 2 20 3 \u00d7 \u00d7 \u00d7 \u00d7 \u00d7 \u00d7 \u00d7 \u00d7 . . . . \u2234 t = 500 sec
Answer: 1900
\nSolution: \u0394 U = 3 \u00d7 1.5 R \u00d7 100 + 2 \u00d7 2.5 R \u00d7 100 = 1900 cal\n4.51 Thermodynamics HINTS AND EXPLANATIONS
Answer: 3120
\nSolution: V A (1200 K) (300 K) 64 atm 1 atm (300 K) B C P Path AB (Adiabatic): P P T T 2 1 1 2 1 1 1 1 3 5 1 64 1200 300 2 = \u22c5 \u239b \u239d \u239c \u239e \u23a0 \u239f = \u00d7 \u239b \u239d \u239c \u239e \u23a0 \u239f = \u2212 \u2212 \u03b3 atm Now, w total = w AB + w BC = \u22c5 \u22c5 \u0394 + \u2212 \u239b \u239d \u239c \u239e \u23a0 \u239f n C T nRT P P V m B C , ln = \u00d7 \u00d7 \u2212 \u2212 \u00d7 \u00d7 \u00d7 1 3 2 300 1200 1 300 2 1 R R ( ) ln = \u20133120 cal
Answer: 0169
\nSolution: \u0394 \u00b0 = \u2212 + \u00d7 = \u2212 + \u00d7 S S S S CH grap H 4 ( ) . ( . . ) 2 186 2 6 0 2 130 6 2 = \u201381 J/K Now, \u0394 G \u00b0 = \u0394 H \u00b0 \u2013 T \u22c5 \u0394 S \u00b0 = \u2013 T \u22c5 \u0394 S univ or, (\u201375 \u00d7 10 3 ) \u2013 300 \u00d7 (\u201381) = \u2013300 \u00d7 \u0394 S univ \u21d2 \u0394 S univ = 169 J/K
Answer: 0300
\nSolution: (\u2013 mgh ) \u00d7 N = (\u2013 \u0394 G \u00b0) or 50 \u00d7 10 \u00d7 2 \u00d7 N = 600 10 2 27 27 300 3 \u00d7 \u00d7 \u00d7 \u21d2 = N
Answer: 2870
\nSolution: (\u2013 \u0394 G ) = \u2013 ( \u0394 H \u2013 T \u22c5 \u0394 S ) = \u2212 \u2212 \u00d7 \u23a1 \u23a3 \u23a2 \u23a4 \u23a6 \u23a5 = ( ) 2808 310 200 1000 2870 kJ
Answer: 0012
\nSolution: 2 + 4 + 6 = 12
Answer: 2349
\nSolution: a = 2 (i, vi) b = 3 (ii, v, vii) c = 4 (iii, iv, viii, ix) d = 9 (all)