window.CHAPTER_SOLUTIONS = {
"equivalent-concept": {
"chapterTitle": "Equivalent Concept",
"chapterNumber": "2",
"sections": {
"A": {
"title": "Section A (Only one Correct)",
"solutions": [
{
"number": 1,
"pageNumber": 110,
"text": "Addition of hydrogen is reduction.",
"html": "Solution: Addition of hydrogen is reduction.
Solution: Sodium amalgam is an alloy (mixture of sodium and mercury).
Solution: 2 4 2 2 0 2 0 2 3 x x + − ( ) + × + × = − ⇒ = +
Solution: KBF 4 1 4 1 0 3 : + ( ) + + − ( ) = ⇒ = + x x
Solution: LiBiO 2 1 2 2 0 3 : + ( ) + + − ( ) = ⇒ = + x x
Solution: N N XeF Xe F : . : . : = = ⇒ 53 5 131 46 5 19 1 6 6 ∴ Oxidation state of Xe = +
Solution: 7. S H3C CH3 O S is more E N than C, but less E N than O.
Solution: Oxidation state of Fe is +2 in both as NO − and NOS − are the ligands.
Solution: CO is a neutral oxide.
Solution: Phosphorus acid: H PO 3 3 , oxidation state of P = +3. Orthophosphoric acid: H PO 3 4 , oxidation state of P = +5. Metaphosphoric acid: HPO 3 , oxidation state of P = +5. Pyrophosphoric acid: H P O 4 2 7 , oxidation state of P = +5.
Solution: Reducing agent undergoes oxidation. In HNO 3 , H and N are in their maximum oxidation state and hence, oxidation is possible only for ‘O’.
Solution: I − is a strong reducing agent.
Solution: Oxidizing agent must undergo reduction. I − can never be reduced.
Solution: In Ca OCl Cl ( ) , the oxidation state of ‘Cl’ is +1 and –1 but in the product Cl 2 , it becomes zero.
Solution: 2 5 4 2 5 2 2 2 4 2 Mn PbO H MnO Pb H O + + − 2+ + + → + +
Solution: AsO OH AsO H O e 3 3 4 3 2 2 2 − − − − + → + +
Solution: ‘Cr’ should be in +6 oxidation state in the product. For basic medium, the product should be CrO 4 2 − .
Solution: N O e + − − + → 5 3 4 Oxidation state of N in the product should be +1.
Solution: CH CH OH H O CH COOH H e 3 2 2 3 4 4 + ⇒ + + + −
Solution: CH CH OH H O CH COOH H e 3 2 2 3 4 4 + ⇒ + + + −
Solution: The oxidation state of ‘Cr’ in ClO x − must be +1 and hence, x =
Solution: 23. 1 mole of AO 3 − should gain 6 moles of electron. AO H e A H O x 3 2 6 6 3 − + − + + + → + From charge conservation, (–1) + ( + 6) + ( − 6) = + x ⇒ x = −
Solution: 24. E A V A V V = ⇒= = 21 14 1 2 As V 1 and V 2 must be integer, A should be 42, 84, 126, etc.
Solution: Mass of oxygen present = 14.9 g and mass of hydrogen combined with oxygen to form water = 16.78 – 14.9 = 1.88 g. ∴ Equivalent weight of oxygen = × = 14 9 1 88 1 008 7 989 . . . .
Solution: M e M A gm N e A 2 2 2 + − + → − ∴ 1 2 1 1 81 10 66 54 22 g → × = × ⇒ = N A A A . . .
Solution: For minimum equivalent weight, the basicity of acid should be maximum.
Solution: In the neutralization reactions, one mole of H O 2 is formed from 1 mole of H + and 1 mole of OH − ions and hence, equivalent weight of water =
Solution: 29. CH COOH Cl CCl COOH H e 3 3 3 3 6 + → + + − + − ∴ E CH COOH 3 60 6 10 = = EXERCISE II (JEE ADVANCED)\n2.36 Chapter 2 HINTS AND EXPLANATIONS
Solution: N H NH n factor 2 6 2 2 3 3 3 2 − = + → (No. of e − involved = 6) ∴ E x NH 3 1 3 = and E x N 2 2 6 = .
Solution: P NaOH H O PH NaH PO 4 2 3 2 2 3 3 3 + + → + The reaction is balanced by the loss and gain of 3 moles of electron per mole of P 4 and hence, E M P 4 = 3 .
Solution: Pb PbO H SO PbSO H O + + → + 2 2 4 4 2 2 2 2 The reaction is balanced by the loss and gain of 2 moles of electron per mole of Pb and hence, E M M H SO 2 4 2 2 = = .
Solution: 3 6 5 3 2 3 2 Cl NaOH NaCl NaClO H O + → + + The reaction is balanced by the loss and gain of 5 moles of electron per mole of NaClO 3 and hence, E M H O 2 3 5 = .
Solution: O H e O H O 3 2 2 2 2 + + → + + − ∴ E O 3 48 2 24 = = .
Solution: The reaction is balanced by the loss and gain of 2 moles of electron per mole of MnO 2 and hence, E M M HCl = = 4 2 2 .
Solution: n n eq eq I H AsO 2 3 4 = or w w 254 2 1 5 10 6 10 3 175 22 23 × = × × ⇒ = . . . g
Solution: n eq = × × = 6 10 6 10 0 001 20 23 .
Solution: 1 g equivalent always reacts with 1 g equivalent of other substance.
Solution: n eq metal oxide = n eq water or 8 6 8 1 8 18 2 35 . . E E M M + = × ⇒ =
Solution: Mass of acid = × = 20 0 5 1 40 . g
Solution: Let the acid be H A n . n eq acid = n eq magnesium salt or 1 0 1 1 301 12 35 54 . . . + = + ⇒ = − − − E E E A A A n n n ∴ Equivalent weight of acid = 1 + 35.54 = 36.54.
Solution: n eq metal oxide = n eq CO 2 or 3 7 8 1 0 44 4 . . E M + = × C CO ° + → ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 4 2 ∴ E M = 32 7 .
Solution: n n n eq -factor = × and n -factor is maximum for IO 3 − + − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 5 1 3 to .
Solution: n n eq eq KMnO FeC O 4 2 4 = ( ) FeC O Fe CO 2 4 3 2 → + + or n n × = × ⇒ = 5 1 3 3 5 .
Solution: n n eq eq KMnO SO 4 3 2 = − SO SO 3 2 4 2 − − → ( ) or n n × = × ⇒ = 5 1 2 2 5 .
Solution: n n n x eq eq R A MnO . . = ⇒ × = × 2 1 2 and n n n y eq eq R A K CrO . . . = ⇒ × = × 2 4 2 3 ∴ x y : : = 3 4
Solution: CHCl COOH H O CO H O Cl H e 2 2 2 2 2 3 2 6 6 + → + + + + + − n n eq eq CHCl COOH O A 2 = . . or n n × = ⇒ = 6 1 2 0 2 . . . Now, n n eq eq CHCl COOH NH 2 3 = or 0 2 1 1 0 2 . . . × = × ⇒ = x x
Solution: M M E E H SO H PO H SO H PO 2 4 3 4 2 4 3 4 98 2 98 1 1 2 : : : : = = =
Solution: 28 2 = ⇒ A M Atomic mass of metal, A M =
Solution: Now, n n eq eq H SO M O 2 4 2 3 = or w w 98 2 4 8 160 6 8 82 × = × ⇒ = . . g
Solution: Approximate atomic mass of metal = = 6 4 0 26 24 62 . . . . Let the metal chloride be MCl V . 24 62 35 5 95 2 . . + × ≈ ⇒ V V \u001f ∴ Exact atomic mass of metal = − × = 95 2 35 5 24 . . Now, n eq metal = n eq H 2 or 1 2 24 2 22 4 2 1 12 2 2 . . . × = × ⇒ = V V H H L L\n2.37 Equivalent Concept HINTS AND EXPLANATIONS
Solution: Approximate atomic mass of metal = = 6 4 0 55 11 63 . . . . Let the metal chloride be MCl V . Then, 2 74 6 11 63 35 5 4 × + × ⇒ ≈ . . . . \u001f V V Again, 2 74 6 4 35 5 7 2 × = + × ⇒ = . . . . A A
Solution: n n eq eq HCl CO = 2 or 200 1000 560 22400 2 0 25 × = × ⇒ = N N N HCL .
Solution: The resulting solution becomes neutral.
Solution: n x n eq eq FeSO NH SO H O KMnO 4 4 2 4 2 4 ⋅( ) ⋅ = or 5 88 284 18 1 20 25 250 3 16 1000 75 100 158 5 6 . . . + × = × ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ × × × ⇒ = x x
Solution: 2 2 2 2 Ag H S Ag S H + + + → ↓ + Cu H S CuS H 2 2 2 + + + → ↓ + ∴ Ratio of amount of H S 2 1 2 = :
Solution: NaAl OH CO HCl NaCl AlCl CO H O ( ) + → + + + 2 3 3 2 2 4 3 n n eq eq NaAl OH CO HCl ( ) = 2 3 or 100 0 1 1000 4 0 25 1000 160 × × = × ⇒ = . . V V ml
Solution: n n eq eq Au SnCl = 2 or 1 97 197 3 0 05 1000 2 300 . . × = × × ⇒ = V V ml
Solution: n n eq eq I Ce − + = 4 or 250 1000 2 20 0 05 1000 1 500 × × = × ⇒ = − M M . I M ∴ Concentration in g L = × = 1 500 127 0 254 .
Solution: n n eq eq U KMnO 4 4 + = or 0 5 238 2 50 1000 5 0 0168 . . × = × × ⇒ = M M
Solution: 2 6 10 6 3 2 2 4 2 4 2 2 2 2 IO H C O C O CO I H O − − + → + + + n n w w KIO H C O g 3 2 2 4 2 6 1 214 1 3 90 1 262 = ⇒ = × ⇒ = .
Solution: n n n eq eq eq H C O MnO I 2 2 4 4 = = − − or w w 90 2 500 1 0 1000 22 5 × = × ⇒ = . . g
Solution: n n eq eq Fe O KMnO 0 9 4 . = or 26 56 66 4 0 7 0 2 1000 5 280 . . . . × = × × ⇒ = V V ml
Solution: n n eq eq KMnO FeC O 4 2 4 = or 25 20 1 200 1000 5 20 1000 3 0 208 2 4 × × ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ × = × × ⇒ = M M FeC O M .
Solution: 20 1 200 0 1 20 0 1 75 0 1 3 30 0 1 3 150 1 1 2 2 3 3 × = × ⇒ = × = × ⇒ = × = × ⇒ = M M M M M M . . . . M M 1 1 150 M Now, n n eq eq K Cr O H O 2 2 7 2 2 = or 15 1 150 1000 6 0 02 1000 2 15 2 2 × × = × × ⇒ = V V . H O ml
Solution: n n eq eq metal H = 2 or 0 1 51 43 9 22400 2 2 . . × = × ⇒ x x \u001f Now, n n eq x eq M MnO + − = 4 or 0 1 51 58 8 0 1 1000 5 . . . × − ( ) = × ⇒ y x y \u001f
Solution: n n eq eq Na S O K BrO 2 2 3 3 = or 40 1000 1 0 1336 167 0 02 2 2 3 × × = ⇒ = M M . . Na S O M
Solution: n n eq eq HCl M PO = ( ) 3 4 2 or 10 0 1 1000 0 0517 31 67 20 03 × = + ⇒ = . . . . E E M M
Solution: n n eq eq HCl CaSO H O = ⋅ 4 2 1 2 or 525 0 1 1000 145 2 3 81 × = × ⇒ = . . w w g
Solution: Cu I Cu I 2 2 + − + + → + n n n eq eq eq Na S O I Cu 2 2 3 2 2 = = + or V V × × = × × ⇒ = 0 4 1000 1 50 0 2 1000 1 25 2 2 3 . . Na S O ml\n2.38 Chapter 2 HINTS AND EXPLANATIONS
Solution: n n n eq eq eq H O I Na S O 2 2 2 2 2 3 = = or 25 1000 2 20 0 3 1000 6 50 2 2 × × = × ⇒ = M M . H O M ∴ Volume strength = × = 6 50 11 35 1 362 . .
Solution: CS O CO SO 2 2 2 2 3 2 x x x mole mole 2 mole + → + H S O H O SO 2 2 2 2 3 2 y y y mole mole mole + → + 3 2 1 97 20 0 082 400 1 2 x y + = × × = . . . (1) Now, n n eq eq SO I 2 2 = or 2 2 250 2 8 1000 2 2 0 7 x y x y + ( ) × = × × ⇒ + = . . (2) From (1) and (2), x = 0 2 . ; y = 0 3 . Now, X x x y CS 2 0 4 = + = .
Solution: n CO 2 formed = = × − 0 9 100 9 10 3 . n NaOH = × = × − 100 0 5 1000 50 10 3 . 2 18 10 2 9 10 2 3 9 10 2 3 3 3 NaOH CO Na CO H O × × × − − − + → + ∴ Final solution contains NaOH = (50 − 18) × 10 –3 = 32 × 10 –3 mole and Na Co mole 2 3 3 9 10 = × − . Now, n n n eq eq eq HCl NaOH Na CO = + 2 3 or V V × × = × × + × × ⇒ = − − 0 5 1000 1 32 10 1 9 10 1 82 3 3 . HCl ml
Solution: Phenolphthalein: x x × × = × × ⇒ = 0 05 1000 1 40 0 05 1000 1 40 . . Methyl orange: y y × × = × × ⇒ = 0 05 1000 1 40 0 05 1000 3 120 . .
Solution: Let M x Na CO M 2 3 = ; M y NaHCO M 3 = Now, V x 1 1 1 100 1000 1 × × = × × (Phenolphthalein) and V x y 2 1 1 100 1000 2 100 1000 1 × × = × × + × × ( M e t h y l orange) ∴ x V = 10 1 and y V V = − ( ) 10 2 2 1
Solution: In presence of phenolphthalein, n -factor =
Solution: In presence of methyl orange, n -factor = 3.
Solution: NO + and ClO 4 − ⇒ , Oxidation state of N = +3, Cl = +7.
Solution: PbO 2 is an oxide.
Solution: (a) Oxidation state of S = +6 in both. (b) Oxidation state of Cr = +6 in both. (c) Oxidation state of P = +5 in both.
Solution: As KMnO 4 reduces, the compound must oxidize. Fe + 3 cannot oxidize.
Solution: – C N Oxidation state of C = +2, N = –3 – O O C C N – N O.S. of C = +4, O = –2, N = –3
Solution: ‘Cl’ should be in intermediate oxidation state.
Solution: KMnO eq 4 15 8 158 5 0 5 = × = . . HCl eq = × = 18 25 36 5 1 0 5 . . . H C O eq 2 2 4 22 5 90 2 0 5 = × = . . SO eq 2 32 69 2 1 0 = × = . FeSO eq 4 38 152 1 0 25 = × = .
Solution: 2 6 2 3 3 3 2 2 2 Cu HNO Cu NO NO NO H O + → + + + ( )
Solution: Equivalent volume is the volume occupied by 1g-equivalent of the gas. (a) 1g-equivalent of CH mole L L 4 1 8 1 8 22 4 2 8 = = × = . .\n2.39 Equivalent Concept HINTS AND EXPLANATIONS (b) 1g-equivalent of O mole L 3 1 2 11 2 = = . (c) 1g-equivalent of H S mole L 2 1 2 11 2 = = . (d) 1g-equivalent of CO mole L 2 1 4 5 6 = = .
Solution: n eq metal nitrate = n eq metal sulphate or 0 5 62 0 43 48 38 . . E E E M M M + = + ⇒ =
Solution: The chemical formula of sulphate is MSO H O 4 2 7 ⋅ . A A A + ( ) × = ⇒ = 222 20 100 55 5 . As the valency of metal is 2, the equivalent weight = = 55 5 2 27 75 . .
Solution: Higher oxide: 80 20 8 32 E E M M = ⇒ = . Lower oxide g Higher oxide g ( ( 4 29 4 77 . ) . ) → Mass of metal = × = 4 77 80 100 3 816 . . g Mass of oxygen = 4.29 – 3.816 = 0.474 g Lower oxide: 3 816 0 474 8 64 4 . . . E E M M = ⇒ =
Solution: Atomic weight (approx.) = = 6 4 0 03 213 33 . . . Now, 10 18 9 62 69 66 E E E M M M = + ⇒ = . . Now, Valency ≈ ≈ 213 33 69 66 3 . . (Integer value) ∴ Atomic weight (exactly) = 69 66 3 208 98 . . × =
Solution: Let molarities of Na CO 2 3 and NaHCO 3 be x M and y M, respectively. Phenolphthalein: 100 1000 1 1 1 10 1 1 × × = × × ⇒ = x V x V Methyl orange: 100 1000 2 100 1000 1 1 1 10 2 2 2 1 × × + × × = × ( ) × ⇒ = − ( ) x y V y V V
Solution: Molarity of oxalic acid solution = + 6 3 90 18 . x Now, 20 6 3 90 18 1000 2 40 0 05 1000 1 2 × + ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ × = × × ⇒ = . . x x E H C O H O 2 2 4 2 2 126 2 63 . = = but E H C O 2 2 4 90 2 45 = = Molarity = = 6 3 126 0 05 . . M Now, n eq acid = n eq KMnO 4 or 100 0 05 1000 2 0 05 1000 5 40 × × = × × ⇒ = . . V V ml.
Solution: (a) 109 % oleum means 100 g oleum (H SO SO 2 4 3 + ) exactly requires 9 g water to produce exactly 109 g of pure H SO 2 4 . H O SO H SO g g 2 18 3 80 2 4 + → ∴ 9 40 g g → ∴ Percentage of free SO 3 40 = % (b) 1 g of oleum contains 0.4 g of SO 3 and hence, 0.6 g of H SO 2 4 . Now, n n n eq eq eq SO H SO NaOH 3 2 4 + = or 0 4 80 2 0 6 98 2 0 5 1000 1 44 49 . . . . × + × = × × ⇒ = V V ml (c) n n n eq eq eq SO H SO Ba OH 3 2 4 2 + = ( ) or 2 0 80 2 3 0 98 2 0 1 1000 111 22 . . . . × + × = × ⇒ = V V ml (d) 100 g of oleum requires 9g of water to give 109 g of H SO 2 4 . Hence, the fi nal solution contains 109 g of H SO 2 4 and 491 g of water. ∴ Molality = × = 109 98 491 1000 2 265 / . . m
Solution: (a) Moles of P O 4 10 5 68 284 0 02 = = . . P O H PO aqueous 4 10 3 4 4 ⎯ → ⎯⎯⎯ ∴ Molarity of H PO 3 4 solution = × × = 0 02 4 250 1000 0 32 . . M (b) 25 0 32 1000 3 0 5 1000 1 48 × × = × × ⇒ = . . V V ml\n2.40 Chapter 2 HINTS AND EXPLANATIONS (c) 15 0 32 1000 3 0 2 1000 2 36 × × = × × ⇒ = . . V V ml (d) 40 0 32 1000 1 0 8 1000 16 × × = × ⇒ = . . V V ml
Solution: SO Cl H O H SO HCl mole mole mole 2 2 0 01 2 2 4 0 01 0 02 2 2 . . . + → + (a) H SO M 2 4 0 01 200 1000 0 05 [ ] = × = . . (b) HCl M [ ] = × = 0 02 200 1000 0 1 . . (c) 20 0 05 1000 2 20 0 1 1000 1 0 2 1000 1 20 × × + × × = × × ⇒ = . . . V V NaOH ml. (d) 100 0 1 1000 143 5 1 435 × = ⇒ = . . . w W AgCl g
Solution: n n n n eq eq eq eq H O KMnO K Cr O O 2 2 4 2 2 7 2 = = = or n n n n KMnO K Cr O KMnO K Cr O 4 2 2 7 4 2 2 7 5 6 × = × ⇒ > or W W W W KMnO K Cr O KMnO K Cr O 4 2 2 7 4 2 2 7 158 5 294 6 × = × ⇒ < or V M V M M M × × = × × ⇒ = KMnO K Cr O KMnO K Cr O 4 2 2 7 4 2 2 7 1000 5 1000 6 6 5
Solution: Moles of S = = 8 0 32 0 25 . . (a) 0 25 2 2 200 1000 5 0 25 4 . . × = × × ⇒ = M M KMnO M (b) M M BaCl 2 0 25 2 100 1000 1 25 = × = . . / (c) M g BaSO 4 0 25 2 234 29 25 = × = . .
Solution: E MnBr 2 215 17 12 65 = \u001f .
Solution: E PbO 2 240 2 120 = =
Solution: 2 17 30 2 2 15 14 2 2 3 4 3 2 3 2 2 MnBr PbO HNO HMnO Pb BrO Pb NO H O + + → + ( ) + ( ) + The reaction is balanced by the loss or gain of 34 electrons. Hence, E HNO 3 30 63 34 55 6 = × = . Comprehension II
Solution: Atomic weight (approx.) = = 6 4 0 08 80 . . n n eq eq Ag = arsenious chloride or 100 108 1 56 35 5 24 98 × = + ⇒ = E E As AS . . Now, Valency = 80 24 98 3 . \u001f ∴ Exact atomic mass = 24.98 × 3 = 74.94
Solution: Molecular mass of arsenious chloride = 6.25 × (22.4 × 1.3) =
Solution: Let the formula be AsCl x 3 ( ) . x x 74 94 3 35 5 182 1 0 . . . + × ( ) = ⇒ = ∴ Molecular formula = AsCl 3\n2.41 Equivalent Concept HINTS AND EXPLANATIONS Comprehension III Let the oxide be M O x 2 and the halide be MX y . For oxide: 0 4 8 . = × w A x (1) For halide: 4 0 . E w A y = × (2) From (1) and (2) , we get: E x y 80 =
Solution: y x E = ⇒ = 2 40
Solution: A E n = ⇒ . As valency should be an integer, A must be an integer multiple of
Solution: Comprehension IV
Solution: n n eq eq n KMnO X 4 = + or 1 5 10 5 2 5 10 5 2 3 3 . . . × × = × × − ( ) ⇒ = − − n n
Solution: E M n = -factor ∴ M n = × − ( ) = 56 5 168 Hence, atomic mass of X = − × = 168 2 35 5 97 . .
Solution: n n eq eq n KMnO X 4 = + or n n × = × ⇒ = 1 1 3 3 Comprehension V Let the sample contains x mole KCl and y mole KClO 3 . KClO KCl mole SO mole 3 10 10 2 y y ⎯ → ⎯⎯ Now, mole of AgCl formed, x y x y 10 10 0 1435 143 5 0 01 + = ⇒ + = . . . (1) For second experiment, n n n eq eq eq KC O O A FeSO l 3 4 + = . . or y y 10 6 37 5 0 08 1000 30 0 2 1000 5 10 3 × + × = × ⇒ = × − . . . From Equation (1), we get: x = × − 5 10 3
Solution: n n y x KClO Kcl 3 1 1 : : : = =
Solution: m x KCl g = × = 74 5 0 3725 . . m y KClO g 3 122 5 0 6125 = × = . . ∴ m moisture g = − + ( ) = 1 0 3725 0 6125 0 015 . . . , i.e., 1.5 %
Solution: Mass percent of KCl = × = 0 3725 1 100 37 25 . . % Comprehension VI
Solution: n n eq eq As O I 4 6 3 = − or 0 1188 396 8 10 1000 2 0 12 3 . . × = × × ⇒ = − M M M I
Solution: n n HCN I = − 3 or 15 1000 5 0 12 1000 0 04 × = × ⇒ = M M . . HCN M
Solution: Mass of HCN g = × ( ) × = 6 0 04 27 6 48 . . Comprehension VII
Solution: 4 20 8 5 4 2 4 4 2 Mn MnO F H MnF H O + − − + − + + + → +
Solution: Let the moles of Mn O 3 4 = x. Hence, moles of Mn 2 + formed = 3 x .\n2.42 Chapter 2 HINTS AND EXPLANATIONS Now, for KMnO 4 solution, molarity = 0 125 5 . = 0.025 M. Now, n n eq eq KMnO Mn 4 2 = + or 30 0 025 1000 4 3 1 10 3 × × = × ⇒ = − . x x Now, mass of Mn O g 3 4 229 0 229 = × = x . ∴ Percentage of Mn O 3 4 0 229 0 458 100 50 = × = . . %
Solution: Normality = × = 0 025 4 0 1 . . . N Comprehension VIII Let the ore contains x mole of FeCr O 2 4 and y mole of Fe O 0 95 1 00 . . . Now, n n n eq eq eq FeCr O Fe O O 2 4 0 95 1 00 2 + = . . or x y x y × + × = × ⇒ + = 7 0 85 280 22400 4 7 0 85 0 05 . . . (1) Now, Moles of K Fe CN 4 6 ( ) ⎡ ⎣ ⎤ ⎦ taken = × = 10 1 1000 0 01 . n n eq eq Fe K Fe CN 2 4 6 + = ( ) ⎡ ⎣ ⎤ ⎦ or 6 0 4 1000 2 0 0012 × = = ⇒ = . . n n . ∴ Moles of K Fe CN 4 6 ( ) ⎡ ⎣ ⎤ ⎦ reacted with Fe 3 0 01 0 0012 + = − . . = 0 0088 . Now, n n eq eq Fe K Fe CN 3 4 6 + = ( ) ⎡ ⎣ ⎤ ⎦ or ( . ) . . . x y x y + × = × ⇒ + = 0 95 3 0 0088 3 0 95 0 0088 (2) From (1) and (2), we get: y = × − 2 10 3 , x = × − 6 9 10 3 .
Solution: Mass per cent of Fe O 0 95 1 00 3 2 10 69 2 2 100 6 92 . . . . % = × × × = −
Solution: Let in Fe O 0 95 1 00 . . , ‘z’ Fe-atom are in +2 state. z z z × + ( ) + − ( ) × + ( ) = ⇒ = 2 0 95 3 2 00 0 85 . . . Hence, per cent of total iron in +2 state = + × + × x y x y 0 85 0 95 100 . . = 97.73 %
Solution: Moles of Prussian blue = moles of Fe 3 + = + = × − x y 0 95 8 8 10 3 . . Comprehension IX
Solution: Moles of NaHSO 3 needed = × 3 moles of NaIO 3 = × = 3 5 94 198 0 09 . . ∴ Mass of NaHSO 3 needed = × = 0 09 104 9 36 . . . g
Solution: Moles of SO 3 − needed in 2nd reaction. = × 1 5 moles of I − formed in 1st reaction = × = × − 1 5 5 94 198 6 10 3 . ∴ Volume of solution required = × = − 6 10 5 94 198 0 2 3 . / . L
Solution: Mass of I 2 produced = × × × = − 6 10 3 254 4 572 3 . g ∴ Mass of I 2 produced per litre of solution = + ( ) = 4 572 1 0 2 3 81 . . . g Comprehension X
Solution: n n n eq eq eq HCl Na CO NaOH = + 2 3 or V V × = × + × ⇒ = 1 1000 1 106 1 1 40 1 34 43 HCl ml .
Solution: n n n n eq eq eq eq HCl Na CO NaHCO NaOH = + + 2 3 3 or V V × = × + × + × ⇒ = 1 1000 1 106 2 1 84 1 1 40 1 55 77 HCl ml .\n2.43 Equivalent Concept HINTS AND EXPLANATIONS
Solution: n n eq eq HCl NaHCO = 3 formed + n eq NaHCO 3 , present initially. or V V × = × + × ⇒ = 1 1000 1 106 1 1 84 1 21 34 HCl ml .
Solution: I − can never be reduced.
Solution: The oxidation state of terminal C-atoms are –3 and middle C-atoms are –2.
Solution: I − can not be reduced and hence, reduction of O 3 occurs. O H e O H O 3 2 2 2 2 + + → + + −
Solution: Oxidation state of Cl = + 3
Solution: n -factor may be fractional.
Solution: V V × × = × × 0 3 2 0 2 3 . .
Solution: In the presence of methyl orange, the colour change appears when Na CO 2 3 converts completely in H CO 2 3 .
Solution: Copper converts from Cu 2 + to Cu + .
Solution: n n eq eq A C = only when n-factor of B is same in both reactions.
Solution: Column I Column II (A) n eq = 1 × 5 (P) n eq = 3 × 1 (B) n eq = 1 × 3 (Q) n eq = 0.5 × 10 (C) n eq = 1 × 2 (R) n eq = 1 × 2 (D) n eq = 1 × 3 (S) n eq = 1.5 × 2
Solution: Column I Column II (A) n n n n n n eq O.A. eq R.A. MnO C O 4 2 4 2 1 2 1 2 3 2 2 3 − ( ) − ( ) = ⇒ × = × ⇒ = : : (B) n n n n n n eq O.A. eq R.A. ClO Fe OH − ( ) ( ) = ( ) ⇒ × = × ⇒ = 3 1 2 1 2 2 3 3 2 : : (C) n n n n n n eq O.A. eq R.A. HO Cr OH 2 3 1 2 1 2 2 4 2 1 − ( ) ( ) = ( ) ⇒ × = × ⇒ = : : (D) n n n n n n eq R.A. eq O.A. N H Cr OH 2 4 2 1 2 2 1 6 2 3 1 ( ) ( ) = ( ) ⇒ × = × ⇒ = : :
Solution: Equivalent volume is the volume of gas corresponding to 1 g-equivalent of the gas.
Solution: (A) n -factor = 2 (B) n -factor = 10 (C) n -factor = × + = 2 10 2 10 5 3 (D) n -factor = × + = 2 2 2 2 1
Solution: (P) n -factor = 1 (Q) n -factor = 1 2 (R) n -factor = 2 (S) n -factor = 6 (T) n -factor = 1 2\n2.44 Chapter 2 HINTS AND EXPLANATIONS
Solution: Milliequivalents: Column I Column II (A) 100 × 0.3 × 2 = 60 for P, Q, R, S (P) 100 × 0.3 × 1 = 30 (B) 50 × 0.6 × 1 = 30 for P, S 50 × 0.6 × 2 = 60 for Q, R (Q) 120 × 0.1 × 5 = 60 (C) 50 × 0.6 × 1 = 30 for P, Q, R, S (R) 60 × 0.1 × 5 = 30 (D) 100 × 0.2 × 3 = 60 for P, Q, S (S) 100 × 0.6 × 1 = 60
Solution: Milliequivalents: Column I Column II (A) 50 × 0.5 × 2 = 50 (P) 50 × 0.5 × 2 = 50 (B) 50 × 0.5 × 1 = 25 (Q) 50 × 0.5 × 1 = 25 (C) 50 × 0.5 × 1 = 25 (R) 25 × 0.5 × 2 = 25 (D) 50 × 0.5 × 1 = 25 (S) 50 × 1.0 × 1 = 50
Solution: AO H e HAO H O n n 4 2 2 9 2 2 4 − + − − + − + → + − ( ) ( ) n n From charge conservation, − ( ) + − ( ) + − ( ) = − ⇒ = n n n 9 2 2 2
Solution: 2. 5 4 2 3 2 2 4 2 AO H AO A H O n + → + + + − + From charge conservation, + ( ) = − ( ) + ⇒ = 4 2 3 2 n n .
Solution: n n eq eq Cl 2 = metal chloride or 1 12 22 4 2 5 55 35 5 . . . . × = + ⇒ E Equivalent weight of metal, E = 20 ∴ Valency = = = A E 40 20 20
Solution: 160 96 2 2 10 = × + ⇒ = ⇒ n n n Oxidation state of Br in unknown product = 5
Solution: E = × + ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = 30 2 2 2 2 30
Solution: M s Cl g Chloride g ( ) + ( ) → ( ) 2 As the volume reduced by one-third, moles of chloride formed = × 2 3 Moles of Cl 2 reacted. Mass of chorine used = × n 71 g. Mass of chloride formed = × × = 2 3 68 75 2 275 3 n n . g ∴ Mass of element used = − = 275 3 71 62 3 n n n g Now, equivalent weight of element, E n n = × = 62 3 71 35 5 31 3 / .
Solution: n n eq eq SO HClO 2 3 = or V V 22 4 2 16 9 84 5 6 13 44 . . . . × = × ⇒ = L Now n PV RT = = × × ( ) × = 2 5 13 44 0 0821 546 3 . .
Solution: n n eq eq N H K CrO 2 4 2 4 = or w w 32 4 24 194 3 2 97 3 × = × ⇒ = . g g \u001f
Solution: n n eq eq KHC O H C O H O NaOH 2 4 2 2 4 2 2 ⋅ ⋅ = or n n × = × ⇒ = 3 30 1 10 Now, n n eq eq KHC O H C O H O KMnO 2 4 2 2 4 2 4 2 ⋅ ⋅ = or 10 4 5 8 × = × ⇒ = n n
Solution: n n eq eq Mn NO H O Na O 3 2 4 6 6 ( )⋅ = Xe or n n × = × × ⇒ = × − 5 62 5 0 04 1000 8 4 10 3 . .\n2.45 Equivalent Concept HINTS AND EXPLANATIONS
Solution: Cl NaClO ClO NaCl 2 2 2 2 2 2 + ⇒ + n n eq eq NaClO ClO 2 2 90 100 × = or 5 2 90 100 9 × × = ⇒ = n n
Solution: n x n n eq eq eq CuSO H O H SO NaOH 4 2 2 4 ⋅ = = or 1 245 159 5 18 2 10 1 1000 1 4 97 5 . . . + × = × × ⇒ = x x \u001f
Solution: CS O CO SO mole mole mole 2 2 2 2 2 2 x x x + → + H S O H O g SO mole mole mole 2 2 2 2 3 2 y y y + → ( ) + From the question, 3 2 7 2 82 1 0 0821 600 12 x y + = × × = . . . and 2 2 75 2 x y + ( ) × = × Hence, x = 2 and y =
Solution: 14. XeF H Xe HF x mole mole + → + x x a a x 2 2 a = 56 22400 and ax x × = × × ⇒ = 1 60 0 25 1000 1 6 .
Solution: n H O n I n Na S O eq eq eq 2 2 2 2 2 3 = = or 12 5 675 11 35 1000 2 24 1000 1 0 5 × × = × × ⇒ = . . . M M Now, n n n eq eq eq O I Na S O 3 2 2 2 3 = = or V V 22 4 2 9 0 5 1000 1 0 504 . . . × = × × ⇒ = L ∴ Percentage of O 3 0 0504 1 100 5 04 5 = × = . . % % \u001f
Solution: H C O H O CO CO mole mole 2 2 4 2 2 x x → + + SCO I O CO I mole mole x x + → + 2 5 2 2 5 5 n n eq eq I Na S O 2 2 2 3 = or x x 5 2 200 0 2 1000 0 1 × = × ⇒ = . . ∴ Mass of H C O g 2 2 4 0 1 90 9 = × = .
Solution: Ammonium Vanadate Z V oxalic acid V KMnO SO x ⎯ → ⎯⎯ ⎯ → ⎯⎯⎯ ⎯ → + ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ + 4 2 5 ⎯ ⎯⎯ ⎯ → ⎯⎯⎯ + + V V KMnO 4 5 4 n Z n n x V N eq eq KMnO = ⇒ × − ( ) = × 4 1 5 n n n V N eq eq V KMnO 4 4 2 1 + = ⇒ × = × From question, V V x 1 2 5 1 0 = ⇒ =
Solution: Let Cu a 2 + = mole and C O b mole 2 4 2 − = . n n b eq eq C O MnO 2 4 2 4 2 22 6 0 02 1000 5 − − = ⇒ × = × × . . n n n a eq eq eq Cu I Na S O 2 2 2 2 3 1 11 3 0 05 1000 1 + = = ⇒ × = × × . . ∴ a : b = 1 : 2
Solution: Moles of Fe O 2 3 0 552 160 = ⇒ . Moles of Fe 2 + formed = × 2 0 552 160 . Now, n n eq eq Fe 2 + = Oxidizing agent or 2 0 552 160 1 17 25 100 0 0167 1000 6 × × = × ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ × × ⇒ . . , n n n f f -factor \u001f
Solution: CH CH COOH CO Na mole mole NaOH excess 3 2 2 2 2 2 ( ) → + ( ) ⎯ → ⎯⎯ + ( ) + ( ) n a a n n n 2 2 3 2 CO mole NaOH left mole + + ( ) x a n In presence of phenolphthalein: x a n × + + ( ) × = × 1 2 2 1 50 1 1000 (1) In presence of methyl orange: x a n × + + ( ) × = × 1 2 2 2 80 1 1000 (2) From (1) and (2), a n + ( ) = 2 2 0 03 . and a n = + 1 16 60 14 . ∴ n = 4\n2.46 Chapter 2 HINTS AND EXPLANATIONS Four-digit Integer Type
Solution: 1 3 2 1 1 5 3 2 0 2 3 + ( ) + + ( ) + − ( ) + ( ) + × − ( ) = ⇒ = x x x ∴ Percentage of ‘ x ’ in +2 state = × = 2 3 2 100 33 33 % . %
Solution: n n eq eq OH − = Metal hydroxide or 0 1 1 8 50 17 68 . . × = + ⇒ = E E
Solution: n n eq eq Ag AgCl = or w w 108 1 2 87 143 5 1 2 16 × = × ⇒ = . . . g ∴ Mass of Cu = − = 2 7 2 16 0 54 . . . g ∴ Percentage of Cu . . % = × = 0 54 2 7 100 20
Solution: n n eq eq H SO CaO 2 4 2 = or V V × × = × ⇒ = 0 25 1000 2 7 2 72 2 400 . . ml
Solution: n n eq eq NaOH Oxalate = or 27 0 12 1000 30 9 15 1000 × = × × . . M y (1) n n eq eq KMnO Oxalate 4 = or 36 0 12 1000 30 9 15 1000 2 × = × × . . M z (2) From charge conservation, x y z + = 2 (3) and molar mass, M x y z n = + + + 39 88 18 (4) Solving (1), (2), (3) and (4), we get: x y z : : : : = 1 3 2 and n =
Solution: 6. 5 8 5 2 200 0 1 1000 4 200 0 1 1000 600 1 0 1000 3 Fe MnO H Fe M M M + × − × + × + + + → + . . . M Mn H O 2 2 4 + + H M final + ⎡ ⎣ ⎤ ⎦ = − × = 600 1000 8 5 20 1000 568 1000
Solution: Moles of KOH used in saponification = × − × × = × − 25 0 4 1000 8 0 0 25 1000 2 6 10 3 . . . ∴ Mass of KOH used = × × = − 6 10 56 0 336 3 . g ∴ Saponification number = × = 0 336 10 1 5 224 3 . .
Solution: n eq Oxalic acid = + n n eq eq MnO KMnO 2 4 or 50 1 0 1000 2 320 0 1 1000 9 10 2 2 3 × = × + × ⇒ = × − . . n n MnO MnO MnO MnO available oxygen 2 → + ’O’ ∴ Percentage of available oxygen = 9 10 16 1 6 100 9 3 × × × = − . %
Solution: n n n V eq eq SeO BrO 3 2 3 1 2 1 40 1000 5 − − = ⇒ × = × × / (1) n n V eq eq AsO BrO 2 3 2 7 5 1 25 1000 2 1 40 1000 6 − − = ⇒ × × = × × . (2) and V V 1 2 20 + = (3) From (1), (2) and (3), we get: n = − 10 3 ∴ Mass of SeO g mg 3 2 3 10 127 127 − − = × =
Solution: Moles of V O 2 5 91 182 0 5 = = . ∴ Moles of V 2 + formed = × = 0 5 2 1 0 . . Now, n n eq eq V I 2 2 + = or 1 0 2 2 254 2 254 . × = × ⇒ = w g
Solution: n n n eq eq eq I C H O Na S O 2 6 8 6 2 2 3 = + or 10 0 025 1000 2 176 2 2 5 0 01 1000 8 0 0264 × × = × + × × ⇒ = . . . . w w ∴ Vitamin C content = × = 0 0264 200 10 132 6 . mg/mL\n2.47 Equivalent Concept HINTS AND EXPLANATIONS
Solution: n eq Mohr salt = + n n eq eq Na CrO K Cr O 2 4 2 2 7 or 1 96 392 1 2 3 40 0 05 1000 . . × = × + × a ∴ Moles of Fe CrO 2 2 4 5 10 ( ) = = × − a ∴ Mass of Cr – present = × × = 2 52 0 2 100 26 a . % %
Solution: n n n n eq eq eq eq CuS Cu S Fe MnO + + = + − 2 2 4 or x x 96 6 10 160 8 200 1 1000 1 900 0 4 1000 5 × + − × + × × = × × . ∴ x =
Solution: ∴ Percentage of CuS = × = x 10 100 80%
Solution: Let the mixture contains x moles of As S 2 3 and y moles of As S 2 5 . n n eq eq As S I 2 3 2 = or x x × = × ⇒ = × − 4 20 0 05 1000 2 5 10 4 . . Now, n n n eq eq eq As S I Na S O H O 2 5 2 2 2 3 2 5 = = ⋅ or x y y + ( ) × = × ⇒ = × − 4 1 24 248 1 1 0 10 3 . . ∴ Mole percent of As S 2 3 100 20 = + × = x x y %
Solution: 3 6 6 3 2 6 3 2 2 2 3 4 Se s Ag NH H O Ag Se Ag SeO NH ( ) + + + → + + + + x mole 45 0 02 1000 9 10 4 × = × − . mole 0 0 0 ( ) 9 10 2 4 × − − x mole 2 x 3 mole x 3 mole From the question, 9 10 2 2 3 10 0 01 1000 4 × − ( ) + × = × = ( ) − + − x x n n . Ag SCN ∴ x = × − 6 10 4 ∴ Mass of Se per ml 6 10 80 2 24 4 × × = − gm mg
Solution: n n eq eq KI KIO = 3 or 20 1000 2 30 1 10 1000 4 × × = × × M ⇒ Molarity of KI solution = 0.3 M Now, moles of KI taken = × = × − 50 0 3 1000 15 10 3 . and moles of KI reacted with KIO 3 3 2 50 1 10 1000 10 10 = × × = × − ∴ Moles of KI reacted with AgNO 3 3 3 3 15 10 10 10 5 10 = × − × = × − − − = Moles of AgNO 3 present ∴ Percentage of AgNO 3 3 5 10 170 1 100 85 = × × × = − %
Solution: Let the original sample contains x moles of Fe O 3 4 and y mole of Fe O 2 3 . As, n n n n eq eq eq eq Fe O Fe O I Na S O 3 4 2 3 2 2 2 3 + = = or x y x y × + × = × ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ × × ⇒ + = × − 2 2 11 2 20 100 0 5 1000 1 2 2 28 10 3 . . (1) Now, moles of Fe 2 + formed = + 3 2 x y As, n n eq eq Fe KMnO 2 4 + =\n2.48 Chapter 2 HINTS AND EXPLANATIONS or 3 2 1 12 8 50 100 0 25 1000 5 3 2 32 10 3 x y x y + ( ) × = × ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ × × ⇒ + = × − . . (2) From (1) and (2), we get: x = × − 4 10 3 and y = × − 10 10 3 . Now, the percentage of Fe O 2 3 in sample = × × = y 160 4 100 40%
Solution: IO I H I H O mole mole 3 1 07 214 5 10 2 15 10 2 3 3 5 6 3 3 − = × − + × − − + + → + . Now, n n eq eq I Na S O 2 2 2 3 = or 15 10 2 50 1000 1 0 6 3 × × = × × ⇒ = − M M .
Solution: C O MnO H CO Mn mole mole 2 4 2 10 0 2 1000 2 10 2 2 10 3 2 2 3 4 2 2 − × = × + × − + − + + → + + . H H O 2 2 3 2 5 4 2 5 2 10 0 8 10 2 2 2 2 10 3 3 3 MnO Mn H O MnO mole mole mol − × × = × + × − − − + + → . e e H + + 4 2 6 5 4 0 8 10 2 2 5 2 0 8 10 2 10 2 1 3 3 3 MnO H H O mole mole mole − × + × × = × = × − − − + + . . 0 0 34 2 2 2 3 2 8 5 − × + → + + g Mn H O O ∴ Mass of H O 2 2 per 100 ml of solution = × × = − 68 10 20 100 340 3 g mg
Solution: In the presence of phenolphthalein, 20 0 1 1000 1 0 05 1000 1 40 1 1 × × = × × ⇒ = . . V V ml In the presence of methyl orange, 20 0 1 1000 3 0 05 1000 1 120 80 2 2 2 1 × × = × × ⇒ = ∴ − = . . V V V V ml ml
Solution: P gas + 15.6 = 53.3 + 76.3 ⇒ P gas = 114 cm Hg = 1.5 atm
Solution: Fractional increase = V V V 2 1 1 − = V V 2 1 1 − = P P 1 2 1 − = H H H + − 7 7 1 = 1 7
Solution: P 2 P 1 44.5 46 P 0 P 0 45.25 45.25 P 0 × 45.25 = P 1 × 46 = P 2 × 44.5 P 1 + 5 sin30° = P 2 P 0 45 25 46 × . + 5 × 1 2 = P 0 45 25 44 5 × . . ⇒ P 0 = 75.4 cm Hg
Solution: P 1 V 1 = P 2 V 2 ⇒ 10 × 2A = (10 + h ) × h A ⇒ h = 1.71 m
Solution: dv dt = V 0 273 = 0.08 ⇒ V 0 = 21.84 L
Solution: V T 1 1 = V T 2 2 ⇒ 1 0 0 . x + = 0 6 100 . ( ) x + − ⇒ x = 250 ⇒ 0 K = –250°C
Solution: V T 1 1 = V T 2 2 ⇒ V T = V V T T + + Δ Δ ⇒ Δ Δ V V T . = 1 T ⇒ y = 1 x
Solution: V T 1 1 = V T 2 2 ⇒ V t 1 1 273 + = 1.1 V t 1 2 273 + ∴ Percentage increase in temperature = t t t 2 1 1 − × 100 = ( ) 10 2730 1 + t %
Solution: Number of SO 2 molecules = N ⇒ Number of atoms = 3N
Solution: n n N O 2 2 = V V N O 2 2 ⇒ m m N O 2 2 28 32 = 1 7 8 ⇒ m m N O 2 2 = 1 1
Solution: V n 1 1 = V n 2 2 ⇒ 4 3 10 2 8 3 π ( ) = 4 3 2 1 3 π d ( ) ⇒ d = 5 cm
Solution: P = P CO 2 + P air = 0 5 0 0821 300 1 1 . . × × + = 13.315 atm
Solution: Weight of fi lled balloon, W = 20 g + 40 × 0.6 = 44 g Weight of displaced air, B = 40 × 1.3 = 52 g ∴ Balloon will lift upward with pay load = 52 – 44 = 8 g B W\n3.45 Gaseous State HINTS AND EXPLANATIONS
Solution: Water will behave like ideal gas on disappearance of intermolecular forces. V = nRT P = 4 5 10 18 3 . × × (22.4 × 10 –3 ) m 3 = 5.6 m 3
Solution: d = m v ⇒ 1.5 = n n n n co co co co × + × + × × 28 44 0 0821 300 1 2 2 ( ) . ⇒ 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 × P total = 7 055 7 055 8 945 760 . . . + × mm = 335.1 mm
Solution: V V water vapour water = 1 0 0821 373 1 18 0 96 × × ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ . . l ml = 1633.24
Solution: V O 2 = 3 2 32 0 0821 310 1 . . × × = 2.5451 L V CO 2 = 8 8 44 0 0821 310 1 . . × × = 5.0902 L
Solution: n CO 2 = 200 0 1 1000 × . = 0.02 ∴ V CO 2 = 0.02 × 22.4 = 0.448 L
Solution: d d d ( ) ρ = M RT = 1.2 × 10 –5 Kg m –3 Pa –1 ⇒ M 8 314 300 . × = 1.2 × 10 –5 ∴ M air = 0.03 Kg/mol = 30 gm/mol Now, 30 = n n n n N O N O 2 2 28 + 32 + × × 2 2 ⇒ n n N O 2 2 : = 1:1
Solution: m T P 1 1 1 = m T P 2 2 2 ⇒ 4 × T P = m T P 2 2 × 2 ⇒ m 2 = 16 gm Hence, (16 – 4) = 12 gm gas should be added.
Solution: 74 5 50 . = 1.49 times
Solution: V 1 d 1 = V 2 d 2 ⇒ 1500 × 1.25 = 3.92 × d 2 ⇒ d 2 = 478.3 kg/mol
Solution: P V T 1 1 1 = P V T 2 2 2 ⇒ P r T × 4 3 1 3 π = P r T 4 4 3 2 3 × π 2 ⇒ r 2 = 2 r 1 ∴ % Increase in radius = r r r 2 1 1 − × 100 = 100 %
Solution: Constant = P 2 V = nRT V V 2 ⇒ T V 2 = Constant ∴ On expansion, temperature will increase.
Solution: r R M = ⇒ r r r n H N e 2 > > 2
Solution: P × 3 = 7 28 0 0821 300 × × . ⇒ 2.0525 atm
Solution: 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 × × . = 900 K.\n3.46 Chapter 3 HINTS AND EXPLANATIONS
Solution: 2Al + 2NaOH + 2H 2 O → 2NaAlO 2 + 3H 2 2 mole 3 mole ∴ 0.15 27 mole 3 2 × 0.15 27 mole ∴ V H 2 = 1 5 0 15 27 0 0831 300 0 831 . . . . × × × = 0.25L
Solution: N 2 → 2N Initial mole a = 1 4 28 . 0 Final mole a – 0.4 a 2 × 0.4 a = 0.6 a = 0.8 a Final total moles = 0.6 a + 0.8 a = 1.4 × 1 4 28 . = 0.07 ∴ P = 0 07 0 0821 1800 5 . . × × \u001e 2.07 atm
Solution: CH 4 (g) + 2O 2 (g) 127 ° ⎯ → ⎯⎯ 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 ⇒ 1 300 = P 2 400 ⇒ P 2 = 1.33 atm
Solution: P c = X c P total ⇒ 10 – (1 + 3) = n C 10 10 × ⇒ n c = 6 ∴ Mass of C = 6 × 2 = 12 gm
Solution: x x M R 4 + 5 = 760 2.4 300 − × × and x R 4 = 19 2.4 15 × × ∴ M = 96
Solution: C 2 H 6 + 7 2 O 2 2CO 2 + 3H 2 O( l ) 1 vol 7 2 vol 2 vol 0 vol ∴ 10 ml 35 ml 20 ml 0 Final volume should be 20 + (40 – 35) = 25 ml but it is 26 ml. Hence, volume occupied by water vapour is (26 – 25) = 1 ml. ∴ Vapour pressure of water = 1 26 × 1 atm = 760 26 = 29.23 mm Hg
Solution: n H 2 O vapour needed = ( . ) . 26 463 24 1 760 0 0821 300 − × × × = 1.32 × 10 –4
Solution: P = 1 2 0 0821 300 18 50 760 . . × × × × = 24.96 mm Hg
Solution: Mass of water lost per day = ∆ P.V RT M × = ( ) . 45 5 760 0 0821 310 18 − × × ×10000 = 372.23 gm
Solution: Vapour pressure is a function of temperature only
Solution: After achievement of equilibrium with its liquid form which will form on continuous injection of vapour, the pressure due to vapours become constant.
Solution: Rate of evaporation will remain constant throughout because neither surface area nor temperature are changing
Solution: r r x y = 1 5 and r r y z = 1 6 ⇒ r r z x = 30 1
Solution: Smaller the rate of diff usion of HX, more closer to the HX end, NH 4 X will form.
Solution: r r N H 2 2 = M M H N 2 2 ⇒ Δ Δ P P t / / 60 = 2 28 ⇒ t = 16.04 min
Solution: M dry air > M moist air
Solution: r r CH HBr 4 = P P CH HBr 4 M M HBr CH 4 ⇒ 1 1 = n n CH HBr 4 81 16 ⇒ n n CH HBr 4 = 0.4 ∴ X CH 4 = n n n CH CH HBr 4 4 + = 0.31
Solution: As HCl will diff use slowly, white fumes will form closer to HCl end.
Solution: In gases, the intermolecular distance is much higher than the size of molecules.
Solution: u u av,2 av,1 = T T 2 1 = 375 250 = 1.22
Solution: u u rms, O rms, O 2 = 3R 2T 16 3RT 32 × 2 1 ⇒ u rms, o = 2 V\n3.47 Gaseous State HINTS AND EXPLANATIONS
Solution: Average speed for a gas depends on temperature and it is independent from the presence of other gas.
Solution: Difference in any two kind of speed, ∆ u K T = × Now, d u dT ( ) Δ = K 2 T ⇒ On increasing temperature, Δ u decreases.
Solution: u u av x av y , , = 2 1 = T T X Y ⇒ 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 × = 4 1 2 1 × = 8 1
Solution: 1 × V = 1 M R T A × × ⇒ M M B A = 4 1 0.5 × V = 2 M RT B × ∴ u u av A av B , , = M M B A = 2 1
Solution: u av, A = u av, B ⇒ 8 RT M A A π = 3 RT M B ⇒ M M B A = 3 8 π Now, u av, A = u av, B ⇒ 8 RT M A A π = 8 RT M B B π ⇒ T T A B = M M A B = 8 3 1 π <
Solution: 1 4 . N *. u av = 1 4 6 10 22 4 10 8 8 314 273 28 10 23 3 3 × × × × × × × × − − . . π = 3.05 × 10 27 m –2 s –1
Solution: u u O O rms rms , , 2 3 = 3 600 32 48 3 300 R R × × × = 3 ⇒ u rms,O 2 = 3 v m/s
Solution: Average translational K.E. per gm = 3 2 RT M
Solution: T M A A = T M B B ⇒ u rms = 3 RT M = Same for both
Solution: 3 2 KT = qV ⇒ 3 2 8 314 6 022 10 23 × × × . . T = 1.602 × 10 –19 × 3 ⇒ T = 23207.2 K
Solution: u 2 rms ≠ u 2 av
Solution: Z w = 1 4 . N *. u av = 1 4 8 × × P N RT RT M A . π ⇒ Z w ∝ 1 T
Solution: u u rms, CH rms, SO 4 2 = 3 16 64 3 300 R T R × × × = 4 1 ⇒ T = 1200 K ∴ Average K.E. per mole = 3 2 RT = 3 2 2 1200 × × = 3600 cal
Solution: u av ∝ T ⇒ u u 2 1 = 432 300 = 1.2
Solution: Mole of gas cannot change.
Solution: Collision number, Z 1 = 2 πσ 2 . u av . N * = 2 πσ 2 . 8 RT M PN RT A π × ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ∴ Z 1 ∝ 1 T Collision frequency, Z 11 = 1 2 πσ 2 u av . N* 2 = 1 2 πσ 2 . 8 2 RT M PN RT A π × ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ∴ Z 11 ∝ 1 3 2 T Mean free path, λ = 1 2 2 πσ N * = RT PN A 2 2 πσ . ⇒ λ ∝ T
Solution: λ = RT PN A 2 2 πσ × = 8 314 300 2 1 5 10 4 1 10 1 013 10 6 022 10 10 2 14 5 23 . ( . ) ( . . ) ( . × × × × × × × × × − − π ) ) = 1.0 × 10 7 m
Solution: Velocity is a vector quality.
Solution: dN N = 4 2 3 2 2 2 2 π π m KT u e du mu KT ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ × × × − = 2 1 3 2 π KT E e dE E KT ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ × × × −\n3.48 Chapter 3 HINTS AND EXPLANATIONS For most probable K.E., d dN N dE ( ) = 0 ⇒ E = 1 2 KT
Solution: Deviation from ideal behavior is maximum at low temperature and high pressure.
Solution: Z > 1 for H 2 at 0°C at all pressure.
Solution: Z < 1 at low pressure and Z > 1 at high pressure.
Solution: V 1 = V – nb ⇒ b = V V n − 1 ⇒ 4 6 3 × × π d N A = V V n − 1 ∴ d = 3 2 1 1 3 ( ) V V nN A − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ π
Solution: P i = P an V + 2 2 ⇒ P = P an V i + ⎛ ⎝ ⎜ ⎜ ⎞ ⎠ ⎟ ⎟ 2 2 Greater the value of ‘ a ’, smaller will be ‘ P ’.
Solution: Gaseous mixture is always homogeneous.
Solution: P 1 V 1 = P 2 V 2 ⇒ 0.5 × 2000 = 100 × V 2 ⇒ V 2 = 10 ml < 13 ml As the real volume is greater than ideal, the volume occupied by the molecule is significant.
Solution: When attractive forces are dominant, V real < V ideal .
Solution: B = b a RT − = 0.03 – 1 344 0 0821 273 . . × = – 0.03 l/mol
Solution: Z = PV RT m = V V b m m − = 10 10 b b b − = 10 9
Solution: P an V V nb + ⎛ ⎝ ⎜ ⎜ ⎞ ⎠ ⎟ ⎟ − 2 2 ( ) = nRT may be expressed as P a d M M d b + ⎛ ⎝ ⎜ ⎜ ⎞ ⎠ ⎟ ⎟ − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ . 2 2 = RT as d = m v = n m v × Now, P + × ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ − ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ 3 6 2 2 44 44 2 2 0 05 2 2 . ( . ) ( ) . . = 0.0821 × 300 ⇒ P = 1.226 atm
Solution: B = b a RT − = –1.0 L/mol Now, PV m = RT 1 + ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ B V m and d = M V m Hence, PM d = RT 1 + × ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ B d M or, 1 40 × d = 0.08 × 262.5 1 1 0 40 + − × ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ ( . ) d ∴ d = 2.005 g/L
Solution: When P → 0, V → ∞ and hence e a/VRT → 1 and ( V – b ) → V . Hence, P = RT V = 0 0821 300 410 5 . . × = 0.06 atm
Solution: For a van der Waals gas, Z = V V b a V RT m m m − − or, 0.8 = 0 5 0 5 0 04 0 5 0 08 300 . . . . . − − × × a ⇒ a = 3.44 atm L 2 /mol 2
Solution: At Boyle’s temperature, dz dp = 0 ⇒ T = 168 0 35 . = 480 K
Solution: Theory based. The initial slope of Z vs. P curve increases with increase in temperature, above Boyle’s temperature, only upto 2 × T B . Then, the slope starts decreasing.
Solution: For van der Waals gas, Z = V V b a V RT m m m − − × At Boyle’s temperature, Z = V V b a V R a Rb m m m − − × × ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = 1 2 + − b V V b m m ( )
Solution: Ideal gas can never be liquified.
Solution: For ideal behavior, Boyle’s temperature should be closer to 600 K.
Solution: T P c c = 8 27 27 2 a Rb a b = 8 b R ∴ T P T P c c c c ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ CO CH 2 4 = b b CO CH 2 4 = 304 72 190 45 = 1 1\n3.49 Gaseous State HINTS AND EXPLANATIONS
Solution: At T > T c , the gas can never be liquified.
Solution: P c V c = 3 8 RT c ⇒ V c = 3 8 0 0821 128 41 05 × × . . = 0.096 L/mol
Solution: Boyle’s law constant = PV = nRT
Solution: P V RT P V RT 0 0 0 0 0 0 + = PV R T PV RT 0 0 0 0 2 × + ⇒ P = 4 3 P 0 and n = 4 3 2 0 0 0 P V R T × × = 2 3 0 0 0 P V RT
Solution: P ƒ = P i V V V n + ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ Δ (a) P ƒ = 24.2 × 10 10 1 1 + ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = 22 atm (b) P ƒ = 24.2 × 10 10 1 2 + ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = 20 atm (c) P η = P 10 10 1 + ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ n ⇒ n = ln ln . η 1 1 (d) P ƒ = 24.2 × 10 10 1 + ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ n atm
Solution: As the average molar mass increases, the molar mass of vapours must be greater than that of N 2 .
Solution: (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 ∝ Change in momentum ∝ M
Solution: At valve – I: P 1 V 1 = P 2 V 2 ⇒ 1 × 60 A = P 2 × 45 A ⇒ P 2 = 1.33 atm < 1.5 atm Hence, valve – I will not open. At valve – II: P 1 V 1 = P 2 V 2 ⇒ 1 × 60 A = P 2 × 30 A ⇒ P 2 = 2 atm < 2.2 atm Hence, valve – II will not open. At valve – III: P 1 V 1 = P 2 V 2 ⇒ 1 × 60 A = P 2 × 20 A ⇒ P 2 = 3 atm > 2.5 atm Hence, valve – III will open fi rst. As the piston will reach at valve – III, the gas will come out till the pressure of gas becomes 2.5 atm. Now, 2 5 20 821 1000 . × × = n × 0.0821 × 300 ⇒ Moles of gas remained, n = 5 3 At valve – IV: P 1 V 1 = P 2 V 2 ⇒ 2.5 × 20 A = P 2 × 15 A ⇒ P 2 = 3.33 atm < 4.4 atm Hence, valve - IV will not open. At valve – V: P 1 V 1 = P 2 V 2 ⇒ 2.5 × 20 A = P 2 × 10 A ⇒ P 2 = 5 atm > 4.8 atm Hence, valve – V will open until the gas pressure becomes 4.8 atm.
Solution: u u rms, A rms, B = 3 300 3 400 R M M R A B × × : = 3 2 ⇒ M A = M B
Solution: On increasing the temperature at constant volume, the average speed of molecules as well as number of molecular collisions at wall increases.
Solution: At very high pressure, P a V + ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 2 \u001e P ⇒ Z = 1+ b P RT . and Z = PV RT ⇒ P RT = Z V ⇒ Z = V V b −\n3.50 Chapter 3 HINTS AND EXPLANATIONS
Solution: T c = 273 + (–177) = 96 K ⇒ T B = 27 8 × 96 = 324K = 51°C (a) Z = PV RT m . = 0 821 9 6 0 0821 96 . . . × × = 1 But at T = T c , Z < 1 at low pressure (b) Z = PV RT m . = 0 821 40 0 0821 400 . . × × = 1 But at T > T B , Z > 1 at all pressure (c) Z = PV RT m . = 82 1 0 310 0 0821 324 . . . × × = 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 . . . × × × − = 10 –3 < 1 But at T = T c and P < 50 atm, Z = 1
Solution: a. 9 8 27 36 8 R a Rb × × = a b. 3 × a b 27 2 × (3 b ) 2 = a c. 3 8 27 36 8 27 2 × × a b a Rb ≠ a d. 27 64 8 27 27 2 2 2 × × ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ R a Rb a b = a
Solution: Real gas may behave ideally at Boyle’s temperature. P = RT V m = R a Rb V m × = a bVm .
Solution: Z = PV RT m = 1 + B ′ .P + C ′ .P 2 + …. (1) Z = PV RT m = 1 + B Vm + C V m 2 + …. (2) Or, P = RT V m (1 + B V m + C V m 2 + ….) Substituting this value in Equation (1), we get: Z = 1 + B ′ . RT V B V C V m m m 1 2 + + + ⎛ ⎝ ⎜ ⎜ ⎞ ⎠ ⎟ ⎟ ⎧ ⎨ ⎪ ⎩ ⎪ ⎫ ⎬ ⎪ ⎭ ⎪ ... + C ′ . RT V B V C V m m m 1 2 2 + + + ⎛ ⎝ ⎜ ⎜ ⎞ ⎠ ⎟ ⎟ ⎧ ⎨ ⎪ ⎩ ⎪ ⎫ ⎬ ⎪ ⎭ ⎪ ... + … = 1 + ′ B RT V m + ′ + ′ B RT B C RT V m . ( ) 2 2 + …. Comparing this with Equation (2), we get: B ′ RT = B and B ′ RT . B + C .( RT ) 2 = C
Solution: Number of strokes = ( ) ( ) ( ) ( ) 8 bar cm 1 bar cm × × × 1000 25 4 3 3 = 80
Solution: F = P.A = 8 10 5 2 × ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ N m × (4 × 10 –4 m 2 ) = 320 N
Solution: m = F g = 320 10 = 32 kg\n3.51 Gaseous State HINTS AND EXPLANATIONS Comprehension – II
Solution: PV m = RT (Let 0K = – x °N) 28 = R (0 + x ) x = 233.33 40 = R (100 + x ) ∴ 0K = –233.33°N
Solution: R = 28 x = 0.12 L – atm/K-mol
Solution: V = nRT P = 2 0 12 66 67 233 33 2 × × + . ( . . ) = 36 L Comprehension – III
Solution: 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 – 20 cm Hg = 76 – 20 = 56 cm Hg V 2 = 30 A cm 3 Now, P 1 V 1 ≠ 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 ≠ P 2 V 2 , only end D is not closed. Hence, both the ends are closed.
Solution: 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 × 20 = P 1 × 30 For CD column: P 0 × 60 = P 2 × 50 As P 1 + 20 cm Hg = P 2 or, 20 30 0 P + 20 cm Hg = 60 50 0 P ⇒ P 0 = 37.5 cm Hg
Solution: D D C A B P 4 P 3 (80 – x ) 20 x P 4 + 20 cm Hg = P 3 or, 60 80 0 P x ( ) − + 20 = 20 0 P x ⇒ x = 13.88 If both ends are open, then mercury will fall down.\n3.52 Chapter 3 HINTS AND EXPLANATIONS Comprehension – IV
Solution: Total moles of product gases = PV RT = 410 5 2 9 0 0821 2000 . . . × × = 7.25 ∴ Moles of gases per 0.04 mole of nitroglycerine = 0.04 × 7.25 = 0.29
Solution: Moles of gases except A = 4 75 0 821 0 0821 250 . . . × × = 0.19 ‘A’ must be H 2 O because it solidifies at –23°C and its mole = 0.29 – 0.19 = 0.10.
Solution: Moles of gases C and D = 2 1 0 821 0 0821 300 . . . × × = 0.07 ∴ Mole of gas ‘B’, which is CO 2 = 0.19 – 0.07 = 0.12
Solution: The gas remained, D must be N 2 and its mole = 1 8 0 821 0 0821 300 . . . × × = 0.06 and gas ‘C’ is O 2 and its mole = 0.07 – 0.06 = 0.01. Comprehension – V
Solution: All H 2 O(g) will solidify in bulb ‘B’.
Solution: n H O 2 + n CO 2 + n N 2 = 570 1 642 760 0 0821 300 × × × . . = 0.05 n CO 2 + n N 2 = 0 21 1 642 0 0821 300 0 21 1 642 0 0821 200 . . . . . . × × + × × = 0.035 ∴ n H O 2 = 0.05 – 0.035 = 0.015
Solution: H 2 O(g) will solidify in ‘B’ as well as ‘C’ but CO 2 (g) will solidify only in ‘C’.
Solution: n N 2 = 22 8 1 642 760 0 0821 1 300 1 200 1 80 . . . × × + + ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ = 0.0125 ∴ n CO 2 = 0.035 – 0.0125 = 0.0225 Comprehension VI
Solution: Let x mole, NH 4 Cl was present initially. NH 4 Cl(s) → NH 3 (g) + HCl(g) x mole x mole Now, PV = nRT 114 × V = 0.01 × R × 300 and 908 × V = (0.01 + 2 x ) × R × 600 ∴ x ≈ 0.015 ∴ Mass of NH 4 Cl = x × 53.5 ≈ 0.8 gm
Solution: P NH 3 = 908 114 2 2 − × 340 mm Hg
Solution: V = 0 01 0 0821 300 760 114 . . × × × = 1.642 L Comprehension – VII
Solution: Water will vaporize till P H O 2 = 0.04 atm Now, PV = nRT ⇒ 0.04 × (40 × 10 3 ) = w 18 × 0.08 × 300 or w = 1200 = 1.2 kg ∴ Percentage of water vaporized = 1 2 5 . × 100 = 24 %
Solution: V = nRT P = 5000 18 0 08 300 0 04 × × . . = 1.67 × 10 5 L Comprehension – VIII
Solution: − dP dt = K ( P – P 0 ) ⇒ − − ∫ dP P P P P 0 1 2 = K dt t 0 ∫ ⇒ ln P P P P 1 0 2 0 − − = Kt or, ln 20 1 1 2 − − P = 0.001 × 3600 = ln38 ⇒ P 2 = 1.5 atm
Solution: Number of balloons = ( . ) 20 1 5 10 1 2 − × × = 92.5 ≈ 92
Solution: ln P P P P 1 0 2 0 − − = Kt ⇒ ln 20 1 2 1 − − = 0.001 × t ⇒ t = 2900 sec\n3.53 Gaseous State HINTS AND EXPLANATIONS Comprehension – IX
Solution: − dP dt = K.P ⇒ − ∫ dP P P 1 5 . atm = K dt t 0 ∫ ⇒ P = (1.5 atm). e – Kt
Solution: A 38 cm 2 A x x 2 The pressure of gas in closed arm, P = 1 atm + 38 3 2 − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ x cm = 114 3 2 − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ x cm Hg Now, 114 3 2 − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ x = 114 × e – kt ⇒ x = 76 (1– e – kt ) cm Hg
Solution: x 2 = 38(1 – e – kt ) cm Hg Comprehension – X
Solution: u mp = 2 RT M ⇒ T = M u R × mp 2 2 = ( ) ( ) 32 10 400 2 8 3 2 × × × − = 320K = 47°C
Solution: u rms – u mp = 400 m/s ⇒ 3 2 RT M RT M − = 400 m/s or, T = 400 3 2 2 − ⎛ ⎝ ⎜ ⎜ ⎞ ⎠ ⎟ ⎟ × M R = ( ) 400 3 2 2 6 2 10 8 2 3 + − × × − = 400 K = 127°C
Solution: c 1 f ( c ) c 2 C 4 π M RT 2 3 2 π ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ × C 1 2 × e MC RT − 1 2 2 = 4 π M RT 2 3 2 π ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ × C 2 2 × e MC RT − 2 2 2 or, C C 1 2 2 2 = e M C C RT ( ) 1 2 2 2 2 − or, M C C RT ( ) 1 2 2 2 2 − = 2 ln C C 1 2 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ∴ T = M C C R C C ( ) .ln 1 2 2 2 1 2 4 − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = ( )( ) ln 28 10 300 600 4 8 300 600 3 2 2 × − × × ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ − = 337.5 K = 64.5°C
Solution: c = ? f ( c ) T n . T C 4 2 3 2 2 2 2 π π M RT c e MC RT ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ × × − = 4 2 3 2 2 2 2 π π M RT n c e MC RT n . . ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ × × − or, n 3/2 = e MC RT n 2 2 1 1 − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ or, 3 2 ln n = MC RT n n 2 2 1 × − ∴ C = 3 1 nRT n M n ln ( ) −\n3.54 Chapter 3 HINTS AND EXPLANATIONS
Solution: f ( c ) T 1 T 2 > T 1 C
Solution: f ( c ) M 1 M 2 > M 1 C
Solution: ( ) ( ) dN dN 1 2 = 4 2 2 4 2 3 2 2 2 2 3 2 π π π π × ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ × × × × × ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ − M RT u e du N M RT M u RT ( ) ( ) mp mp 2 2 2 2 2 × × × × − × u e du N M u RT mp mp = 4 × e Mu RT mp − − 2 2 4 1 ( ) =
Solution: e –3 Comprehension – XI
Solution: n A = m 2 , n B = m 16 , n C = m 32
Solution: Z W = 1 4 × u av × N * = 1 4 8 RT M π × N * ⇒ Z W ∝ N M *
Solution: λ = 1 2 2 πσ N * ⇒ λ A : λ B : λ C = 1 1 2 1 2 16 1 2 32 2 2 2 × × × m m m : : = 1 : 2 : 4
Solution: E total = 3 2 nRT and n max for A
Solution: Z = 1 for all (Ideal behaviour)
Solution: Z 1 = 2 π σ 2 u av × N * Z A : Z B : Z C = 1 2 × 1 2 2 × m : 2 2 × 1 16 16 × m : 2 2 × 1 32 32 × m = 1 2 2 1 64 1 128 2 : :
Solution: u av ∝ 1 M Comprehension – XII For critical point, dP dV = 0 and d P dV 2 2 Now, dP dV = 0 ⇒ − − + RT V b a T V ( ) . 2 3 2 = 0 ⇒ RT V b a T V ( ) . − = 2 3 2 (1) and d P dV 2 2 = 0 ⇒ 2 6 3 4 RT V b a T V ( ) . − − = 0 ⇒ 2 3 3 4 RT V b a T V ( ) . − = (2) From (1) ÷ (2) : V – b = 2 3 V ⇒ V C = 3 b Eq 1 : RT b b ( ) 3 2 − = 2 3 3 a T b .( ) ⇒ T C = 8 27 a Rb\n3.55 Gaseous State HINTS AND EXPLANATIONS and P C = RT V b a T V − − . 2 = aR b 216 3
Solution: T C = 8 27 a Rb
Solution: P C = aR b 216 3
Solution: Avogadro’s hypothesis is valid only for gases due to large intermolecular distance.
Solution: d = PM RT but M is independent from d , P or T .
Solution: H 2 and Cl 2 are reactive gases.
Solution: Escaping tendency increases only on increasing the energy of molecules.
Solution: Graham’s law is valid for ideal as well as non-ideal gases.
Solution: Volume of ideal gas should be the total volume minus the volume occupied by gas molecules.
Solution: As the average K.E. is same, increase in mass decreases their speed.
Solution: 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.
Solution: f ( c ) T 1 T 2 > T 1 C
Solution: Excluded volume is ‘nb’.
Solution: T C < T B and hence, attractive forces are dominant.
Solution: PV is constant at constant temperature.
Solution: Above Boyle’s temperature, gases show positive deviation.
Solution: K.E. of molecules is the function of T.
Solution: Boyle’s law : PV = K ⇒ dP dV T ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = − K V 2 = − P V And d PV dP T ( ) ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ = 0 Charle’s law V T = K ⇒ dV dT P ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = K = V T Avogadro’s law : V n = K = RT P ⇒ dV dn P T ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ , = RT P Graham’s law r = − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ dP dt ∝ 1 d
Solution: Average translational K.E. per mole = 3 2 RT Average translational K.E. per gram = 3 2 RT M
Solution: T C = 8 27 a Rb a b X ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = 120, a b Y ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = 333.33, a b Z ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = 171.4 V C = 3 b P C = a b 27 2 , a b X 2 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = 4800, a b Y 2 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = 11111.11, a b Z 2 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = 4898
Solution: A. P V = P nRT P ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = 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
Solution: Moles of water vapour formed, n = PV RT = 22 8 827 6 760 0 0821 300 . ( ) . × − × × = 1
Solution: Pressure correction = a n V . 2 2 = 4 5 10 2 2 × = 1 atm Ideal volume = V – nb = 10 – 5 × 0.05 = 9.75 L Volume occupied by molecules = nb 4 = 0 25 4 . = 0.0625 L Volume correction = nb = 0.25L
Solution: (A) λ = 1 2 2 πσ N * = RT PN A 2 2 πσ At constant volume, N * = constant ⇒ λ λ 2 1 = 1 At constant pressure, λ ∝ T ⇒ λ λ 2 1 = 2 (B) Z 1 = 2 πσ 2 × u av × N * = 2 πσ 2 × 8 RT M PN RT A π ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ At constant volume : Z 1 ∝ T ⇒ Z Z 1 2 1 1 , , = 2 At constant pressure : Z 1 ∝ 1 T ⇒ Z Z 1 2 1 1 , , = 1 2 (C) Z 11 = 1 2 2 2 πσ . . * u N av = 1 2 8 2 2 πσ π RT M PN RT A ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ At constant volume : Z 11 ∝ T ⇒ Z Z 11 2 11 1 , , = 2 At constant pressure : Z 11 ∝ 1 3 2 ( ) T ⇒ Z Z 11 2 11 1 , , = 1 2 2
Solution: 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 × × . . = 1.5 atm
Solution: T C = 273 + (–177) = 96 K = –177°C T B = 27 8 96 × = 324 K = 51°C A. V i = 0 2 0 08 96 20 10 3 . . × × × = 76.8 ml But V real < V ideal in given condition ⇒ V r < 76.8 ml B. Z = 1 ⇒ V r = V i = 0 2 0 08 324 6 48 10 3 . . . × × × = 800 ml C. Above Boyle’s temperature, Z > 1 ∴ V r > V i = 0 2 0 08 350 7 10 3 . . × × × = 800 ml D. Below Boyle’s temperature, Z < 1 ∴ V r < V i = 0 2 0 08 300 6 10 3 . . × × × = 800 ml E. T = T B but P > 50 atm ⇒ Z > 1 ∴ V r > V i = 0 2 0 08 324 64 8 10 3 . . . × × × = 80 ml
Solution: V T 1 1 = V T 2 2 ⇒ 45 300 = 42 2 T ⇒ T 2 = 280 K = 7°C\n3.57 Gaseous State HINTS AND EXPLANATIONS
Solution: 1 2 P 1 = 1 atm + hm water = (10 + h ) m water V 1 = 4 3 π (1 mm) 3 A = π r 2 = π mm 2 ∴ r = 1 mm P 2 = 1 atm = 10 m water V 2 = 2 π mm 3 Now, P 1 V 1 = P 2 V 2 ⇒ (10 + h ) × 4 3 π = 10 × 2 π ⇒ h = 5 m Hence, water holding capacity of pool, V = 1 3 π r 2 h = 1 3 π (10 m) 2 × 5 m = 500 3 π m 3
Solution: Number of cosmic events = Number of Ar-atoms = 1 911 10 22 7 6 . . × − l l × 6 × 10 23 = 5.05 × 10 16
Solution: For lifting of balloon, B > W or, ( V × ρ outside air × g ) > ( V × ρ inside air + m additional ) g or, V ( ρ outside air – ρ inside air ) > m additional or, 91 1 29 0 08314 290 1 29 0 08314 10 8 314 × × − × × ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ > . . . T ∴ T > 293.22 Hence, diff erence in temperature, Δ T > 3.22 ≈ 4 K.
Solution: n released gas = n taken – n remained or, m 22 4 1 25 . . × = P p × × − − × × 3 0 0821 273 0 8 3 0 0821 273 . ( . ) . ⇒ m = 3 gm
Solution: 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 × × + + × × 600 2 600 ( ) = 4 300 × × V R ⇒ P = 3 atm
Solution: Let the mixture contains a moles C x H 8 and b moles C x H 10 . Now, a × (12 x + 8) + 6 × (12 x + 10) = 28.4 (1) a + b = PV RT = 2 46 5 0 082 300 . . × × = 0.5 (2) and 28.4 × 84 5 100 . = a × 12 x + b × 12 x (3) On solving, x ≈ 4
Solution: 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 → nX Initial a mole o Final a – 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 − − ⇒ n = 6
Solution: 2H 2 + O 2 → 2H 2 O 2a a 0 Final 2a – 1.6a a – 0.8a 1.6a = 0.4a = 0.2a Now, P nT = R V = Constant ⇒ P n T 1 1 1 = P n T 2 2 2 ⇒ 4 5 3 330 . a × = P a 2 2 2 400 . × ∴ P 2 = 4 atm
Solution: n total = n n O N 2 2 + or, 1 1 30 . × RT = Po RT RT 2 30 0 9 10 × + × . ⇒ P O 2 = 0.8 atm
Solution: P H O 2 = 40 100 × (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 ⇒ 40 100 × (V.P.) × 10 = (V.P.) × V 2 ∴ V 2 = 4 ml
Solution: H 2 O(l) → H 2 (g) + 1 2 O 2 (g) a mole 0 0 Final 0 a mole 0.5 a mole Δ P.V = Δ n .RT or, (1.86 – 0.96) × 20 = (1.5a) × 0.08 × 300 ⇒ a = 0.5 ∴ Mass of water present initially = 0.5 × 18 = 9 gm
Solution: Initial total moles = PV RT = 24 63 3 0 0821 300 . . × × = 3 ∴ Initial mole of H 2 = 3 – 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 = × ⎛ ⎝ ⎜ ⎜ ⎞ ⎠ ⎟ ⎟ or, 1 2 2 1 4 2 = × ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ n ⇒ n = 4
Solution: For critical point, dP dV m = 0 and d P dV m 2 2 0 = On solving, b =
Solution: But b ≠ 0 from question. Hence, the gas does not have critical condition.
Solution: Boyle’s temperature: T a Rb B = = × = 4 105 0 0821 0 1 500 . . . K Hence, at 500 K, the gas will behave ideally. Not, d PM RT = = × × 2 164 2 0 0821 500 . . = 8 g/L = 8 kg/m 3 Four-digit Integer Type
Solution: l mm 760 750 l – 750 800 770 l – 770 P 760 l – 760 (760 – 750) × ( l – 750) = (800 – 770) × ( l – 770) = ( P – 760) × ( l – 760) ∴ P = 775 mm Hg
Solution: Mass of LNG = 10 m 3 × 416 kg/m 3 = 416 × 10 4 gm ∴ Moles of CH 4 = 916 10 16 4 × = 26 × 10 4 Now, V = nRT P = × × × 26 10 0 021 300 1 692 4 . . = 3.9 × 10 6 L = 3900 m 3
Solution: V n T V n T V n V n T 1 1 1 2 2 2 2 303 1 6 1 2 = ⇒ × = × . . ⇒ T 2 = 404 K
Solution: V initial = a = nRT P = × × × 64 0 08 300 64 3 . = 8 L V initial = b = nRT P = − × × × ( ) . 64 8 0 08 300 64 3 = 7 L P = nRT V = × × × = 64 0 08 300 64 7 24 7 . atm\n3.59 Gaseous State HINTS AND EXPLANATIONS
Solution: P 1 320 K P 0 P 2 T K P 0 P ′ 2 P ′ 1 P 2 + P 0 = P 1 P ’ 2 + P 0 = P ’ 1 P 0 = P 1 – P 2 = P ’ 1 – P ’ 2 or, n R V n R V n R T V n R T V × × ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ − × × ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = × × ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ − × × ⎛ ⎝ ⎜ 320 5 320 4 5 4 3 4 ⎞ ⎞ ⎠ ⎟ ∴ T = 450 K
Solution: Mass of gas used = 28.8 – 23.2 = 5.6 kg Volume of gas used up, V = nRT P = × × × × ( . ) . 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 = ⇒ − = − ( . . ) ( . . ) ⇒ P 2 = 21 atm
Solution: n NO = × × 1 6 0 75 0 08 3 00 . . . . = 0.05 n O 2 1 2 0 25 0 08 3 00 = × × . . . . = 0.0125 2NO + O 2 → 2NO 2 → N 2 O 4 0.05 0.0125 0 0 Final − 0 025 0 025 . . − 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 × 1000 = 25 Now, P = 0 025 0 08 200 0 75 0 25 . . ( . . ) × × + = 0.4 atm
Solution: 76 70 6 76 l = ? 3 P 1 = (6 – 1) = 5 cm Hg P 2 = ? V 1 = 6 A cm 3 V 2 = 3 A cm 3 P 2 = 5 6 3 × A A = 10 cm Hg Hence, fi nal total pressure in table above mercury = 10 + 1 = 11 cm Hg ∴ Barometer reading = 76 – 11 = 65 cm
Solution: Mass of water vapour present initially, m V R 1 756 100 24 760 300 18 = × ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ × × × Mass of water vapour fi nally remained, m V R 2 8 4 760 280 18 = × × × . ∴ Fraction of water condensed = m m m 1 2 1 − = 0.5
Solution: 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 = × ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ × × ( ) × × . 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 = × ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ × × ( ) × × . 18 gm = 6.25 t gm From question, m 1 + 200 kg × 36 100 = m 2 ⇒ t = 19200\n3.60 Chapter 3 HINTS AND EXPLANATIONS
Solution: At a depth of 10 m, P 1 = 1 atm + 10 m water = 1 + 1 013 1000 1000 1 013 10 6 . . × × × = 2 atm V 1 = 24 ml n 1 = 2 24 10 0 08 300 3 × × × − . = 2 × 10 − 3 At surface, P 2 = 1 atm, V 2 = ?, n 2 = 2 × 10 − 3 − 0.05 × 10 − 3 × 10 = 1.5 × 10 − 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 = ⇒ × × = × × ⇒ − − . = 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 × = × − × ⇒ = − − − r r mol/min
Solution: N 2 O 4 → 2NO 2 Initial mole 20 × V RT 0 Final mole 20 10 × − 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 :
Solution: 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 ∴ Pressure of gas in chamber-I = 15 1 2 V RT R T V × × . = 18 mm and pressure of gases in chamber-II = 15 1 2 3 V RT R T V × × . + 30 = 36 mm
Solution: 5 2 n ≤ 0.01 ⇒ n ≥ 12.28
Solution: 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
Solution: 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 ∴ P P n n N A A N 4 1 5 − = / / = 3 ⇒ N = 15
Solution: r r M M t t M H air air H air 2 2 100 26 2 = ⇒ = / / and r r M M t t M M gas air air H air gas 2 = ⇒ = 100 130 / / ∴ M gas = 50
Solution: 3 2 kT = mgh ⇒ 3 2 × 8 4 . N A × 300 = ( ) 40 10 3 × − N A × 10 × h ∴ h = 9450 m
Solution: At Boyle’s temperature, the second virial coefficient is zero. B = a + b . e c T / 2 = 0 or, e c T − / 2 = − a b ⇒ e T − 950 2 / = − − 0 02 0 22 . . ⇒ T = 100 k\n3.61 Gaseous State HINTS AND EXPLANATIONS
Solution: P c = 73.89 atm = a b 27 2 and T c = 300 k = 8 27 a Rb ∴ 300 73 89 . = 8 b R ⇒ b = 0.0416 l mol –1 b = 4 × Volume occupied by molecules per mole ∴ Volume occupied by molecules in 24 moles = b 4 × 24 = 0.25 L = 250 ml
Solution: d d dP ( ) = 4 + 0.02 × 2P + … ∴ d d dP P ( ) ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ → 0 = 4 = M RT ⇒ M = 4 × 0.08 × 300 = 96
Solution: U n f R T = × × 2 For larger U , n , f , T , should be higher.
Solution: w P dv a V b dv a V V b V V V V V V = − ⋅ = − + ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = − − − ∫ ∫ 1 2 1 2 2 1 2 1 ln ( )
Solution: q = m.s. Δ T ⇒ 10 × 10 6 = 80 × (4.2 × 10 3 ) × Δ T ⇒ Δ T = 29.76 K = 29.76° C
Solution: Heat lost by water = Heat gained by ice or, 500 75 6 18 20 9 6000 18 × × = × × . ( ) N ⇒ N = 14
Solution: Let T 1 > T 2 . Now, heat lost by gas (1) = Heat gained by gas (2) or, n 1 ⋅ C m ⋅ ( T 1 – T f ) = n 2 ⋅ C m ⋅ ( T f – T 2 ) or, PV RT T T P V RT T T f f 1 1 1 1 2 2 2 2 ⋅ − = − ( ) ( ) ⇒ T T T PV P V PV T P V T f = + + 1 2 1 1 2 2 1 1 2 2 2 1 ( )
Solution: w = – P ext ( V 2 – V 1 ) = − − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ P nRT P nRT P 2 2 1 = − − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = − − − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ nRT P P nRT P P 1 1 1 2 1 1 1 = − × nRT P 1 1\n4.37 Thermodynamics HINTS AND EXPLANATIONS Now, w total = w 1 + w 2 + … + w f = − × ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ + − × − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ + + − × ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ nRT P nRT P nRT 1 1 1 1 2 \u001d = − + − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = = − ∑ nRT P i i i P 1 1 1 1
Solution: 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 = − ⋅ = − ⋅ = + − ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ = − ∫ ∫ = − = − = − 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 = − ⋅ = − ⋅ = − = − × ∫ ∫ ln .
Solution: As Δ T = 0, Δ U = 0
Solution: Area = P 2 × Δ V ⇒ P 2 × 4 = 49.26 L -atom Now, correct work, w = − ⋅ = − ⋅ nRT V V P V ln ln 2 1 2 2 4 2 = –49.26 × 0.693 = – 34.137 L -atom
Solution: PV x = Constant ⇒ = ⇒ = ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ⇒ = ⇒ = 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 = + − = + − = − 1 1 3 2 1 3 2 2
Solution: Dulong and Petit’s law is applicable only for solid element. (Molar heat capacity ≈ 6.4 cal/K-mol = 26.8 J/K-mol).
Solution: PV P V 1 1 2 2 γ γ = ⇒ P d P d 1 1 2 2 γ γ = ⇒ P P d d 2 1 2 1 7 5 32 128 = ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = = γ ( ) /
Solution: U U T T rms rms , , 2 1 1 2 1 4 2 1 = = ⇒ Now, T.V r – 1 = Constant ⇒ T T V V V V r 2 1 1 2 1 1 2 7 5 1 1 4 = ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ⇒ = ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ − − / ∴ V 2 = 32 V 1
Solution: w w A B = × 2 But Δ U A = Δ U B , Hence q A > q B or, ( C A ⋅ Δ T ) > ( C B ⋅ Δ T ) ⇒ C A > C B
Solution: q = n ⋅ C m ⋅ Δ T = 1 × (0.22 × 32) × (273 × 1.1 – 273) × 4.2 J
Solution: q q n C T n C T V P V m P m = ⋅ ⋅ Δ ⋅ ⋅ Δ = , , 1 γ ⇒ q V = ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ × = 1 74 3 66 793 660 . J ( C P m , . = × = 743 5 2 74 3 ⇒ C V m , = 74.3 – 8.3 = 66.0)
Solution: Isothermal: P.V = P V n P n P i i × ⇒ = ⋅ Adiabatic: P.V r = P V n P n P a a × ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ⇒ = ⋅ γ γ ∴ P P n P n P n i a = ⋅ ⋅ = − γ γ 1
Solution: Δ U = n C T T n R P V nR P V nR PV V m ⋅ ⋅ − = ⋅ − ⋅ ⋅ − ⋅ ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = − , ( ) 2 1 1 2 1 γ γ
Solution: K.E. = Δ U ⇒ 1 2 40 1000 100 8 314 1 5 1 2 × × × = × − × Δ ( ) ( ) . ( . ) n n T ∴ Δ T = 12.03 K
Solution: For minimum pressure, compression should be irreversible. Δ = ⇒ ⋅ − ⋅ − = − − = − − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ U w n R T T P V V P nRT P nRT P ext γ 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 − − = − − ⋅ ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ⇒ − − = − − × ⎛ ⎝ ⎜ ⎞ γ . ⎠ ⎠ ⎟ ∴ P 2 = 362.5 kPa
Solution: q n C T n C T V m Ne V m SO = ⋅ ⋅ Δ + ⋅ ⋅ Δ ( ) ( ) , , 3 or, 12 × 10 3 = 2 × 3 × ( T f – 300) + 3 × 6 × ( T f – 400) ⇒ T f = 875 K Now, P nRT V final atm = = × × = 5 0 08 875 10 35 .
Solution: Free expansion is isothermal.
Solution: C R R N R N R V m , ( ) ( ) = × + × + − × = − 3 1 2 3 1 2 3 6 3 3 ∴ γ = = + = + − C C R C N P V V m 1 1 1 3 3 ,
Solution: C C P dV dT C P C RT V C R T V V m V m V m V m V m = + ⋅ = + = + ⋅ = + + , , , , ( ) α α α α 0 = + C RT V P m , 0 α Now, q C dT C dV C RT V dV m T T m V V P m V V = ⋅ = ⋅ ⋅ = ⋅ + ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ∫ ∫ ∫ 1 2 1 2 1 2 0 ( ) , α α = ⋅ − + ⋅ α C V V RT V V P m , ( ) ln 2 1 0 2 1
Solution: w = –25 J = – nR ⋅ Δ T Δ = ⋅ ⋅ Δ = ⋅ ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ⋅ Δ = U n C T n R T J V m , 6 2 75 ∴ q = Δ U – w = 100 J
Solution: C C R x R R R m V m = + − = + − = , 1 3 2 1 5 2 5 6 ∴ q n C T R J m = ⋅ ⋅ Δ = × × = ⋅ 1 5 6 26 180 14
Solution: PV K dP dV x P V x x x = ⇒ = − ⋅ ⇒ − = − × ⇒ = 1 4 2 1 2 ∴ C C R x R R R m V m = + − = + − = , . 1 3 2 1 1 2 3 5
Solution: V 0 2 V 0 V P 2 1 3
Solution: V 1 V 2 V Isobaric Isothermal Adiabatic P Δ E adiabatic = Negative Δ E isothermal = 0 Δ E isobaric = Positive
Solution: Boyle temperature, T B = 20 + 273 = 293 K Inversion temperature, T i = 2 × T B = 586 K = 313° C > 50° C
Solution: 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 ⋅ = ⋅ / / Diatomic : P V P V 1 0 7 5 2 0 7 5 3 4 ⋅ = ⋅ ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ / / ∴ V V 1 0 21 25 3 4 = ⎛ ⎝ ⎜ ⎞ ⎠ ⎟\n4.39 Thermodynamics HINTS AND EXPLANATIONS
Solution: V 0 2 V V P P 2 V = Constant Isothermal reversible Adiabatic irreversible Adiabatic reversible
Solution: For greater heat exchange, heat capacity should be high. C C R x m V m = + − , 1 for PV x = Constant
Solution: q ABC = 600 + 200 = 800 J w AB = 0 and w N m m BC = − × × ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ × − × − − 8 10 5 10 2 10 4 2 3 3 2 ( ) = –240 J ∴ Δ U AC = Δ U ABC = q ABC + w ABC = 800 + (–240) = 560 J
Solution: 3 60° 30° 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 : = − + 1 3 2 P V C 0 0 2 1 3 6 = − ⋅ + (3) 3 1 3 0 2 P V C B = − ⋅ + (4) From equation (1), (2), (3) and (4), V V B = 9 4 0 Now, T T P V P V B A = ⋅ ⋅ = 3 9 4 27 4 0 0 0 0
Solution: F = P 0 = P P 0 P i P f dw = F ⋅ dx = ( P 0 – P ) A ⋅ dx = ( P 0 – P ) ⋅ dV ∴ w P nRT V dV P V V RT V V V V = − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = − − ⋅ ⋅ ⋅ ∫ 0 0 η η η ( ) ln P 0 V ( η -1) − RT.ln η = RT [ η -1-ln η ]
Solution: T 0 ( V 0) P 0, P 0, P 1, P 2, T 0 ⋅ ( V ) η ( V 0) ( V ) Work performed on the piston = − ⋅ ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ ⋅ + ∫ ∫ ⋅ P P dV dV V V V V 1 2 0 0 η and ( V + h ⋅ V ) = 2 V 0 = ⋅ + P V 0 0 2 1 4 ln ( ) η η
Solution: Isothermal : P A ⋅ V = P ⋅ (2 V ) ⇒ P A = 2 P Adiabatic : P B ⋅ V 1.5 = P ⋅ (2 V ) 1.5 ⇒ P A = 2 2 P Isobaric : P C = P ∴ P A : P B : P C = 2 : 2 2 : 1
Solution: 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 ⋅ = ⋅ γ γ\n4.40 Chapter 4 HINTS AND EXPLANATIONS ∴ V V T T B B = ⇒ = 4 9 3 2 0 0 and V V V V T T A A = − = ⇒ = 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 = Δ + Δ = ⋅ ⋅ − + ⋅ ⋅ − , , ( ) ( ) 0 0 = ⋅ ⋅ ⋅ − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ + − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ = ⋅ 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
Solution: h 1st step Final position x = ? mg A mg A P 0 P 1 After 1st step, the process is irreversible adiabatic. Hence, Δ U = w n C T T P V V V m ⋅ ⋅ − = − − , ( ) ( ) 2 1 2 1 ext or, n R P V nR PV nR P V V A H ⋅ ⋅ − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = − − + ⋅ 3 2 1 2 1 1 1 2 1 [ ( )] ∴ V 2 = V 1 + 0.4 H.A ⇒ x = 0.4 H (The final pressure of gas after 2nd step will remain same as initial, beginning of processes.)
Solution: Smaller the heat capacity larger is Δ T.
Solution: Δ = ⋅ ⋅ − = × × − = H n C T T P m , ( ) ( ) 2 1 1 40 500 300 8000 J ∴ Δ U = Δ H – P ⋅ Δ V = 8000 – 2(40 – 30) × 100 = 6000 J
Solution: q = 0 ⇒ Δ U = w = – P ext ⋅ ( V 2 – V 1 ) = –4 × (30 – 40) = 40 l -bar Now, Δ H = Δ U + Δ ( PV ) = 40 + (4 × 30 – 2 × 40) = 80 L -atom = 8000 J
Solution: Δ U = 0 Δ H = Δ U + Δ ( PV ) = 0 + B ( P 2 – P 1 ) = B RT V B RT V B ⋅ − − − 2 1 = × × − − − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = − 2 8 314 400 1 22 2 1 12 2 332 56 . . J
Solution: Δ = ⋅ ⋅ − U n C T T V m 1 2 1 , ( ) Δ H 1 = Δ U 1 + V ⋅ Δ P = 1 × C V,m × ( T 2 – T 1 ) + V 1 ( P 2 – P 1 ) Now, Δ U 2 = w 2 = – P ext ( V 2 – V 1 ) = – P 3 ( V 2 – V 1 ) Δ H 2 = Δ U 2 + Δ ( PV ) = – P 3 ( V 2 – V 1 ) + ( P 3 V 2 – P 2 V 1 ) ∴ Δ H total = Δ H 1 + Δ H 2 = C V ( T 2 – T 1 ) + V 1 ( P 3 – P 1 )
Solution: η = − 1 T T C H 1 6 1 = − T T C H and 1 3 1 65 390 = − − ⇒ = T T T C H H K = 117° C
Solution: q q T T q C H rej abs rej cal = ⇒ = × = 390 600 120 78
Solution: 1 1 − − Δ ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ > − + Δ ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ T T T T T T C H C H
Solution: T V T V H C ⋅ = ⋅ − − 2 1 3 1 γ γ ⇒ T T V V C M = ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = − − 2 3 1 1 4 1 1 2 75 1 1 5 γ . . . ∴ η = − = 1 1 3 T T C H
Solution: 2 1 3 4 V T 2 T 3 T 4 T 1 P T V T V 2 1 1 3 2 1 ⋅ = ⋅ − − γ γ and T V T V 1 1 1 4 2 1 ⋅ = ⋅ − − γ γ ∴ T T T T T T T T T T 2 1 3 4 2 1 1 3 4 4 = ⇒ − = − @TheBookCorner\n4.41 Thermodynamics HINTS AND EXPLANATIONS η = − = − ⋅ − ⋅ − = − = − ⎛ ⎝ ⎜ 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 , , ( ) ( ) ⎞ ⎞ ⎠ ⎟ − γ 1 = − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = − − 1 1 10 0 6 7 5 1 .
Solution: Δ S unit = Δ S Source + Δ S Heat engine + Δ S Sink = − × + + × = + 40 10 500 0 30 10 300 20 3 3 J/K
Solution: Δ = ⋅ ⋅ S n C T T P m , ln 2 1 = × × = − 3 2 32 14 900 1000 0 14 . ln . c al/K
Solution: Δ = ⋅ S nR V V ln 2 1 = × × = 2 8 314 2 34 58 3 3 . ln ( ) . a a J/K
Solution: Δ = ⋅ ⋅ S n C T T P m , ln 2 1 = × × = 1 5 2 1000 250 7 0 R ln . cal/K
Solution: Δ = ⋅ ⋅ S n C T T V m , ln 2 1 S R K 500 46 2 1 3 2 500 250 − = × × . ln ∴ S 500 K = 48.3 Ccal/K-mol
Solution: Δ = ⋅ S nR V V ln 2 1 ⇒ − = × × × ⇒ = − 5 0 10 15 10 300 15 5 4 5 3 2 2 . ( ) ln . V V L
Solution: Δ = − × = − S Surr J/K 1 5 10 300 5 3 . Now, Δ S unit = Δ S Sys + Δ S Surr = 5.51 + (–5) = + 0.51 J/K Hence, the process is irreversible.
Solution: Δ = ⋅ ⋅ + ⋅ 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 = × × + × ⇒ = n R nR P P ln ln bar
Solution: Δ S = Δ S adiabatic + Δ S isobaric = 0 2 1 + ⋅ ⋅ n C T T P m , ln = × × = − 1 6 4 5 2 1 3 2 2 . ln . R cal/K
Solution: Δ = ⋅ ⋅ + ⋅ S n C T T nR V V V m , ln ln 2 1 2 1 = ⋅ ⋅ + ⋅ ⋅ n C P P n C V V V m P m , , ln ln 2 1 2 1 = × − × + × − × 2 1 5 1 1 4 2 1 5 1 5 1 2 R R . ln . . ln = –11.64 J/K
Solution: S S n C T T nR V V V m 2 1 2 1 2 1 − = ⋅ ⋅ + ⋅ , ln ln = × × + × × 1 2 3 2 1 2 1 2 2 . l n . ln R R = –0.84 cal/K
Solution: Δ = ⋅ ⋅ + ⋅ S n C T T nR V V V m , ln ln 2 1 2 1 = × − × + × × ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ − 1 1 1 2 1 1 2 1 1 R T T R T T n γ ln ln / (as T.V n -1 = Constant) R T T n ⋅ − − − ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ ln 2 1 1 1 1 1 γ = − − − ⋅ ( ) ( )( ) ln n R n γ γ τ 1 1
Solution: Δ = ⋅ ⋅ + ⋅ ⋅ S n C P P n C V V V m P m , , ln ln 2 1 2 1 = × × + × × 2 3 2 2 2 5 2 2 R R ln ln = + 11.2 cal/K
Solution: dS n C dT P dV T V m = ⋅ ⋅ + ⋅ , For maximum entropy, dS dV = 0\n4.42 Chapter 4 HINTS AND EXPLANATIONS or, n C dT dV P V m ⋅ + = , 0 (1) Now, P RT V P V dT dV R P V = = − ⇒ = − 0 0 1 2 α α ( ) (2) From (1) and (2), 1 1 1 2 0 0 0 × − × − + − = R R P V P V γ α α ( ) ( ) ∴ V P = ⋅ + γ α γ 0 1 ( )
Solution: dS C dT T P dV T a dT C T dT V m V m = ⋅ + ⋅ = ⋅ + ⋅ ⋅ , , 1 or, R V dV a dT R V V a T T V V T T 0 0 0 0 ∫ ∫ ⋅ = ⋅ ⇒ ⋅ = − ln ( ) ∴ T = T 0 + R a V V ⋅ ln 0
Solution: dS C dT T T dT T S aT S T T 0 0 3 3 0 3 ∫ ∫ ∫ = ⋅ = ⋅ ⇒ =
Solution: Δ = Δ + Δ = − + = + S S S A B 12000 600 12000 400 10 J/K
Solution: Heat lost by alloy = Heat gained by water or 4 × 4 × (800 – T ) = 4 × 1.0 × ( T – 300) ⇒ T = 700 K (As date is not given for vaporization of water) Now, Δ S mix = Δ S alloy + Δ S water = 4 × 4 × ln 700 800 + 4 × 1 × ln 700 300 = 1.0 K cal/K
Solution: Final temperature of both blocks = T T 1 2 2 + ∴ Δ S = Δ S 1 + Δ S 2 = C T T T C T T T ⋅ + + ⋅ + ln ( ) / ln ( ) / 1 2 1 1 2 2 2 2 = ⋅ + C T T T T ln ( ) 1 2 2 1 2 4
Solution: Δ S = – R [ n 1 ⋅ ln x 1 + n 2 ⋅ ln x 2 ] = – R [0.8 × ln 0.8 + 0.2 × ln 0.2] = + 0.96 Cal/K.
Solution: Larger molar mass, greater is the molar entropy.
Solution: Greater the number of atoms, greater is the molar entropy.
Solution: nC( s ) + (n + 1) H 2 ( g ) → C n H 2n + 2 ( g ) with increase in n , the decrease in entropy increases.
Solution: H 2 O ( l , 1 atm, 100°C) ( ) 1 ⎯ → ⎯ H 2 O ( g , 1 atm, 100°C) ( ) 2 ⎯ → ⎯ H 2 O ( g , 5 atm, 100°C) Δ G 1 = 0 and Δ G 2 = nRT ln P P 2 1 = 5 × 2 × 373 × ln 5 1 = 3730 ln 5 Cal
Solution: q = Δ U – w = 0 – − ⋅ ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ nRT P P ln 1 2 = nRT P P ⋅ ln 1 2 = – Δ G ∴ Δ G = – q = –(–1200) = +1200 cal
Solution: Δ G ° = – R T ⋅ ln K eq ⇒ –1743 = – 8.3 × 300 × ln K eq ∴ K eq = 2 Now, (a) K eq = × 3 3 6 (b) K eq = × 6 3 3 2 (c) K eq = × 6 3 3 2 (d) K eq = × 3 3 6 2
Solution: H O C Pa H O C Pa H O 2 2 2 ( , , . ) ( , , . ) ( l s G G g − ° ⎯ → ⎯ − ° ↓ Δ = ↑ Δ = 10 0 28 10 0 26 0 0 1 3 , , , . ) ( , , . ) − ° ⎯ → ⎯⎯ − ° Δ 10 0 28 10 0 26 2 C Pa H O C Pa 2 G g Δ G 2 = nRT ln P P 2 1 = 1 × R × 263 × ln 0 26 0 28 . . ∴ Δ G = Δ G 1 + Δ G 2 + Δ G 3 = 263 R ln 13 14
Solution: Free expansion is isothermal Δ G = n RT ln P P 2 1 = nRT ln V V 2 1 = 10 5 × (1.2 × 10 –3 ) × ln 1 2 2 4 . . = –84 J\n4.43 Thermodynamics HINTS AND EXPLANATIONS
Solution: Δ G 1 ° = – RT ⋅ ln K 1 and Δ G 2 ° = – RT ⋅ ln K 2 Now, Δ G 2 ° – Δ G 1 ° = – RT [ln K 2 – ln K 1 ] = – RT [ln e 4 ] = –2 × 300 × 4 = –2400 cal
Solution: At 0.04 atom, the system is in equilibrium.
Solution: For isolated system, there should not be any mass and energy transfer with surroundings.
Solution: 2 V V V Given 0.5 P Isothermal P P 0.5 P T P P
Solution: The internal energy of real gas may change on changing the volume of gas. Change in physical state also changes the physical state.
Solution: Option (c) should be changed with (c) adiabatic free expansion of any gas is also isothermal.
Solution: w rev – w irr = (– P ⋅ dV ) – (– P ext ⋅ dV ) = ( P ext – P ) ⋅ dV = negative, always and q rev – q irr = positive, always
Solution: q = 0 w = Δ U = n C T T R V m ⋅ ⋅ − = × × − = − , ( ) ( ) cal 2 1 4 3 2 290 320 360 Δ H = γ ⋅ Δ U = 5 3 360 600 × − = − ( ) cal
Solution: Process BC : P T P T P P B B C C B B = ⇒ = ⇒ = 500 1 250 2 bar and Δ = ⋅ − = × × − = − U n C T T BC V m C B 1 2 1 5 250 500 750 ( ) . R ( ) R Process CD : Δ = ⇒ ⋅ − = − − 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 × × − = − − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ⇒ = 1 5 450 . ( ) K and Δ = ⋅ ⋅ − = × × − = H n C T T R CD P m D C 1 2 2 5 450 250 1000 ( ) . ( ) R
Solution: PV K P K V dP dV K V P V γ γ γ γ γ = ⇒ = ⋅ ⇒ = ⋅ − = − ⋅ − − − 1 1 1 1 ( ). The gas having higher γ will have higher magnitude of slope of P vs. V curve. Now, n R dT P dV nRT V dV ⋅ − ⋅ = − ⋅ = − ⋅ γ 1 or, dV dT V T = − − ⋅ 1 1 γ\n4.44 Chapter 4 HINTS AND EXPLANATIONS Gas having higher γ will have lower magnitude of slope of V vs. T curve. Now, n C dT P dV nRdT V dP V m ⋅ ⋅ = − ⋅ = − − ⋅ , [ ] or, n C dT V dP n R dT nRT P dP P m ⋅ ⋅ = − ⋅ ⇒ ⋅ − ⋅ = ⋅ , γ γ 1 ∴ dP dT P T = − ⋅ γ γ 1 Gas having higher γ will have lower magnitude of slope of P vs. T curve.
Solution: γ V V 2 V CH4 SO2 O2 Ne P Here, γ 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, Δ T = 0 ⇒ Δ U total = 0 or, Δ U I + Δ U II = 0 ⇒ (0 + w I ) + ( q II + 0) = 0 ∴ q II = – w I = maximum for CH 4
Solution: C C R P m V m , , − = ⇒ S S R M P V − = = 0 04545 . ∴ M = 44 gm/mol
Solution: q = 0 ⇒ Δ U = w = – P 0 (4 V 0 – V 0 ) = –3 P 0 V 0 Now, Δ H = Δ U + Δ ( PV ) = (–3 P 0 V 0 ) + ( P 0 ⋅ 4 V 0 – 2 P 0 ⋅ V 0 ) = – P 0 ⋅ V 0
Solution: In reversible cycle, heat rejected is minimum. For reversible cycle, q T T q C H rej abs J = × = × = 400 500 100 80
Solution: dS q T = rev and \u001e ∫ dS = 0
Solution: In rusting, moles of gas decreases.
Solution: Δ U = 0 ⇒ q = – w
Solution: For a process to be spontaneous at low temperature and non-spontaneous at high temperature, Δ H = negative and Δ S = negative.
Solution: ( ) , Δ = × = S J K K Vap atm mol 350 1 3 35 10 350 100 On increasing pressure at constant temperature entropy decreases. ( ) , Δ = G K Vap atm 350 1 0 On increasing pressure at constant temperature energy increases.
Solution: Δ = ⋅ ⋅ − = × × − − = − U n C T T R V m 1 2 1 4 5 2 50 0 1000 ( ) ( ) cal
Solution: Δ H = γ ⋅ Δ U = 7 5 1000 1400 × − = − ( ) cal
Solution: w = 0 ( V = Constant)\n4.45 Thermodynamics HINTS AND EXPLANATIONS Comprehension II
Solution: q U n C T n C T C C V m V m m V m = −Δ ⇒ ⋅ ⋅ Δ = − ⋅ ⋅ Δ ⇒ = − , , ,
Solution: C C R x m V m = + − , 1 ⇒ − = + − ⇒ − ⋅ − = − C C R x R R x V m V m , , 1 2 1 1 γ ∴ x = + γ 1 2 Now, T ⋅ V x –1 = Constant ⇒ T ⋅ V ( γ –1)/2 = Constant
Solution: T T V V T T 2 1 1 2 1 2 2 5 3 1 2 2 300 1 8 150 = ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ⇒ = ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ⇒ = − − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ γ / / K ∴ w nR T T x = − − − = − × − − + ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = − ( ) ( ) / 2 1 1 1 2 150 300 1 5 3 1 2 900 cal Comprehension III
Solution: w = – nR ⋅ Δ T = –1 × 8.314 × 72 = –598.6 J = – 0.6 kJ
Solution: Δ U = q + w = 1.6 + (–0.6) = 1.0 kJ
Solution: γ = Δ Δ = = H U 1 6 1 0 1 6 . . . Comprehension IV
Solution: V 2 = 4 V 0 ⇒ P 2 = 4 P 0 As PV T P V T 1 1 1 2 2 2 = ⇒ P V T P P T 0 0 0 0 0 2 4 4 = ⋅ ⇒ T T 2 0 16 = Now, Δ = ⋅ ⋅ Δ = × − × − = − = − U n C T n R T T P V V V m , ( . ) γ γ α γ 1 16 15 1 15 1 0 0 0 0 0 2
Solution: w nR T T x nR T V = − − = × − − = ( ) ( ) 2 1 0 0 2 1 15 1 1 15 2 α
Solution: C C R x R R R m V m = + − = − + − − = + − 1 1 1 1 1 1 2 1 γ γ γ ( ) ( ) ( ) Comprehension V
Solution: P = a ⋅ T a = a ⋅ PV nR ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ α ⇒ P V ⋅ = − α α 1 Constant ∴ w nR T x R T R T = ⋅ Δ − = × ⋅ Δ − − = − ⋅ Δ 1 1 1 1 1 α α α ( )
Solution: C C R x R R R m V m = + − = − + − − = − + − ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ , ( ) . 1 1 1 1 1 1 1 γ α α γ α
Solution: 1 1 1 0 γ α − + − < ( ) ⇒ α γ γ > − 1\n4.46 Chapter 4 HINTS AND EXPLANATIONS Comprehension VI
Solution: U V n C T V m = ⋅ = ⋅ ⋅ α α , ⇒ T V ⋅ = − α Constant As T V x ⋅ = − 1 Constant ⇒ x – 1 = – a Now, w n R T x U = ⋅ ⋅ Δ − = − ⋅ Δ − 1 1 ( ) γ α
Solution: q U w U U U = Δ − = Δ + − ⋅ Δ = Δ + − ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ ( ) γ α γ α 1 1 1
Solution: C R R x R R m = − + − = − + γ γ α 1 1 1 Comprehension VII
Solution: PV P V 1 1 2 2 γ γ = ⇒ P 2 7 5 1 320 10 128 = × ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = / atm
Solution: PV T P V T 1 1 1 2 2 2 = ⇒ 1 320 300 128 10 2 × = × T ⇒ T 2 1200 = K
Solution: w n C T V m = ⋅ ⋅ Δ = × × × × ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ × − = , . . . ( ) . 1 0 32 0 082 300 5 2 8 314 1200 300 243 3 3 J Comprehension VIII
Solution: PV P V P 1 1 2 2 2 5 3 1 8 32 γ γ = ⇒ = × = ( ) / atm
Solution: 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 = ⇒ T 2 = 1638 K
Solution: Δ = ⋅ ⋅ Δ = × × × × ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ × − H n C T A P m , . . . ( . ) 1 22 4 0 082 27 3 5 2 2 1638 27 3 = 80535 cal Comprehension IX
Solution: Δ U = 0 Δ H = Δ U + V ⋅ Δ P = 0.9 L × 1 532 760 − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ atm = 0.27 L -atm = 27 J
Solution: Δ = → U 1 2 0 Δ = Δ → U U 2 3 for temperature increase + Δ U for vaporization of water. = m . s . ⋅ Δ T + ( q + w ) = × × ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ + × ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ + − × × ⎛ ⎝ ⎜ ⎞ ⎠ 900 4 2 1000 20 450 18 40 450 18 8 1000 373 . ⎟ ⎟ = 1001 kJ
Solution: Δ = Δ + = × + × × ⎛ ⎝ ⎜ ⎜ ⎜ ⎞ ⎠ ⎟ ⎟ ⎟ → → → H H q 1 3 1 2 2 3 27 450 18 40 4 2 1000 20 J+ kJ 900 kJ . = 1075.573 kJ
Solution: w 1 2 0 → = w 2 3 450 18 8 1000 373 74 6 → = − × × = − . kJ\n4.47 Thermodynamics HINTS AND EXPLANATIONS Comprehension X
Solution: T A > T B < T C < T D = T A From question : T T A B = 4 and T A = 800 K ⇒ 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 ⋅ = ⋅ − − γ γ 1 1 ⇒ T T V V C B B C r = ⋅ ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ − 1 = × ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = − 200 8 2 4 400 5 3 1 K
Solution: Δ U BC = C T T R V m C B , ( ) ( ) . ⋅ − = × × − = 1 3 2 400 200 2 4942 kJ
Solution: q = 0 because Δ U = 0 and w = 0
Solution: Enthalpy of ideal gas is independent from pressure.
Solution: For non-ideal gas, U = f ( T , V )
Solution: In adiabatic free expansion, Δ T = 0
Solution: V 1 V 2 V Isothermal Adiabatic P P 1 Pi Pa
Solution: V 1 V 2 V Isothermal Adiabatic P Pa Pi P 2
Solution: Magnitude of work in adiabatic process depends on change in temperature.
Solution: Magnitude of work in adiabatic process depends on change in temperature.
Solution: q n C T P P m = ⋅ ⋅ Δ ,
Solution: Endothermic reactions may also be spontaneous.
Solution: At high temperature, process may become entropy driven.
Solution: At low temperature, process may become enthalpy driven.
Solution: If Δ G = negative and Δ S = negative, Δ H must be negative and Δ G = Δ H – T ⋅ Δ S.
Solution: Δ S sys = 0 but Δ S univ = +ve\n4.48 Chapter 4 HINTS AND EXPLANATIONS
Solution: For ideal gas, H = f ( T ) but in general, H = f ( T , P )
Solution: U f T V dU U T dT U V dV V T = ⇒ = ∂ ∂ ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ⋅ + ∂ ∂ ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ⋅ ( , ) ∂ ∂ ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = ⋅ ∂ ∂ ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = ⋅ ∂ ∂ ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ − U T n C U V T P T P V V m T V , and
Solution: N 2 ( g ) + O 2 ( g ) → 2NO ( g ) ; Δ H = Positive, Δ S ≈ 0 2 KI ( aq ) + HgI 2 ( aq ) → K 2 [HgI 4 ]( aq ) ; Δ H = Negative, Δ S ≈ Negative PCl 5 ( g ) → PCl 3 ( g ) + Cl 2 ; Δ H = Positive, Δ S ≈ Positive NH 3 ( g ) + HCl ( g ) → NH 9 Cl ( s ); Δ H = Negative, Δ S = Negative
Solution: (A) Δ H = 0, Δ U = 0, Δ S total = 0, Δ S Sys = Positive (B) q = 0, Δ S Sys = 0, Δ S total = 0, Δ S surr = 0 (C) q = 0, Δ S surr = 0, Δ S total = Positive.
Solution: (A) Δ H = Negative, Δ V = ± V e (B) Δ H = ± V e, Δ S = − V e, Δ G = Δ H – T , Δ S = + V e, if Δ H = + V e = ± V e, if Δ H = − V e (C) Δ H = Ea f – Ea b = 10 kJ / mol = + V e Δ S = + V e ∴ Δ G = Δ H – T Δ S = − V e, at high temperature (D) Δ H = + V e, Δ S ≈ 0 ⇒ Δ G ≈ Δ H
Solution: (A) Solid \u001f Liquid ; Δ G = 0, Δ S = Positive, Δ V ≈ 0 ⇒ Δ H ≈ Δ U (B) Liquid \u001f Vapour : Δ G = 0, Δ S = Positive, (C) Triple point is equilibrium condition. (D) Melting at boiling point is spontaneous.
Solution: dG = V ⋅ dP – S ⋅ dT ⇒ ( dG ) T = V ⋅ dP and ( dG ) P = – S ⋅ dT
Solution: w = – P ext ( V 2 – V 1 ) = – P ext [ V 0 (1 + γ ⋅ t 2 ) – V 0 (1 + γ ⋅ t 1 )] = – P ext ⋅ V 0 ⋅ γ ⋅ ( t 2 – t 1 ) − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ × × × ° × ° − 10 18 10 0 0002 10 5 2 6 3 N m ( ) . m C C = 0.0036 J
Solution: q = Δ U – w = 0 – − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ nRT V V ln 2 1 or, 420 = 1 × 2 × 300 × ln V 2 1 0 082 300 8 21 × × ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ . . ⇒ V 2 = 6 L
Solution: P = K ⋅ l For initial condition, 1 bar = K × 1 m ⇒ K = 1 bar/m And, V r l = = 4 3 6 3 3 π π ⇒ dV l dl = ⋅ π 2 2 Now, w P dV K l l dl K l l l l V V = − ⋅ = − ⋅ ⋅ ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = − ⋅ − ∫ ∫ ( ) π π 2 2 4 2 2 4 1 4 1 2 1 2 = − × × − − × 10 2 4 1 4 1 10 5 2 4 4 4 7 N/m m J π ( ) m \u001c
Solution: – w = mgh ⇒ P ext ( V 2 – V 1 ) = mgh = × ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ × − × = × × ⇒ = − 4 10 8 2 10 40 10 6 5 2 3 3 N m m ( ) m h h
Solution: q U n C T n C T Q Q Q m V m Δ = ⋅ ⋅ Δ ⋅ ⋅ Δ = − , 2 ⇒ C R m 3 2 2 = ⇒ C m = 6 cal/K mole\n4.49 Thermodynamics HINTS AND EXPLANATIONS
Solution: q w n C T nR T P m = ⋅ ⋅ Δ − ⋅ Δ , ⇒ q R R − = − 2 7 2 ⇒ q = 7 J
Solution: q = q 1 + q 2 = Δ U 1 + Δ H 2 = n C n C V m P m ⋅ ⋅ − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ + ⋅ ⋅ − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ , , 300 2 300 300 300 2 = ⋅ ⋅ = × × = n R 300 2 10 2 300 2 3000 cal
Solution: State I State II Isothermal Isochoric ⎯ → ⎯⎯⎯⎯ ⎯ → ⎯⎯⎯ T 1 = 273 K T 2 = 273 K T 3 = 5 × 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 ⋅ + ⋅ ⋅ − ln ( ) 2 1 3 2 1 or, 80 × 10 3 = 3 × 8.314 × 273 × ln5 + 3 × C V m , × 4 × 273 ∴ C V m , ≈ 21 J/K-mol = 5 cal/K-mol
Solution: PT = Constant ⇒ P ⋅ V 1/2 = Constant Now, C C R x f f m V m = + − ⇒ = × + − ⇒ ≈ 1 1 29 8 314 2 8 314 1 1 2 3 . .
Solution: u u T T 2 1 2 2 2 300 1200 = = ⇒ = K ⇒ T 2 = 1200 K Now, q n C T T V V m = ⋅ ⋅ − = × × ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ × − = , ( ) ( ) 2 1 56 28 5 2 2 1200 300 9000 cal
Solution: Z u N RT M PN RT w aV A = ⋅ ⋅ = ⋅ ⋅ = 1 4 1 4 8 * π Constant or, P T = ⇒ ⋅ = − Constant P V Constant 1 ∴ C C R x R R R m V m = + − = + − − = = 1 1 5 2 1 1 3 6 ( ) cal/K mol
Solution: Δ = ⋅ Δ = × ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ − × = − G V P 13 0 78 10 3001 1 10 5000 6 3 5 . ( ) m N m J 2
Solution: Δ G 1 = Δ H – T 1 ⋅ Δ S and Δ G 2 = Δ H – T 2 ⋅ Δ S ∴ (– Δ G 2 ) – (– Δ G 1 ) = ( T 2 – T 1 ) ⋅ Δ S = ( T 2 – T 1 ) × Δ − Δ H G T 1 1 = (302 – 298) × ( ) ( ) − − − = 5737 6333 298 8 kJ
Solution: Graphite \u001f Diamond; Δ G ° = 5.0 kJ P = 1 bar Δ G = 0 P = ? Now, Δ G 2 – Δ G 1 = ( V P – V G ) ( P 2 – P 1 ) or, 0 5000 12 3 6 12 2 4 10 10 3 10 6 2 5 9 − = − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ × ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ × − ⇒ × − . . ( ) P N m 2
Solution: Δ G ° = – RT ⋅ ln K eq = –2 × 300 × ln ( e –10 ) = + 6000 cal Now, Δ ° = Δ ° − Δ ° = − = S H G T 7500 6000 300 5 cal/k-mol\n4.50 Chapter 4 HINTS AND EXPLANATIONS Four-digit Integer Type
Solution: w nRT V nb V nb an V V = − − − + − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ ln 2 1 2 2 1 1 1 = − × × × − × − × ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ + × × × − 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 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ × ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ ⎥ − = –5713.16 J
Solution: Δ H = Δ U + Δ ( PV ) = 30 + (4 × 5 – 2 × 3) = 44 L -atm
Solution: Calcite → Aragonite Δ H = Δ U + P ⋅ Δ V Δ V = 210 J + 2 7 10 100 3 100 2 7 10 5 6 3 . . × ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ × − N m m 2 = 209 J
Solution: Δ U = ( q + w ) path I = ( q + w ) path II or, 10 × 10 3 × 4.2 J + 0 = (11 × 10 3 × 4.2 J) + (–0.5 w max ) ∴ w max = 8400 J
Solution: H = U + PV = 2.5 PV ∴ Δ H = 2.5 × 10 200 100 10 25 5 3 3 N m m kJ 2 × − × = − ( ) 0 10 25 3 m kJ × − × = − ( )
Solution: q = Δ U – w = 1.5 nR ⋅ Δ T + P ext ⋅ A ⋅ Δ l 42 = 1.5 × 1 × 8.314 × 2 + 100 × 10 3 × 8.5 × × 10 -4 × Δ l ∴ Δ l ≈ 0.2 m
Solution: Δ H – Δ U = P ( V D – V G ) ⇒ − = − ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ × − 1000 12 3 6 12 2 4 10 6 P . . ∴ P = 6000 × 10 5 Pa
Solution: Solid (70 C) Liquid (70 C) Liquid (450 C) Vapour (450 ° ⎯ → ⎯ ° ⎯ → ⎯ ° ⎯ → ⎯ 1 2 3 ° ° C) q = q 1 + q 2 + q 3 = 30 × 10 + 10 × 0.215 × 380 + 10 × 45 = 1567 cal
Solution: q = × × = 12 0 5 1805 6 1805 . J w = – P ⋅ ( V g – V l ) = – P ⋅ V g = – nRT = − × × = − 0 9 18 8 314 373 155 . . J ∴ Δ U = q + w = 1805 + (–155) = 1650 J (for 0.9 g ) = × × = − 1650 0 9 18 10 33 3 . kJ
Solution: q = Δ U – w = 0 ⇒ Δ U = w = – P ext ( V 2 – V 1 ) = – 100(–1) = 100 bar-ml Now, Δ H = Δ U + Δ PV = 100 + (100 × 99 – 1 × 100) = 9900 bar-ml = 990 J
Solution: q = Δ U – w = 0 ⇒ Δ U = w = × × 1 2 10 1000 bar-ml = 500 J V 990 1000 1001 bar 1 bar (ml) 2 1 P
Solution: For adiabatic process: T T V V 2 1 1 2 1 1 4 1 300 2 400 = ⋅ ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = × = − ⋅ − γ ( ) K Now, w = w 1 + w 2 = − ⋅ ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ + ⋅ − nRT V V nC T T V m ln [ ( ) , 2 1 2 1 = [–1 × 8.3 × 300 × ln 2] + 1 8 3 1 4 1 400 300 × − × − ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ . . ( ) = 332 J
Solution: q n C T V m = ⋅ ⋅ Δ , or, 50 × 166 × t = 1 8 21 10 2 9 10 0 0821 290 5 8 3 2 20 3 × × × × × × × × . . . . ∴ t = 500 sec
Solution: Δ U = 3 × 1.5 R × 100 + 2 × 2.5 R × 100 = 1900 cal\n4.51 Thermodynamics HINTS AND EXPLANATIONS
Solution: 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 = ⋅ ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = × ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = − − γ atm Now, w total = w AB + w BC = ⋅ ⋅ Δ + − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ n C T nRT P P V m B C , ln = × × − − × × × 1 3 2 300 1200 1 300 2 1 R R ( ) ln = –3120 cal
Solution: Δ ° = − + × = − + × S S S S CH grap H 4 ( ) . ( . . ) 2 186 2 6 0 2 130 6 2 = –81 J/K Now, Δ G ° = Δ H ° – T ⋅ Δ S ° = – T ⋅ Δ S univ or, (–75 × 10 3 ) – 300 × (–81) = –300 × Δ S univ ⇒ Δ S univ = 169 J/K
Solution: (– mgh ) × N = (– Δ G °) or 50 × 10 × 2 × N = 600 10 2 27 27 300 3 × × × ⇒ = N
Solution: (– Δ G ) = – ( Δ H – T ⋅ Δ S ) = − − × ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ = ( ) 2808 310 200 1000 2870 kJ
Solution: a = 2 (i, vi) b = 3 (ii, v, vii) c = 4 (iii, iv, viii, ix) d = 9 (all)
Solution: N O NO Brown 2 4 2 2 \u001f On heating, colour deepens means reaction is endothermic Δ H ve = + ( ) .
Solution: 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.
Solution: Δ Δ Δ Δ Δ H H H H 2 1 2 1 2 1 3 2 0 − − = ( ) = ( ) − + ( ) = ⇒ = T T C x x x P
Solution: I 2 (s) → I 2 (g); Δ Δ H cal/gm at K H at K 1 1 2 2 24 473 523 = = = = T T ? Δ Δ Δ H H H 2 1 2 1 2 2 2 24 523 473 0 055 0 031 − − = ( ) − ( ) ⇒ − − = − T T C I g C I s P P , , . . ∴ Δ H c al/gm 2 25 2 = .
Solution: For direct measurement, reaction must occur directly in the conditions to measure heat.
Solution: Greater the mass per cent of hydrogen, greater is the calorific value.
Solution: For the reaction: Δ Δ Δ U H RT kJ = − ⋅ = − ( ) − − ( ) × × = − n g 72 3 1 8 314 1000 298 69 8 . . . As HCl is limiting reagent, for the given amount, Δ U kJ = × − ( ) = − 2 69 8 139 6 . .\n5.36 Chapter 5 HINTS AND EXPLANATIONS
Solution: HAuBr 4 + 4HCl → HAuCl 4 + 4HBr; Δ H = (–28) – (–36.8) = 8.8 kcal ∴ Percentage reaction = × = 0 44 8 8 100 5 . . %
Solution: (a) C + O CO; 1 2 9 0 4 5 2 → . . Δ H = –75 kcal Heat evolved = 75 × 9 = 675 kcal (b) C + O CO 2 2 2 2 → ; Δ H = –95 kcal Heat evolved = 95 × 2 = 190 kcal (c) 4C O CO 3CO + → + 3 5 2 2 . Heat evolved = 75 × 1 + 95 × 3 = 360 kcal (d) C + O CO 2 2 2 5 2 5 → . . Δ H = –95 kcal Heat evolved = 95 × 2.5 = 237.5 kcal
Solution: CH3 NO2 NO2 O2N l O g CO g H O l N g ( ) + ( ) → ( ) + ( ) + ( ) 21 4 7 5 2 3 2 2 2 2 2 Δ H kJ/mol k = × − ( ) + × − ( ) − ( ) = − = − × 7 395 5 2 285 65 3542 5 3542 5 227 1 816 . . . J J/mol kJ/mol MJ/L = − = − 28 34 28 34 . .
Solution: For H A 2 , enthalpy of ionization = × − ( ) = 2 13 5 13 1 . kcal/mol For B OH 2 ( ) , enthalpy of ionization = × − ( ) = 2 13 5 10 7 . kcal/mol ∴ Required Δ = × − − = − H kcal 2 13 5 1 7 19 .
Solution: If BaSO 4 were water soluble, then Δ = × − ( ) = − H kJ expected 2 57 114
Solution: Required Δ = − − × ( ) = − H cal 13700 400 0 9 13340 .
Solution: Aerobic oxidation results release of energy and hence, it is biologically benefical by (2880 + 2530 = 5410 kJ/mol)
Solution: (a) Si H g H g SiH g H kcal 2 6 2 4 ( ) + ( ) → ( ) Δ = − 2 11 7 ; . (b) SiH g SiH g H g H kcal 4 2 2 ( ) → ( ) + ( ) = + ; . Δ 239 7 (c) 2Si s H g Si H g H kcal 2 ( ) + ( ) → ( ) = + 3 80 3 2 6 ; . Δ Required thermochemical equation is Si s H g SiH g 2 2 ( ) + ( ) → ( ) From (b) a) (c) H kcal/mol + + = + 1 2 1 2 274 ( : Δ
Solution: Required thermochemical equation is Dy s Cl g DyCl s 2 3 ( ) + ( ) → ( ) 3 2 From (ii) + 3 × (iii) – (i), we get: Δ H kJ/mol = − ( ) + − ( ) − − ( ) = − 699 43 3 158 31 180 06 994 3 . . . .
Solution: Δ − ( ) + − ( ) + − ( ) + − ( ) ⎡ ⎣ ⎤ ⎦ − − ( ) + − H= 188 84 22 05 2 22 1 2 17 63 70 97 8 5 . . . . . . 6 6 0 2 68 32 74 18 ( ) + + − ( ) ⎡ ⎣ ⎤ ⎦ = − . . kcal
Solution: H aq OH aq H O l 2 + − ( ) + ( ) → ( ) Δ = Δ − Δ + Δ ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ ( ) ( ) ( ) + − H H H H f H O l f H aq f OH aq 2 or, − = − ( ) − + Δ ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ − ( ) 57 32 285 84 0 . . f OH aq H ∴ Δ = − − ( ) f OH aq H kJ/mol 228 52 .
Solution: CH CH COOH l O g CO g H O l 3 2 2 2 2 7 2 3 3 ( ) + ( ) → ( ) + ( ) Δ = × Δ + × Δ ⎡ ⎣ ⎤ ⎦ − Δ ( ) ( ) ( ) 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 ( ) ( ) + × Δ ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ 7 2 2 or, 3 3 94 3 68 3 2 3 2 × Δ = − ( ) + − ( ) ⎡ ⎣ ⎤ ⎦ − Δ ⎡ ⎣ ⎤ ( ) ( ) f CH CH COOH f CH CH COOH H H +O l l ⎦ ⎦ ∴ Δ = − ( ) f CH CH COOH H kcal/mol 3 2 121 5 l .\n5.37 Thermochemistry HINTS AND EXPLANATIONS
Solution: phOH(solution II) → phOH(solution I) Δ = − × ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ − − × ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = − H kcal/mol 0 02 0 47 94 0 03 1 410 94 2 . . . .
Solution: Δ = − ( ) − − ( ) ⎡ ⎣ ⎤ ⎦ − − [ ] = + H kcal/mol required 14 7 2 7 13 75 1 75 . . . .
Solution: 2 1 2 197 2 65 2 3 FeO O Fe O H 2 + → = − ( ) − − ( ) ; Δ = − 67 kcal Initial Final 2 2 2 a a a x a x − + 2 2 1 2 3 5 a x a x x a − + = ⇒ = and heat released = 67 x kcal ∴ Heat released per mole of initial mixture = = 67 3 13 4 x a . kcal
Solution: 3C H OH 25H O 3C H OH 25H O 2 5 2 2 5 2 + → ( ) ; Δ Δ H kcal H cal theo exp = − ( ) + − ( ) = − = − ( ) = − 1120 2 1760 4640 3 1650 4950 As experimentally, more heat is released means the mixing is exothermic by (4950 – 4640) = 310 cal.
Solution: Heat absorbed in solubility = Heat released from solution = Δ = + ( ) × × = m.s. T J 200 25 4 2 3 2835 . ∴ Δ H J = + × = + 2835 7 45 74 5 28350 . .
Solution: Heat released by reaction = Heat gained by ice = = × = m.L cal 0 2 80 16 . Δ − = − × − H= cal 16 10 16 10 3 3
Solution: let C 2 H 6 = x L, then CH 4 = (4 – x )L Volume of CO 2 produced, 2 x + (4 – x ) = 6 ⇒ x = 2 ∴ Total heat evolved = − × + × ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ × × = 1 2 1573 1 2 890 1 0 0821 300 50 . kJ
Solution: C O CO H cal + → Δ = − × = − 2 2 2400 0 3 12 96000 ; . C O CO H cal + → Δ = − × = − 1 2 1400 0 6 12 28000 2 ; . Now, CO O CO H cal + → Δ = − − ( ) − − ( ) = − 1 2 96000 28000 68000 2 2 ; ∴ Heat produced cal = × = 68000 28 0 7 1700 .
Solution: Heat liberated from propane = Heat absorbed by water or, n n × × × = × × × ⇒ = 500 10 40 100 160 10 1 50 40 3 3
Solution: q = v . i . t = 15 × 0.125 × (14 × 60) J = 1575 J ∴ Δ = × = H J/mol 1575 0 1 1 15750 .
Solution: H g O g H O g H 240 kJ 2 2 2 1 1 2 ( ) + ( ) → ( ) Δ = − ; 2H g O g H O H 672 kJ ( ) + ( ) → ( ) Δ = − + ( ) = − 1 2 240 432 2 2 2 g ; ∴ Δ Δ = − − = H H 2 1 672 240 2 8 .
Solution: Δ = = × × × ( ) = × − m E C kg 2 3 8 2 12 103 10 4 2 3 10 4 8 10 . .
Solution: − = − ( ) + + ( ) ⎡ ⎣ ⎤ ⎦ − − ( ) + − ( ) + + − ( ) ⎡ ⎣ ⎤ ⎦ 12 75 3 9 11 8 5 17 5 . . . x ∴ X = − 22 1 . kcal/mol
Solution: Dulong and petit’s law : Atomic mass × Specific heat ≅ 6.4 for greater temperature rise, heat lost should be high.\n5.38 Chapter 5 HINTS AND EXPLANATIONS
Solution: g O g CO g H O l ( ) + ( ) → ( ) + ( ) 9 2 3 3 2 2 2 Δ = × Δ + × Δ ⎡ ⎣ ⎤ ⎦ − Δ ( ) ( ) 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 + × Δ ( ) ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ 9 2 2 = × − ( ) + × − ( ) ⎡ ⎣ ⎤ ⎦ − ( ) + ( ) + × ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ 3 394 3 286 33 20 9 2 0 { } = − 2093 kJ/mol
Solution: w = − = − = − × × = − P.V n RT cal H 2 1 5 2 298 894 .
Solution: w = − = − × × = − nRT cal 1 2 353 706 ∴ Δ = + = + − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = U kcal/mol q w 7 4 706 1000 6 694 . .
Solution: C H g +5O g 3CO g +4H O l 3 8 2 2 2 ( ) ( ) → ( ) ( ) Δ = × Δ + × Δ ⎡ ⎣ ⎤ ⎦ − Δ + × Δ ( ) ( ) ( ) ( ) 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 ( ( ) ⎡ ⎣ ⎤ ⎦ = − ( ) + − ( ) ⎡ ⎣ ⎤ ⎦ − − ( ) = − 3 393 5 4 285 8 103 8 2219 9 . . . . kJ Δ Δ Δ Δ Δ 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 = − + ⎡ ⎣ ⎤ ⎦ + + ⎡ ( ) ( ) ( ) ( ) ⎣ ⎣ ⎤ ⎦ = − − ( ) + − ( ) ⎡ ⎣ ⎤ ⎦ + − ( ) + − ( ) ⎡ ⎣ ⎤ ⎦ = − 1560 0 890 0 2219 9 285 8 55 7 . . . . . kJ J
Solution: The required thermochemical equation is K s Cl g KCl s ; H 2 ( ) + ( ) → ( ) = 1 2 Δ ? From (iv) + (iii) – (v) + (i) – (ii): we get, Δ H Kcal = − ( ) + − ( ) − ( ) + − ( ) − − ( ) = − 116 5 39 3 4 4 13 7 68 4 105 5 . . . . . .
Solution: From 1 3 2 2 3 × + × + × ( ) ( ) ( ) : i ii iii we get, Δ = − ( ) + ( ) + − ( ) = − H kJ Required 1 3 46 4 2 9 0 2 3 41 24 8 . . .
Solution: For the reaction, 1 2 1 2 2 2 H g I s HI g ( ) + ( ) → ( ) ; Δ = − ( ) − − ( ) + − ( ) − − ( ) + − ( H Required 1 2 44 20 1 2 52 42 17 31 19 21 13 74 . . . . . ) ) − − ( ) = 13 67 5 94 . . kcal
Solution: The required thermochemical equation is I s O g I O s 5 2 2 2 5 2 ( ) + ( ) → ( ) From 2 × (ii) + 6 × (v) + 5 × (vii) – (i) – 6 × (iii) – 6 × (iv) – (vi) – 10 × (viii) – 10 × (ix), we get, Δ = − ( ) + − ( ) + − ( ) − ( ) − − ( ) − − ( ) − − H Required 2 322 6 100 5 255 4 0 6 44 6 57 22 . 4 4 10 92 10 75 169 ( ) − − ( ) − − ( ) = − kJ
Solution: Given thermochemical equations are (i) H g H g H kJ 2 ( ) → ( ) = 2 218 ; Δ (ii) Cl g 2Cl g H kJ 2 ( ) → ( ) = ; Δ 124 (iii) 1 2 3 2 46 N g H g NH g H kJ 2 2 3 ( ) + ( ) → ( ) = − ; Δ (iv) 1 2 2 1 2 314 N g H g Cl g NH Cl s H kJ 2 2 2 4 ( ) + ( ) + ( ) → ( ) = − ; Δ (v) H g H g e H kJ ( ) → ( ) + = + − ; Δ 1310 (vi) Cl g e Cl g H kJ ( ) + → ( ) = − − − ; Δ 348 (vii) NH Cl s NH g Cl g H kJ 4 ( ) → ( ) + ( ) = + − 4 683 ; Δ Required thermochemical equations are NH g H g NH g H 3 ( ) + ( ) → ( ) = + + 4 ; ? Δ From ( ) ( ) ( ) ( ) ( ) ( ) ( ) vii i v iii ii i v vi + − − − − − 1 2 1 2 Δ = ( ) + − ( ) − − ( ) − × ( ) − × ( ) − ( ) − − H required 683 314 46 1 2 124 1 2 218 1310 348 8 718 ( ) = − kJ/mol\n5.39 Thermochemistry HINTS AND EXPLANATIONS
Solution: In such polymerization, one sigma bond is formed on cleavage of one pi bond. Δ H B.E. B.E. required C C bond C C bond = ( ) − ( ) = − ( ) − ( ) = − − − π σ 590 331 331 7 72 kJ/mole
Solution: Required thermochemical equation is C S 2H g O g CH OH l 2 2 3 ( ) + ( ) + ( ) → ( ) 1 2 Δ H kJ = + × + [ ] − × + + [ ] − = − 715 4 218 249 3 415 356 463 38 266
Solution: 3 3 C s H g C H g H 53kJ 2 3 6 exp ( ) + ( ) → ( ) = ; Δ Δ H 3 715 6 218 3 356 6 408 63kJ theo = × + × [ ] − × + × [ ] = − ∴ Strain energy H H kJ theo = − = Δ Δ exp 116
Solution: 2C g 6H g C H g 2 6 ( ) + ( ) → ( ) Δ H B.E B.E kJ C C C C = − = + [ ] − + × [ ] ⇒ = − − 2839 0 0 6 412 367 and, 2C g 4H g C H g 2 4 ( ) + ( ) → ( ) Δ H B.E B.E. kJ C C C C = − = + [ ] − + × [ ] ⇒ = = = 2275 0 0 4 412 627 Now, 6C g 6H g C H g 6 6 ( ) + ( ) → ( ) Δ H R.E. = − = + [ ] − × + × + × [ ] − 5506 0 0 3 367 3 627 6 412 ∴ R.E. kJ/mol = 52
Solution: C H S C H g S g C H S S C H g 2 5 2 5 2 5 2 5 − − ( ) + ( )→ − − − ( ) Δ H kJ = − ( ) − − ( ) + ⎡ ⎣ ⎤ ⎦ = − 202 143 222 276 B.E. kJ/mol s s − = 276
Solution: CH g CH g H g 4 3 ( ) → ( ) + ( ) 103 103 2 18 33 5 = + ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ − − [ ] ⇒ = ( ) ( ) Δ Δ f CH g f CH g H H kcal/mol 3 3 .
Solution: Δ H required = × + × + + × [ ] − × + + + + + 6 414 2 348 580 2 610 3 414 348 580 354 462 11 18 2 580 140 2 462 + × + + × ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ = −348 Κ J
Solution: Δ H kJ = − = − 50 70 20
Solution: Δ H B.E. B.E. bond in C C bond inC C = ( ) − ( ) = − ( ) − ( ) = − = − π σ 835 610 348 123 k kJ
Solution: n HCHO g HCHO 2 ( ) → ( ) Δ H n n = − = × − ( ) − − ( ) ⇒ = 72 134 732 6 ∴ Molecular formula HCHO C H O 6 6 12 6 = ( ) =
Solution: There is 2.5 B-B bond per B atom. Hence, Δ H = –2.5 × 300 = –750 kJ/mole of Boron.
Solution: Let the enthalpy of combustion of gauche form be – x kcal/mol. Now, 690 0 7 2 0 2 0 06 3 0 04 5 5 = × − ( ) + × + × + ( ) + × + ( ) . . . . . x x x x ∴ x = 691
Solution: M s X g MX s H 1.5 H eg g ( ) + ( ) → ( ) Δ = × Δ ( ) 2 2 ; x From Born–Hafer Cycle, we get: Δ = Δ + Δ + Δ + Δ + × Δ + Δ ( ) ( ) ( ) ( ) ( ) 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 . . . . . × − ( ) = Δ + × Δ + × Δ + × × Δ ( ) ( ) ( ) sub M s sub M s sub M s s H H H u ub M s sub M s H H ( ) ( ) ( ) + − ( ) + × − × × Δ ⎡ ⎣ ⎤ ⎦ 2 96 5 0 8 1 2 . . Δ = ( ) sub M s H kcal/mol 41 38 .\n5.40 Chapter 5 HINTS AND EXPLANATIONS
Solution: Combustion is exothermic. Decomposition or elimination are endothermic. Graphite is more stable form.
Solution: Conversion of liquid into gas is endothermic.
Solution: Δ ° = f H 0 for elements in their reference state.
Solution: Endothermic compounds have +ve Δ ° f H .
Solution: For Δ = Δ Δ = H E, n g 0
Solution: One mole of the substance should burn completely.
Solution: Δ = + ( ) f NO g H ve
Solution: Heat released in reaction = Heat gained by calorimeter system = × = 1 5 1 4 2 1 . . . kJ n H SO eq 2 4 100 0 5 1000 0 05 ( ) = × = . . n NH OH Limiting reagent eq 4 200 0 2 1000 0 04 ( ) = × = . . ( ) Δ neut NH OH H By strong acid kJ/eq kJ/mole 4 ( ) . . . . = − = − = − 2 1 0 04 52 5 52 5 Δ = − ( ) − − ( ) = diss NH OH H kJ/mol 4 52 5 57 4 5 . . Δ = − ( ) − = diss CH COOH H kJ/mol 3 57 48 1 4 5 4 4 . . .
Solution: (a) Δ = + × ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ − − = r H kJ 436 1 2 495 242 925 5 ( ) . (b) Δ = × + × ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ − − = r H kJ 1 2 436 1 2 495 42 423 5 ( ) . (c) Δ = × = f H(g) H kJ mol 1 2 436 218 / (d) Δ = f OH(g) H kJ/mol 42
Solution: Resonance occurs in 1, 3-Butadiene and N 2 O.
Solution: Resonance occurs in product but not in reactant.
Solution: (a) Δ = × − ( ) ⎡ ⎣ ⎤ ⎦ − − ( ) + − ( ) ⎡ ⎣ ⎤ ⎦ = − r H kJ 2 1263 2238 285 3 (b) − = Δ − × Δ 3KJ H H C -maltose C glucose α 2
Solution: (a) n C s ( ) = × = 1 2 1000 12 100 . ∴ Maximum obtainable heat = 100 × 94 = 9400 cal (b) Heat released = × + × = 100 68 100 68 13600 cal (c) Heat released = × ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = 13600 100 1200 30 5440 cal
Solution: C H COOH s O g CO g H O l 2 6 5 2 2 15 2 7 3 ( ) + ( ) → ( ) + ( ) Δ = − ( ) + − ( ) ⎡ ⎣ ⎤ ⎦ − − [ ] = − H kJ/mol 7 393 3 286 408 3201 and Δ = Δ − Δ = − ( ) − − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ × × = − U H n RT kJ/ g . . . 3201 7 15 2 8 314 1000 300 3199 75 m mol
Solution: Δ H kcal = = × − ( ) = − q 3 35 105 Δ Δ Δ U H n RT kcal g = − = − ( ) − − ( ) × × ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ × = − . . 35 2 3 2 1000 300 3 103 2 and w u q = − = − ( ) − ( ) = 103 2 105 1 8 . . kcal
Solution: From − × − × + × − × 3 4 1 4 1 4 9 4 a b c d , Δ = − × − − × + × − = − H kcal required 3 4 76 1 4 240 1 4 36 9 4 68 147 ( ) ( ) ( ) ( )
Solution: From 3 4 1 4 1 4 1 4 × + × − × + × a bc d , Δ = × − + × − × + = H kcal/mol required 3 4 76 1 4 240 1 4 36 1 4 68 11 ( ) ( ) ( ) ( )
Solution: Given data based\n5.41 Thermochemistry HINTS AND EXPLANATIONS Comprehension II
Solution: Δ = − × + ⋅ = − ⋅ H kJ 75 5 5 1 8 55 15
Solution: Δ = − × + ⋅ = − ⋅ H kJ 75 10 10 1 8 63 56
Solution: Δ = − + ⋅ = − + ⋅ ∞ = − H n kJ 75 1 1 8 75 1 1 8 75
Solution: Δ = − − − = − H kJ ( . ) ( . ) . 63 56 55 15 8 41
Solution: Δ = − − − = − H kJ ( ) ( . ) . 75 63 56 11 44 Comprehension III
Solution: Δ = − ⋅ = − ⋅ H kJ/mol H 1 2 483 636 2 241 818 Δ = − ⋅ = − ⋅ H kJ/mol H 2 2 868 2 3 289 4 Δ = − H kJ/mol H 3 2 347 33 .
Solution: Δ = − ⋅ = − ⋅ H kJ/gm O 1 2 483 636 32 15 11 Δ = − ⋅ = − ⋅ H kJ/gm O 2 3 868 2 48 18 09 Δ = − ⋅ = − ⋅ H kJ/gm H O 3 2 2 347 33 34 10 22
Solution: Δ = − ⋅ = − ⋅ H kJ/gm reactant 1 483 636 36 13 43 Δ = − ⋅ = − ⋅ H kJ/gm reactant 2 868 2 54 16 08 Δ = − ⋅ = − ⋅ H kJ/gm reactant 3 347 33 36 9 65 Comprehension IV
Solution: Heat released = 0.25 × 320 = 80 cal ∴ Molar enthalpy of solution = − ⋅ × = − 80 0 98 98 8000 cal
Solution: Heat released = 0.80 × 320 = 256 cal ∴ Δ − ⋅ × = − r H = cal 256 0 49 98 51200 Comprehension V
Solution: 2 3 2 1 2 2 2 3 C(S)+ H g N g CH CN(g); ( ) ( ) + → Δ H B.E. B.E. C C C N = = × + × + × − × + + → − ≡ 88 2 719 3 2 435 1 2 948 3 414 1 [ ] [ ] 3 2 3 8 C(s) 4H g C H g + → ( ) ( ) Δ = − = × + × − × + × → − 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 − ≡ = = 335 899 5 . Now, CH CN(g) 2H g CH CH NH (g), 3 2 3 2 2 + → ( ) Δ = × + + + × − × + + + × H [ . ] [ ] 3 414 335 899 5 2 435 5 414 335 378 2 426 = − 288 5 . kJ/mol\n5.42 Chapter 5 HINTS AND EXPLANATIONS Comprehension VI
Solution: C < C p,m,N (g) p,m,H O(g) 2 2
Solution: H g O g H O g 2 2 2 1 2 ( ) ( ) ( ); + → Δ = − ⋅ Δ = − ⋅ 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 × = × mol N 2 is also present. Now, Heat released from reaction = Heat gained by H O(g) 2 and N g 2 ( ) 56.0 × 10 3 = x × 6.2 × ( T 2 – 300) + 2 x × 4.9 × ( T 2 – 300) ∴ T 2 = 3800 K
Solution: 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 = ⇒ + + ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ × = + × ( ) ∴ p 2 = 10.86 atm
Solution: q = 0, w = 0 ⇒ ∆ E = 0 Comprehension VII
Solution: C H l O g CO g H O(g) 18 8 2 2 2 25 2 8 9 ( ) ( ) ( ) + → + ∆ c H = [8 × (−94) + 9 × (−58)] − [−74] = −1200 kcal/mol
Solution: ∆ H required = [8 × (−26.5) + 9 × (−58)] − [−74] = −660 kcal/mol
Solution: 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 × + ⋅ − × − × × × + ⋅ − × ⋅ × + ⋅ × 1200 0 1 660 1 1000 8 8 500 8 0 1 7 0 500 0 9 6 6 0 500 87 3 ⋅ × = ⋅ ] ∴ x = 0.05 Moles of CO 2 formed = 0.05 × 8 = 0.4
Solution: Moles of H 2 O formed = 9 x + (0.1− x ) × 9 = 0.9
Solution: w p v v p n RT p n RT p R n T n T = − − = − − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = − − ( ) ( ) 2 1 2 2 1 1 2 2 1 1 = − × ⋅ × + ⋅ − ⋅ × + ⋅ × − ⋅ × + ⋅ − ⋅ × 2 0 05 8 0 1 0 05 8 0 9 800 0 05 25 2 0 1 0 05 17 2 [{ ( ) } ( ) { { } × − 300 2090 ] = cal Comprehension VIII
Solution: CO 2H CH OH rd mol + ⎯ → ⎯⎯ 2 2 3 3 1000 = × = 3 2 1000 1500 mol In reformer, CO and H 2 is forming in 1 : 3 ratio.
Solution: CO = 1500 −1000 = 500 mole H 2 = 4500 − 2000 = 2500 mole
Solution: Heat produced in 1 min = 1000 × 100 R × 60 = 1.2 × 10 7 cal\n5.43 Thermochemistry HINTS AND EXPLANATIONS Comprehension IX
Solution: Let x mole C be converted into CO. Hence, x × 26 + (1 – x ) × 94 = 53.2 ⇒ x = 0.6 Hence, moles of C formed = 0.6
Solution: O consumed +(1 ) ]32 gm 2 2 22 4 = − ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ = ⋅ x x Comprehension X Given thermochemical equations are (a) H 2 S (g) → H (g) + H S (g); ∆ H = 376.0 kcal (b) H 2 (g) + S(s) → H 2 S (g); ∆ H = −20.0 kcal (c) S (s) → S (g); ∆ H = 277.0 kcal (d) H 2 (g) → 2H (g); ∆ H = 436.0 kcal
Solution: 1 2 2 H S(s) HS(g) ( ) g + → From a b d we get: + − × 1 2 , Δ = + − − × = H kJ mol required 376 20 1 2 436 138 ( ) /
Solution: HS(g) → H(g) + S(g) From (d) − (a) − (b) + (c) ∆ H required = 436 − 376 − (−20) + 277 = 357 kJ/mol Comprehension XI
Solution: Δ ° = Δ ° + Δ ⋅ = + × × = H E n RT kcal g 2 1 2 2 1000 298 3 292 . . Now, Δ ° = Δ ° − ⋅ Δ ° = − × = − G H T S kcal 3 292 298 1000 20 2 668 . .
Solution: Spontaneous as ∆ G° = −ve
Solution: H 2 SO 4 is dibasic but HCl is monobasic.
Solution: Information based
Solution: Heat liberated will be four times but as quantity is also four times, the change in the temperature will be same.
Solution: Solubility is exothermic but all gases are not highly soluble in all liquid.
Solution: If | ∆ Hydration H| < | ∆ lattice H|, the salt dissolves partially and the extent depends on the difference in two values.
Solution: H HDiamond + O2 Hgraphite + O2 HCO2
Solution: H H1 – Extent + O2 H2 – Extent + O2 HCO2 + H2O
Solution: Theory based\n5.44 Chapter 5 HINTS AND EXPLANATIONS
Solution: C → CO, ∆ H ≠ ∆ H combustion
Solution: (A) ∆ n g = 0 (B) ∆ n g = −1 (C) ∆ n g = 1 (D) ∆ n g = − 2
Solution: (A) ∆ H = (−57.3) + 15 = −42.3 kJ (B) ∆ H = −42.3 − 70.7 + 20 = −93.0 kJ (C) ∆ H = −70.7 + 15 = −55.7 kJ (D) ∆ H = 0
Solution: (A) Mg (s) + Cl2 (g) –(1110 + 790) (2360 – 1110) Mg 2+ (aq) + 2Cl – (aq) Mg 2+ (g) + 2Cl – (aq) + 1110 Mg 2f (aq); +Cl2 (g) or = 1110 – (1110 + 790) + (2360 – 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 ( ) ( ); ( ) ( ) ( ) → Δ = − + + = − − + + K KJ/mol (C) Mg 2+ (g) +2Cl − (aq) → Mg 2+ (aq) +2Cl − (aq); ∆ H = −790 − 1110 = −1900 kJ (D) Mg 2+ (g) + 2Cl − (g) → MgCl 2 (s); ∆ H = − 640 − 1870 = − 2510 kJ
Solution: Defi nition based
Solution: 3O 2 (g) → 2O 2 (g)
Solution: ∆ n g = 0 ⇒ ∆ H = ∆ U ∆ n g = + ve ⇒ ∆ H > ∆ U ⇒ If ∆ H = −ve, then | ∆ H| < | ∆ U| ⇒ If ∆ H = +ve, then | ∆ H| > | ∆ U| ∆ n g = − ve ⇒ ∆ H < ∆ U ⇒ If ∆ H = −ve, then | ∆ H| > | ∆ U| ⇒ If ∆ H = +ve, then | ∆ H| < | ∆ U|
Solution: Defi nition based (10) (A) Δ = Δ + − × − = + − × × 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 − = + ) kJ/mol (B) Δ = Δ + − × − = + − × × 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 − = + ) kJ/mol (C) ∆ H 300 = 3 × 25 − 100 − 52 = − 77 kJ/mol (D) Δ = Δ + − × × − = − + − × × 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 × − = − ( ) kcal/mol
Solution: For HCl : 13.7 × 0.05 = c × 411 (1) For HCOOH : q × 0.05 = c × 321 (2) From (2) ÷ (1) ⇒ q = 10.7 kcal ∴ Enthalpy of ionisation of HCOOH = 13.7 − 10.7 = 3.0 kcal/mol
Solution: (C 6 H 10 O 5 ) x + 6 x O 2 (g) → 6 x CO 2 (g) + 5 x H 2 O(l) ∆ H = − 4.6 × 162 x = [6x( −94.2) + 5 x (−68.4)] − Δ f C H O H 6 10 5 ( ) x + ⎡ ⎣ ⎤ ⎦ 0 ∴ Δ f C H O H 6 10 5 ( ) x = −162 kcal/mol = −1kcal/gm
Solution: 1 800 3120 800 3120 × = × ⇒ = a a L/hr Butane C H O CO H O 2 4 10 2 2 13 2 4 5 + → + ∴ Rate of Oxygen Supply L/hr = × × = 800 3120 13 2 3 5
Solution: Δ = × − = H kcal/mol required 100 75 13 7 12 2 2 ( . . )\n5.45 Thermochemistry HINTS AND EXPLANATIONS
Solution: Total moles of gases = × × = 1 192 1 642 0 0821 298 0 08 . . . . Now, n n CH CH 4 4 210 10 1260 0 667 0 004 3 × × = × ⇒ = . . ∴ Volume per cent of CH 4 = ⋅ ⋅ × = 0 004 0 08 100 5%
Solution: Heat released by 6 3 64000 ⋅ mole haemoglobin = 25 4.2 J × × = 0 03 3 15 . . ∴ Heat released per mole haemoglobin = ⋅ × ⋅ = 3 15 64000 6 3 32000 J ∴ Heat released per mole O 2 = = 32000 4 8000 J
Solution: Heat released = × × × − = 300 1 0 1 0 26 25 300 . . ( ) cal Now, n HA = × ⋅ = ⋅ 200 0 4 1000 0 08 n NaOH = × ⋅ = ⋅ 100 0 5 1000 0 05 Hence, NaOH is a limiting reagent. ∴ Δ = − ⋅ × = − neut H cal/mol 300 0 05 1 6000
Solution: There is 3 H-bond per NH 3 molecule because for each bond two NH 3 molecules are required. ∴ Strength of H-bond = − = 30 4 15 4 3 5 0 . . . kcal/mol
Solution: C(s)+ O g CO(g); H kcal/mol 2 1 1 2 7 5 3 12 30 ( ) → Δ = − ⋅ × = − C(s)+O g CO (g); H kcal/mol 2 2 ( ) → Δ = − × = − 2 32 4 12 96 Now, CO (g) CO(g)+ O g H H H kcal/mol 2 1 2 2 1 2 66 → Δ = Δ − Δ = + ( ); For 4 gm CO H kcal 2 ⋅ Δ = × = 0 66 44 4 6 ,
Solution: ( ) 1 6900 3 4 − × + × = ⇒ = a a a 2900 3900 ∴ n n eq(HA) eq(HB) : ( ) : : : = − = = 1 1 4 3 4 1 3 a a Four-digit Integer Type
Solution: C 2 H 6 + H 2 → 2 CH 4 ; ∆ H = −65.2 kJ C 3 H 8 + 2H 2 → 3 CH 4 ; ∆ H = −87.4 kJ Hence, for CH 4 (g) + C 3 H 8 (g) → 2 C 2 H 6 (g); ∆ H = (−87.4) −2 × (−65.2) = + 43 kJ
Solution: Moles of O 2 consumed = × − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ × × ⎧ ⎨ ⎩ ⎫ ⎬ ⎭ × ⋅ × = 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 + → + Δ − ; ∴ Heat produced in body per hr = × = 3100 6 24 31 400 kJ
Solution: Number of glycogen units oxidized per day = × × × × = 150 60 60 24 432 10 30 3
Solution: Moles of C = = 15 12 1 25 . Moles of O 2 = × ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ × ⋅ × = 20 19 100 8 21 0 0821 380 1 . 1 25 0 5 2 2 . . C + O CO +0.75 CO → ∴ Heat produced = × + × = 0 5 26 0 75 96 85 . . kcal\n5.46 Chapter 5 HINTS AND EXPLANATIONS
Solution: C 2 H 5 OH(l) + O 2 (g) → CH 3 COOH(g) + H 2 O(l) ∆ H = [(−118)+(−68)] − [(−66)+0] = −120 kcal Hence, rate of heat removal = × × × = 120 46 2 3 10 40 100 2400 3 . kcal/mol
Solution: C 6 H 12 O 6 (s) + 6 O 2 (g) → 6 CO 2 (g) + 6 H 2 O(l) ∆ H = [6 × (−395) + 6 × (−285)] − [(−1280) + 0] = −2800 kJ Moles of CO 2 released per astronaut = × = 6 2800 2100 4 5 . ∴ Mass of LiOH required = 4.5 × 2 × 24 = 216 gm
Solution: 16 1 322 100 10 500 3 3 . . × × = × ⇒ = = V 10000 V 0 5m L
Solution: Total heat absorbed = × + × ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ × = 8 9 45 1 9 72 2 5 120 . KJ
Solution: For banana: q = c × 3.0 (1) For benzoic acid: 800 122 0 305 0 × = × . . c 4 (2) From (1) and (2), q = 1.5 kacl for 2.5 gm banana ∴ Heat obtained per banana = ⋅ ⋅ × = 1 5 2 5 125 75 kcal
Solution: H O H O(l); H kJ 2 2 2 298 1 2 286 ( ) ( ) g g + → Δ = − H O(l) H O(g); H kJ 2 2 → Δ = ⋅ 398 40 8 Δ = Δ + Δ ⋅ Δ = − × − = H H C T 40.8+ kJ 398 p 298 33 4 75 4 1000 298 398 45 . . ( ) ∴ H O(g) H O H 2 2 → + Δ = − − 2 298 1 2 45 286 ( ) ( ); [ ] g g = 241 kJ
Solution: Δ − Δ = Δ ⋅ ∫ H H C dT p T T 1 2 1 2 or, 0 T dT T T − − = × = × − − − ∫ ( ) ( ) ( ) 4000 2 10 2 10 2 300 2 300 2 2 2 ∴ T K = 700
Solution: 3C(s) + 3H 2 (g) → C 3 H 6 (g) ∆ H theo = (3 × 715 + 6 × 218) − (3 × 356 + 6 × 408) = −63 kJ ∆ H exp = [3 × (−393) +3 × (−285)] − [3 × (−697)] = 57 kJ ∴ Strain energy = 57 − (−63) = 120 kJ/mol
Solution: KF.CH COOH s K g F CH COOH g H 734 kJ 3 3 ( ) → ( ) + ( ) Δ = + − . ; CH COOH l CH COOH g H 20 kJ 3 3 ( ) → ( ) Δ = ; KF s K aq F aq H kJ ( ) → ( ) + ( ) Δ = + − ; 35 K g K aq H kJ + + ( ) → ( ) Δ = − ; 325 F g F aq H kJ − − ( ) → ( ) Δ = − ; 389 KF s CH COOH l KF CH COOH s H kJ ( ) + ( ) → ( ) Δ = − 3 1 3 25 ; Required: F g CH COOH g F CH COOH g − − ( ) + ( ) → ( ) 3 3 ; Δ = − ( ) + − − + − ( ) + − ( ) = − H kJ/mol 389 734 20 35 325 25 60
Solution: B s H BH g ( ) + ( ) → ( ) 3 2 2 3 g Δ = = + × ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ − × ( ) − H B E. B H 100 565 3 2 436 3 . ∴ B E. kJ/mol B H . − = 373 2B(s) + 3H g B H g 2 ( ) ( ) → 2 6 Δ = = × + × [ ] − × + × [ ] − H B E. 3c 2e 36 2 565 3 436 4 373 2 . ∴ B E. kJ/mol 3c 2e . − = 455\n5.47 Thermochemistry HINTS AND EXPLANATIONS
Solution: XeF Xe F F F H kcal 4 + → + + + Δ = × ( ) + + − ( ) + − ( ) = − 2 4 34 279 85 38 292
Solution: (a) KF CH COOH s K ACOH F ACOH CH COOH l kJ . . 3 3 3 ( ) → ( ) + ( ) + ( ) = + − (b) KF s K ACOH F ACOH H 35 kJ ( ) → ( ) + ( ) Δ = + − ; (c) F CH COOH g F g CH COOH g H 46 kJ . ; − − ( ) → ( ) + ( ) Δ = 3 3 (d) KF CH COOH s K g F CH COOH g H 734 kJ . . ; 3 3 ( ) → ( ) + ( ) Δ = + − (e) KF s K g F g H 797 kJ ( ) → ( ) + ( ) Δ = + − ; Required: CH COOH l CH COOH g 3 3 ( ) → ( ) From (c) (a)+(d) e b − − + ( ) ( ), we get: Δ = − − ( ) + − + = H 46 kJ/mol 3 734 797 35 21
Solution: (l) + 3H2 (g) (g) N NH ∆ H = (−50) − [ ∆ f H py(l) + 0] = (40 + 125) + [2 × {(−156) − (−37)} + {(−18) − 44}] ∴ ∆ f H py(l) = 85 kJ/mol
Solution: 1 2 5 2 847 2 2 5 I F g IF g H kJ s ( ) + ( ) → ( ) Δ = − ; − = × + ( ) + × − × − 847 B E. I F 1 2 62 149 5 2 155 5 . B E. kJ/mol I F . − = 268 1 2 3 2 470 2 2 3 I s F g IF g H kJ ( ) + ( ) → ( ) Δ = − ; − = + ( ) + × − × + ⎡ ⎣ ⎤ ⎦ − ( ) 470 1 2 62 149 3 2 155 2 268 B E. I F eq . B E. kJ/mol I F eq . − ( ) = 272
Solution: H g O g H O l H kJ 2 2 2 1 2 286 ( ) + ( ) → ( ) Δ = − ; − + × ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ − × − − − 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 ( ) + ( ) → ( ) Δ = − ; − + ( ) − × + − − − − 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 − = − . 142
Solution: (a) Ag + (aq) +Br - (aq) → AgBr(s); ∆ H = −84.54 kJ (b) Ag(s) → Ag + (aq); ∆ H = −8 x kJ (c) 1 2 9 Br l Br aq H kJ 2 ( ) → ( ) = − ; Δ x (d) Ag s Br l Ag Br s H kJ 2 ( ) + ( ) → ( ) = − 1 2 99 54 ; . Δ As (a) + (b) + (c) = (d), we get: (−84.54) + (−8 x ) + 9 x = −99.54 ⇒ x = −15 ∴ ∆ f H Ag + (aq) = −8 x = 120 kJ/mol
Solution: K eq for the reaction in backward direction = = × × = − − − − − − K K s b f 2 1 10 3 9 10 53 846 3 1 1 5 1 1 . . . L mol s L mol
Solution: Stability constant, K K K f b = = × × = × − 1 45 10 1 22 10 1 1885 10 13 9 17 . . .
Solution: K K K B A f b eq = = [ ] [ ] 2 ⇒ 1 5 10 100 10 10 10 3 2 5 . ( / ) ( / ) × = − − K b ⇒ K b = 1.5 × 10 –11 M –1 S –1
Solution: For pent hydrate to be efflorescent, Q < K p or, P H O 2 atm 2 4 2 10 < − ⇒ P H O 2 atm mm < = − 10 7 6 2 .
Solution: Q P P K P P = × = × = < 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.
Solution: Q K P K P P p < ⇒ < H O 2 2 ⇒ 40 760 100 1 21 10 2 4 × ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ < × − R H . . ∴ R.H. < 20.9 %
Solution: H 2 (g) + I 2 (s) \u001f 2HI(g) ; K p = 6.4 × 10 –4 atm I 2 (s) \u001f I 2 (g) ; K p = 1.6 × 10 –4 atm ∴ H 2 (g) + I 2 (g) \u001f 2HI(g) ; K P = × × = − − 6 4 10 1 6 10 4 4 4 . .
Solution: Net rate of reaction of HI, − ⋅ = − = − − 1 2 1 2 1 2 2 d dt r r b f [ ] [ ] [ ][I ] HI K HI K H
Solution: α = − − ⋅ = − − × = M M n M 0 1 208 5 124 2 1 124 0 681 ( ) . ( ) .
Solution: PCl g PCl g 3 2 5 ( ) Cl ( ) ( ) + g \u001f ⇀ \u001f ↽ \u001f \u001f Initial partial pressure P 0 P 0 0 Equilibrium partial pressure P 0 – 0.75 P 0 P 0 – 0.75 P 0 0.75 P 0 –0.25 P 0 –0.25 P 0 Now, K P P P P = × PCl PCl Cl 5 3 2 ⇒ 2 0 75 0 25 0 25 0 0 0 = × . . . P P P ⇒ P 0 = 6 atm ∴ Initial total pressure of mixture = 2 P 0 = 12 atm
Solution: N g 3H g NH g 2 2 3 2 ( ) ( ) ( ) + \u001e ⇀ \u001e ↽ \u001e \u001e Initial moles 1 3 0 Moles at equilibrium 1 – x 3 – 3 x 2 x Total moles of gases = (1 – x ) + (3 – 3 x ) + 2 x = 4 – 2 x Equilibrium partial pressure 1 4 2 3 3 4 2 2 4 2 − − × − − × − × x x P x x P x x P Now, K x x P x x P x x P x P = − × ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ − − × ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ − − × ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = 2 4 2 1 4 2 3 3 4 2 4 27 1 2 3 2 ( − − × − ≈ × × x x P x P ) ( ) 4 2 2 2 2 4 2 4 16 27 ∴ x K P P K P P = ⋅ = ⋅ 27 64 3 3 8 2
Solution: X 2 + Y 2 \u001f 2 XY Initial moles 2 3 0 Final moles 2 – x 3 – x 2 x\n6.37 Chemical Equilibrium HINTS AND EXPLANATIONS [ ] . XY = = 2 5 0 7 x ⇒ x = 1.75 ∴ [X ] . 2 2 5 0 05 = − = x M and [Y ] . 2 3 5 0 25 = − = x M
Solution: N 2 + O 2 \u001f 2NO Equilibrium moles 1 – x 1 – x 2 x 0 09 2 1 1 2 . ( ) ( )( ) = − − x x x ⇒ x = 0.13
Solution: N 2 + O 2 \u001f 2NO Initial moles 4 a a 0 Equilibrium moles 4 a – x a – x 2 x Now, 0 0004 2 4 4 4 2 2 . ( ) ( )( ) = − − ≈ ⋅ x a x a x x a a ⇒ x a = 0 02 . ∴ Per cent of NO = 2 5 100 0 8 x a × = . %
Solution: N 2 + 3H 2 \u001f 2NH 3 Initial moles 1 5 0 Moles at equilibrium 1 – x 5 – 3 x 2 x Total moles = (1 – x ) + (5 – 3 x ) + 2 x = 6 – 2 x From question, 2 6 2 0 4 x x − = . ⇒ x = 6 7 K x x x x P P = − × − × − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = × − − ( ) ( ) ( ) . 2 1 5 3 6 2 2 6 10 2 3 2 4 2 atm
Solution: N 2 + 3H 2 \u001f 2NH 3 Initial moles 4 16 0 Moles at equilibrium 4 – x 16 – 3 x 2 x Total moles = (4 – x ) + (16 – 3 x ) + 2 x = 20 – 2 x From question, 20 9 10 20 2 × =− x ⇒ x = 1 Now, K x x x V C = − − ⋅ = × − − ( ) ( )( ) . 2 4 16 3 6 07 10 2 3 2 4 2 M
Solution: 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 + + = × + × = .
Solution: 2SO 2 + O 2 \u001f 2SO 3 Initial moles 2 1 0 Moles at equilibrium 2 – 2 x 1– x 2 x From question n eq SO 2 = n eq MnO 4 − . or, (2 – 2 x ) × 2 = 0.4 × 5 ⇒ x = 0.5 ∴ K x x x C = − × − = − ( ) ( ) ( ) 2 2 2 1 2 2 2 1 M .
Solution: 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 – x 1 – x x = = 44 88 0 5 . x Case II 120 60 2 = mole 46 46 1 = mole 0 0 Moles at Equ. 2 – y 1 – y y y K x x x x y y y y eq = ⋅ − ⋅ − = ⋅ − ⋅ − ( ) ( ) ( ) ( ) 1 1 2 1 ⇒ y = 2 3 ∴ Mass of CH 3 COOC 2 H 5 at equilibrium = 2 3 88 × =
Solution: R 1 OH + CH 3 COOH \u001f CH 3 COOR 1 + H 2 O Initial moles 1 1 0 0 Equ. moles 1 – x 1 – ( 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 – y 1 – ( x + y ) y x + y From question, x + y = 0.8 and x y = 3 2 ∴ 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 = ⋅ + − − + = × × = ( ) ( )[ ( )] . . . . .
Solution: 2NO( g ) + Cl 2 ( g ) \u001f 2NOCl( g ) Initial partial pressure 2P 0 P 0 0 Equ. partial pressure 2P 0 – 2x P 0 – x 2x From question, (2 P 0 – 2 x ) + ( P 0 – x ) + 2 x = 1 ⇒ 3 P 0 – x = 1 (1) and 2 1 4 0 x P x = − ( ) (2) From (1) and (2), P 0 = 9 x and x = 1 26 ∴ K x P x P x P = − − = − ( ) ( ) ( ) 2 2 2 13 256 2 0 2 0 1 atm
Solution: S 8 ( g ) \u001f 4 S 2 ( g ) Initial partial pressure 1 atm 0 Equ. partial pressure 1 – 0.3 4 × 0.3 = 0.7 atm = 1 .2 atm ∴ K P = = ( . ) . . 1 2 0 7 2 96 4 3 atm
Solution: HCl( ) O Cl H O 2 g g g g + + 1 4 1 2 1 2 2 2 ( ) ( ) ( ) \u001e ⇀ \u001e ↽ \u001e \u001e Initial partial pressure 730 8 100 × 730 92 100 × = 58.4 mm = 671.6 mm Equilibrium partial pressure 58.4 – 58.4 × 0.08 671 6 58 4 0 08 4 . . . − × = 670.432 mm
Solution: H 3 BO 3 + Glycerin \u001f complex Initial concent. 0.1 a M 0 Equ. Concert 0.1 – 0.06 ( a – 0.06) M 0.06 M = 0.04 M Now, K a eq = = × − 0 9 0 06 0 04 0 06 . . ( . ) ( . ) ⇒ a = 1.73 M
Solution: 2A( g ) \u001f A 2 ( g ); K P = 8 × 10 8 atm –1 Initial partial pressure 1 atm 0 Partial pressure on complete reaction 0 0.5 atm Equilibrium partial pressure 2 x atm 0.5 – x ≈ 0.5 atm Now, 8 10 0 5 8 2 × = . P A ⇒ P A = 2.5 × 10 –5 atm
Solution: K eq = 3.8 × 10 –7 10 6 3 2 − − × [ ] [ ] HCO CO ⇒ [ ] [ ] . HCO CO 3 2 0 38 − =
Solution: A(g) \u001f nB(g) Initial mole 1(say) 0 Equilibrium mole 1 – a n a Total moles = 1 – a + n a = 1 + a ( n – 1) Now K P P n n P n P n P P B n A n n n = = + − ⋅ ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ − + − ⋅ ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ = ⋅ − α α α α α 1 1 1 1 1 1 ( ) ( ) ( ) ( ( ) [ ( )] 1 1 1 1 − ⋅ + − − α α n n
Solution: A + B \u001f C + D Initial moles a a 0 0 Equilibrium moles a – x a – x x x From question, [ A ] = 2[ C ] ⇒ a – x = 2 x ⇒ a = 3 x Now, K K K x x a x a x f b eq = = ⋅ − ⋅ − ( ) ( ) ⇒ 2 10 2 2 3 × = ⋅ ⋅ − K x x x x b ∴ K b = 8 × 10 –3 mol –1 L S –1\n6.39 Chemical Equilibrium HINTS AND EXPLANATIONS
Solution: (Cl 2 CHCOOH) 2 \u001f 2Cl 2 CHOOH Initial Conc. 0 0129 258 100 1000 . / / 0 = 5 × 10 –4 M Equ. Conc. 5 × 10 –4 – x 2 x Now, K eq = 5 × 10 –4 = ( ) ( ) 2 5 10 2 4 x x × − − ⇒ x = 1.95 × 10 –4 ∴ [Cl 2 CHOOH] = 3.90 × 10 –4 M
Solution: NH 2 CONH 4 ( s ) \u001f 2NH 3 ( g ) + CO 2 ( g ) Initial moles 1 0 0 Equ. moles 1 – a 2 a a From question, 3 α = ⋅ P V RT ∴ Percentage dissociation of solid = 100 a % = ⋅ 100 3 PV RT %
Solution: P H O eq 2 , = ( K P ) 1/4 = (8.1 × 10 –7 ) = 0.03 atm P H O eq actual 0.04 atm 2 30 4 760 , , . = = ∴ Mass of water vapour absorbed = ( . . ) . . 0 09 0 03 1 642 0 0821 300 18 − × × × = 0.012 gm
Solution: Addition of CO will shift second reaction backward. Decrease in Cl 2 will shift the fi rst reaction forward.
Solution: P g P P g P H O HCl(g) H O HCl(g) 2 2 new ( ) ( ) , 2 2 2 = × ⇒ P P HCl (g), new HCl (g) = × 2
Solution: Ionic form of the reaction is NH H O NH OH H 4 2 4 + + + + \u001e ⇀ \u001e ↽ \u001e \u001e
Solution: K A B AB K AB AB B 1 2 2 = = + − − − [ ][ ] [ ] [ ] [ ][ ] and Now, [ ] [ ] [ ] A AB K K B + − − = ⋅ 2 1 2 2
Solution: 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 ′ atm Now, P P P P 2 2 × = × ′ ⇒ P ′ = 0.25 P
Solution: N 2 + 3H 2 \u001f 2NH 3 Equ. partial pressure 100 mm 400 mm 1000 mm New equ. partial pressure 100 – a + x 400 + 3 x = 700 mm 1000 – 2 x = 800 mm ∴ x = 100 mm Now, K P P N = × = × 1000 100 400 800 700 2 3 2 3 2 ⇒ 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
Solution: 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 – 0.25 3 – 0.25 × 2 0.25 = 0.75 = 2.5 Total moles = 0.75 + 2.5 + 0.25 = 3.5 Now, K P P = × × ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = × − − 0 25 0 75 2 5 3 5 6 23 10 2 2 3 . . ( . ) . . ∴ P = 10.24 bar
Solution: A \u001f B + C; K 1 = 10 6 Initial moles 1 0 0 Equilibrium moles 1 – x + y x – y x B + D \u001f A; K 2 = 10 –6 x 1 1 Equilibrium moles x – y 1 – y 1 + y – x\n6.40 Chapter 6 HINTS AND EXPLANATIONS As K 1 >> 1, we may assume x ≈ 1 Now, K y x x y y y y y 2 1 1 1 1 = + − − − ≈ − ⋅ − ( ) ( )( ) ( ) ( ) As K 2 << 1, we may assume y << 1 K y y y y 2 1 1 = − − ( )( ) \u001a ∴ [A] = 1 – x + y ≈ y = 10 –6 M
Solution: A \u001f B Initial a M b M Equilibrium ( a – x ) M ( b + x ) M Now, K K K b x a x eq = = + − 1 2 ∴ x K a K b K K = − + 1 2 1 2
Solution: r b = K b ⋅ P C(g)
Solution: Δ G ° = –2.303 RT ⋅ ln K p ° Δ H ° – T ⋅ Δ S ° or –2.303 × 8.314 × T × ln 1.0 = 240 × 10 3 – T × 50 ∴ T = 4800 K
Solution: Δ G ° = –2.303 RT ⋅ ln K eq ⇒ –2.303 × 10 3 = –2.303 × 2 × 500 × ln K eq ∴ K eq = 10 Now, K P P P eq = × HI H I 2 2 1 2 1 2 / / ⇒ 10 0 001 2 1 2 1 2 P H / / ( . ) × ∴ P H atm 2 1000 =
Solution: Δ ° = − ⋅ = − ⋅ = − ⋅ G RT B RT RT ln [ ] [ ] ln ln . α 64 36 1 78
Solution: There is no net change at equilibrium.
Solution: K eq at 27°C, K 1 3 4 2 10 2 10 4 = × × = − − and K eq at 127°C, K 2 2 3 8 10 4 10 20 = × × = − − Now, ln K K H R T T 2 1 1 2 1 1 = Δ − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ or, 2 303 20 4 1 300 1 400 . log = Δ − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ H R ∴ Δ H = 2.303 × 8.314 × 1200 log(5) J/mol
Solution: n A \u001f A n Initial moles 1 0 Equilibrium moles 1 – x x n Now, K x n V x V x V x n x V n x C n n n n = ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = ⋅ − ⋅ ≈ ⋅ << − − / ( ) ( ) 1 1 1 1 1 as Now, total moles = (1 – x ) + x n = 1 + x ⋅ 1 1 n − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = + ⋅ ⋅ − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = + − ⋅ − − 1 1 1 1 1 1 n KC V n n n K V n C n ( )
Solution: α = − − ⋅ = − − ⋅ = − M M M M M M M M mix mix mix mix mix ( ) ( ) n 1 2 1 1 = ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ − = − M dRT P PM dRT 1 1
Solution: n RT RT NO = × = 0 4 250 100 . n RT RT O 2 0 8 100 80 = × = . 2 NO + O 2 → 2 NO 2 \u001f N 2 O 4 Initial moles 100 RT 80 RT 0 0 Final moles 0 30 RT 100 RT x − x 2 From question, 30 100 2 0 3 350 RT RT x x RT + − + = × . ∴ 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 = − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ × × ⎛ ⎝ ⎜ ⎜ ⎞ ⎠ ⎟ ⎟ − . . = 3.5 atm
Solution: N 2 + 3H 2 \u001f 2NH 3 Initial moles 1 3 0 Equilibrium moles 1 – x 3 – 3 x 2 x From question, 2 4 2 x x a − =\n6.41 Chemical Equilibrium HINTS AND EXPLANATIONS ⇒ x a a = + 2 1 Now, K x x x P x x x x P P = − − ⋅ − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = ⋅ − − ⋅ − ( ) ( )( ) ( ) ( ) 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 = ⋅ − − ⋅ = ⋅ + ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ⋅ − + ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ⋅ − 2 4 2 27 1 2 2 1 4 4 1 27 1 2 1 2 ( ) ( ) + + ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ⋅ a P 2 = ⋅ − ⋅ 32 27 1 2 a a P ( ) ∴ a a P ( ) 1 2 − α
Solution: K P P = ⋅ − α α 2 2 1 ⇒ ( . ) ( ) . 0 3 1 1 0 3 0 1 1 2 2 2 2 × − − = × − α α ⇒ a = 0.973
Solution: K P = P 2 Now, ln K K H R T T 2 1 1 2 1 1 = Δ − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ or, ln ( ) P 2 2 3 2 7 10 3360 2 1 300 1 400 × = − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ − ⇒ P 2 = 1.4 × 10 –2 atm
Solution: CO(g) + H 2 (g) \u001f CO 2 (g) + H 2 (g) Initial 1 5 0 1 Equilibrium 1 – x 5 – x x 1 – x Now, K x x x x eq = = ⋅ + − ⋅ − 1 3 1 1 5 ( ) ( ) ( ) ⇒ x = 1 2
Solution: 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 × = . 22 5 5 4 . = 22 2 5 5 2 × = . Now, K p = × × × = × 4 12 4 2 27 2 3 1 2 10 5 ( ) ( ) ( ) / .
Solution: 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 = ⋅ = ⋅ ( ) ( ) 2 3 0 2 0 0 2 ⇒ P P = 4 9 0 Now, 3 3 31 27 0 0 P P P + =
Solution: Δ H ° = Δ E ° + Δ n g ⋅ RT = (+30) + (3 – 2) × 2 1000 300 30 6 × = + . K cal Now, Δ G ° – RT ⋅ ln K eq = Δ H ° – T ⋅ Δ S ° or, – 2 × 300 × ln K eq = 30.6 × 10 3 – 300 × 100 ⇒ ln K eq = ∴ K eq = 1 e
Solution: trans \u001f Cis ; Δ G ° = 22.112 – 30.426 = – 8.314 KJ Now, Δ ° = − ⋅ G RT Cis trans ln [ ] [ ] ⇒ – 8.314 × 10 3 = – 8.314 × 300 × ln [ ] [ ] Cis trans ∴ [ ] [ ] Cis trans = 28 1
Solution: CO( g ) + H 2 O( g ) \u001f CO 2 ( g ) + H 2 ( g ) Initial moles 2 5 0 2 Equilibrium moles 2 – x 5 – x x 2 + x Now, K x x x x eq = = ⋅ + − − 3 0 2 2 5 . ( ) ( )( ) ⇒ x = 1.5 ∴ [ ] . H M 2 2 2 1 75 = + = x
Solution: NH 4 HS(s) \u001f NH 3 (g) + H 2 S(g) X 2 bar X 2 bar ∴ Δ G ° = – RT ⋅ ln K P ° = – RT ln X X RT 2 2 2 2 , (ln ln ) ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = − × −\n6.42 Chapter 6 HINTS AND EXPLANATIONS
Solution: 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 × 0.2 = 0.5 × P ⇒ P = 0.08
Solution: 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 – a ) 2 a P 0 α P 0 2 and α = − − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ⋅ = − M D D M D D 2 5 2 1 2 2 3 Total equilibrium pressure = P 0 (1 – a ) + 2 a P 0 + α α P P 0 0 2 1 3 2 = + ⎛ ⎝ ⎜ ⎞ ⎠ ⎟
Solution: Le Chatelier’s principle
Solution: Vapour pressure of a particular liquid system depends only on temperature.
Solution: Cl 2 (g) \u001f 2Cl(g) T ↑ P ↓
Solution: Addition of insert gas at constant pressure shifts the equilibrium in the direction of increase in moles of gases.
Solution: Decrease in pressure favors the reaction is the direction of increase in moles of gas and hence, B should be monomer.
Solution: Δ ° > > f G : NO N O NO 2 2 5
Solution: At 300 K : Δ G ° = (–41) – 300 × (–0.04) = –29 KJ/mol Hence, the reaction is spontaneous in forward direction. At 1200 K : Δ G ° = (–33) – 1200 × (–0.03) = + 3 KJ/mol Hence, the reaction is spontaneous in backward direction.
Solution: S A e H RT = ⋅ −Δ / ⇒ ln s = ln A H RT − Δ Positive slope represents that Δ H = negative.
Solution: 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 = ⋅ = × = × − Cl H O(g) 2 atm ( ) . ( . ) P H O(g) 2 = Vapour pressure of ice.
Solution: PCl 5 (g) \u001f PCl 3 (g) + Cl 2 (g) Initial moles 5 0 0 Moles at equilibrium 5 – x x x From question, ( ) . . 5 4 4 8 112 0 0821 546 − + + + = × × x x x ⇒ x = 3 ∴ α = = x 5 0 6 . and K x x x P = ⋅ − ⋅ ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = 5 4 8 12 1 8 . . atm
Solution: 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 = × = × 5 10 0 5 0 4 4 12 2 2 4 ( . ) ( . ) ( ) ⇒ x = 5 × 10 –4
Solution: Δ ° = × Δ ° − Δ ° = G G G f g f g 2 0 2 4 NO N O 2 ( ) ( ) ⇒ K P ° = 1 Now, Δ = Δ ° + ⋅ = + ⋅ G G Q RT P P RT ln NO N O 2 4 0 2 2 ln = ⋅ RT ln 10 10 2 = positive.\n6.43 Chemical Equilibrium HINTS AND EXPLANATIONS
Solution: AB 2 (g) + A( s ) \u001f 2 AB(g) Initial partial pressure 0.7 bar 0 Equ. partial pressure (0.7 – x ) bar 2 x bar Second equ. partial pressure y bar (0.4 – y ) bar From question, (0.7 – x ) + 2 x = 0.95 ⇒ x = 0.25 ∴ K x x P = − = = ( ) ( . ) ( . ) . 2 0 7 0 5 0 45 5 9 2 2 Now, 5 9 0 4 2 = − ( . ) y y ⇒ y = 0.13 ∴ At second equilibrium, the volume per cent of AB 2 0 13 0 4 100 32 5 = × = . . . %
Solution: PCl 5 ( g ) \u001f PCl 3 ( g ) + Cl 2 ( g ) Initial moles 1 1 0 Equ. moles 1 – x 1 + x x ≈ 1 ≈ 1 = 0.004 ∴ K C = × × ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = 1 0 004 1 1 10 0 0004 . . M
Solution: Δ ° = − ⋅ ° G RT K P ln ⇒ –1743 = – 8.3 × 300 × ln K P ° ∴ K P ° = 2
Solution: K K K 1 2 3 1 1 0 24 = × = . As Δ n g = 0, [ A ] + [ B ] + [ C ] = 1 M
Solution: Addition of water will shift the reaction in the direction of increase in mole of aq species.
Solution: For SrCl 2 ⋅ 2H 2 O(s), P H O 2 = (2.56 × 10 –10 ) 1/4 = 0.004 atm For Na 2 HPO 4 ⋅ 7H 2 O P H O 2 = (2.43 × 10 –13 ) 1/5 = 0.003 atm For Na 2 SO 4 (s), P H O 2 = (1.024 × 10 –27 ) 1/10 = 0.002 atm As P H O 2 is minimum for Na 2 SO 4 (s), it is the best dehydrating agent.
Solution: 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 . . . . × ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ < R H ⇒ R . H . < 5 %
Solution: Na 2 HPO 4 ⋅ 7H 2 O(s) to be deliquescent, P H O 2 > 0.003 atm or, 0 04 100 0 003 . . . . × ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ > R H ⇒ R . H . > 7.5 % Comprehension II
Solution: CO( g ) + 2H 2 ( g ) \u001f CH 3 OH( g ) Initial moles 0.2 a (say) 0 Moles at equ. 0.2 – x a – 2 x x = 0.1 = a – 0.2 = 0.1 Total moles = 0.1 + ( a – 0.2) + 0.1 = 7 5 2 463 0 0821 750 . . . × × ⇒ a = 0.3 Now, K P = × × ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = − − 0 1 0 1 0 1 7 5 0 3 0 16 2 2 2 . . ( . ) . . . atm
Solution: K K RT C P n g = = × = Δ − − ( ) . ( . ) 0 16 0 0821 750 607 2 2 M
Solution: P = + × × = ( . . ) . . . 0 2 0 3 0 0821 750 2 463 12 5 atm\n6.44 Chapter 6 HINTS AND EXPLANATIONS Comprehension III
Solution: 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 – x a – x x x From question, x a 2 0 333 1 3 = = . ⇒ x a = 2 3 Now, K x x a x eq a x = ⋅ − = ⋅ − ( ) ( ) 4
Solution: 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 − 2 3 a x x x Now, K x x a x a x eq = = ⋅ − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 4 3 2 3 ⇒ x = 0.2833 a ∴ Fraction of alcohol reacted = x a / . 3 0 85 =
Solution: Solution of 0.7 = x a × = 100 66 67 . % Comprehension IV
Solution: 2HI( g ) \u001f H 2 ( g ) + I 2 ( g ) Initial moles 1(say) 0 0 Equilibrium moles 1 – 0.2222 = 0.7778 0.1111 0.1111 ∴ K eq = × = ≈ 0 1111 0 1111 0 7778 1 49 0 02 2 . . ( . ) .
Solution: In the presence of I 2 ( g ), the extent of dissociation of HI will decrease.
Solution: Addition of He( g ) will not affect thequilibrium. Comprehension V
Solution: 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 ∴ x = 100 and P = 525
Solution: K P = ( P + x ) ⋅ x = 625 × 100 mm 2 ∴ ′ K P (required) = 1 1 6 10 5 K P = × − . mm –2 .
Solution: P P K P NH H S mm 3 2 250 = = =
Solution: Minimum mass of NH 4 HS ( s ) needed. = × × × 250 760 5 0 0 0821 300 51 . . gm\n6.45 Chemical Equilibrium HINTS AND EXPLANATIONS Comprehension VI
Solution: K A A e e e eq f b H RT = ⋅ = − − × × = −Δ / ( . ) . 24 942 10 8 314 300 3 10
Solution: K K K e b f eq = = − 1 10 Δ H = Ea f – Ea b = Ea f – 3 2 Ea f ⇒ Ea f = 2 ⋅ (– Δ H) Now, K A e e e f f Ea RT f = ⋅ = × = − − × × × − / . . 1 2 24 942 10 8 314 300 20 3 and K b = e –30 Comprehension VII
Solution: 2SO 3 \u001f 2SO 2 + O 2 Equilibrium moles 1 – a a α 2 Now. K K P P = ⋅ − ⋅ + ⎛ ⎝ ⎜ ⎜ ⎞ ⎠ ⎟ ⎟ ′ α α α α 2 2 2 1 1 2 ( ) ⇒ α = 2 3
Solution: 2NH 3 \u001f N 2 + 3H 2 Equilibrium moles 1 − α α 2 3 2 α = 1 3 = 1 3 = 1 Now, K P = × ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ × ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ⎛ ⎝ ⎜ ⎜ ⎞ ⎠ ⎟ ⎟ = 1 3 1 1 3 50 5 3 2700 3 2 2 2 atm
Solution: n initial × (17 + 28 + 2 + 20) = 134 ⇒ n initial = 2 2NH 3 \u001f N 2 + 3H 2 Initial moles 2 2 2 Moles at equilibrium 2 – 2 x 2 + x 2 + 3 x = 1.0 = × 134 0 5224 28 . = 3.5 ∴ x = 0.5 Now, K P P = = × × ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 2700 2 5 3 5 1 0 9 3 2 2 . ( . ) ( . ) ⇒ P = × × 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 – x – y 13 P – 3 x – 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 – y – x 13 P – 2 y – 3 x y From question, P x P NH 3 2 0 = = ⇒ x P = 0 2 P P x y P H 2 13 3 2 2 0 = − − = and P total = (9 P – x – y ) + (13 P – 3 x – 2 y ) + 2 x + y = 7 P 0 ∴ 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 = ⋅ = × = NH N H ( ) ∴ K P (required) = 20 0 2 P
Solution: 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 = ⋅ = ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ × = N H N H ( )\n6.46 Chapter 6 HINTS AND EXPLANATIONS
Solution: Direction of shifting of equilibrium will depend on relative values of a and b .
Solution: Equilibrium opposes the changes.
Solution: K K RT P C n g = ⋅ Δ ( )
Solution: Exothermic direction is favoured on lowering temperature.
Solution: NaCl( s ) \u001f Na + ( aq ) + Cl – ( aq )
Solution: On decreasing the volume, moles of A( g ) as well as B( s ) will increase.
Solution: K K RT P C n g = ⋅ Δ ( )
Solution: Le Chatelier’s principle
Solution: Le Chatelier’s principle
Solution: Le Chatelier’s principle
Solution: Le Chatelier’s principle
Solution: Le Chatelier’s principle
Solution: Le Chatelier’s principle
Solution: P K P CO atm 2 2 463 = = . ∴ n CO 2 at equilibrium = × × = 2 463 15 0 0821 900 0 5 . . . (A) % of CaCO 3 decomposed = × = 0 5 1 0 100 50 . . % ( ) Eqn (B) % of CaCO 3 decomposed = × = 0 5 0 5 100 100 . . % ( ) Eqn (C) % of CaCO 3 decomposed = 100% (non ) -Eqn
Solution: Le Chatelier’s principle
Solution: 2H 2 S(g) \u001f 2H 2 (g) + S(g); K e = 10 –6 Initial moles 0.1 0 0 Equilibrium moles 0.1 – x x x 2 \u001a 0 1 . Now, K x x C = = × × ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ − 10 2 0 1 1 0 4 6 2 2 ( . ) . ⇒ x = 2 × 10 –3 ∴ Percentage dissociation = × × = − 2 10 0 1 100 2 3 . %
Solution: V V CF CO ml 4 2 500 300 200 = − = = ∴ V COF ml 2 500 2 200 100 = − × = Hence, P P CF CO atm 4 2 200 500 10 4 = = × = P COF atm 2 100 500 10 2 = × = K p = × = 4 4 2 4 2
Solution: 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 – 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 ⋅ + − ⋅ + ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = + − × ( . ) ( . ) . ( . ) ( . ) 0 2 0 3 0 5 0 2 0 3 or, 8 0 2 0 3 0 0821 546 4 48 = + − × × x x x ( . ) . . . ⇒ x = 0.2 ∴ Final pressure = + × × = ( . ) . . 0 5 0 0821 546 4 48 7 x atm
Solution: 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 − 2 a a − 2 a 2 a 2 ∴ K a a a a eq = × × = 2 2 2 2 1
Solution: 2HI \u001f H 2 + I 2 Equilibrium moles 1 – 0.8 0.4 0.4 = 0.2 ∴ K eq = × = 0 4 0 4 0 2 4 2 . . ( . ) Now, H 2 + I 2 \u001f 2HI Initial moles 2 2 0 Equilibrium moles 2 – x 2 – x 2 x K x x x eq = = − − 1 4 2 2 2 2 ( ) ( )( ) ⇒ x = 0.4 Now, n eq of I 2 = n eq of Na 2 S 2 O 3 or, (2 – x ) × 2 = V × (1.6 × 1) ⇒ V = 2 L
Solution: Cl 2 CHCOOH + C 5 H 10 \u001f Cl 2 CHCOOC 5 H 11 I: Initial moles 1 4 0 Equilibrium moles 1 – x 4 – x x = 0.5 II: Initial moles 1 a 0 Equilibrium moles 1 – y a – y y = 0.6 Now, K a eq = × × = × − × 0 5 0 5 3 5 0 7 0 6 0 4 0 6 0 72 . . . . . . ( . ) . ⇒ a = 5\n6.48 Chapter 6 HINTS AND EXPLANATIONS
Solution: n CO2 at equilibrium = 0.05 ∴ Minimum mass of CaCO 3 needed = 0.05 × 100 = 5 gm
Solution: Ag + (aq) + Fe 2+ (aq) \u001f Fe 3+ (aq) + Ag(s) Initial moles 500 0 9 1000 0 45 × = . . 500 1 0 1000 0 50 × = . . 0 0 Equilibrium moles 0.45 – x 0.50 – x x x Now, n eq Fe 2+ = n eq MnO 4 − or, ( . ) . 0 50 1000 30 1 25 0 06 1000 5 − × × = × × x ⇒ x = 0.25 ∴ K x x x eq M = − − = − ( . )( . ) 0 45 0 5 5 1
Solution: Sb2S3(s) + 3H2(g) \u001f 2Sb(s) + 3H2S(g) Initial moles 0.01 0.01 0 0 Equ. moles 0.01 – x 0.01 – 3 x 2 x 3 1 19 238 x = . = 5 × 10 –3 = 5 × 10 –3 Now, K c = × × = − − ( ) ( ) 5 10 5 10 1 3 3 3 3
Solution: H 2 O + D 2 O \u001f 2HDO Initial moles 28 28 0 Equ. moles 28 – 14 = 14 28 – 14 = 14 2 × 14 = 28 K C = × = ( ) 28 14 14 4 2
Solution: 3A 2 (g) \u001f A 6 (g), K p 1 1 6 2 = − . atm Initial partial pressure 2 P 0 0 Equilibrium partial pressure 2 P 0 – 3 a – b a A 2 (g) + C (g) \u001f A 2 C (g), K x p 2 1 = − atm Initial partial pressure 2 P 0 P 0 0 Equilibrium partial pressure 2 P 0 – b – 3 a P 0 – b b From question, a = 0.2, P P P A A A 6 2 2 3 3 1 6 0 2 1 6 = ⇒ = . . . ⇒ P P a b A 2 0 5 2 3 0 = = − − . and (2 P 0 – 3 a – b ) + a + ( P 0 – b ) + b = 1.4 ⇒ 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 = − − − = × = − ( )( ) . . . . atm
Solution: Initial partial pressure of IBr (g) = × × = 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 – 2 x x x = 4 ∴ K p = × = 4 4 2 4 2 ( )
Solution: H 2 (g) + I 2 (g) \u001f 2HI (g) I: Initial moles 1 3 0 Equilibrium moles 1 2 − x 3 2 − x x II: Initial moles 3 3 0 Equilibrium moles 3 – x 3 – x 2 x\n6.49 Chemical Equilibrium HINTS AND EXPLANATIONS Now, K x x x x x x eq = − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = − − ( ) ( ) ( )( ) 2 2 1 2 3 2 2 3 3 ⇒ x = 3 2 ∴ K eq = 4
Solution: 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 – x x – 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 – y y x + y From question, [ ] . O 2 2 2 5 = + = x y ⇒ x + y = 5 And K x y x y y C 1 2 5 5 4 1 2 5 2 5 1 1 2 = = − × − × = − × − × . ( ) ( ) ( ) ⇒ y = 2 ∴ [N O] 2 2 1 = = y M
Solution: In left chamber, P H e = 2 atm ∴ P P NH H 3 2 atm = = − = 3 4 2 2 1 ∴ K p = × = 1 1 1 2 atm Four-digit Integer Type
Solution: P-xyloquinone + M.W \u001f P-xylohydroquinone + M.B. Initial conc. 0.012 M 0 0.24 M 10 –3 M Equ. con. 0.012 + 4 × 10 –5 4 × 10 –5 M 0.24 – 4 × 10 –5 10 4 100 10 3 3 − − − × ≈ 0.012 ≈ 0.24 M 0.96 × 10 –3 m ∴ K eq = 0 24 0 96 10 0 012 4 10 480 3 5 . . . × × × × = − −
Solution: 6HCHO \u001f C 6 H 12 0 6 ; K eq = 6.4 × 10 19 Initial conc. 0 1 M Equ. con. 6 x 1 – x = 1 M Now, K eq = 6.4 × 10 19 = 1 6 [ ] HCHO ⇒ [HCHO] = 5 × 10 –4 M = 5 × 10 –4 × 30 g/L = 15 mg/L
Solution: 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 – y 2 + x – y From question, (2 + y) + (2 – y ) + 2 + ( x – y ) = 2 × 6 or, x – y = 6 and K c = 2 2 2 1 2 2 2 1 2 × × = − × + − + × V y x y y V ( ) ( ) ( ) Or, 4 = ( ) ( ) 8 8 4 − × − x x ⇒ x = 20 3
Solution: K ° = eq 1 ⇒ ∆ G ° = 0 ⇒ T = Δ Δ H S ° ° = × − × ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 223 10 223 33 1520 10 3 3 = 1784 K
Solution: Graphite \u001f Diamond; ∆ G° = (3.0 – 0) kJ/mol; P 1 = 1 bar ∆ G = 0 P 2 = P Now, ∆ ( ∆ G ) = ∆ V ∙ ∆ P or, ( ∆ G – ∆ G °) = ( V Dia – V Gra ) ( P 2 – P 1 ) or, (0 – 3.0 × 10 3 ) = 12 3 6 12 2 4 10 10 6 5 2 . . − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ × ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ − ⎡ ⎣ ⎤ ⎦ − P ∴ P 2 = 1.8 × 10 9 Pa = 1.8 × 10 4 bar
Solution: K P P P ° = = CO CO 2 10 400 6 Now, ∆ G ° = –5320 – 5.6 T = –RT ln K P °\n6.50 Chapter 6 HINTS AND EXPLANATIONS = –2 × T × ln 10 400 6 ∴ T = 532 k
Solution: 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 – 4 x 4 x 2 x x From question, P 0 – 4 x = 2 atm and ( P 0 – 4 x ) + 4 x + 2 x + x = 30 atm ∴ P 0 = 18 atm and x = 4 atm Now, K x x x P x p o = × × − = ( ) ( ) ( ) 4 2 4 2 4 2 4 20 3 atm and K K RT c p n g = ) = × = Δ ( ( . ) 2 0 08 400 32 20 3 M 3
Solution: Initial: P NOcl = P 0 bar and P N 2 = (1 – P 0 ) bar 2NOCl \u001f 2NO + Cl 2 Initial partial pressure P 0 0 0 Eqn. partial pressure P 0 – 2 x 2 x x = 1.2–1.0 = 0.2 Par. pre. on adding Cl 2 P 0 – 2 x 2 x x + (8.3 – 1.2) New Equ.partial pre. P 0 – 2 x + 2 y 2 x – 2 y 7.1 + x – y From question, y = 8.3 – 8.2 = 0.1 Now, K x x P x x y x y P x y p o = × − = − × + − − + ( ) ( ) ( ) ( . ) ( ) 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 . . ( . ) . . ( . ) . × − = × − ⇒ = P P P and K p = 3.2
Solution: Initial equilibrium: A + 2B \u001f C Initial partial pressure P 0 2 P 0 O Equ. partial pressure P 0 – x 2 P 0 – 2 x x From question, ( P 0 – x ) + (2 P 0 – 2 x ) + x + 3 P 0 = 5 6 6 0 × 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 – y 4 P 0 – 2 y y – 2 z 2C + D \u001f 2F y 6 P 0 0 Equ. partial pressure y – 2 z 6 P 0 – z 2 z From question, 2 P 0 – y = y – 2 z Now for the first reaction, K x P x P x y z P y P y p o = − − = − − − ( )( ) ( )( ) 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 = − ⇒ = = ( ) and Total equilibrium pressure: First equilibrium = 5 P 0 Second equilibrium = (2 P 0 – y ) + (4 P 0 – 2y) + ( y – 2 z ) + (6 P 0 – z ) + 2 z = 3.5 P 0
Solution: A 2 + B 2 \u001f 2AB Initial moles 0.2 0.2 0 Equ. moles 0.2 – x 0.2 – x 2 x = 0.3 = 0.05 = 0.05 K eq = K 1 = ( . ) . . 0 3 0 05 0 05 36 2 × = After adding C 2 : A 2 + B 2 \u001f 2AB Initial moles 0.2 0.2 0 Equ. moles 0.2 – y – z 0.2 – y 2 y = 0.24 = 0.08 – z = 0.08 A 2 + C 2 \u001f 2AC Initial moles 0.2 0.1 0 Equ. moles 0.2 – z – y 0.1 – z 2 z Now, K z z 1 2 36 0 24 0 08 0 08 0 06 = = − × ⇒ = ( . ) ( . ) . . ∴ K eq = K 2 = ( . ) . . 0 12 0 02 0 04 18 2 × =
Solution: Δ G = Δ G ° + RT. ln Q = –RT. 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 × 300 × ln 1 4 e = –2400 cal
Solution: A + B \u001f C; K 1 = 4 × 10 10 Initial moles 5 5 0 Equ. moles 5 – ( x + y ) 5 – x x A + D \u001f C; K 2 = 10 10 Initial moles 5 5 0 Equ. moles 5 – ( x + y ) 5 – 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 = = − × − ⇒ = (2) From (1) and (2), x = 10 3 ∴ Moles of B at equilibrium = 5 – x = 5 3
Solution: Br 2 (l) + Cl 2 (g) \u001f 2Br Cl (g); K p = 1 atm Initial moles x 10 0 Equ. moles ≈ 0 10 – x 2 x Br 2 (l) \u001f Br 2 (g); K p = 0.25 atm Initial moles y 0 Equ. moles ≈ 0 y From question: y y × × = ⇒ = 0 082 300 164 0 25 5 3 . . and ( ) . . 10 2 0 082 300 164 2 00 10 3 − + × × = ⇒ = x x x ∴ Minimum mass of Br 2 (l) = ( x + y ) × 160 gm = 800 gm
Solution: 2SO 3 \u001f 2SO 2 + O 2 Initial moles 1 (say) 0 0 Equ. moles 1 – 0.4 = 0.6 0.4 0.2 ∴ M av = 1 80 1 2 × . Now, d = PM RT p p ⇒ = × × ⇒ = 16 80 1 2 0 0821 920 0 0821 216 . . . atm ∴ K p = × × = ( . ) . ( . ) . 0 4 0 2 0 6 216 1 2 16 2 2 atm
Solution: A 2 \u001f 2A; K 1 = x atm Initial partial pressure 1 atm 0 Equ. partial pressure 1 – ( x + z ) 2 x B 2 \u001f 2B; K 2 = y atm Initial partial pressure 1 atm 0 Equ.partial pressure 1 – ( y + z ) 2 y A 2 + B 2 \u001f 2AB; K 3 = 2 Initial partial pressure 1 1 0 Equ. partial pressure 1 – ( x + z ) 1 – ( y + z ) 2 z = 0.5 (1) From question, [1 – ( x + z )] + 2 x + [1 – ( y + z )] + 2 y + 2 z = 2.75 ∴ x + y = 0.75 (2) Now, K 3 = ( . ) ( . )( . ) . . 0 5 0 75 0 75 2 0 25 0 50 2 − − = ⇒ = x y x or y = 0.50 or 0.25 ∴ 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 = − + − + = × − × − ( ) ( ) ( ) ( ) ( ) ( . ) ( ) ( . ) = = 1 8 1 or 8
Solution: [ ] NH M 2 30 15 10 10 − − − = = ∴ Number of NH 2 − ions per ml = × × × = × − 10 1 1000 6 10 6 10 15 23 5 ( )
Solution: On increasing temperature, the dissociation of water will increase. It will result increase in [H + ] and as well as in [OH – ] and hence, decreases in P H and as well as P OH .
Solution: For maximum dissociation, [H + ] = [OH – ].
Solution: [H + ] = 10 –2 M ⇒ n H + = × = × − − 200 10 1000 2 10 2 3 [OH – ] = 10 –2 M ⇒ n OH − = × = × − − 300 10 1000 3 10 2 3 ∴ Moles of excess OH – remained = 1 × 10 –3 [OH – ] = 1 10 500 1000 2 10 3 3 × × = × − − M ∴ P OH = – log (2 × 10 –3 ) = 2.7 ⇒ P H = 11.3
Solution: [OD – ] excess = 80 0 1 20 0 2 100 0 04 × − × = . . . M ∴ P OD = – log (0.04) = 1.4 Now, P Kw of D 2 O = P D + P OD = 13.6 + 1.4 = 15 ∴ Kw = 1 × 10 –15
Solution: [OT - ] excess = 400 0 2 100 0 4 500 0 08 × − × = . . . M ∴ P OT = – log(0.08) = 1.1 Now, PT = P Kw – P OT = 2 × 7.60 – 1.1 = 14.1
Solution: K Kw Ka b = = × = × − − − 10 2 10 5 10 14 10 5
Solution: NH + H O NH OH 3 2 \u001f ⇀ \u001f ↽ \u001f \u001f 4 + − + Δ H ° = (–52.21) + (54.70) = 2.49 kJ Δ S ° = 1.6 + (–76.3) = – 74.7 J/K Now, Δ H ° = – RT. ln K eq or, 2490 – 300 × (–74.7) = –8.3 × 300 × ln K eq ∴ K eq = e –10
Solution: [ ] ] H [H HCOOH CH COOH 3 + + = or, 2 4 10 0 6 1 8 10 8 4 5 . . . × × = × × ⇒ = − − C C M ∴ Moles of CH 3 COOH added = 100 8 1000 0 8 × = .
Solution: K a ( HA ) = K b ( A – ) = Kw = − 10 7 Now, [ ] . . H M P H + − − = × = ⇒ = 10 0 1 10 4 0 7 4
Solution: NH +OH K M S K NH +H O 4 + f b 3 2 − − − = × = 3 4 10 10 1 1 . ? Given : NH NH H K M 4 + 3 \u001f ⇀ \u001f ↽ \u001f \u001f + = × + − ; . 1 10 5 6 10 and H O H OH K M 2 \u001f ⇀ \u001f ↽ \u001f \u001f + − − + = × ; . 2 14 2 1 0 10 ∴ NH OH NH H O K K K 4 + 3 2 eq + + = − \u001f ⇀ \u001f ↽ \u001f \u001f ; 1 2 Now, 3 4 10 5 6 10 10 6 07 10 10 10 14 5 1 . . . × = × ⇒ = × − − − K K S b b
Solution: CH COOH CH COO H 3 3 0 1 0 1 0 1 . . . − + + ≈ + − + + x x y y x \u001f ⇀ \u001f ↽ \u001f \u001f Cl CHCOOH Cl CHCOO H 2 2 0 1 0 1 0 1 . . . − + + ≈ + − + + y x y y y \u001f ⇀ \u001f ↽ \u001f \u001f 0 15 0 1 0 1 0 05 . ( . ) ( . ) . = × + − ⇒ = y y y y ∴ [H + ] = 0.1 + x + y ≈ 0.1 + y = 0.15 M ∴ P H = – log(0.15) = 0.82
Solution: [ ] . . / / . OH M − = × = 0 4 100 4 25 17 250 1000 0 004 P OH = – log(0.004) = 2.4 ∴ P H = 14 – 2.4 = 11.6
Solution: [ ] . . OH K C M b − − − = × = × × = × 1 6 10 0 0025 4 10 6 9 P P OH H = − × = ⇒ = − log . . 4 10 4 2 9 8 9 EXERCISE II (JEE ADVANCED)\n7.42 Chapter 7 HINTS AND EXPLANATIONS
Solution: [ ] / O HSaC M = × = × − − 4 10 200 1000 2 10 4 3 and P H = 3.0 ⇒ [H + ] = 10 –3 M Now, 2 10 10 2 10 4 10 12 3 3 12 × = × × ⇒ = × − − − − − − [S ] [ ] aC SaC M
Solution: [ ] . . HA M O = × = 20 0 5 50 0 2 [ ] . . HB M O = × = 30 0 2 50 0 12 HA H A 0 2 . − + + − + x x y x \u001f ⇀ \u001f ↽ \u001f \u001f HB H 0 12 . − + + − + y x y y B \u001f ⇀ \u001f ↽ \u001f \u001f Now, 2 10 0 2 0 2 4 × = + ⋅ − ≈ + ⋅ − ( ) x . ( ) . x y x x y x ∴ ( x + y ) ⋅ x = 4 × 10 –5 (1) and, 5 10 0 12 0 12 5 × = + ⋅ − ≈ + ⋅ − ( ) ( . ) ( ) . x y y y x y y ∴ ( x + y ) ⋅ y = 6 × 10 –6 (2) From (1) and (2), [H + ] = x + y = 6.78 × 10 –3 M
Solution: 10 0 01 10 5 8 − − = × ⇒ = K K a a . Now, [ ] . . . OH P OH − − − = × = ⇒ = 10 0 1 10 4 5 8 4 5 ∴ P H = 9.5
Solution: RNH +H O RNH + OH 2 2 0 01 3 10 4 . − + − + − x x x \u001f ⇀ \u001f ↽ \u001f \u001f 2 10 10 0 01 10 6 4 4 × = + − ⇒ = − − − x x x x ( ) . ∴ [OH – ] = 2 × 10 –4 M
Solution: As on adding HCl, [H + ] is not changing and will remain unchanged.
Solution: Co +H O HCo H 2 2 Decrease Shiftleft ↓ ← − + + \u001f ⇀ \u001f\u001f\u001f\u001f ↽ \u001f \u001f\u001f\u001f\u001f 3 As [H + ] decreases, P H increases.
Solution: [ ] . / / . W H M O 2 4 0 16 32 500 1000 0 01 = = ∴ ∝ = 4 10 0 01 0 02 2 6 × = − . . % or
Solution: P H = – log (2 × 10 –6 ) = 5.70
Solution: [OH–] = 6.67 × 10 –3 + 6 67 10 2 0 10 3 2 . × + ≈ − − M ∴ P OH = – log (10 –2 ) = 2.0 ⇒ P H = 12.0
Solution: en + H O enH OH M M M b 2 0 09 5 1 8 1 10 ( . ) ( ) ( ) ;K . − + − − + − + = × x x y x y \u001f ⇀ \u001f ↽ \u001f \u001f enH + H O enH OH M M M b + − + − + − + = × 2 2 2 8 2 7 0 10 ( ) ( ) ;K . x y y x y \u001f ⇀ \u001f ↽ \u001f \u001f Now, 8 1 10 0 09 0 09 2 7 10 5 3 . ( )( ) ( . ) . . × = − + − ≈ ⋅ ⇒ = × − − x y x y x K x x x and 7 0 10 7 0 10 8 8 . ( ) ( ) . × = ⋅ + − ≈ ⋅ ⇒ = × − − y x y x y y x x y ∴ [en H + ] = ( x – y ) ≈ x M = 2.7 × 10 –3 M [enH ]= = 7.0 10 M 2 2+ y × − 8 ∴ [OH – ] = ( x + y ) = x = 2.7 × 10 –3 M and P OH = – log (2.7 × 10 –3 ) = 2.56 ⇒ P H = 11.44
Solution: K K H H S a a 2 1 2 2 2 5 ⋅ = + − [ ] [ ] [ ] or, ( . ) ( . ) ( . ) [ ] . [ ] . 1 4 10 1 0 10 0 1 5 0 2 5 2 8 10 7 14 2 2 2 20 × × × = × ⇒ = × − − − − − M
Solution: [ ] . H M + − − = × × = × 0 2 2 10 2 10 5 3 Now, ( ) ( ) ( ) ( ) [ ] [ ] 2 10 5 10 4 10 2 10 5 9 12 3 3 3 × × × × × = × × − − − − + A H A 3 ∴ = × − − [ ] [ ] A H A 3 3 17 5 10
Solution: [ ] . H M + − − = × = 0 1 10 10 5 3 Now, K a 3 3 2 3 2 13 3 10 10 10 10 = ⇒ = = + − − − − − − − [ ][ ] [ ] [ ] [ ] H A HA A HA ∴ P X = 10
Solution: OH ACOH ACO H O m mol Final 0 m mol 1 m mol m mol − − + + 2 3 0 2 2 \u001f ⇀ \u001f ↽ \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
Solution: Millimoles of ACOH = 6 × 0.1 = 0.6 Millimoles of ACO – =
Solution: × 0.1 = 1.2 ∴ = + = p H 4 75 1 2 0 6 5 05 . log . . .
Solution: 1st solution will finally have 1 mole of CH 3 COOH. ∴ = P P H Ka 1 1 2 and for 2nd solution, P P H Ka 2 =
Solution: 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 ∴ = y x 3 98 .
Solution: 4 0 5 0 0 5 0 05 1 1 . . log . . = + ⇒ = C C M 6 0 5 0 0 5 5 0 2 2 . . log . . = + ⇒ = C C M Now, fi nal P 0.05 + 5.0 0.5 + 0.5 H = + × × × × = 5 0 5 7 . log . V V V V
Solution: For maximum β αβ α , [ ] H P P H K a + = ⇒ = 0
Solution: BOH H B C Final C-0.1 V 0.1 40 40 40 40 40 0 0 0 1 40 × + + + × + + + V M V V V M V \u001f ⇀ \u001f ↽ \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 − 14 =P +log 0.1 5 40C 0.1 5 K b − × − × 10 (1) 14 =P +log 0.1 40C 0.1 K b − × − × 9 20 20 (2) ∴ K b = 2 × 10 –5
Solution: For maximum buffer capacity: [ [ [ ACOH] NaOH] NaOH]= M M = ⇒ = 2 1 2 2 1 ∴ Mass of NaOH added = × × = 500 1 1000 40 20 gm
Solution: n n HA OH M M= 80 = ⇒ = × ⇒ − 0 28 35 0 1 1000 . .
Solution: Fe H O Fe(OH H M 2 M 3 0 9 2 0 1 + + + + + . . ? ) x x \u001f ⇀ \u001f ↽ \u001f \u001f 9 10 0 1 0 9 0 081 1 08 3 × = × ⇒ = ⇒ = − + + . [ ] . [ ] . . x x H H P H
Solution: In fi nal solution: [HA] = [A – ]
Solution: [ ] NH K K K C = 8.33 10 M w a b 3 4 = ⋅ × × −
Solution: For equivalence point, 2 5 2 5 2 15 . × = × V Hcl ∴ 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 . . . × × − x x \u001f ⇀ \u001f ↽ \u001f \u001f 2 2 aq O; K = = − − 10 10 10 12 14 2 100 0 1 2 7 10 2 = − ⋅ ⇒ = × = − + . . ( ) x x x x M H
Solution: 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 ⋅ ⋅ = + − [H ][ ] [ ] or 7 5 10 10 10 10 4 8 12 10 3 3 . ( ) [ ] [ ] × × × = × − − − − − A H A 3 ∴ = = × − − − [ ] [ ] . . H A A 3 3 6 7 10 7 5 1 33 10
Solution: 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 − + − − + = × . . . ; \u001f ⇀ \u001f ↽ \u001f \u001f For HCO 3 − solution, [ ] , . H K K M + − = = × a a 1 2 1 4 10 8 Now, K H O HCO a 2 3 2 3 = + − − [ ][C ] [ ] ∴ = × × × = − − − − [ ] . . CO M 3 2 11 8 3 4 10 0 35 1 4 10 10\n7.44 Chapter 7 HINTS AND EXPLANATIONS
Solution: [ ] . . . LaC M − = × = 0 0 125 0 5 2 0 5 Now, P P C OH K a = − + 7 1 2 ( log ) 5 6 7 1 2 0 5 . ( log . ) = − + P K a ∴ = ⇒ = × − P K K a 3 1 8 10 4 . a
Solution: As HA is stronger acid, it will react first. For first equivalent point, V V ml NaOH NaOH × = × ⇒ = 0 2 50 0 05 12 5 . . . At fi rst equivalent point, [ ] . . . A M − = × = 50 0 05 62 5 0 04 [ . . . HB]= M 50 0 08 62 5 0 064 × = Now, A HB B HA ; K = K (HB) K (HA eq a a − − − − + + 0 04 0 04 0 064 0 064 . . . . x x x x \u001b \u001b \u001f ⇀ \u001f ↽ \u001f \u001f ) ) = = = × − − − − 10 10 10 4 10 8 2 3 8 4 4 5 . . . 4 10 0 04 0 064 3 2 10 5 4 × = ⋅ × ⇒ = × − − x x x . . . Now, K (HA H A HA H a ) [ ][ ] [ ] . [ ] . . = ⇒ × = × × + − − + − 1 6 10 0 04 3 2 10 4 4 ∴ = × ⇒ = + − [ ] . . H P H 1 28 10 5 9 6
Solution: CrO +H O HCrO +OH ; K Kw K 4 2 2 4 h − − − − − = = × 0 005 8 2 2 10 . x x x a \u001f ⇀ \u001f ↽ \u001f \u001f 2 10 0 005 0 005 10 8 2 5 × = ⋅ − ≈ ⇒ = − − x x x x x . . ∴ = = − h 10 0 005 0 002 5 . .
Solution: P and P K K a a 1 2 2 40 9 60 = = . . ∴ Required pH = + = 1 2 2 40 9 60 6 00 ( . . ) .
Solution: HA H A Red Blue \u001f ⇀ \u001f ↽ \u001f \u001f + − + [ ] [ ] [ ] H K HA A + − = ⋅ a ∴ = − = − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ + + + [ ] [ ] [ ] H required H H K a 2 1 75 25 25 75 = 8 × 10 –5 M
Solution: 5 5 4 75 . . log [ ] [ = + − ACo ACOH] O O ∴ = − [ ] [ . ACo ACOH] O O 5 62 1
Solution: n n CO H 2 and used = = = × = + 224 22400 0 01 30 1 1000 0 03 . . CO +OH HCO 2 mole (L.R. 3 mole or less − − ⎯ → ⎯ 0 01 0 01 . ) . Hence, moles of H + used should be 0.01 or less and titration of HCO 3 − and H+ should not be detected by phenolphthalein. Hence, OH – must be in excess. CO + 2OH CO H O 2 mole mole 3 mole 2 0 01 0 02 2 0 01 . . . − − ⎯ → ⎯ + Thus, 0.01 mole of CO 3 2 − 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 – must be present in excess. Hence, total moles of OH – used = 0.02 + 0.02 = 0.04, ∴ = = [ ] . . NaOH M used 0 04 1 0 04
Solution: PbSO g/100ml 4 = × × × − − 2 10 304 10 1 36 10 9 3 \u001b . ZaS g/100ml = × = × − − 10 97 10 9 7 10 22 11 . AgBr g/100ml = × × = × − − 4 10 188 10 1 19 10 13 5 . CuCo g/100ml 3 8 3 10 123 10 1 23 10 = × = × − − .
Solution: Hg 4Cl HgC 2+ M 100% ram 0 Eqn. 1.6 M M 0.5M 0.5M 0 1 10 0 9 17 . . × − − + \u001f ⇀ \u001f ↽ \u001f \u001f l l M 0.1M 4 2 0 0 1 − . ∴ = × × = − K form 0 1 1 6 10 0 5 10 17 4 17 . . ( . )\n7.45 Ionic Equilibrium HINTS AND EXPLANATIONS
Solution: 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 . . . ) + + − − − \u001f ⇀ \u001f ↽ \u001f \u001f B Br − 0 0 1 . K eq = × × × = × − ⇒ = − 4 10 1 6 10 0 1 0 1 0 2 0 325 13 12 2 . . . ( . ) . a a
Solution: Tl S Tl S SM 2 2 2 2 (S) \u001f ⇀ \u001f ↽ \u001f \u001f + − + x S H O HS OH Kw K M SM SM h a 2 2 2 − − − + + = x \u001f ⇀ \u001f ↽ \u001f \u001f ; K 10 10 2 10 2 10 4 10 14 14 6 6 12 − − − − − = × × × ⇒ = × x x ∴ = × × × × = × − − − K sp ( ) . 2 2 10 4 10 6 4 10 6 2 12 23
Solution: 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 + × × − = × − × × − = − − − − − + x x . . . . . ) 0 10 2 0 1 5 10 5 3 × + = × − − M M M M(SCN \u001f ⇀ \u001f ↽ \u001f \u001f x ∴ = × × × × = × − − − K f 1 5 10 5 10 1 10 3 10 3 4 5 5 .
Solution: SrCO s) Sr CO M 3 2 2 10 3 2 2 10 4 4 ( \u001f ⇀ \u001f ↽ \u001f \u001f + × − × − − − + x CO H O HCO OH M M M 3 2 2 10 2 3 4 10 4 6 − × − − − × − − + + ( ) x x \u001f ⇀ \u001f ↽ \u001f \u001f Now, 10 5 10 4 10 2 10 0 01 51 14 11 6 4 − − − − × = × × × − ⇒ = x x x ( ) . ∴ = × × × − = × − − − K sp ( ) ( ) 2 10 2 10 4 51 10 4 4 8 x
Solution: MnS(S) Mn S SM S M \u001f ⇀ \u001f ↽ \u001f \u001f 2 2 + − − + ( ) x S H O HS OH S M M M 2 2 − − − − + + ( ) x x x \u001f ⇀ \u001f ↽ \u001f \u001f 10 10 14 14 10 − − − = ⋅ − × = ⋅ − x x x x ( ) ( ) S and 2.5 10 S S ∴ S = 6.3 × 10 –4 M
Solution: AgCl(s) Br AgBr(s) Cl aq M M 0 1 0 075 0 075 0 075 . . . . (aq) ( − − − + + x \u001f ⇀ \u001f ↽ \u001f \u001f ) ) K Br K (AgCl) K (AgBr) Br eq sp sp = = ⇒ = × × − − − − − [Cl ] [ ] . [ ] 0 075 2 10 4 10 10 13 3 ∴ = × − − [ ] . Br M 1 5 10 4
Solution: [ ) ] . Ag(CN M 2 0 01 − = K Ag CN Ag CN Ag diss = ⇒ × = × × + − + − − + − [ ][ ] [ ( ) ] [ ] ( . ) . 2 2 20 7 2 1 10 2 5 10 0 01 ∴ = × + − [ ] . Ag M 1 6 10 9
Solution: [ ] . CO M 3 2 2 0 − = Now, K (CaCO K (CaF CO F F F sp sp 3 2 3 2 2 3 3 4 2 8 ) ) [ ] [ ] [ ] ( ) = ⇒ = ⇒ = − − − − x y y x
Solution: BaF s O BaC O s M M 2 2 4 2 0 1 2 4 2 2 ( ) C (aq) ( ) F (aq) ( . ) + ⇒ + − − − x x K F C O K BrF K O eq 2 sp = = = = − − − − [ ] [ ] ( ) (BrC ) 2 4 2 2 4 6 10 4 10 10 10 sp ∴ x ≈ 0.1 ⇒ [F – ] = 0.2 M ∴ [ ] ( . ) . Ba M 2 6 2 5 10 0 2 2 5 10 + − − = = ×
Solution: S Zn(OH) Zn(OH) Zn Zn(OH) Zn(OH) = + + + + + + − − [ (aq)] [ ] [ ] [ ] [ ] 2 2 3 4 2 = + ⋅ + ⋅ + + − − − − 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 [ ] [ ] [ ] ] = + × + × × + × × + × − − − − − − − 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 × × − ( . ) = 10 –6 +10 –12 + 10 –15 + 10 –4 + 10 –4 ≈ 2 × 10 –4 M
Solution: For molecular solubility of CaCl 2 , Δ H solution = 209.2 + (–33.5) = 175.7 KJ > 30 KJ For ionic solubility of CaCl 2 , Δ H solution = 209.2 + 1004.2 + 1715.4 – 1598.3 – 719.6 – 711.2 = – 100.3 KJ For molecular solubility of HgCl 2 , Δ H solution = 83.7 – 66.9 = 16.8 KJ < 30 KJ For ionic solubility of HgCl 2 , Δ H solution = 83.7 + 460.2 + 2815.8 – 1845.1 – 719.6 – 711.2 = 83.3 KJ > 30 KJ Hence, CaCl 2 is ionic and HgCl 2 is molecular solubility.
Solution: [ ] . C O M 2 4 2 5 6 0 001 5 250 2 6 10 − − = × × × = × ∴ = × = × − − K sp ( ) . 6 10 3 6 10 5 2 9\n7.46 Chapter 7 HINTS AND EXPLANATIONS
Solution: Sr NO M Eqn. 0.001 M 0.05 M 2 0 001 0 001 75 100 3 0 05 0 05 + − = × − − + . . . . x x \u001b \u001f ⇀ \u001f \u001f ↽ \u001f \u001f Sr(NO 3 0 ) + x ∴ x = 0.00025 Now, K f = × × × = − − 2 5 10 7 5 10 0 05 20 3 4 4 . . .
Solution: ACOAg(s) H Cl ACOH AgCl(s) Mole M M 0 1 0 1 0 1 . . . + + + + − \u001f ⇀ \u001f ↽ \u001f \u001f K K K eq sp sp ACOH H Cl ACO ACO Ag Ag ACOAg] ( = × × = + − − − + + [ ] [ ][ ] [ ] [ ] [ ] [ ] [ A AgCl) × K a = × = ⇒ − − − 10 10 10 10 8 10 5 7 Almost complete reaction ∴ = + − [ . , [H . ACOH] M ] =10 M \u001b 0 1 0 1 10 7 4 and [ ] [ ] [ ] . ACO Ka ACOH H M − + = × = 0 01
Solution: A B (s) A B x y y x x y \u001f ⇀ \u001f ↽ \u001f \u001f + − + K x y s s K x y x y x y x y x y sp sp = ⋅ ⋅ ⇒ = ⋅ ⎛ ⎝ ⎜ ⎜ ⎞ ⎠ ⎟ ⎟ + + 1 As K sp << 1, greater the value of ( x + y ), greater is s.
Solution: s = + = + + − − − [ ] [ [ ] [ ] Zn Zn(OH) K OH K OH sp f 2 4 2 2 For maximum or minimum S, d d OH S [ ] − = 0 or, − + = ⇒ = ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = − − − − 2 2 0 10 3 1 4 4 K OH K OH OH M sp f sp [ ] [ ] [ ] K K f ∴ P H = 10 and S min = 2.4 × 10 –9 M
Solution: Al(OH) s) OH Al(OH) From question M 3 4 33 10 3 8 10 1 ( ; . ? + = × − − − − \u001f ⇀ \u001f ↽ \u001f \u001f K 6 6 10 50 34 × = − 50 10 2 10 9 30 3 5 = ⇒ = × ⇒ = − − − − [OH ] [ ] . OH M P H As the calculated [OH – ] is minimum OH – , P H is minimum. Al(OH) (s) Al OH K 3 M \u001f ⇀ \u001f ↽ \u001f \u001f 3 33 10 3 3 8 10 + − − − + = × ; 8 10 10 2 10 4 30 33 3 3 10 × = × ⇒ = × ⇒ = − − − − − [ ] [ ] . OH OH M P H As the calculated [OH – ] is maximum OH – , P H is maximum.
Solution: From the question, [ . Cu(CN) M 4 3 0 1 − = and [CN – ] = 0.2 M ∴ = ⋅ = × × = × + − − − − [ ] [ ] [ ] . . ( . ) 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 − + − − − = = × × = × K ∴ = × = × × × = + − − − − [ ] [ ] [ ] . . . 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
Solution: S M ppm = × = × = × × × = × − − − 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.
Solution: During precipitation, the concentration of both Ba 2+ and SO 4 2 − ions will decrease.
Solution: K sp AgCl) ( = × = − − − 10 10 10 4 6 10 K sp (Ag CrO 2 4 4 2 4 12 10 8 10 8 10 ) ( ) = × × = × − − − After precipitation of AgCl, fi nd the concentration of Cl – . [ ] . Cl M final − − − − = × − × × × 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 − − = [ ] ( ) ( ) [CrO ] . CrO final final 4 2 7 2 12 10 2 4 2 2 10 8 10 10 3 2 1 − − − − − × = × ⇒ = × 0 0 5 − M\n7.47 Ionic Equilibrium HINTS AND EXPLANATIONS Hence, moles of Ag 2 CrO 4 precipitated = × − × = × − − − 8 10 3 2 10 7 68 10 4 5 4 . .
Solution: To prevent precipitation of AgCl, the concentration of Ag + needed in solution = K sp AgCl) Cl ( ( ) − = × << − 1 8 10 0 16 1 8 10 . . . M Hence, almost all Ag + ion must form complete with CN – ions. Ag CN Ag(CN) M CM M + × − − − + = × 1 8 10 0 16 2 1 8 17 10 2 6 4 10 . . . ; . \u001f ⇀ \u001f ↽ \u001f \u001f K f
Solution: [CO ] ( ) [ ] [ ] 3 2 2 3 2 − + = ⋅ K overall H CO H a To prevent precipitation of MCO 3 , or, [ ][ ] M CO K sp 2 3 2 + − ≤ or K K sp , [M ] [H CO ] [H ] 2 2 3 2 + + ⋅ ⋅ ≤ a ∴ ≥ ⋅ + + [H ] [ ] [ ] M K H CO K a sp 2 2 3 For MgCO 3 : [H ] . . . + − − − ≥ × × × × = × 0 1 5 10 0 05 9 10 2 5 10 17 8 6 M ∴ P H ≤ 5.6 For SrCO 3 : [H ] . . + − − − ≥ × × × × = × 0 1 5 10 0 05 9 10 5 10 3 17 10 5 M ∴ P H ≤ 4.78 For precipitation of SrCO 3 without any precipitation of MgCO 3 , the P H range should be 4.78 to 5.6
Solution: 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 ] . . . . [ ] + − − + − × < × × ⇒ < × 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 + − = ⋅ K a 3 3 = × × = × > × − − − 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 × = × × − − . [CH COONa] 3 O ∴ [CH 3 COONa] O = 1.25 M
Solution: Mg(OH) S NH Mg NH OH 2 4 2 4 2 2 ( ) + + + + \u001f ⇀ \u001f ↽ \u001f \u001f To re-dissolve Mg(OH) 2 , Q ≤ K eq or, [ ][ ] [NH ] Mg NH OH sp 2 4 2 4 2 2 + + ≤ 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 . . . . . . . . ( × ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ × ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ≤ × − n . . ) 0 10 5 2 × − ∴ n ≥ 0.035 Hence, minimum mass of (NH 4 ) 2 SO 4 needed = × = 0 035 2 132 2 31 . . gm
Solution: Ag Cl M Eq 10 M + × = × × − + − × = × − − + 500 0 01 1000 5 10 5 250 0 02 1000 5 3 3 . ( ) . x y 1 10 5 10 3 3 − − × − M M AgCl(S) ( ) x \u001f ⇀ \u001f ↽ \u001f \u001f Ag Br AgBr( M 10 M M M + = × × − + − = × × − − − − − + 5 10 5 5 10 5 10 3 3 3 3 ( ) ( ) x y y \u001f ⇀ \u001f ↽ \u001f \u001f S S) As both reactions will tend towards completion, ( x + y ) = 5 × 10 –3 Now, [Ag ]( ) [ ] + − − + − × − = ⇒ ⋅ = 5 10 10 10 3 10 10 x y Ag (1) and [Ag ]( ) + − − × − = × 5 10 5 10 3 13 y (2) From (1) ÷ (2), y y y 5 10 200 1 201 3 × − = ⇒ = − ∴ = × − ≈ × − − − [ ] . Br M 5 10 2 5 10 3 5 y\n7.48 Chapter 7 HINTS AND EXPLANATIONS
Solution: (a) Complete neutralization ⇒ P H = 7.0 (b) [ ] . . . . H M P final H + = × − × = ⇒ = 55 0 1 45 0 1 100 0 01 2 0 (c) OH – is in excess. (d) [ ] . . H M P final H + = × − × = ⇒ = 75 1 5 25 1 5 100 0 1 1 0
Solution: For basic solution: [ ] [ ] ] H OH and [H Kw + − + < < ∴ P H > P OH or P P or P P H kw OH kw > < 2 2
Solution: HA H A C(1 M C M C M − + − + α α α ) \u001f ⇀ \u001f ↽ \u001f \u001f K C C C C C K C a a = ⋅ − = ⋅ − ≈ ⋅ ⇒ = α α α α α α α ( ) 1 1 2 2 Now, K C C K K a a a = ⋅ − = ⋅ − ⇒ = + + + + [ ] ( ) [ ] [ ] H H H α α α α α 1 1 = + = + + − 1 1 1 1 10 [ ] ( ) H P P Ka H K a
Solution: Dilution results in increased degree of dissociation but decrease in concentrations of all active components.
Solution: Relation is valid only for conjugate pairs.
Solution: P H may decrease only on increasing [H + ].
Solution: CH COOH CH COO H M 0.1 M M M 0.1 M 3 3 0 1 0 1 ( . ) ( . ) − + − + + x x x \u001b \u001b \u001f ⇀ \u001f ↽ \u001f \u001f Now, 1 8 10 0 1 0 1 1 8 10 5 5 . . . . × = × ⇒ = × − − x x and α = = × − x 0 1 1 8 10 4 . . Now, [ ] [ ] [ ] H OH Kw H M T from water acid = = = − + − 10 13
Solution: RNH g) H O(l) RNH aq OH aq bar M M 2 1 2 3 ( ( ) ( ) + + + − \u001f ⇀ \u001f ↽ \u001f \u001f x x 10 1 10 3 0 11 0 6 3 − − = ⋅ ⇒ = ⇒ = ⇒ = x x x P P OH H . .
Solution: NH OH(aq) NH OH 4 4 \u001f ⇀ \u001f ↽ \u001f \u001f + − + Addition of solid NH 4 OH will increase NH 4 OH(aq) concentration and hence, [OH – ] will increase.
Solution: (c) H O H O H O OH ve 2 2 3 + + Δ ° = + + − \u001f ⇀ \u001f ↽ \u001f \u001f ; H (d) HA OH A H O Final a a a a 2 2 0 2 2 + + − − \u001f ⇀ \u001f ↽ \u001f \u001f P P P H K K a a = + = log / / a a 2 2
Solution: [ ][ ] H C O + − >> 3 2
Solution: NH Cl NaOH NH OH NaCl For buffer: ( 4 4 0 0 a a b b a b b − ≈ ⇒ > + + ) \u001f ⇀ \u001f ↽ \u001f \u001f CH COONa HCl CH COOH NaCl For buffer: ( 3 3 0 0 a a b b a b b − ≈ ⇒ > + + ) \u001f ⇀ \u001f ↽ \u001f \u001f
Solution: KCN is a salt of weak acid (HCN) and strong base (KOH).
Solution: BOH H B mole th run a mole Equivalent point a a b a 1 5 5 0 0 0 0 − + + + \u001b \u001b \u001f ⇀ \u001f ↽ \u001f \u001f 5 5 2 a mole H O + Now, for 1/5th reaction, P P B BOH] OH K b = + − log [ ] [ or, ( ) log / / 14 9 5 4 5 − = + P K b a a ∴ = ⇒ = × − P K K b b 5 6 2 5 10 6 . . At equivalent point: P P C) H K b => − + 1 2 ( log or, 4 5 1 2 5 6 . ( . log => − + ⇒ = C) C 0.25 M Now, n n HCl used B formed = + or, V 0.5 V V ml HCl × = + × ⇒ = 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 = × ⇒ = ∴ Percentage purity of base = × = 2 25 2 5 100 90 . . %
Solution: CO H O HCO OH K Kw K M M M h a 1 2 3 2 2 3 0 5 2 1 − − − − − + + + = = × ( . ) ( ) ( ) ; x x y x y \u001f ⇀ \u001f ↽ \u001f \u001f 0 0 4 − HCO H O H CO OH K Kw K M M M 2 h a 3 2 3 9 2 1 2 5 10 − − − − + + + = = × ( ) ( ) ; . x y y x y \u001f ⇀ \u001f ↽ \u001f \u001f Now, 2 10 0 5 0 5 10 4 2 × = − ⋅ + − ≈ ⋅ ⇒ = − − ( ) ( ) ( . ) . x y x y x x x x and 2 5 10 2 5 10 9 9 . ( ) ( ) . × = ⋅ + − ≈ ⋅ ⇒ = × − − y x y x y y x x y Now, h x = = 0 5 0 02 . . P P OH H = − + ≈ − = ⇒ = − log( ) log( ) . . x y 10 2 0 12 0 2 and [H 2 CO 3 ] = y = 2.5 × 10 –9 M
Solution: N H CH COOH N H CH COO NH CH P P K a 1 K a2 + = + − = ⎯ → ⎯⎯⎯⎯ ⎯ → ⎯⎯⎯⎯ 3 2 2 22 3 2 9 78 3 2 . . C COO − 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 = ⇒ = × ⊕ − + ⊕ − − ⊕ [NH ][H ] [NH ] . [NH . CH COO CH COOH C 2 2 H H COOH 2 ] ∴ = = × ⊕ − − [NH ] . . 3 5 78 6 10 1 7 10 CH COOH M 2 % of glycine in cationic form = × × = − 1 7 10 0 01 100 0 017 6 . . . %
Solution: At equivalent point, the solution should be acidic.
Solution: Sodium acetate solution is basic.
Solution: For precipitation of Fe(OH) 2 , [ ] . . . min max min OH P P OH H − − − = × = × ⇒ = ⇒ = 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 − − − = × ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = × ⇒ = ⇒ = 4 10 0 05 2 10 8 7 5 3 28 1 3 9
Solution: K x x x = × = = − ⇒ = 1.5 10 Dimer] [Monomer] 2 2 2 0 1 2 5 120 [ ( . ) ∴ [ ] [ . Dimer Monomer] = − = x x 0 1 2 5 2
Solution: K x x x = × = = − ⇒ = × − − 3.6 Dimer Monomer 10 0 1 2 3 6 10 2 2 4 ( ) [ ] ( . ) . ∴ (D ) [Monomer] . . imer = − ⇒ = x x x 0 1 2 0 1 9 2500
Solution: [ ] . H M + − − ≈ × × = × 0 1 2 10 2 10 5 3 (Dimerization is negative as Q. 2) ∴ P H = 2.85\n7.50 Chapter 7 HINTS AND EXPLANATIONS Comprehension II
Solution: [ ] . CH COOH M O 3 3 3 8 0 7 10 10 10 7 10 = × × = × − − The solution is so dilute that we may assume almost complete dissociation of acid. ∴ ≈ × + − [ ] H M acid 7 10 8 Now, H O H OH M M 2 7 10 8 \u001f ⇀ \u001f ↽ \u001f \u001f + × + − − + ( ) x x 10 7 10 7 10 14 8 8 − − − = × + ⋅ ⇒ = × ( ) x x x ∴ = − × + = − P H log( ) . 7 10 6 85 8 x
Solution: CH COOH CH COO H Eqn M 7 10 M M 3 3 8 7 10 8 14 10 8 . y x \u001f ⇀ \u001f ↽ \u001f \u001f − + × × − + = × − − + Now, 2 0 10 7 10 14 10 5 8 8 . × = × × × − − − y ∴ y = 4.9 × 10 –10 M Comprehension III
Solution: K a = × = × − − ( ) . . 8 10 0 2 3 2 10 3 2 4
Solution: K a = × = × × − + 3 2 10 1 0 0 8 0 2 4 . [ ] ( . . ) . H ∴ [H + ] = 8 × 10 –5 = ⇒ P H = 4.1 Comprehension IV
Solution: p p NH NH NH H Ka O = + = + = + + ( ) log [ ] [ ] . log . . . 4 3 4 9 3 0 8 0 2 9 9
Solution: 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 . . . . . + + + ≈ + \u001f ⇀ \u001f ↽ \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
Solution: H + added in excess. Final [H + ] = 1.0 – 0.8 = 0.2 M ∴ P H = – log(0.2) = 0.7 Comprehension V
Solution: No hydrolysis ⇒ p H = 7.0
Solution: 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 + − + + + + ( . ) ) x x x \u001f ⇀ \u001f ↽ \u001f \u001f 1 4 10 0 25 0 25 1 87 10 5 2 3 . ( . ) . . ] × = ⋅ − ⇒ = × − − + x x x x x \u001b M = [H
Solution: SO H O HSO OH M 4 2 0 5 2 4 − − − − + + ( . ) x x x \u001f ⇀ \u001f ↽ \u001f \u001f 10 1 25 10 0 5 0 5 6 32 10 19 2 2 7 − − − × = ⋅ − ⇒ = × . ( . ) . . x x x x x \u001b M ∴ = × = × + − − − [ ] . . H M 10 6 32 10 1 58 10 14 7 8
Solution: 1 4 10 1 25 10 0 25 0 5 0 25 0 5 5 2 2 . . ( . )( . ) . . × × = ⋅ − − × − − x x x x x \u001b ∴ x = 1.18 × 10 –2 Now, 1 4 10 1 18 10 0 25 5 2 . . [ ] . × = × × − − + H ∴ [H + ] = 2.97 × 10 –4 M Al SO H O Al(OH HSO M M M 3 0 25 0 25 4 2 0 5 2 2 0 4 + − − + − + + + . Eqn. ( . ) . ) x x \u001f ⇀ \u001f ↽ \u001f \u001f 0 0 x (0.5 – x )M\n7.51 Ionic Equilibrium HINTS AND EXPLANATIONS Comprehension VI
Solution: PuO H O PuO (OH H 2 2 0 01 0 01 2 2 1 6 10 4 + − ≈ + + = × + + − . . . ) x x x \u001f ⇀ \u001f ↽ \u001f \u001f ∴ = ⋅ − = × − K x x x x h 0 01 0 1 2 56 10 2 6 . . . \u001b
Solution: K Kw K K a b b = ⇒ = × − 3 9 10 9 . Comprehension VII
Solution: p P P H K K a 2 a = + = + = 1 2 1 2 8 13 1 0 5 3 ( ) ( ) .
Solution: For 2nd equivalent point, n n NaoH H PO = × 2 3 4 V V 40 ml × = × × ⇒ = 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 − + ≈ − + \u001f ⇀ \u001f ↽ \u001f \u001f ∴ = + = P H 8 5 5 8 0 log .
Solution: S K OH M sp = = × = × − − − − [ ] . ( ) . 2 30 6 2 18 4 0 10 10 4 0 10 Comprehension VIII
Solution: P P [HCO H CO H K O O a = + = + = − log ] [ ] . log . 3 2 3 6 4 8 1 7 3
Solution: 7 4 6 4 1 10 3 2 3 2 3 3 . . log [ ] [ ] [ ] [ ] = + ⇒ = − − HCO H CO H CO HCO O O O O
Solution: More CO 2 should dissolve in solution. Comprehension IX
Solution: pH P C)=7+ a 2 = + + + = 7 1 2 1 2 10 6 1 12 3 ( log ( . log ) . K
Solution: CO H HCO M Final M M M 3 2 50 1 75 25 75 25 1 75 3 0 25 75 0 − × + × − + ≈ \u001f ⇀ \u001f ↽ \u001f \u001f pH CO HCO a 2 O O = + = + = − − P K log [ ] [ ] . log / / . 3 2 3 10 6 1 3 1 3 10 6
Solution: CO H HCO M Final M M 3 2 50 1 100 0 50 1 100 3 0 50 100 0 − × ≈ + × − + ≈ \u001f ⇀ \u001f ↽ \u001f \u001f ∴ = + = + = pH a 1 a 1 2 1 2 5 4 10 6 8 0 2 ( ) ( . . ) . P P K K
Solution: CO H HCO M Final M M M 3 2 50 1 125 0 75 1 125 25 125 3 0 50 125 − × + × − + \u001f ⇀ \u001f ↽ \u001f \u001f HCO H H CO M Final M M M 3 50 125 25 125 25 125 0 2 3 0 25 125 − + = + \u001f ⇀ \u001f ↽ \u001f \u001f ∴ = + = + = − pH [HCO H CO a O O P K log ] [ ] . log / / . 3 2 3 5 4 25 125 25 125 5 4
Solution: [H CO ] M 2 3 = × = 50 10 150 1 3 ∴ = + = + = 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
Solution: N H CH COOH N H CH 3 3 K K 3 3 b + = × = × + − − ⎯ → ⎯⎯⎯⎯⎯ ← ⎯ ⎯⎯⎯⎯⎯ − − − 2 12 1 3 2 5 10 4 10 . a − − ⎯ → ⎯⎯⎯⎯⎯ ← ⎯ ⎯⎯⎯⎯⎯ − − − = × = × − − − 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 = = × × − 1 6 25 10 10 5 .
Solution: ∴ = + = + = pH a 1 a 1 2 1 2 2 4 9 8 6 1 2 ( ) ( . . ) . P P K K Comprehension XI
Solution: Moles of Cu reacted = 6 35 10 63 5 10 3 4 . . × = − −
Solution: ln . K G RT K eq eq = Δ − = − × − × = ⇒ >>> ° 120 10 8 0 300 50 1 3 ∴ = + [ ] Ag 2 × Mole of Cu reacted = 2 × 10 –4 M
Solution: K sp Ag M = = × = × + − − − [ ][BRO ] ( ) 3 4 2 8 2 2 10 4 10 Comprehension XII
Solution: S K AgCN CN M sp = = × = × − − − [ ] [ ] . . . 1 0 10 0 02 5 0 10 16 15
Solution: AgCN(s) CN Ag(CN S M SM eq sp + = × = − − − ( . ) ) 0 02 2 15 \u001f ⇀ \u001f ↽ \u001f \u001f K K K f 15 0 02 0 3 16 1 875 10 2 = − ⇒ = = × − s s s M . . .
Solution: s [Ag ] [Ag(Cu Cu Cu M sp sp = + + ⋅ ⋅ + − − − ) ]; [ ] [ ] 2 5 K K K f For minimum solubility: ds d[Cu ] − = 0 or, − + ⋅ = ⇒ = = × − − − K K K f sp sp f Cu Cu K M [ ] [ ] . 2 9 0 1 2 58 10
Solution: For acidic solution, pH < 7.0 at 25° C.
Solution: If there were no common ion effect, P H should lie in between 7.0 and 7.3
Solution: [ ] ] . H M [H M HCl HCOOH + − + − = < = × 10 3 16 10 4 2
Solution: Dilution results decrease in concentration of BOH (aq), B + (aq) as well as OH – (aq).
Solution: 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.
Solution: pH of buff er containing H A and A – may be less than, greater than or equal to 7.0.
Solution: K K a b ( ( ) CH COOH) NH OH 3 = 4 and hence, CH 3 COONH 4 solution is also neutral. But, it undergoes hydrolysis.
Solution: Reaction occurs but at equivalent point, pH will be less than
Solution: 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
Solution: 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.
Solution: (P) [OH–] to just start precipitation = K sp Mg [ ] 2 + = × × = − − − 2 10 2 10 10 6 3 1 5 . ∴ p OH = 1.5 ⇒ P H = 12.5 (Q) [ ] [ ] . max / / OH Al M sp − + − − = ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = K 3 1 3 28 1 3 9 10 0 1 10 ∴ = ⇒ = 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 . . . . − × ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ − × \u001f ⇀ \u001f ↽ \u001f \u001f 1 1 M H + + ? Now, K a = = × ⇒ = − + + − 10 1 11 0 1 11 10 5 6 [H ] . [ ] H M ∴ p H = 6.0 (S) [ ] . . H C M pH a + − − = ⋅ = × = ⇒ = K 10 0 001 10 5 0 7 5 (T) A H O HA OH M M M − × − − − − + + = × ( ) ; 6 10 2 6 5 2 10 x x x a Kw K \u001f ⇀ \u001f ↽ \u001f \u001f 2 10 6 10 1 10 6 5 5 × = ⋅ × − ⇒ = × − − − x x x x ( ) ∴ P OH = 5 ⇒ pH = 9
Solution: (A) pH [H A H A] a 2 O O = + = − P K 1 3 4 0 log ] [ . (B) pH [HA H A ] a O O = + = − − P K 2 2 2 8 0 log ] [ . (C) pH [A HA ] a O O = + = − − 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
Solution: (A) K K a b ( ( [ ] . H O) H O)= Kw H O 2 2 2 14 16 10 1000 18 1 8 10 = = ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = × − − (C) On increasing temperature, Kw increases and hence, P Kw decreases.
Solution: [ ] log[ ] D K M P D W D + − − + = = = ⇒ = − = 10 10 8 16 8
Solution: [ . HCOOH] M C O = × = = 1 15 10 46 25 3 HCOOH HCOOH HCOOH HCOO C C − − + − + + x x x x \u001f ⇀ \u001f ↽ \u001f \u001f 2 From given data: x = = − K M 10 3 ∴ Percentage of HCOOH molecules converted into HCOO – = × = × = × − − x C 100 10 25 100 4 10 3 3
Solution: [ ] / / . NH M O 3 3 10 17 100 0 85 10 5 = × =\n7.54 Chapter 7 HINTS AND EXPLANATIONS [ ] ) OH C Kw K (NH C M a − + − − − = ⋅ = ⋅ = × × = K b 4 14 10 2 10 5 10 5 10 ∴ = = = + − − − − [ ] [ ] H O Kw OH M 3 14 2 12 10 10 10
Solution: [ ] . H K C M a + − − = ⋅ = × × = 4 10 0 0025 10 10 6 ∴ = − = − P H log10 6 6
Solution: [ . . H SO ] M 2 3 O = = 1 28 64 0 02 H SO H HSO 2 3 M M M ( . ) 0 02 3 − + − + x x x \u001f ⇀ \u001f ↽ \u001f \u001f Now, K x x x x x a M pH = = ⋅ − ⇒ = ⇒ = − = − 10 0 02 0 01 2 2 ( . ) . log
Solution: pH HC H O [H C H O a 4 4 O O = + = + = − P K 1 6 2 4 4 6 3 3 18 8 18 8 30 150 3 log [ ] ] . log . / . /
Solution: Buff er capacity, β = − Δ = − − = + [ ] . / . ( . ) H added H P 0 05 0 2 0 05 5
Solution: HA OH A millimole 0 millimole Equ.point a \u001b \u001b \u001f ⇀ \u001f ↽ \u001f \u001f + − − × 36 12 0 1 0 0 3 . . . .612 2 millimole H O + A H HA mmole Final mmole mmole − + + × 3 612 1 806 18 06 0 1 0 0 1 . . . . . \u001b \u001f ⇀ \u001f ↽ \u001f \u001f 8 806 mmole ∴ = + = + = − pH A HA] a O O P K log [ ] [ log . . 5 1 806 1 806 5
Solution: ∴ = + = + = pH a a 2 1 2 1 2 2 28 9 72 6 1 ( ) ( . . ) P P K K
Solution: For appearance of only ln + colour, log [ln ] [ln . . . log + = − = = OH] 4 6 3 4 2 0 6 4 ∴ = + [ln ] [ln OH] 4 Four-digit Integer Type
Solution: 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 . ( . ) . − + = + + − x x \u001b \u001f ⇀ \u001f ↽ \u001f \u001f − − + CM (C ) M CM x \u001b Now, K C b = ⇒ × = × + − − − [C H NH ][OH ] C H NH . 6 5 3 6 5 2 10 8 4 10 10 0 2 ∴ C = 8 × 10 –3 M Now, mass of NaOH added = × × × = = − 8 10 1000 500 40 0 16 160 3 . gm mg
Solution: K a = ⇒ × = × × + − − − − [ ][CH COO ] [CH COOH . [CH COO ] . H ] 3 3 5 4 3 1 8 10 4 10 0 2 ∴ = × − − [CH COO ] 3 3 9 10 M ∴ Mass of CH 3 COONa added 9 10 500 1000 82 0 369 3 × × × = = − . gm 369 mg
Solution: H SO H HSO 2 3 CM Final 0 CM CM \u001b \u001f ⇀ \u001f ↽ \u001f \u001f + − + 0 4 0 H A OH HA mmole Final 0 mmole mmole 2 20 0 09 30 0 06 0 1 8 + − × × + + . . . \u001b \u001b \u001f ⇀ \u001f ↽ \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 − − + + − + ( ( ) x x x \u001f ⇀ \u001f ↽ \u001f \u001f Now, 1 2 10 0 01 6 11 2 . . × = × − ⇒ = − x x x C C and C + x = 0.01 ⇒ C M = × 0 01 11 17 .
Solution: P P NaHCo [H Co H K O O a = + log [ ] ] 3 2 3 or, 7 4 6 1 10 2 80 . . log = + × × ⇒ = V 5 V ml
Solution: [ ] . / / . HSO M O 4 1 8 120 100 1000 0 15 − = = HSO H SO M M M 4 0 15 4 2 − − + − + ( . ) x x x \u001f ⇀ \u001f ↽ \u001f \u001f 4 10 0 15 6 10 60 2 2 × = ⋅ − ⇒ = × = − − x x x x . M millimole/L
Solution: 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 + − = + = − + + + x x x \u001f ⇀ \u001f ↽ \u001f \u001f 0 0 3 − ≈ M C 0.1M ∴ Mass of Al(OH) 3 added = 400 × 0.1 × 133.5 = 5.34 gm = 5340 mg
Solution: 9 18 60 40 . log = + P K a (1) 9 00 100 . log = + − P x x K a (2) ∴ x = 50
Solution: I I I M Eqn. (0.05 M M 2 12 7 254 0 05 0 1 0 1 3 0 . . ) . . = − − − − + x x x \u001f ⇀ \u001f ↽ \u001f \u001f = = 0 254 254 0 001 . . ∴ x = 0.049 Now, K x x x c = − − = × = ( . )( . ) . . . 0 05 0 1 0 049 0 001 0 051 960
Solution: Concentration of Ca(OH) 2 in its saturated solution = ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = × ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = − 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 + − + . . ( ) \u001f ⇀ \u001f ↽ \u001f \u001f 0.01 M (0.02 + 0.8) M = 0.82 M Equ (0.01 – x )M = 0 (0.82 – 2 x ) M = 0.8M ∴ = × = × << + − − [ ] . ( . ) . Ca left 2 5 2 5 3 2 10 0 8 5 10 0 01 ∴ Ca(OH) 2 precipitated = 0.01 mole = 0.01 × 74 = 0.74 gm = 740 mg
Solution: S M = = − 0 0055 550 100 1000 10 4 . / / ∴ K sp of Ca(pam) 2 = 4S 3 = 4 × 10 –12 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 + × = − − + / . ( \u001f ⇀ \u001f ↽ \u001f \u001f s s) ∴ Ca(pam) 2 participated = 10 –3 × 10 = 0.01 mole = 0.01 × 550 = 5.50 gm = 550 mg On adding NaOH
Solution: (I) Cu cannot reduce Pb (II) Pb can reduce Ag (III) Ag cannot reduce Cu. Hence, reducing power: Pb > Cu > Ag
Solution: E E n = ° − ⋅ 0 06 . log [R] [O] ⇒ 0 24 0 36 0 06 1 . . . log [ [ = − R] O] ∴ [ [ O] R] = 1 100
Solution: For the complex ion to get oxidised, its reduction potential should be low.
Solution: Ag NH Ag(NH) E + + + ° = − = 2 0 79 0 37 0 42 3 3 \u001e ⇀ \u001e ↽ \u001e \u001e ; . . . V Now, E n K eq ° − ⋅ 0 06 . log ⇒ 0 42 0 06 1 . . log = ⋅ K f ⇒ K f = 10 7
Solution: Given: Co 3+ + e – Co 2+ ; E° = 1.81 V; Δ ° = − × × G F 1.81 1 1 Co(CN) e Co(CN) 6 3 6 4 + − − + ⎯ → ⎯ E° = –0.83 V; Δ ° = − × × − G F 2 1 0 83 ( . ) Co 6CN Co(CN) 2 6 4 + − − + ⎯ → ⎯ ; K f = 10 19 ; Δ ° = − × G RT ln10 3 19 Required Co CN Co(CN) 3 6 3 6 + − − + ⎯ → ⎯ K f = ?; Δ ° = − × G RT lnK f Now, Δ ° = Δ ° − Δ ° + Δ ° G G G G 1 2 3 or, − ⋅ = − − + − RT lnK F) F RTln10 f ( . . ( ) 1 81 0 83 19 or, RT ln K F ⋅ = − 10 2 64 19 f . ⇒ log . . . . 10 2 64 2 303 2 64 0 06 19 K F RT f = − = − ∴ K f = 10 63
Solution: E E Cu E Cu Cu Cu Cu Cu Cu Cu 2 2 2 0 06 2 1 0 03 2 2 + + + = ° − ⋅ = ° + + + | | | . log [ ] . log[ ] ] = 0.34 + 0.03 × log(0.1) = 0.31 V ∴ E Cu/Cu 2+ = − 0.31 V
Solution: E E RT F Cd Ag cell cell = ° − ⋅ + + 2 2 2 ln [ ] [ ] As CN– will form complex with Ag+ ion in the cathodic compartment, E cell will decrease.
Solution: −Δ = ⋅ = ° − ⋅ = ° − + + G nF E nF E RT nF Zn Cu nFE cell cell Cell Cell [ ln [ ] [ ] 2 2 RT T C C ⋅ ln 1 2
Solution: E E RT F K X Ag X|Ag Ag |Ag sp ° = ° − ⋅ − + | ln 1
Solution: Cell reaction: H + (cathode, P H = 3) H + (anode, P H = ?) E H anode H P P cell anode H catho = − ⋅ = − + + 0 0 059 1 0 059 . log [ ] [ ] cathode . d de H ⎡ ⎣ ⎤ ⎦ or, 0 272 0 059 3 . . [P ] = − anode H ⇒ P anode H = 7.6
Solution: Cl 2 + H 2 O \u001f Cl – + ClO – + 2H + ; E V cell ° = − = − 1 36 1 63 0 27 . . . Now, E E H cell cell = ° − + 0 06 1 2 . log[ ] or, 0 0 27 0 06 1 2 2 25 = − + ×= . . . P H
Solution: E E E cell quinohydrone calomel = − 0.210 = E quinohydrone – 0.279 ⇒ E quinohydrone = 0.489 V EXERCISE II (JEE ADVANCED)\n8.41 Electrochemistry HINTS AND EXPLANATIONS Quinohydrone electrode is + 2H + + 2e – (Quinone, Q) (Hydroquinone, H2Q) O O OH OH E E H E P H = ° − ⋅ = ° − + 0 06 2 1 0 06 . log [ ] . or, 0.489 = 0.699 – 0.06.P H ⇒ P H = 3.5
Solution: Anode: Ag(s) Ag + (aq) + e – 1 × 6 Cathode: Cr O aq H aq Cr aq H O(l) 2 7 2 3 2 14 6 2 7 − + − + + + ⎯ → ⎯ + ( ) ( ) e ( ) Net: 6 14 6 2 7 2 3 2 Ag(s) Cr O aq H aq Ag aq) 2Cr aq)+7H O(l) + + ⎯ → ⎯ + − + + + ( ) ( ) ( ( E E n Ag Cr Cr O H cell cell = ° − ⋅ + + − + 0 06 6 3 2 2 7 2 14 . log [ ] [ ] [ ][ ] = (1.33 – 0.80) – 0 06 6 0 1 0 4 1 6 0 1 0 46 6 2 14 . log ( . ) ( . ) . ( . ) . ⋅ × × = V
Solution: Net cell reaction: Zn – Hg(C 1 M) Zn – Hg(C 2 M) E C C V cell = − = − = 0 0 059 2 0 059 2 1 10 0 0295 2 1 . log . log .
Solution: E K cell eq ° = ⋅ 0 06 2 . log ⇒ 0 75 1 50 3 1 68 1 3 1 0 03 . . . . log − × − × − = K eq ∴ K eq = 10 –22
Solution: [ ] . . H Ka C C M left 1 + − = ⋅ = × × = 1 8 10 0 1 5 [ ] . . H K Kb C C M right w + − − = ⋅ = × × = 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 = − ⋅ = − 0 0 06 1 0 465 1 2 . log .
Solution: E V V V ° = × + × − × = + + 2 3 1 0 616 1 0 439 1 0 799 0 256 | . . . . ∴ E V V V ° = − + + 3 2 0 256 | .
Solution: E E n H P cell cell H = ° − ⋅ + 0 06 2 2 . log [ ] or, 0 70 0 28 0 0 06 2 1 2 . ( . ) . log [ ] = − − ⋅ + H ⇒ P H = 7.0
Solution: Net cell reaction: Ag C M) Ag C K M sp + + = ⎯ → ⎯ = ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ( . / 1 2 1 3 0 1 2 4 Now, E C C cell = − ⋅ 0 0 06 1 2 1 . log ⇒ 0.162 = − ⋅ 0 06 2 0 1 1 3 . log ( ) . / K sp ∴ K sp = 4 × 10 –12
Solution: E E V Tl Tl Pb Pb + + − = − | | . 2 0 444 or, E E Pb Tl Tl Tl Pb Pb + + − ( ) − ⋅ = − + + | | . log [ ] [ ] . 2 0 06 2 0 444 2 2 or, [( . ) ( . )] . log . . . − − − − ⋅ ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = − 0 336 0 126 0 03 0 1 0 1 0 444 2 K sp ∴ K sp = 4 × 10 –6
Solution: E cell = E Cu – E Zn = (E Cu – E calomel ) – (E Zn – E calomel ) From question E calomel – E Zn = 1.083 V and E calomel – E Cu = – 0.018 V
Solution: The potential of hydrogen electrode at H 2 (1 bar) may be expressed as E = – 0.059 P H Now, E P Ka 1 0 059 = − + ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ . log x y and E P Ka 2 0 059 = − + ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ . log y x ∴ P E E ) Ka 1 2 = − + ( . 0 118\n8.42 Chapter 8 HINTS AND EXPLANATIONS
Solution: For the given reaction, E E E cell given hydrogen ° = ° − ° ∴ (–0.84) – 0 = ⋅ 0 06 1 . log K eq ⇒ K eq = 10 –14
Solution: Net cell reaction may be written as H 2 + Zn 2+ \u001f 2H + + Zn E E n H P cell cell H = ° − ⋅ + + 0 06 2 2 2 . log [ ] [Zn ] or, ( . ) ( . ) . log [ ] . − = − − ⋅ × + 0 61 0 76 0 06 2 1 0 4 2 H ∴ [H + ] = 2 × 10 –3 M Now, K H SO HSO a 2 3 2 3 3 2 2 10 6 4 10 0 4 = = × × × + − − − − [ ][ ] [ ] ( ) ( . ) . = 3.2 × 10 –4
Solution: Cu NH Cu(NH K f 2 3 3 4 2 12 4 10 + + + = \u001e ⇀ \u001e ↽ \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 = × . ( . ) x ⇒ x = = × − − 10 16 6 25 10 12 14 . Now, cell reaction: Zn + Cu 2+ \u001f Zn 2+ + Cu E E Zn Cu cell cell = ° − ⋅ + + 0 06 2 2 2 . log [ ] [ ] = − − ⋅ × = − [ . ( . )] . log . . . 0 34 0 76 0 06 2 1 0 6 25 10 0 704 14 V
Solution: Net cell reaction: Zn + 2H + \u001f Zn 2+ + H 2 E E Zn P H cell cell H = ° − ⋅ + + 0 06 2 2 2 2 . log [ ] [ ] or, 0 70 0 0 76 0 06 2 0 01 1 2 . [ ( . )] . log . [ ] = − − − ⋅ × + H ⇒ [H + ] = 0.01 M Moles of HCl in RHS = 500 0 01 1000 5 10 3 × = × − . ∴ Mass of NaOH needed = 5 × 10 –3 × 40 = 0.2 gm
Solution: 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 = × = . . and C K Cl K K M sp sp sp 2 0 1 50 100 0 05 = = × = − [ ] . . Now, E C C cell = − 0 0 06 1 2 1 . log or, 0 42 0 06 1 0 05 0 04 . . log / . . = − K sp ⇒ K sp = 2 × 10 –10
Solution: Δ G cell = – nFE cell ⇒ – 965 × 3 × 10 3 = –12 × 96500 × E cell ∴ E cell = 2.5 V
Solution: Theoretical efficiency = −Δ ° −Δ ° G H ⇒ 0 84 285 . = −Δ ° G ∴ Δ G° = – 0.84 × 285 KJ = –nF ⋅ E° cell or, 0.84 × 285 × 10 3 = 2 × 96500 × E° cell ⇒ E° cell = 1.24 V
Solution: Net cell reaction: Ag(s) + H + + Cl – \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 = ° − ⋅ ⋅ + + 0 06 1 2 1 2 . log [ ] [ ] / = − − ⋅ ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ × = − − [ . ] . 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
Solution: x y y x Ag NH Ag(NH + + + 3 3 \u001e ⇀ \u001e ↽ \u001e \u001e ) aM bM 0 b >> a Eqn. ? bM a x M K / Ag f x y a x b = ⋅ + [ ] ⇒ [Ag ] / + = ⋅ ⋅ ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ a x b f y x K 1 Cell reaction: Ag + (Right) Ag + (Left)\n8.43 Electrochemistry HINTS AND EXPLANATIONS ∴ E cell = 0 0 059 1 − ⋅ + + . log [ ] [ ] Right Ag Left Ag Case-I: 0 118 0 059 4 10 4 10 4 2 1 . . log / = − ⋅ × × ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ − − x ⇒ x = 1 Case-II: 0 118 0 059 0 1 1 . . log . y/ = − ⋅ ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ x ⇒ y = 2
Solution: E K V cell eq ° = ⋅ = × = − 0 06 1 0 06 1 1 667 10 0 3732 6 . log . log . . Now, E E V Cu Cu Cu Cu ° − ° = − + + + 2 0 3732 | | . (1) and E E E V Cu Cu Cu Cu Cu Cu ° = × ° + × ° + = + + + + 2 2 1 1 1 1 0 3376 | | | . or, E E Cu Cu Cu Cu ° + ° = + + + 2 0 6752 | | . V (2) From (1) and (2), E V Cu Cu ° = + | . 0 5242
Solution: Zn + Ni 2+ \u001f Zn 2+ + Ni E K cell eq ° = 0 06 2 . log ⇒ (–0.24) – ( –0.75) = 0.03 log K eq ∴ K eq = 10 17 ⇒ It means that Ni 2+ will react almost completely and [Zn 2+ ] ≈ 1.0 M Now, 10 1 0 17 2 = + . [Ni ] ⇒ [Ni 2+ ] = 10 –17 M
Solution: Assuming the cell as concentration cell, the cell reaction may be written as Ag C M Ag C M + − − + − − = × = × ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ⎯ → ⎯ = × = 1 13 10 2 10 9 4 10 0 001 4 10 2 10 0 2 10 . . ⎛ ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ Now, E V cell = − ⋅ × = − − − 0 0 06 1 10 4 10 0 024 9 10 . log .
Solution: 0 03 0 06 2 0 3 0 5 2 2 2 . . log [ ] [ ] . log . [ ] lower = ⋅ = ⋅ + + + Cu Cu Cu higher lower r ∴ [Cu 2+ ] lower = 0.05 M
Solution: Cell reaction: H + (C 1 , HA 1 ) H + (C 2 , HA 2 ) E C C Ka Ka cell = − ⋅ ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = ⋅ 0 0 059 1 0 059 2 1 2 1 . log . log = − = 0 059 2 0 059 1 2 . ( ) . V P P K K a a
Solution: Au + + 2CN – \u001f Au ( ) CN 2 − , Δ G° 1 = –RT ⋅ ln x O 2 + 2H 2 Ο + 4e – \u001f 40H – ; Δ G° 2 = –4 × F × 0.41 Au 3+ + 3e – \u001f Au; Δ G° 3 = –3 × F × 1.50 Au 3+ + 2e – \u001f Au + ; Δ G° 4 = –2 × F × 1.40 From Δ ° + Δ ° − Δ ° + Δ ° Δ ° = − + G G G G G RT F required 1 2 3 4 1 4 1 29 , ln . x
Solution: P H of right electrode will increase due to formation of OH – ion.
Solution: At low [Cl – ], O 2 becomes anode product
Solution: In the electrolysis of aq. KNO 3 , neither K + are NO 3 − participate in electrode reaction.
Solution: Na will react with water. S will not conduct electricity.
Solution: w E Q F = ⇒ 0 635 63 5 2 0 965 3600 96500 . . . × = × × i ⇒ i = 2 3 6 . A ∴ % . . . % error = − = 2 3 6 0 5 2 3 6 10
Solution: Detonating gas is mixture of H 2 and O 2 n eq S n = n eq H 2 = n eq O 2 ⇒ 1 2 120 2 2 4 2 2 . × = × = × n n H O ∴ n and n H O 2 1 0 01 0 005 = = . .\n8.44 Chapter 8 HINTS AND EXPLANATIONS ∴ Vol. of mixture of H 2 and O 2 = 0.015 × 22400 = 336 ml
Solution: n eq of Li OH = Q F ⇒ w 24 1 2 5 4825 0 8 96500 × = × × . .
Solution: n eq Cu deposited at cathode = Q F or, w 63 5 2 12 4 4825 96500 . . × = × ⇒ 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. ∴ % of Cu = 19 05 20 100 95 25 . . % × = Now, Q F n Cu n Fe eq eq = + (oxidised at anode) or, 12 4 4825 96500 19 05 63 5 2 56 2 . . . × = × + × w ⇒ w = 0.56 gm ∴ Percentage of Fe = 0 56 20 100 2 8 . . % × =
Solution: Theoretical n eq of NaOH formed = n eq Cu = 3 18 63 6 2 0 1 . . . × = Actual n eq of NaOH formed = × = 60 1 1000 0 06 . ∴ Percentage yeild = 0 06 0 1 100 60 . . % × =
Solution: 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 = ⇒ 200 240 2 10 96500 × = × t ⇒ t = 16083.33 sec
Solution: Q F of I N a S O eq eq 2 2 3 = = − n n or, i × × = × × 2 3600 96500 72 1 0 1000 1 . ⇒ i = 0.965 A
Solution: C 14 H 10 + 2H 2 O C 14 H 8 O 2 + 6H + + 6e – n eq C 14 H 8 O 2 = Q F ⇒ w 208 6 1 40 60 0 965 96500 × = × × × . ∴ w = 0.832 gm
Solution: Number of coulombs required = 1000 0 00033 . per Kg Cu Energy required = × = 1000 0 00033 0 33 10 6 . . J ∴ Cost of electricity = × × = 4 10 3600 10 1 11 3 6 . Rupee
Solution: n eq CH 3 Coo – oxidised = Q F or, n × = × × × 1 0 5 482 5 60 0 8 96500 . . . ⇒ n = 0.12 ∴ Moles of (C 2 H 6 + CO 2 ) produced = 3 2 0 12 0 18 × = . . and total volume = 0.18 × 22.4 = 4.032 L
Solution: Δ ° = ⋅ = E V 0 059 2 10 0 177 6 . log .
Solution: Back EMF = ⋅ = × − 0 06 2 0 12 0 08 5 4 10 3 . log . . . V
Solution: Equivalent of charge used = 0 4825 10 3600 96500 0 18 . . × × = F Cell reaction during charge: Cu Zn Cu Zn + ⎯ → ⎯ + + + 2 2 100 1 1000 0 1 × = . mole 100 1 1000 0 1 × = . mole = 0.2 eq = 0.2 eq Final 0.2 – 0.1 8 = 0.2 + 0.18 = 0.02 eq = 0.38 eq = 0.01 mole = 0.19 mole ∴ Final [ ] . . Zn M 2 0 01 100 1000 0 1 + = × = and [ ] . . Cu M 2 0 19 100 1000 1 9 + + × = 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 = ° − ⋅ + + 0 06 2 2 . log [ ] [ ] = − − − ⋅ = [ . ( . )] . log . . . 0 34 0 76 0 06 2 0 1 1 9 1 1084 V
Solution: Reactions involved are 2 2 2 2 H O H O 2 electrolysis ⎯ → ⎯⎯⎯⎯ + 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 = × × = × 3 1152 10 80 4 32 10 3 4 . Now, n eq H Q F 2 = ⇒ 4.32 × 10 4 × 2 = i × × 24 3600 96500 ∴ i = 96500 A/day
Solution: n eq HC 3 H 5 O 3 = n OH Q F − = or, w 90 1 50 10 1158 96500 3 × = × × − ⇒ w = 0.054 gm ∴ % of lactice acid = 0 054 1 100 5 4 . . % × =
Solution: 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 = ⇒ n i × = × = × × 2 102 34 2 3600 0 5 96500 . [ ] 2 2 4 2 2 8 2 SO S O e − − − ⎯ → ⎯ + ⇒ i = 321.67 A
Solution: 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 × = × × − . . ⇒ w = 6.3 × 10 –5 gm
Solution: w E Q F = ⇒ 52 2 87 2 19 3 2 3600 96500 . . × = × × × η ⇒ η = 0.8333 or 83.33 %
Solution: n eq Cu 2+ reduced = Q F ⇒ n × 2 = 2 10 19 3 60 96500 3 × × × − . ∴ n = 1.2 × 10 –5 ∴ [ ] ( . ) . CuSO M 4 0 5 5 1 2 10 2 250 1000 9 6 10 = × × × = × − −
Solution: w E Q F = ⇒ 12 3 123 6 . × = × Q 0.5 F ⇒ Q = 1.2 F + 6H + + 6e – + 2H2O NH2 NO2 and Energy consumed = 1.2 F × 3.0 V = 347.4 KJ
Solution: 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 . . . . × = × + × w ∴ w = 38.8 gm
Solution: Initial mass of H 2 SO 4 , w 1 = 3600 × 1.5 × 40 100 2160 = gm Final mass of H 2 SO 4 , w 2 = 3600 × 1.1 × 10 100 396 = gm ∴ Moles of H 2 SO 4 consumed = w w 1 2 98 18 − = Now, n eq H 2 SO 4 = Q F ⇒ 18 × 1 = ( ) amp-hr × 3600 96500 ∴ Number of ampere-hr = 482.5
Solution: ∧ ∧ = = ⋅ ⋅ ⋅ ∗ ∗ m m NaCl KCl NaCl NaCl KCl NaCl KCl /C) /C) R G C R G ( ) ( ) ( ( κ κ 1 1 1 ⋅ ⋅ = ⋅ ⋅ 1 C R C) R C) KCl KCl NaCl ( ( or, ∧ = × × m NaCl) ( . . 120 200 0 1 6400 0 003 ⇒ ∧ m(NaCl) = 125 Ω –1 cm –1 mol –1 2 2 4 2 2 8 2 SO S O e − − − → + ⎡ ⎣ ⎤ ⎦\n8.46 Chapter 8 HINTS AND EXPLANATIONS
Solution: ∧° = × ° + + − m m m NH CrO NH CrO [( ) ] ( ) ( ) 4 2 4 4 4 2 2 λ λ = (2 × 6.6 × 10 –8 + 5.4 × 10 –8 ) × 96500 = 0.01795 Ω –1 m 2 mol –1
Solution: ∧ = ⋅ ∧° = m m C α κ or, 0.9 × 4.25 × 10 –2 = 382 5 3 . C 10 × ⇒ C = 0.1 M
Solution: ∧ = m K C ⇒ 1.5 × 10 –2 = 3 06 10 2 56 10 3 3 . . × − × − − C ∴ C mol m mol l V = = × ⋅ = − − − 1 30 1 30 10 585 58 5 3 3 1 / . ∴ V = 3 × 10 5 L
Solution: ∧ = = ⋅ ∗ m C G G C κ ⇒ 100 1 0 5 1 5 0 1 10 3 = × × − R . . . ⇒ R ohm = 100 3 Now, V = IR ⇒ I V R A = = = 5 100 3 0 15 / .
Solution: Ionic mobility, μ λ ° = = ° speed of ion Pot. gradient F m or, speed 19 3 5 50 96500 . ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = ⇒ speed = 2 × 10 –3 cm/s
Solution: ∧ eq = C κ ⇒ 150 3 4 10 1 6 10 5 6 6 × × − × − − . . ∴ S = 1.2 × 10 –8 mol cm –3 = 1.2 × 10 –5 M
Solution: K = G ⋅ G* ⇒ κ 1 4 1 280 1 50 . / / = ⇒ κ = 0.25 s m –1 for 0.5 M Now, ∧ = = × = × − − m C s m mol κ 0 25 0 5 10 5 10 3 4 2 1 . .
Solution: Ag A S Ag A ( ) ( )M \u001e ⇀ \u001e ↽ \u001e \u001e + + − + x y x M Ag B S Ag B M ( ) ( )M \u001e ⇀ \u001e ↽ \u001e \u001e + + − + x y y Now, ( x + y ) ⋅ x = 3 × 10 –14 and ( x + y ) ⋅ y = 1 × 10 –14 ∴ [Ag ] , [A ] . ; [B ] . + − − − − − = + = × = = × = = × x y x y 2 10 1 5 10 0 5 10 7 7 7 M M M Now, κ solution = × = × × × + × × × + × − − − − − 3 75 10 2 10 10 60 1 5 10 10 80 0 5 10 8 7 3 7 3 . . . − − − ° × 7 3 10 λ B ∴ λ B ° − − = 135 ohm cm mol 1 2 1
Solution: ∧ ° = ∧° + ∧° − ∧ ° eq eq eq eq [Be (Po ) ] [BeCl ] [K Po ] [K ] 3 4 2 2 3 4 Cl = + − = − − 160 140 100 200 1 2 1 ohm cm eq Now, n eq = ⇒ = × − κ C C 200 1 2 10 5 . ⇒ C = 6 × 10 –8 eq cm –3 = 6 × 10 –5 N = 10 –5 M Now, K sp = = × = × − − 108 108 10 1 08 10 5 5 5 23 S ( ) .
Solution: Net cell reaction is spontaneous in electrochemical cell but non-spontaneous in electrolytic cell. Cathode is +ve in electrochemical cell but –ve in electrolytic cell.
Solution: For the cell: Ag(s) | Ag Cl (s) | Cl− || Ag + | Ag(s), the net cell reaction is Ag + + Cl− \u001f Ag Cl (s).
Solution: Left electrode: Ag(s) + Cl − (aq) → Ag Cl (s) + e − 1 × 2 Right electrode : Hg 2 Cl 2 (s) + 2e − → 2Hg(l) + 2 Cl − (aq) Net reaction: 2Ag(s) + Hg 2 Cl 2 (s) → 2Ag Cl(s) + 2Hg (l)\n8.47 Electrochemistry HINTS AND EXPLANATIONS
Solution: n n n n eq eq eq eq Cu Mg Na Al = × = = × = = × = = 63 5 63 5 2 2 24 24 2 2 11 5 23 1 0 5 . . ; . . ; 9 9 27 3 1 × =
Solution: ∧ ° eq = 60 + 80 = 140 ohm –1 cm 2 eq –1 ∧ ° m = 140 × 6 = 840 ohm –1 cm 2 eq –1
Solution: Salt bridge does not change standard potential of any electrode.
Solution: Moles of electron involved = 0 25 9 65 3600 96500 0 09 . . . × × = ∴ Mass of Zn involved = 0 09 2 65 4 2 943 . . . × = gm Mass of MnO 2 involved = 0.09 × 87 = 7.83 gm Mass of NH 4 + involved = 0.09 × 18 = 1.62 gm
Solution: Net cell reaction is Cd(s) + 2AgCl(s) \u001f 2Ag(s) + Cd 2+ (aq) + 2Cl – (aq) Δ ° = − ° = − × × = − ° G 50 2 96500 0 6 115800 C nFE J . Δ ° = − ° = − × × = − ° G 0 2 96500 0 7 135100 C nFE J . Δ ° = ⋅ ° − ° − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = × × − = − S T T nF E E J/K 2 1 2 1 2 96500 0 6 0 7 50 386 . . Δ H ° = Δ G ° + T ⋅ Δ S ° = (–135100) + 273 × (–386) = –240478 J
Solution: Anode: 2H 2 O(l) O 2 (g) + 4H + (aq) + 4e – Cathode: 2H 2 O(l) + 2e – H 2 (g) +2OH – (aq) n eq H + produced = n eq OH – produced = Q F or, n n H OH + − × = = × × = 1 1 25 965 60 96500 0 75 . . Anode: HPO H H PO 4 2 2 4 − + − + \u001e ⇀ \u001e ↽ \u001e \u001e 1.0 M 0.75 M 1.0 M Final 0.25 M 1.75 M ∴ P P HPO H PO H K = + = + = − − 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 − − − + + \u001e ⇀ \u001e ↽ \u001e \u001e 1.0 M 0.75 M 1.0 M Final 0.25 M 0 1.75 M ∴ P P HPO H PO H K = + = + = − − a log [ ] [ ] . log . . . 4 2 0 2 4 2 15 1 75 0 25 3 0
Solution: V, Fe and Hg will be oxidised by NO 3 −
Solution: Resistance and heat capacity depends on quantity.
Solution: 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 × 1.26 × 40 100 504 = gm Final mass of H 2 SO 4 , w 2 = (1260 – 98 x + 18 x ) × 28 100 = (352.8 – 22.4 x ) gm From reaction, 504 – 98 x = 352.8 – 22.4 x ⇒ x = 2
Solution: E° Cell = E E H H Zn Zn ° − ° + + / / 2 2 because E H H ° + / 2 was higher or, 0.76 = 1.00 – E Zn Zn ° + 2 / ⇒ E Zn Zn ° + 2 / = 0.24 V
Solution: E° Cell = E E Cu Cu H H ° − ° + + 2 2 / / because E Cu Cu ° + 2 / was higher or, 0.34 = E Cu Cu ° + 2 / – 1.00 ⇒ E Cu Cu ° + 2 / = 1.34 V
Solution: E° Cell = E E Cu Cu Zn Zn ° − ° + + 2 2 / / = 1.34 – 0.24 = 1.10 V\n8.48 Chapter 8 HINTS AND EXPLANATIONS Comprehension II
Solution: E E Ag K Ag Ag Ag Ag sp + + = ° − ⋅ = − + / / . log [ ] . . .log 0 06 1 1 0 80 0 06 1 1 = + 0.314 V
Solution: E E K V I |Ag I|Ag Ag Ag sp ° = ° − ⋅ = − − + / . log . 0 06 1 0 172
Solution: E E I I |Ag I|Ag I /Ag I Ag − − = ° − ⋅ − / . log[ ] 0 06 1 = –0.172 – 0.06 ⋅ log 0.04 = –0.088 V Comprehension III
Solution: As(0.40) > (–0.87), reduction of Ni 2 O 3 (s) will occur.
Solution: E° cell = (0.40) – (–0.87) = 1.27 V
Solution: Net cell reaction is independent from OH – (aq)
Solution: – Δ G ° = nFE° cell = 2 × 96500 × 1.27 = 245110 J Comprehension IV
Solution: H 3 O + is only needed in balancing cathode reaction: NO H O e HNO H O 3 3 2 2 3 2 4 − + − + + ⎯ → ⎯ +
Solution: Moles of electron needed = 2 × moles of HNO 2 formed
Solution: n eq HNO Q F 2 = ⇒ 0 1 2 10 96500 . × = × t ⇒ t = 1930 sec Comprehension V
Solution: Δ G° = – n FE° ⇒ –237.39 × 10 3 = –2 × 96500 × E° cell ∴ E °cell = 1.23 V
Solution: Moles of H 2 needed = 23 739 237 39 0 1 . . . = ∴ Volume of H 2 needed = 0.1 × 22.7 = 2.27 L
Solution: E cell is independent from [OH – ]
Solution: Δ ° = Δ ° − Δ ° = − × − − × S H G T ( . ) ( . ) 285 8 10 237 39 10 298 3 3 = –162.4 J/K –1
Solution: η = −Δ ° −Δ ° = = ( ) ( ) . . . G H 237 39 285 8 0 8306 or 83.06% Comprehension VI
Solution: Refer theory given is passage.
Solution: E E E V O H O H Fe Fe ° = ° − ° = − − = + + 2 2 2 1 229 0 447 1 676 / , / . ( . ) .
Solution: E E Cu Cu Pb Pb ° > ° + + 2 2 / / and hence Cu will not oxidise easily.
Solution: n eq Fe Q F = ⇒ w 56 2 0 5 1 0 3600 96500 × = × × . . ⇒ w = 0.522 gm
Solution: Theoretical Comprehension VII
Solution: n eq Ni Q F = ⇒ w 58 7 2 15 3600 0 6 96500 . . × = × × ⇒ w = 9.85 gm
Solution: V = A × t ⇒ 9 85 8 9 4 0 2 . . ( . ) = × × t ⇒ t = 0.138 cm\n8.49 Electrochemistry HINTS AND EXPLANATIONS
Solution: n eq H Q F 2 = ⇒ V H 2 22 4 2 15 3600 0 4 96500 . . × = × × ⇒ V L H 2 2 5 = .
Solution: Anode produced is only O 2 gas. n eq O Q F 2 = ⇒ w 8 15 3600 96500 = × ⇒ w = 4.477 gm Comprehension VIII
Solution: E° cell = 0.8 – 0.05 = 0.75 V Now, E RT nF K cell ° = ⋅ ln ⇒ 0 75 1 2 38 92 . . ln = × ⋅ K ⇒ ln K = 58.38
Solution: The oxidation reaction of glucose contains H + and E E H = ° − ⋅ + 0 0592 2 2 . log[ ] ⇒ E – E° = 0.0592 P H = 0.6512 V
Solution: Standard potential is independent from ammonia concentration. Comprehension IX
Solution: Left electron: H 2 (g) 2H + (aq) + 2e – Right electrode: 2AgCl(s) + 2e – 2Ag(s) + 2Cl – (aq) ∴ Net reaction: H 2 (g) + 2AgCl(s) 2Ag(s) + 2H + (aq) + 2Cl – (aq)
Solution: Δ ° = ⋅ − − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = × × − = − S nF E E T T J/K 2 1 2 1 2 96500 0 21 0 23 20 193 . . Δ G° = –nFE° = –2 × 96500 × 0.23 = – 44390 J at 15°C Now, Δ H ° = Δ G ° + T ⋅ Δ S ° = (–44390) + 288 × (–193) = –99974 J = –49987 J/mole AgCl
Solution: Δ S° = –193 J/K = –96.5 J/K per mole AgCl
Solution: Δ G ° 298 = Δ H ° – T ⋅ Δ S ° = (–49987) – 298 × (–96.5) or, –1 × 96500 × E° = – 21230 ⇒ E° = 0.22 V = E Cl AgCl/Ag ° − / or, E E K Cl AgCl/Ag Ag Ag sp ° = ° + − + / / . log 0 058 1 or, 0 22 0 80 0 058 1 . . . log = + K sp ⇒ K sp = 1 × 10 –10 Hence, Solubility, S = K M sp = − 10 5 Comprehension X
Solution: Add Ag(s) in both sides of given reaction to get cell reaction. Now, for E° cell Δ G ° = –nFE° ⇒ [–109] – [77 + (–129)] × 10 3 = –1 × 96500 × E° cell ∴ E° cell = 0.59 V
Solution: E K cell eq ° = 0 059 . log n ⇒ 0 59 0 059 1 1 . . log = ⋅ K sp ∴ K sp = 10 –10
Solution: Zn(s) + 2Ag + (aq) ¡ Zn 2+ (aq) + 3Ag(s); E° = 0.80 – (–0.76) Now, E K eq ° = ⋅ 0 059 . log n ⇒ 1 56 0 059 2 2 2 . . log [ ] [ ] = ⋅ + + Zn Ag ∴ log [ ] [ ] . Zn Ag 2 2 52 88 + + =
Solution: Moles of Zn added = 6 539 10 65 39 10 2 3 . . × = − − Moles of Ag + present = 10 1000 100 10 5 6 − − × = (L.R.) ∴ Moles of Ag precipitated = 10 –6\n8.50 Chapter 8 HINTS AND EXPLANATIONS Comprehension XI
Solution: For KCl: ∧ = = ⋅ ∗ m C G G C κ ⇒ G* = ∧ m ⋅ C/G or, G* = ∧ m ⋅ C ⋅ R = 200 × (0.02 × 10 –3 ) × 100 = 0.4 cm –1
Solution: κ water G G cm = ⋅ = × = × Ω ∗ − − − 1 10000 0 4 4 10 5 1 1 .
Solution: For NaCl: ∧ = = ⋅ ∗ m C G G C κ or, 125 1 8000 1 10000 0 4 585 58 5 = − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ × ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ . / . V ⇒ V = 1.25 × 10 8 cm 3 = 1.25 × 10 5 L Comprehension XII
Solution: E E Cl Cl I ° > ° − − 2 2 / /I
Solution: E E Mn Mn O H O, H ° > ° + + + 3 2 2 2 / / Comprehension XIII
Solution: Net reaction: M + (1M) M + (0.05 M); Higher Conc. Lower Conc. E cell > 0, Δ G cell < 0
Solution: 70 mV = E RT F ° − ⋅ ln . 0 05 1 (1) E E RT F E RT F req = ° − ⋅ = ° − ⋅ ln . ln ( . ) 0 0025 1 0 05 1 2 (2) and E° =
Solution: Hence, E req = 140 mV Comprehension XIV
Solution: Δ G cell = –nFE cell = –2 × 96500 × 0.059 = –11387 J
Solution: E E M M cell cell Left Right = ° − ⋅ + + 0 059 2 2 2 . log [ ] [ ] or, 0 059 0 0 059 2 4 0 001 1 3 . . log ( / ) . / = − ⋅ K sp ⇒ K sp = 4 × 10 –15 Comprehension XV
Solution: H + + Cl – + (NaOH) Na + + Cl – + 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.
Solution: CH 3 COOH + (NaOH added) CH 3 COO – + 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 – ion, number of OH – ions increases and hence slope becomes greater.
Solution: H + + Cl – + (NH 4 OH added) NH Cl H O 4 2 + − + + 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.
Solution: HCl will neutralize fi rst followed by CH 3 COOH.
Solution: Ionic mobilities of Ag + and K + ions do not differ largely
Solution: First number of ions increases and then remain atmost constant. added\n8.51 Electrochemistry HINTS AND EXPLANATIONS
Solution: CuCl 2 Cu + Cl 2
Solution: E E informative Zn Zn Ag Ag ° < ° + + 2 / / ( )
Solution: Reason is different charges on ions.
Solution: Negative reduction potential means greater tendency to get oxidised.
Solution: Number of ions increases considerably only for weak electrolytes.
Solution: It is due to very high over voltage potential of hydrogen at mercury cathode.
Solution: E E RT nF Q = − ⋅ ° log
Solution: Cl – will combine with Ag + to precipitate AgCl.
Solution: At Cathode: 2H O(l) + 2e H (g) + 2OH aq 2 2 − − ⎯ → ⎯ ( ) At Anode: 2 2 2 3 2 6 2 CH COO aq C H CO e − − ( ) ( ) ( ) ⎯ → ⎯ + + g g
Solution: E E E V Fe Fe Fe Fe Fe Fe 3 3 2 2 1 2 1 2 0 037 + + + + ° ° ° = × + × + = − | , | . E V H ( H O OH ) . . . 4 4 4 2 0 40 1 23 0 83 → + ° + − = − = − E E E Cu Cu Cu Cu (Cu Cu ) | Cu | . . . . 2 2 2 0 34 0 52 2 1 0 52 0 + + + + + + → ° ° ° = − = − − − = − 7 70 V E V Cr Cr 3 2 3 0 74 2 0 91 3 2 0 4 + + ° = × − − × − − = − , ( . ) ( . ) .
Solution: For concentration cell, both half cell must have same configuration. P. Ag Cl AgCl sponteneous, K K eq sp + − + ⎯ → ⎯ = >> 1 1 Q. Ag Br + Cl AgCl Br Cl Non-Spontene eq sp sp − − ⎯ → ⎯ + = << ; (Ag Br) (Ag ) , K K K 1 o ous R. Ag Ag + + ⎯ → ⎯ ( . M) ( . M) 1 0 0 1 Higher to lower concentration, spontaneous S. Cl Cl − − ⎯ → ⎯ − ( . M) ( . M) Non spontaneous 0 1 1 0
Solution: E = E P H ° ⋅ + → + + − − 0 06 2 2 2 2 2 . log [H ] (assuming H e H ) n − = − ⋅ ⇒ = + + − 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 + − + + + ( )M C x x \u001e ⇀ \u001e ↽ \u001e \u001e h x c = = = − 10 0 04 4 3 1 40 . % or\n8.52 Chapter 8 HINTS AND EXPLANATIONS
Solution: E E = − ⋅ ° + 0 06 2 1 . log [Cu ] 0 31 0 34 0 06 2 1 0 1 2 2 . . . log [Cu ] [Cu ] . M = − ⋅ ⇒ = + + ∴ [OH ] . − + − − = = = ⇒ = K sp H Cu p 2 19 9 10 0 1 10 5
Solution: E E E Fe Fe Fe Fe Fe Fe 3 2 3 2 3 2 3 2 3 0 04 2 0 44 + + + + ° ° ° = × − × − = × − − × − | | | ( . ) ( . ) 1 1 0 76 = . V Now, E E Fe Fe Fe Fe 3 2 3 2 0 06 1 2 3 + + + + = − ° + + | | . log [Fe ] [Fe ] or, 0 7 8 0 76 0 06 5 2 3 2 3 . . . log [Fe ] [Fe ] [Fe ] [Fe ] 1 = − ⇒ = + + + +
Solution: 2Fe 3+ + 2I – → 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; ° = − = 0 77 0 53 0 24 . . . K eq = 10 8 Now, 10 0 5 1 0 5 10 8 2 2 2 5 = × ⇒ = × − ( . ) ( . ) C C M
Solution: Assuming the cell as concentration cell, net cell reaction is Ag C Ag C sp sp(AgI) + − + = ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ⎯ → ⎯ = C K K 1 2 (AgCl) [ l ] ( ) and E C C cell O = − 0 06 1 2 1 . log or, 0.102 = – 0.06 ⋅ log . . [Cl ] [Cl ] 8 1 10 1 8 10 4 10 17 10 4 × × ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ⇒ = × − − − − − M
Solution: n n eq eq Pb Tl = + Q F ⇒ × × + × × = × 5 0 70 100 208 2 5 0 30 100 204 1 1 1 96500 . . . t ∴ t = 3597.4 sec ; 1 hr
Solution: n Q F x x eq r I = ⇒ × = × × ⇒ = 0 36 192 0 075 2 3600 96500 3 . . Now, x + 6(–1) = y ⇒ y = –3
Solution: 2 2 2 2 2 NaCl + 2H O NaOH H Cl electrolysis ⎯ → ⎯⎯⎯⎯ + + 2 2 2 NaOH + Cl NaCl NaClO H O ⎯ → ⎯ + + n NaClO formed = n Cl 2 produced from electrolysis or, ( . ) . . . 10 10 1 0 7 45 100 74 5 2 2 5 96500 3 × × × × = × t ( n -factor of Cl 2 in electrolysis) ∴ t = 7.72 × 10 5 sec = 8.93 days ≈ 9 days
Solution: Cathode: Cu 2+ + 2e – Cu Anode 2H 2 O O 2 + 4H + + 4e – Moles of e – used = Q F = × × × − 0 161 5 60 96500 5 10 4 . \u001a Eq. of Cu 2+ present = 500 0 1 1000 2 0 1 × × = . . ( ) excess ∴ Moles of H + produced = Moles of e – = 5 × 10 –4 ∴ [H + ] fi nal = 5 10 500 1000 10 4 3 × × = − − M ⇒ P H = 3.0
Solution: cathode: Cu 2+ + 2e – Cu 250 0 1 1000 × . 5 1351 96500 × = 0.025 mole = 0.07 mole As Cu + will not remain fi nally in solution, no complex formation.
Solution: n eq metal = n eq Cl 2 ⇒ 52 8 9 08 22 7 2 . . . E = × ⇒ E = 66 At. wt. (approx) = 6 4 0 032 200 . . = ∴ valency At wt Eq.wt = = . . 200 66 3 \u001a\n8.53 Electrochemistry HINTS AND EXPLANATIONS
Solution: Initial mass of H 2 SO 4 , w 1 = 2000 1 1 16 100 352 × × = . gm Final mass of H 2 SO 4 , w 2 = 2000 1 42 40 100 1136 × ×= . gm Now, n eq H 2 SO 4 produced = Q F or, ( ) 1136 352 98 1 965 9 3600 96500 − × = × × ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ i ⇒ i = 2A
Solution: Ionic mobility, μ λ = ° = m F Speed of ion Potential gradient or, 7 5 10 96500 1 93 0 12 3 . / . / . × = × − distance 20 3600 ⇒ distance = 0.09 m = 9 cm
Solution: n eq Ag oxidised = Q F ⇒ w 108 1 9 65 1 3600 96500 × = × × . ∴ w = 38.88 gm ∴ Mass of anode dissolved = 38 88 100 60 64 8 . . × × gm ∴ Final mass of anode = 72.8 – 64.8 = 8 gm
Solution: Cathode: 2H 2 O(l) + 2e – H 2 (g) + 2OH – (aq) Anode: 2ce – (aq) Cl 2 (g) + 2e – n eq OH– produced = Q F ⇒ n OH − × = × = − 1 9 65 10 96500 10 3 . ∴ [OH – ] fi nal = 10 100 10 3 5 − − = M ⇒ P H = 9.0 Four digit integer type
Solution: Λ° = ° + ° − m m m (AgCl) (AgCl) (Cl ) λ λ = × + × = × − − − − − 6 19 10 7 81 10 14 00 10 3 3 3 1 2 1 . . . ohm m mol Now, Λ m C S = ⇒ × = × − − κ 14 10 2 8 10 3 4 . ⇒ S = 0.02 mol m –3 = 2 × 10 –5 M = 287 × 10 –5 g/L
Solution: [S ] K [H S] [ ] . ( ) 2 1 2 2 2 8 13 3 16 10 10 0 1 10 10 − + − − − − = ⋅ ⋅ = × × = a a K H M ∴ [Ag ] [S ] + − − − − = = × = × K M sp 2 48 16 16 4 10 10 2 10 Now, E Ag /Ag + = − × − 0 80 0 06 1 1 2 10 16 . . log = –0.142 V ∴ Required potential = 142 mV
Solution: 2Hg(l) + 2Fe 3+ \u001f Hg Fe 2 2 2 2 + + + excess 10 –3 M Equ. 10 –3 – x x 2 x = × − 10 100 10 3 M ∴ x = 9 × 10 –4 Now, K eq = × − = × × × × = × − − − − − 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+ ° = ° − ° = 0 06 2 . log or, 0 7724 0 06 2 9 10 2 3 4 . . log − ° = ⋅ × − E Hg /Hg 2 2+ ∴ E V Hg /Hg 2 2+ ° = 0 815 .
Solution: The cell reaction is 6 14 6 2 7 3 3 Fe Cr O H Fe Cr H O 2+ 2 7 2 2 + + → + + − + + + E E n cell cell = ° − ⋅ + + + + + 0 06 3 6 3 2 2 6 2 7 2 8 . log [Fe ] [Cr ] [Fe ] [Cr O ][H ] = − − × × × ( . . ) . log ( . ) ( ) ( . ) ( ) 1 35 0 77 0 06 6 0 75 4 0 75 2 1 6 2 6 8 = 0.571 V
Solution: Cell reaction: H 2 (g) + 2Ag + → 2H + + 2Ag(s) E E P cell cell H 2 = ° − ⋅ + + 0 06 2 2 2 . log [H ] [Ag ]\n8.54 Chapter 8 HINTS AND EXPLANATIONS or, 0.50 = 0.80 – 0.03 log 1 1 2 2 × + [Ag ] ⇒ [Ag + ] = 10 –5 M ∴ Mass of Ag in alloy = × ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ × = × − − 250 10 1000 108 2 7 10 5 4 . gm ∴ % of Pb in alloy 2 7 10 2 7 10 2 7 10 100 90 3 4 3 . . . % × − × × × = − − −
Solution: E K cell eq ° = 0 06 . log n ⇒ (0.2 – 0.08) = 0 06 1 . log K eq ⇒ K eq = 100
Solution: Left electrode: Mn(s) → Mn 2+ + 2e – Right electrode: 2H 2 O(l) → O 2 (g) + 4H + + 4e – ∴ Net cell reaction: 2Mn (s) + 2H 2 O(l) → 2Mn e+ + O 2 (g) + 4H + Now, E E Mn P cell cell 2+ O = ° − ⋅ ⋅ + 0 06 4 1 2 4 2 . log [ ] [H ] = − − − × × [ . ( . )] . log ( . ) ( . ) . 1 229 1 185 0 06 4 0 001 0 01 0 25 1 2 4 = 2.633 V
Solution: Δ S nF E T P = ⋅ ∂ ∂ ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = × × = 2 96500 0 001 193 . J/k
Solution: Δ Σ Δ Σ Δ r f f G nFE G G ° = − ° = ° − ° Products Reactants or, − × × = × ° ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ − × + × + × − + × − 12 9600 2 5 1000 4 4 0 3 0 6 280 4 4 . [ ( ) (OH) Δ f G Al ( ( . )] − 156 25 ∴ Δ f G ° = − − Al KJ/mol (OH) 4 1300
Solution: n eq MnO 2 = Q F ⇒ 8 7 87 1 3 99 10 96500 3 . . × = × × − t ∴ t = 2.418 × 10 6 sec ≈ 28 days
Solution: SnCl 2 Sn 2+ + 2Cl – Cathode: Sn 2+ + 2e – Sn Anode: 2Cl – Cl 2 + 2e– 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 – (0.01 + 0.01) = 0.08 ∴ m m SnCl SnCl 2 4 0 08 190 0 01 261 1520 261 = × × = . .
Solution: n eq Au = Q F ⇒ × × × × = × − 80 8 0 10 19 7 3 197 2 4 96 500 4 . . . t ∴ t = 772 sec
Solution: AgBr(s) \u001f Ag + + Br – (10 –7 + x )M x M Now, (10 –7 + x ) x = 12 × 10 –14 ⇒ x = 3 × 10 –7 Final solution: [Ag + ] = 4 × 10 –7 M, [Br – ] = 3 × 10 –7 M; [ ] NO M 3 7 10 − − = Now, κ solution = λ ° m (Ag + ) × [Ag + ] + λ ° m (Br – ) × [Br – ] + λ °m ( ) [ ] NO NO 3 3 − − × = 6 × 10 –3 × (4 × 10 –7 × 10 3 ) + 8 × 10 –3 × (3 × 10 –7 × 10 3 ) + 7 × 10 –3 (10 –7 × 10 –3 )
Solution: C 60 + 60O 2 60 CO 2 Moles of O 2 needed = 60 60 96 60 12 8 60 × = × × = n C n eq O 2 = n eq azobenzene ⇒ 8 × 4 = w 182 8 × ⇒ w = 728 gm + 8H + + 8e – + 4H2O N N NO2 2
Solution: Λ° eq[Ba (PO ) ] 3 4 2 = 160 + 140 – 100 = 200 Ohm –1 cm 2 eq –1 Now, ∧ ° eq = ∧ eq = κ C ⇒ 200 1 2 10 5 5 = × − . ∴ S = 6 × 10 –8 eq/cm 3 = 6 × 10 –5 N = 10 –5 M ∴ K sp = 108 S 5 = 10 8 × 10 –25 M 5
Solution: a b c ≠ ≠ = = = ° , α β γ 90 ⇒ Orthorhombic
Solution: Volumeof metal taken cm 3 = = = m d 100 6 25 16 . Volumeof each unit cell cm cm 3 = × ( ) = × − − 4 10 64 10 8 3 24 ∴ Number of unit cells = × = × − 16 64 10 2 5 10 24 23 .
Solution: d Z M N V = ⋅ ⋅ A ⇒ 45 16 27 6 10 4 10 23 8 3 = × × ( ) × × ( ) − Z ⇒ Z = 4 Hence, Al crystal is FCC. For FCC: 2 4 a r = ⇒ r = × = 2 4 0 4 1 414 . . Å
Solution: d Z M N V FCC A 3 gm/cm = ⋅ ⋅ = × × ( ) × × ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = − 4 56 6 10 4 125 10 2 8 45 23 10 3 . d Z M N V BCC A 3 gm/cm = ⋅ ⋅ = × × ( ) × × ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = − 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
Solution: Packing fraction = × ( ) × × × ( ) ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = − 6 10 4 3 144 10 108 10 6 0 7 23 10 3 π . . 3 36 Hence, the crystal should be FCC.
Solution: d Z M N V = ⋅ ⋅ A ⇒ 10 5 4 198 5 10 8 3 . = × × × ( ) − N A ⇒ N A = × 6 034 10 23 .
Solution: Void space per unit cell = 0.26 × V unit cell = × ( ) = 0 26 4 16 64 3 3 . . Å Å
Solution: d Z M N V = ⋅ ⋅ A ⇒ 12 5 4 6 10 4 100 2 10 2 23 10 3 . gm cm gm cm 3 3 = × × ( ) × × × ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ − M ∴ M = 120 ⇒ Metal is Sn.
Solution: 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 . . / = × × ( ) × × − Z Z M π ⇒ M = 8 π
Solution: Fraction of edge covered by atoms = = = 2 2 4 2 0 707 r a r r / . Hence, fraction of edge not covered = 0.293
Solution: d Z M N V = ⋅ ⋅ A ⇒ 8 3 4 6 10 5 10 23 8 3 = × × ( ) × × ( ) − M ⇒ M = 50 gm/mol At 0°C, the substance will exist as gas and its density = = 50 22 4 2 23 gm L g/L . .
Solution: M HCl MCl 1 2 H 2 + → + 1 1 2 mole mole = A gm = × ° 1 2 22 7 0 1 . L at C and bar ∴ (7.68 × 4.5) gm 11 35 7 68 4 5 4 54 . . . . A × × ( ) = ∴ A = 86.4 Now, d Z M N V = ⋅ ⋅ A ⇒ 4 5 86 4 6 10 400 10 23 10 3 . . = × × ( ) × × ( ) − Z ∴ Z = 2 ⇒ Unit cell is BCC.
Solution: Percentage of occupied space in water = × = x x 0 99 0 96 33 32 . . ∴ Percentage of empty space in water is 100 33 32 − x .
Solution: (Volume of crystal containing one mole metal) × 70 100 = × ( ) × × × ( ) − 6 10 4 3 0 2 10 23 7 3 π . cm ∴ V crystal cm = 64 7 3 π ∴ Density gm/cm = ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = 32 64 7 3 5 3 π π .
Solution: 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.
Solution: 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
Solution: Packing is FCC for which 2 4 a r = . ∴ r = × = 2 10 4 2 5 2 Å Å . Now, density of metal atom = m V atom atom = × ( ) × × × ( ) = − 60 22 6 022 10 4 3 2 5 2 10 0 54 23 8 3 . . . . π gm/cm 3
Solution: P.F. particle unit cell = = × × × ( ) ⎧ ⎨ ⎪ ⎩ ⎪ ⎫ ⎬ ⎪ ⎭ ⎪ × = V V r r r 3 4 3 6 3 4 2 2 3 3 3 2 π π
Solution: 1 2 4 3 3rd layer 2nd layer 1st layer 1 4 3 2
Solution: A B A 0.V 0.V 0 h 3 h 4 h 2 h 4
Solution: Number of NaCl formula units in 1 gm = × × ( ) 1 58 5 6 10 23 . Each unit cell contains 4 NaCl formula units and hence, volume of crystal = × × × × = − 6 10 58 5 4 4 7 10 0 12 23 23 . . . ml
Solution: d Z M N V = ⋅ ⋅ = × × ( ) × × ( ) = − A 3 gm/cm 4 58 5 6 10 600 10 1 8 23 10 3 . .
Solution: The structure is simple cubic for both metals.
Solution: ‘B’ should occupy all the tetrahedral voids and hence, its C.N. =
Solution: 25. P.E. A ( ) = × + × ( ) ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = 4 4 3 8 4 3 0 225 4 2 0 76 3 3 3 π π r r r . . P.E. B ( ) = × + × ( ) ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = 4 4 3 4 4 3 0 414 4 2 0 79 3 3 3 π π r r r . . P.E. C ( ) = × + × ( ) ( ) = 1 4 3 1 4 3 0 732 2 0 72 3 3 3 π π r r r . .
Solution: Z Z x Z O X Y 2 2+ 3+ − = = × = × = 4 8 100 4 50 100 2 ; ; For neutrality of crystal, 4 2 8 100 2 2 3 0 × − ( ) + × + ( ) + × + ( ) = x ⇒ x = 12.5
Solution: r r x Rb Cl pm + − + = = 328 5 . r r y r r z r r w K Cl Na Br K Br pm pm pm + − + − + − + = = + = = + = = 313 9 298 1 329 3 . . . ∴ r r x w y Rb Br pm + − + = + − = 343 9 .\n9.32 Chapter 9 HINTS AND EXPLANATIONS
Solution: r r + − = 0 414 . ⇒ r − = = 200 0 414 483 1 . . pm
Solution: Z Z Z A B C = = × = = × = 4 2 1 2 1 4 1 4 1 ; ; ∴ Formula = A 4 BC
Solution: Z Z Z O Metal I M Metal II N = = × = = × = ( ) ( ) 4 1 8 8 1 1 2 4 2 ; ∴ Formula = MN 2 O 4 For neutrality, M Zn and N Al 3+ = = + 2
Solution: Z M 3+ If M are at corners = × = ( ) + 8 1 8 1 3 Z X If F are at face centres − = × = ( ) − 6 1 2 3
Solution: Octahedral voids in FCC = 4, but only one is occupied by ‘ x ’ and one by ‘ y ’.
Solution: (II) r r + − = = − ( ) ⇒ = 20 95 0 21 0 155 0 225 3 . . . C.N.
Solution: 0.V. T.V. T.V. Fraction of body diagonal covered = + × + × × ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = 2 2 0 414 4 0 225 3 4 2 0 76 r r r r . . . ∴ Fraction, not covered = 1 – 0.76 = 0.24
Solution: For electrical neutrality, A = bivalent and B = trivalent. Now, one octahedral void is occupied by ‘A’ and one by ‘B’.
Solution: Frenkel defect does not change the density.
Solution: ln f f H R T T 2 1 1 2 1 1 = − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ Δ or, ln / / 1 2 10 1 10 1 1100 1 1200 9 10 × ( ) ( ) = − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ Δ H R ⇒ Δ H = 176 8 . KJ/mol
Solution: 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
Solution: Shortest distance between two T.V. is a 2 .
Solution: Both have same packing efficiency and C.N.
Solution: There are the voids per sphere in hexagonal close packing.
Solution: For octahedral void, the orientation of both tetrahedral voids should be opposite to each other.
Solution: a r r = + ( ) + − 2 Na Cl
Solution: Informative\n9.33 Solid State HINTS AND EXPLANATIONS
Solution: (a) P P c K b H = + + ( ) 7 1 2 log ⇒ 5 0 7 1 2 4 4 . . log = − + ( ) c ∴ c M M = = ⇒ = 0 4 80 2 100 . / (b) h K K c h b = × = × × = × − − − 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 = + ( ) ⇒ = × + ( ) = + − . pm (d) d M N V = ⋅ = × × × × ( ) = − Z gm/cm A 3 1 100 6 10 400 10 2 6 23 10 3 .
Solution: (a) Z Z Z C A B 3 2 4 4 1 2 8 4 − + + = = = × = , , Crystal is electrically neutral and hence, it is possible. (b) Z Z Z B A C 2 3 6 6 1 2 12 6 + + − = = = × = , , Crystal is electrically neutral and hence, it is possible. (c) Z Z Z A B C + + − = × = = × = = 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 + − + = = = , , Crystal is negatively changed and hence, it is not possible.
Solution: CaF Al O 2 2 3 8 4 6 4 : , : ( ) ( )
Solution: Metal defi cient defect can occur with extra anion present in the interstitial voids, but it is very rare.
Solution: Δ H = + ve and hence, the surroundings must lose heat.
Solution: O CCP 2 − = Fe O.V. 2 1 4 1 + = × = Fe in T.V. and 1 in O.V. 3 1 + =
Solution: 3 4 2 3 5 0 2 4 33 a r r = ⇒ = × = . . Å
Solution: Next nearest neighbours are at ‘ a ’ distance.
Solution: Number of next nearest neighbours = 6
Solution: d Z M N V = ⋅ ⋅ = × × ( ) × × ( ) = − A 3 gm/cm 2 39 6 10 5 10 1 04 23 8 3 .
Solution: Fractional space occupied by atoms = = V V V V atoms liquid atoms/mol liquid/mol = × ( ) × × × × ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ( ) = − 6 10 4 3 3 5 10 4 39 0 9 0 5883 23 8 3 π / . . ∴ Percentage of empty space = (1 – 0.5883) × 100 = 41.17 %\n9.34 Chapter 9 HINTS AND EXPLANATIONS Comprehension II
Solution: 2 4 2 0 5 2 4 0 125 a r r = ⇒ = × = . . nm Now, size of octahedral void = 0.414 × 0.0125 = 0.052 nm
Solution: Size of tetrahedral void = 0.225 × 0.0125 = 0.028 nm Comprehension III
Solution: Volume Mass Density cm 3 = = × × ( ) × = × − 6 24 6 10 1 92 1 25 10 23 22 . .
Solution: Volume occupied by particles = × π 3 2 V unit all or, 6 4 3 3 2 1 25 10 1 5625 10 3 22 8 × = × × ⇒ = × − − π π r r . . cm
Solution: Height of unit cell = ⋅ = 4 2 3 5 r Å
Solution: Number of nearest neighbours = 12 Comprehension IV
Solution: Number of T.V. per particle = 2 Number of O.V. per particle = 1
Solution: T.V. are smaller than O.V. Comprehension V
Solution: a a r r r r r r r r r r KCl NaCl K Cl Na Cl K Na Cl Na Cl N = + ( ) + ( ) = + + + − + − + + − + − 2 2 1 a a + = + + = 1 0 7 1 0 5 1 1 0 5 1 143 . . . .
Solution: d d a a NaCl KCl NaCl KCl = = × ( ) = 58 5 74 5 58 5 74 5 1 143 1 172 3 3 3 . . . . . . Comprehension VI
Solution: Centre is octahedral void.
Solution: Number of O.V. = 4, but only one is occupied.\n9.35 Solid State HINTS AND EXPLANATIONS Comprehension VII
Solution: C.N. of Zn 2+ = 4
Solution: C.N. of S 2 − = 4
Solution: Zn 2+ is tetrahedrally linked with 4 S 2 − ions.
Solution: r r + − > 0 225 . Comprehension VIII
Solution: Z Mn = × = 8 1 8 1 Z F = × = 12 1 4 3 ∴ Formula = MnF 3
Solution: a r r = + ( ) = + ( ) = + − 2 2 0 65 1 35 4 00 3 Mn F . . . Å
Solution: d = × + × ( ) × ( ) × × ( ) = − 1 55 3 19 6 10 4 10 2 92 23 8 3 . gm/cm 3 Comprehension IX
Solution: V m d formula unit 3 cm = = + ( ) × × = × − 132 5 35 5 6 10 3 5 8 10 23 23 . . .
Solution: d Z M N V A = . . ⇒ 3 5 1 168 6 10 23 3 . = × × ( ) × a ⇒ a = × − 4 3 10 8 . cm Hence, nearest Cs – Cs distance = a = 4.3 Å
Solution: Nearest Cs Cl distance − = = 3 2 3 72 a . Å Comprehension X
Solution: N N e e O E RT = = = × − − × × × − 2 46 10 2 2 1000 5 3 1 0 10 .
Solution: In NaCl, all O.V. are occupied. Hence, the available voids are only tetrahedral. ∴ N i = 2 × N o Now, N N N e o o E RT = × ⋅ − 2 2 / ∴ N N e e o E RT = × = × = × − − × × × − 2 2 1 41 10 2 73 6 1000 2 2 1000 8 / . . Comprehension XI
Solution: d Z M N V = ⋅ ⋅ A = × × ( ) × ⋅ × ( ) = = − 4 6 023 6 023 10 2 10 1 200 5 23 1 3 7 3 . . / Y Y gm/cm kg/m 3 3
Solution: d observed >> d theoretical Such large diff erence is possible due to impurity defect.\n9.36 Chapter 9 HINTS AND EXPLANATIONS Comprehension XII
Solution: For diamond crystal, 3 8 a r = and Z = 8 Now, d Z M N V = ⋅ ⋅ A ∴ 3 6 8 12 6 10 8 3 23 3 . = × × ( ) × ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ r ∴ r = 0.76 × 20 –8 cm
Solution: For SiC, 3 4 a r r c si = + ( ) and Z = 4 Now, d Z M N V r r c si = ⋅ ⋅ ⇒ = × × ( ) × + ( ) ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ A 3 2 4 40 6 10 4 3 23 3 . d Z M N V r r c si = ⋅ ⋅ ⇒ = × × ( ) × + ( ) ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ A 3 2 4 40 6 10 4 3 23 3 . ∴ r si = × − 1 12 10 8 . cm
Solution: Interchange between C and si atoms will neither change ‘ Z ’ nor ‘ a ’.
Solution: For the same volume, m m dia sic 3 6 3 2 . . = or, n n n n c sic sic c × = × ⇒ = 12 3 6 40 3 2 1 3 75 . . .
Solution: Packing efficiency of SiC is greater due to unequal size of particles.
Solution: Voids are named according to the orientation of spheres constituting the voids.
Solution: Relative increase in packing efficiency is high when larger voids are occupied.
Solution: BCC : 2 3 2 r a = FCC : 2 2 2 r a =
Solution: Density depends on mass and size of particles but not the packing efficiency.
Solution: Packing efficiency of diamond is only 0.34.
Solution: Δ Δ H S = + = + ve ve , Hence for –ve Δ G , the temperature should be high.
Solution: Theoretical (Cubic crystals)
Solution: Informative (Ionic solids)
Solution: Theoretical (Classification of solids)
Solution: Theoretical (Classification of solids)\n9.37 Solid State HINTS AND EXPLANATIONS
Solution: Theoretical (Classification of solids)
Solution: CaCl : 3 2 a r r = + ( ) + − CaF and 2 3 4 2 4 : . a r r a r = + ( ) = + − + Diamond : 3a r = 8 Nacl: + + – – = √ 2 a 2 a √ 2
Solution: Theoretical (Ionic solids)
Solution: Informative (Basic crystal system)
Solution: Informative (Basic crystal system)
Solution: CaF and 2 2 2 2 2 : d a d a F F Ca Ca − − + − − − = = NaCl Na Na : d a + − − = CsCl Cs Cs : d a + + − = d a T.V. from corner = 3 4 Na O and 2 Na Na O O 2 : d a d d a + + − − − = − = 2 2 2
Solution: SC Nearest Next nearest : , = = a a 2 BCC : = = 3 2 a a FCC : = = 2 2 a a
Solution: Volume cm cm , . . l m d l 3 3 58 5 2 167 27 3 = = = ⇒ =
Solution: d Z M N V Z Z = ⋅ ⋅ ⇒ = × × × × ( ) ⇒ = − A 4 72 6 10 0 493 10 4 23 7 3 .
Solution: d d l s CH CH A Z M N V 4 4 ( ) ( ) = = ⋅ ⋅ or, 0 5 16 6 10 0 6 10 4 23 7 3 . . = × × × × ( ) ⇒ = − Z Z
Solution: d Z M N V = ⋅ ⋅ A ⇒ 0 92 18 6 10 2 3 4 4 53 10 7 41 10 23 8 2 8 . . . = × × × × × × ( ) × × ⎧ ⎨ ⎪ ⎩ ⎪ ⎫ ⎬ ⎪ ⎭ ⎪ − − Z ∴ Z = 4
Solution: d Z M N V = ⋅ ⋅ A ⇒ 2 4 12 6 10 6 3 4 10 2 3 10 23 8 2 8 . = × × × × ⋅ × ( ) × × ⎧ ⎨ ⎪ ⎩ ⎪ ⎫ ⎬ ⎪ ⎭ ⎪ − − Z x ∴ x 2 200 108 =
Solution: 2 4 a r = ⇒ 2 2 2 1 r a = = nm
Solution: + + – – r 2 2 186 2 214 2 400 r = + ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = pm pm\n9.38 Chapter 9 HINTS AND EXPLANATIONS
Solution: d Z M N V = ⋅ ⋅ A = × × × × ( ) = − 4 58 5 6 10 500 10 3 12 23 10 3 . . gm/cm 3 ∴ Percentage vacancy = − × = 3 12 2 964 3 12 100 5 . . . %
Solution: (ii), (iv), (v), (vi) , (vii) are true statements.
Solution: d theo 3 gm/cm = × × × × ( ) = − − 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 ∴ Moles of metal missing per litre = − = 1670 1607 5 31 25 2 . .
Solution: Z Z ZnS ZnS but due to defect, = = 4
Solution: 12. Octahedron has eight triangular faces. Hence, truncated octahedron will have eight hexagonal faces.
Solution: BCC: Fraction of edge covered by atom = 2 r a . ∴ Fraction of edge uncovered = − = − × = 1 2 1 2 3 4 0 134 r a . From question: 0.134 a = 67 pm ⇒ a = 500 pm Now, d Z M N V = ⋅ ⋅ = × × × × = − A 3 gm/cm 2 75 6 10 500 10 2 23 10 3 ( ) ( )
Solution: 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 − × + × = × ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ × x N y N N A A A or, 30 x – 11 y = 360 (1) and ( ) ( . ) . 100 30 11 1 0 001 0 001 3000 30 11 999 − × × = − ⇒ − = x N y N x y A A (2) From (1) and (2), y x × = 100 2%
Solution: In truncated octahedron, all corner of octahedron become square faces and hence, its number =
Solution: And, the number of c-atoms per unit cell in diamond =
Solution: Four-digit Integer Type
Solution: m d N r × = × 0 74 4 3 3 . A π or, 197 19 7 0 74 6 10 4 3 23 3 . . × = × × × π r ⇒ r = × = − 1 43 10 143 8 . cm pm
Solution: P.E. of diamond and here silicon is 0.34.
Solution: d Z M N V = ⋅ ⋅ A ⇒ 1 5 10 6 6 10 12 5 8 0 3 0 10 3 23 9 3 . . . . × = × × ( ) × × × × ( ) ⎧ ⎨ ⎪ ⎩ ⎪ ⎫ ⎬ ⎪ ⎭ ⎪ − M ⇒ M = 45 Kg/mol
Solution: Let the percentage of fayalite be x . V V V olivine fayalite fosterite = + or, 100 3 9 4 2 100 3 3 . . . = + − x x ⇒ x = 71.79
Solution: d Z M N V = ⋅ ⋅ A ⇒ 2 14 4 426 18 6 10 1 26 10 23 7 3 . . = × + ( ) × ( ) × × ( ) − x ⇒ x = 12\n9.39 Solid State HINTS AND EXPLANATIONS
Solution: For diamond: 3 4 a d = × − C C Now, d Z M N V = ⋅ ⋅ A ⇒ 2 3 8 12 6 10 4 3 23 3 = × × ( ) × × ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ − d C C ∴ d C C cm pm − − = × = 1 55 10 155 8 .
Solution: The maximum backing efficiency of identical spheres in 2D is 0.90. Hence, 40 0 90 10 2 2 2 ( ) × = × × ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ . N π ⇒ N = 18
Solution: 4 2 520 . r Cl − = × ⇒ r Cl pm − = 182
Solution: Z Ti = × = 1 2 1 1 2 Z O = 1 (assume) ∴ Formula = Ti O T O i 1 2 1 2 / ≅ Now, Ti = + × = 48 48 32 100 60% and oxidation state of Ti = +4
Solution: For one litre crystal, m m m m initial Bremoved Cadded final − + = or, 4800 30 1 15 4795 − × + × = x ⇒ x = 2 3 ∴ Percentage of C-atoms which replaced B-atoms = × = x 1 100 67%
Solution: Percentage of body diagonal covered = + + ( ) ⎡ ⎣ ⎤ ⎦ × − + − 2 4 3 2 100 r r r r B C A B % = + × + ( ) × = − − − − 2 4 0 225 2 3 0 414 100 59 2 r r r r B B B B . . % . %
Solution: Out of 8 Fe 2+ in original crystal, 1 is missing. ∴ Percentage of cation vacancy = × = 1 8 100 12 5 . %
Solution: Percentage occupied space = × = × × × ( ) ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ × − V V particles solid 100 6 10 4 3 1 54 10 40 4 1 23 8 3 % . / π π 0 00 27 % % = ∴ Empty space = 73 %
Solution: For cubic void, r a = 0 732 . ⇒ a = = 366 2 0 732 250 / . pm Now, d Z M N V = ⋅ ⋅ = × × ( ) × × ( ) = − A 3 gm/cm 1 125 6 10 250 10 100 7 5 23 10 3 .
Solution: d Z M N V = ⋅ ⋅ = × × ( ) × × × × ( ) − A 2 28 6 10 667 500 280 10 23 30 = 1 gm/cm 3 = 1000 kg/cm 3
Solution: P K X N H N 2 2 = ⋅ ( ) Solution or 5 0 8 1 0 10 10 10 10 5 5 2 2 2 × = × × + × . ( . ) n n n N N N \u001f ∴ n N 2 4 10 4 = × −
Solution: p K X K n n m P V H H = ⋅ ≈ ⋅ ⇒ ⋅ gas liq gas liq α ∴ = ⋅ ⋅ ⇒ = × × ⇒ = 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 = ⋅ = ⋅ ⋅ gas liq liq ∴ Volume of gas dissolved, V RT K n H = ⋅ liq (Volume of gas dissolved is independent of pressure of gas) ∴ = ⇒ = ⇒ = 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
Solution: P X P Hg Hg = ⋅ = × × + × = × − − − total t 0 8 10 200 0 8 10 200 50 4 28 720 1 6 10 3 3 3 . . . . o orr
Solution: 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. ∴ + = − ⇒ = 20 200 10 100 33 33 x x x .
Solution: Y P P X P P A A A A = = ⋅ ° 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 = ⋅ ° ⋅ ° + − ° ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = ⋅ ° ° + ° − ° ° ( )
Solution: n n n n P P P Q P Q P Q n ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ⋅ ° ° ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 2 nd condense initial Where n = number of condensation steps. Now, n n P Q ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = × ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = final 1 1 300 100 9 1 2 ∴ = + = X P 9 9 1 0 90 .
Solution: The mole fraction of A in distillate, ′ = = ⋅ ° = × × + × = 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 = ′ ⋅ ° + ′ ⋅ ° = × + × = 5 17 100 12 17 80 85 88 . m m Kg
Solution: Let the final composition: liquid (10 mole): A = x mole, B = (10 – x ) mole Vapour (10 mole): A = (10 – x ) mole, B = x mole Now, Y Y X X P P x x x x A B A B A B = ⋅ ° ° ⇒ − = − ⋅ 10 10 200 100 ∴ x = 4.14 Now, p X P X P x x A A B B = ⋅ ° + ⋅ ° = × + − × 10 200 10 10 100 = 141.4 torr
Solution: 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 = ° + ° = + = ⇒ = . . . .
Solution: If the solution were ideal, P total mm kg = × + × = 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.
Solution: For ideal behavior, P total = 0.25 × 512 + 0.725 × 344 = 386 mm Hg < 600 mm Hg Hence, the solution shows positive deviation ⇒ Δ H mix = positive
Solution: For increase in temperature, the solution shows negative deviation ( Δ H = negative).
Solution: n n P P Chlorobenzene water Chlorobenzene water = ° ° or, x x x / . ( ) / . . . 112 5 100 18 9 031 10 7 031 10 7 031 10 64 4 4 4 − = × − × × ⇒ =
Solution: As the available surface area for solvent molecules decreases, the rate of vaporization decreases.
Solution: P P P n n m m m m ° − = ⇒ = ⇒ = 1 2 1 2 2 1 5 95 0 3 57 10 / / . M M
Solution: P P X P ° − = ⋅ ° 1 10 = 0.2 × P ° (1) 20 = X 1 × P ° (2) From (1) and (2): X 1 = 0.4 ⇒ X 2 = X solvent = 0.6
Solution: P P P n n ° − = ⇒ − = ⇒ = 1 2 3000 2985 2985 5 100 18 179 1 / / . M M
Solution: P P P n n ° − = ⇒ − = ⇒ = 1 2 0 85 0 845 0 845 0 5 39 78 169 . . . . / / M M\n10.32 Chapter 10 HINTS AND EXPLANATIONS
Solution: P X P = ⋅ ° = + × = 2 1000 18 1 1000 18 12 3 12 08 . . K Pa
Solution: P X P = ⋅ ° 2 2 8 90 18 30 90 18 . = + × ° M P and 2 9 108 18 30 108 18 . = + ⋅ ° M P ∴ = M 23
Solution: P P X P ° − = ⋅ ° = + × = 1 1 1 1000 18 760 13 44 . m m kg
Solution: P P P n n ° − = ⇒ − = ⇒ = 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 × = . and the number of H-atoms in each molecule = 178 5 6 100 1 10 × = . ∴ Hydrocarbon is C 14 H 10 .
Solution: Loss is mass of solvent, w 1 a ( P ° – P ) and gain is mass of absorbent, w 2 a P ° ∴ = ° − ° = ⇒ = + ⇒ = w w P P P X 1 2 1 0 05 2 05 40 40 100 18 288 . . M M M
Solution: C C M M M M M 1 2 10 20 6 67 30 1 3 = ⇒ + = + ⇒ = A B A B A B M .
Solution: As solution have same concentration, mixing will not change the total molar concentration.
Solution: Isopiestic refers to the same pressure. Blood is isotonic with 0.9 % ( w / v ) NaCl solution. ∴ Osmolarity = × ≈ 0 9 58 5 100 1000 2 0 31 . / . / . M
Solution: π = = × ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ × × = CRT atm 2 5 58 5 100 1000 2 0 0821 300 21 05 . / . / . .
Solution: Δ = ⋅ ⇒ = × ⇒ = T K m b b 0 104 0 52 2 98 1000 104 . . / / M M
Solution: Clausius–Clapeyron equation: dT dP RT H P = Δ = × × = 2 2 3 2 350 4 9 10 50 . ( ) . K/atm
Solution: Δ = ⋅ = × = T K m b b 0 52 72 180 0 208 . . K ∴ B.P. of solution = 373.208 K
Solution: Δ = ⋅ ⇒ − = × × T K m b b ( . . ) . . / 354 11 353 23 2 53 1 8 90 1000 M ∴ M = 57.5
Solution: If molality of solution is ‘ m ’, then P P P n n m m ° − = ⇒ − = ⇒ = 1 2 760 750 750 1000 18 20 27 / Now, Δ = ⋅ = × = T K m b b 0 52 20 27 0 385 . . K ∴ B.P. of solution = 100.385° C
Solution: K H b = Δ = × = 0 002 0 002 320 80 2 56 2 2 . ( ) . ( ) . T Cal/gm K/m o vap Now, Δ = ⋅ ⇒ = × ⇒ = T K m x x b b 0 32 2 56 6 4 32 200 1000 8 . . . / / ∴ Molecular formula of sulphur = S x = S 8
Solution: Δ = ⋅ = × + ≈ ° T K X b x b , / . solute C 32 1 128 1 128 94 94 0 25 ∴ B.P. of solution = 110.75 + 0.25 = 111° C
Solution: Δ = ⋅ ⇒ = × ⇒ = T K m m m b b 1 04 0 52 2 . . Now, P P P n n P P ° − = ⇒ ° − = ⇒ ° = 1 2 750 750 2 1000 18 777 / torr
Solution: Δ = ⋅ ⇒ = × ⇒ = T K m f f 0 93 1 86 7 93 1000 150 5 . . / / . M M\n10.33 Liquid Solution HINTS AND EXPLANATIONS
Solution: Δ = ⋅ ⇒ = × ⇒ = 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 ⇒ x = 2 ∴ Molecular formula = C 2 H 4 O 2
Solution: Δ = ⋅ ⇒ = × ⇒ = T K m m m f f 15 1 86 8 06 . .
Solution: Δ Δ = ⇒ Δ = ⇒ Δ = ° T T K K T T f b f b f f 0 78 1 86 0 52 2 79 . . . . C ∴ F.P. of solution = –2.79° C
Solution: Δ + Δ = + ⋅ T T K K m f b f b ( ) or 4.76 = (1.86 + 0.52) × w w / / . 342 100 1000 68 4 ⇒ = gm
Solution: Δ = ⋅ ⇒ = × ⇒ = T K m w w f f 5 8 5 120 425 1000 30 . / / gm
Solution: Δ Δ = ⇒ Δ = ⇒ Δ = ° T T K K T T f b f b b b 0 7 5 17 5 0 2 . . . C ∴ B.P. of solution = 90.2°C
Solution: If the molarity of solution is m , then P P P n n m m ° − = ⇒ = ⇒ = 1 2 2 100 1000 78 10 39 / Now, Δ = ⋅ ⇒ = × ⇒ = T K m K K f f f f 1 3 10 39 5 07 . . K/m
Solution: K K T H T H H H T T K K f b f b f b b f = ° Δ ° Δ ⇒ Δ Δ = ° ° ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ × 2 2 2 / / fus vap fus vap = ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ × = 280 350 2 5 5 6 2 7 2 . .
Solution: Δ = ⋅ ⇒ = × ⇒ = T K m K K f f f f 2 0 0 25 8 . . K/m
Solution: Δ = ⋅ T K m f f For cane sugar solution: 273 15 272 85 5 342 95 1000 . . − ( ) = ⋅ K f / / (1) For glucose solution: 273 15 5 180 95 1000 . / / − ( ) = ⋅ T K f f (2) From (1) and (2), T f = 272.58 K
Solution: Δ = ⋅ T K m f f For AB 2 solution: 2.55 = 5 1 1 2 20 1000 . / ( ) / × + x y (1) For AB 4 solution: 1.7 = 5 1 1 4 20 1000 . / ( ) / × + x y (2) ∴ Atomic mass of A = x = 50 Atomic mass of B = y = 25
Solution: Δ = ⋅ ⇒ = × ⇒ = T K m m m f f 0 744 1 86 0 4 . . . Now, π = CRT = 0.4 × 0.0821 × 300 = 9.852 atm
Solution: ϕ = Δ ⋅ = × = T K m f f 0 93 1 86 0 4 1 25 . . . .
Solution: Δ = ⋅ ⇒ = × ⇒ = T K m m m f f 1 0 1 80 1 1 8 . . . P X P = ⋅ ° = + × = 2 1000 18 1 1 8 1000 18 24 24 24 . . mm Hg
Solution: Δ = ⋅ T K m f f ∴ = × = × 0 2 100 1000 1000 . / / K x K x y f f and 0.25 ∴ Mass of ice separated out = 100 – y = 20 gm
Solution: 2 KI(aq) HgI K HgI aq particles added 2 4 particles ( ) ( ) ( ) ( ) 4 2 3 + ⎯ → ⎯ As the number of ions in solution decreases, osmotic pressure decreases.
Solution: Osmolarity of both solution should be equal ∴ × = × + ⇒ = 0 1 2 0 1 1 2 0 5 . . ( ) . α α
Solution: Osmolarity = 0.2 × 3 = 0.6 M
Solution: Δ = ⋅ ⇒ = × × ⇒ = T K m n n f f 3 72 1 86 1 0 2 . . . ∴ From each particle, two ions should form.
Solution: Δ = ⋅ ⇒ = × + − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ T K m K b b b 1 43 1 1 0 9 1 2 1 . . ∴ K b = 2.6 K/m
Solution: Δ = ⋅ + − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ T K m n f f 1 1 1 α ∴ = × + − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ ⇒ = 1 96 4 9 2 122 25 1000 1 1 2 1 0 78 . . / / . α α\n10.34 Chapter 10 HINTS AND EXPLANATIONS
Solution: HA H A M M =0.01M M ( . ) 0 1 − + + x x x \u001f ⇀ \u001f ↽ \u001f \u001f − ∴ Osmolarity = (0.1 – x ) + x + x = 0.11 M Now, π = CRT = 0.11 RT
Solution: Δ = ⋅ = × × + × = ° T K m f f 1 86 0 1 2 0 025 2 0 465 . [ . . ] . C ∴ F.P. of solution = – 0.465° C
Solution: The eff ective molality will be in the range of 0.2 to 0.3 and Δ T f = K b ⋅ m will be in the range of 1.86 × 0.2 = 0.372° C to 1.86 × 0.3 = 0.558° C.
Solution: Let the mixture contain x mole KCl and y mole NaCl. Then x × 74.5 + y × 58.5 = 3.125 (1) and ( ) . . . x y T K f f + × = Δ = = 2 0 186 1 86 0 1 (2) ∴ x y = 1 3
Solution: 2 2 0 2 A A Initial conc. nM ( M M Equilibrium conc. n x x − ) \u001f ⇀ \u001f ↽ \u001f \u001f Now, Δ = ⋅ = ⋅ − + ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ T K m K n x x b b b 2 ∴ x n T K b b = − Δ 2 2 Now, K A A x n x n T K T K n c b b b b = = ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ − = − Δ Δ − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ [ ] [ ] ( ) 2 2 2 2 2 2 = ⋅ − Δ Δ − ⋅ K n K T T n K b b b b b ( ) ( ) 2 2
Solution: Δ Δ = ⋅ ⋅ = × × = 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
Solution: Colloidal solutions have low value of any colligative property than the true solution of same composition.
Solution: If the complex dissociates into n ions, then Δ = ⋅ ⇒ = × × ⇒ ≈ T K m n n f f 0 0054 1 86 0 001 3 . . [ . ]
Solution: XCl X Cl M M 3 3 3 3 (s) \u001f ⇀ \u001f ↽ \u001f \u001f + − + S S P P P n n ° − = ⇒ − = ⇒ = × − 1 2 2 17 25 17 20 17 20 4 1000 18 4 04 10 . . . / . S M S
Solution: (a) Δ = ⋅ = × = ° T K m f f 1 86 1 1 86 . . C ∴ F.P. of solution = – 1.86° C (b) Δ = ⋅ = × = ° T K m b b 0 52 1 0 52 . . C ∴ B.P. of solution should be 100.52° C. As the solute dissociates completely above 100.26° C, its actual Δ = × = ° T b 0 52 2 1 04 . . C and hence, B.P. = 101.04° C. (c) Δ = ⋅ ⇒ = × T K m m f f 7 44 1 86 1 2 . . / solvent (as solute dimerizes) ∴ m Solvent left = 0.125 kg ∴ Percentage of water separated as ice = (1 – 0.125) × 100 = 87.5 % (d) Δ = ⋅ ⇒ = × T K m m b b 2 08 0 52 2 . . solvent (Complete dissociation) ∴ m Solvent left = 0.5 kg Percentage of water evaporated = (1 – 0.5) × 100 = 50 %
Solution: At constant temperature, the vapour pressure may be changed by changing the composition.
Solution: 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
Solution: A, C, D shows negative deviation.
Solution: (a) Mass percent of A = 50 ⇒ n A : n B = 1 : 2 ⇒ Azeotrope Hence, vapour will have the same composition of liquid. (b) Mass percent of A > 50 ⇒ n A : n B = 1 : 2 L L V 0.0 1.0 V ° A ρ ° B ρ mole-fraction of B ρ 2 3 In this case, the vapour must be more rich in A than liquid. (c) X B = = > 3 4 0 75 2 3 . ⇒ Pure A cannot be obtained. (d) X B = = < 3 5 0 60 2 3 . ⇒ Pure A cannot be obtained in traces.
Solution: (a) On changing the solvent, K f will change.
Solution: F. P. and V. P. will become lower for X .
Solution: (a) P P P n n n n ° − = ⇒ − = ⇒ = 1 2 2 2 760 740 740 1 37 ∴ Moles of water separated as ice = 200 – 37 = 163 (b) Δ = ⋅ = × × = × T K m f f 2 0 1 37 18 1000 2000 37 18 . ( ) / K ∴ = − × ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ T K 273 2000 37 18 (c) For original solution: Δ = ⋅ = × × T K m f f 2 1 200 18 1000 ( ) / ∴ F. P. = 0 10 18 − Δ = − ° T C f (d) For final solution: P P P X ° − ° = = + = 1 1 1 37 1 38
Solution: 0.0 1.0 ° T A B T ° T B
Solution: Y X X P P X P P X A A A A B A A B A − = ⋅ ° ° + ⋅ ° − ° − ( ) = ° − ° − ° − ° ° + ⋅ ° − ° = 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 − − = 0 ∴ X P P P P P A A B B A B = ° ⋅ ° − ° ° − °
Solution: P P X P P P P B A A B A B total = ° + ° − ° = ° ⋅ ° ⋅ ( )\n10.36 Chapter 10 HINTS AND EXPLANATIONS Comprehension II
Solution: 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 = ° + ° = + ⇒ = > / . / . . . As the applied pressure is less than equilibrium pressure, the system must be 100 % vapour.
Solution: First drop of liquid will form at 0.5 bar.
Solution: P 1 V 1 = P 2 V 2 ⇒ 0.3 × 10 = 0.5 × V 2 ⇒ V 2 = 6.0 dm 3
Solution: P X P X P X X A A B B A A total = ⋅ ° + ⋅ ° ⇒ = × + − × 0 5 0 4 1 0 6 . . ( ) . ∴ X A = 0.5
Solution: Liquid composition: A ≈ 2 mole, B ≈ 3 mole ∴ P total bar = × + × = 2 5 0 4 3 5 0 6 0 52 . . .
Solution: Y X P P A A A = ⋅ ° = × = total 2 5 0 4 0 52 4 13 . .
Solution: P X P X P X X A A B B B A total = ⋅ ° + ⋅ ° ⇒ = × + − × 0 51 0 4 1 0 6 . . ( ) . ∴ X A = 0 45 . Now, Y X P P A A A = ⋅ ° = × = 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 = + = = − − + − = 0 45 2 2 3 6 17 . ( ) ( ) and ∴ n A (liquid) = x = 12 11 and n A (vapour) = 2 – x = 10 11
Solution: Final total moles of liquid = 5 20 100 1 × = and total moles of vapour = 5 – 1 =
Solution: Let the liquid contain x mole A . P x x x total = × + − × = − 1 0 4 1 1 0 6 0 6 0 2 . . . . (1) and Y X P P x x x A A A = ⋅ ° ⇒ − = × − total 2 4 1 0 4 0 6 0 2 . . . (2) From (2): x = 0.48 ∴ From (1): P total = 0.504 bar Comprehension III
Solution: Δ G mix, m = RT [ X 1 ⋅ ln X 1 + X 2 ⋅ ln X 2 ] = × ⋅ + ⋅ ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ 2 300 1 3 1 3 2 3 2 3 ln ln = –380 cal/mol
Solution: Δ = − Δ = − − = S G T m m mix, mix, . / 380 300 3 8 3 cal K-mol and Δ S mix = 3 3 8 3 3 8 × = . . / cal K Comprehension IV
Solution: P X P X P B B T T total mm Hg = ⋅ ° + ⋅ ° = × + × = 10 20 100 10 20 40 70
Solution: Y X P P A B B = ⋅ ° = × = = total 0 5 100 70 5 7 0 714 . .
Solution: 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 = ° + ° = + ⇒ = . . .
Solution: X Y P P B B B = ⋅ ° = × = total 0 5 57 14 100 0 286 . . .
Solution: 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 = ⋅ ° + ⋅ ° = × + − × 10 100 10 10 40 or, P total = 40 + 6 x (1) Y X P P x x x x B B B = ⋅ ° ⇒ − = × + ⇒ = 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 – x ) mole B = ( n B – y ) mole = ( n A + n B) 3 4 From question: 700 = + × ° + + × ° n n n P n n n P A A B A B A B B (1) 600 = + × ° + + × ° 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 − + = 3 4 0 75 ( ) . (5)
Solution: n B : n A = 29.51
Solution: P A ° = 807 4 . mm
Solution: P B ° = 511 1 . mm Comprehension VI
Solution: P RT V A m ° = = × × × = 0 08 300 100 1 25 3800 1000 760 60 . . torr
Solution: P RT V B m ° = = × × × = 0 08 300 50 1 00 7600 1000 760 48 . . torr
Solution: 1 1 54 60 1 48 5 9 P Y P Y P Y Y Y A A B B A A A total = ° + ° ⇒ = + − ⇒ = Comprehension VII
Solution: Δ = ⋅ = ⋅ T K m K f f f 50 M ∴ Δ Δ = = T A T B f f B B ( ) : ( ) M : : M 3 1
Solution: Average molar mass of solute in S 1 M(S M M M 1 2 3 2 3 11 5 ) = × + × + = 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 + × + = ∴ Δ Δ = ( ) ( ) = T S T S M S M S f f ( ) : ( ) : : 1 2 2 1 9 11 Comprehension VIII
Solution: Δ = ⋅ = × × × = T K m f f 2 0 0 1 0 9 46 1000 4 8 . . . . K ∴ Freezing point of solution = 155.7 – 4.8 = 150.9 K
Solution: P X P = ⋅ ° = × = 2 0 9 40 36 . mm kg
Solution: Δ = ⋅ = × × × = T K m b b 0 52 0 1 0 9 18 1000 3 2 . . . . K ∴ B.P. of solution = 373 + 3.2 = 376.2 K Comprehension IX
Solution: Increase in mass of absorber a P ° and decrease in mass of pure solvent a ( P ° – P ). ∴ P P P X x x x ° − ° = = = + − 0 02 0 24 180 180 100 18 1 . .\n10.38 Chapter 10 HINTS AND EXPLANATIONS ∴ Mass percent of glucose, x = 1000 21 %
Solution: AlCl Al Cl 3 3 1 0 8 0 2 0 8 3 0 8 2 4 3 − = × = + − + . . . . . \u001f ⇀ \u001f ↽ \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 °. ∴ P P X m ° = = + = = Δ 2 17 17 3 4 5 6 0 18 . . absorber ∴ Increase in mass of absorber = 0.216 gm
Solution: As P ° > P , some vapour above the solution from saturated moist air coming will condense and hence, the mass of solution will increase.
Solution: Both have same Δ T f
Solution: KCl will dissociate.
Solution: Relative lowering of V.P. is also independent of solvent.
Solution: Deviation may occur in non-ideal solution.
Solution: (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
Solution: Higher the B.P. of solvent, normally higher is its K b value.
Solution: (A) 2 = 1 + a (2 – 1) ⇒ a = 1.00 (B) 2 = 1 + a (3 – 1) ⇒ a = 0.50 (C) 2 = 1 + a (5 – 1) ⇒ a = 0.25 (D) 2 = 1 + a (4 – 1) ⇒ a = 0.33
Solution: (A) i = 1 (B) i = 1 + 1(2 – 1) = 2 (C) i = 1 + 1 (3 – 1) = 3 (D) i = 1 + 1(4 – 1) = 4
Solution: (P) Eff ective conc. = 0.1 × 3 = 0.3 M = 0.3 m (Q) Eff ective conc. = 0.14 × 2 = 0.28 M = 0.28 m (R) Eff ective conc. = 0.1 [1+0.9(3 – 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 − + − + x x x \u001f ⇀ \u001f ↽ \u001f \u001f K x x x x a = = ⋅ − ⇒ = 0 81 0 1 0 09 . . . ∴ Eff ective conc. = (0.1 + x ) = 0.19 M = 0.19 m
Solution: V gas a n Solvent but independent of pressure. ∴ ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ⇒ = ⇒ = V V V V V V 2 1 2 1 2 2 4 0 5 1 2 gas Solvent ml ml .
Solution: m solution = m water + m ethanol or, V × 0.9344 = 50 × 1.000 + 50 × 0.7939 ∴ V ≈ 96 ml < 100 ml Solution is non-ideal with negative deviation.\n10.39 Liquid Solution HINTS AND EXPLANATIONS
Solution: Ideal gas can never be liquefied.
Solution: P = X 2 ⋅ P ° = 0.8 × 233.5 =
Solution: 8 torr = P exp ∴ Solution is ideal.
Solution: π π π = + + = × + × + = 1 1 2 2 1 2 2 4 2 4 2 2 3 V V V V V V V V ( ) . . atm
Solution: π = CRT = ρ g h or 0 2 100 1000 0 0821 300 1 013 1000 0 2463 1 013 10 6 . / / . . . . M × × = × × × ∴ M = 2 × 10 5
Solution: π = CRT = ρ g h or n 1 0 08 298 1 013 1000 7 45 1 013 10 6 × × = × × × . . . . ∴ n = × − 25 10 80 3 ∴ Millimoles in 320 gm = 25 80 320 20 5 × =
Solution: V V C C T T 2 1 1 2 1 1 2 2 500 283 105 3 298 5 = = = ≈ π π / / / . /
Solution: Δ = ⋅ ⇒ = × × ⇒ ≈ T K m n n f f 0 29 1 86 1 04 267 100 1000 4 . . . / /
Solution: Δ = ⋅ T K m f f For KCN solution: 0.80 = K f × 0.2 × 2 (1) Hg(CN) mCN Hg(CN mole Final 0 mole (0.2 m mole 2 0 1 0 2 0 1 . . . ) ) + − − \u001f ⇀ \u001f ↽ \u001f \u001f m m m + − 2 0 0 1 . mole Final eff ective molality = (0.2 – 0.1 m) + 0.1 + 0.2 = 0.5 – 0.1 m Now, 0.60 = K f × (0.5 – 0.1 m) (2) From (1) and (2): m = 2 Four Digit Integer Type
Solution: P X P = ⋅ ° 2 20 180 18 6 180 18 = + × ° / M P (1) and 20.02 = 11 6 11 M + × ° P (2) ∴ M = 54
Solution: Mole fraction of solvent is same in both. ∴ 90 18 10 90 18 95 18 5 180 95 18 380 M M + = + ⇒ = X
Solution: P P P X ° − ° = = − = 1 2400 2300 2400 1 24 1 mole solution Urea mole gm Water mole = = × = = = 1 24 1 24 60 2 5 23 24 23 2 . 4 4 18 17 25 × = ⎧ ⎨ ⎪ ⎪ ⎩ ⎪ ⎪ . gm ∴ Volume of 1 mole solution = + = 2 5 17 25 1 185 50 3 . . . ml Now, π = = × ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ × × = CRT 1 24 50 3 1000 0 08 300 60 / / . atm
Solution: Δ = ⋅ T K m f f or, 30 1 86 62 795 1000 795 = × ⇒ = . / / w w gm
Solution: Δ = ⋅ T K m f f or, 6 1 86 50 62 1000 250 = × ⇒ = . / / w w water (final) gm ∴ Mass of water separated as ice = 375 – 250 = 125 gm
Solution: Δ = ⋅ T K m f f Naphthalene solution: 13 5 38 4 128 185 1000 . . / / = × K f (1) Unknown substance solution: 9 0 11 6 185 1000 . . / / = × K f M (2) ∴ M = 58
Solution: Δ = ⋅ = × + ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ × = ° T K m f f 1 86 3 6 180 3 6 60 200 1000 0 744 . . . . C ∴ F. P. of solution = – 0.744°C\n10.40 Chapter 10 HINTS AND EXPLANATIONS
Solution: Δ = ⋅ T K m f f ( . . ) . / M / 26 84 25 64 8 2 4 10 100 10 1000 3 3 − = × × × ⇒ = − − M 160
Solution: Δ = ⋅ ⇒ = × × + − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ T K m f f 1 60 4 88 2 122 26 1000 1 1 2 1 . . / / α ∴ a = 0.96 or 96 %
Solution: π = ⇒ = + × ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ × × CRT x 4 92 200 0 05 2 0 08 300 . . . ∴ x = 20
Solution: P P X P n n n P n n P ° − = ⋅ ° = + ° ≈ ⋅ ° 1 1 1 2 1 2 , Urea solution: 0 03 0 1 1000 18 . . = ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ × ° P KCl solution: 0 0594 0 1 1 2 1 1000 18 . . [ ( )] = + − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ × ° α P ∴ a = 0.98 or 98 %
Solution: Loss in weight of solution a P Loss in weight of water a ( P ° – P ) Now, P P P n n ° − = ⇒ = + − 1 2 0 01 0 98 1 25 90 1 3 1 49 18 . . . [ ( )] α ∴ a = 0.50 or 50 %
Solution: Δ = ⋅ ⇒ = × × + − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ T K m f f 7 14 75 2 94 1 1 2 1 . α ∴ a = 0.75 or 75 %
Solution: 100 gm solution (Say) Water = 100 – (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 ⋅ 1 [1 + 0.8(2 – 1)] + 0.1 [1 + 0.6 (3– 1)] = 0.4 Now, Δ = ⋅ = × = T K m b b 0 785 0 4 78 5 1000 4 . . . / K ∴ B. P. of solution = 373 + 4 = 377 K
Solution: Δ = ⋅ ⇒ = × T K m m f f 0 40 1 86 . . ∴ Eff ective molality of NaCl solution = 20 93 m Now, π = = × × = × CRT 20 93 0 0821 279 60 0 0821 . . atm
Solution: Negative sign is for reactants and positive for products.
Solution: r d dt K K rxn = − ⋅ = − 1 2 2 1 2 2 2 2 4 [ ] [ ] [ ] NO NO N O ∴ Rate of disappearance of NO 2 is given by, − = − d dt K K [ ] [ ] [ ] NO NO N O 2 1 2 2 2 2 4 2 2
Solution: P n RT V A A = ⇒ dP dt RT V dn dt A A = ⋅ or, ( ) ( ) − ⋅ = − ⋅ K P RT K C A n A n 1 2 ∴ K K RT P C K RT RT RT A A n n n 2 1 1 1 = ⋅ ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = ⋅ = − ( ) ( )
Solution: − ⋅ = − 1 2 2 2 1 2 2 d dt K K [ ] [ ] [ ][ ] HI HI H I
Solution: Order = 1 ←⎯ with respect to A ⇒ a = 1 Order = 2 ←⎯ with respect to B ⇒ b = 2 Hence, reaction is A + 2B P. r d A dt d B dt = − = − [ ] [ ] 1 2
Solution: A + 2B C + D t = 0 0.6 atm 0.8 atm t = t 0.6 – x 0.8 – 2 x = 0.3 atm = 0.2 atm ⇒ x = 0.3 ∴ r r K K t 0 2 2 0 3 0 2 0 6 0 8 1 32 = × × × × = . ( . ) . ( . )\n11.45 Chemical Kinetics HINTS AND EXPLANATIONS
Solution: For no change in temperature, Δ H net = 0 and hence, for 3 moles of B reacted, 4 moles of Q should form.
Solution: 2.82 = (2) x ⇒ x = 3 2 9 = (3) y ⇒ y = 2 ∴ overall order = x + y = 7 2
Solution: r 1 = 0.0068/65 = 1.046 × 10 –4 gm/min r 2 = 0.0031/120 = 2.583 × 10 –5 gm/min r 3 = 0.0032/60 = 5.333 × 10 –5 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
Solution: From (2) and (3) : order with respect to I – = 1 From (1) and (3) : order with respect to ClO – = 1 From (3) and (4) : order with respect to OH – = 1
Solution: (1 + K 2 ⋅ C A ) ≈ 1
Solution: For steady state, + = d R dt [ ] 0 or, K 1 [ A ] – K 2 [ R ][ B ] – K 3 [ R ][ C ] = 0 ∴ [ ] [ ] [ ] [ ] 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 [ ][ ] [ ][ ] [ ] [ ]
Solution: r K = 3 3 [ ][ ] O O For 1st step, K K 1 2 2 3 = [ ][ ] [ ] O O O ∴ r = K 3 [O][O 3 ] = K K K 3 1 3 2 2 2 [ ] [ ] O O
Solution: r d X dt d Y dt = − = + [ ] [ ] and rate decreases with time.
Solution: A 2B t = 0 0.1 M 0 t = 1 min 0.1 – x 2 x For zero order reaction: [ A 0 ] – [ A ] = Kt or, 0.1 – (0.1 – x ) = 0.01 × 1 ⇒ x = 0.01 ∴ [ B ] = 2 x = 0.02 M
Solution: 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 = − × ×= . P N atm 2 0 1 10 1 = × = . P H atm 2 3 0 1 10 3 = × × = . ∴ P total = 1 + 1 + 3 = 5 atm
Solution: For − = d A dt K A n [ ] [ ] and n ≠ 1 [ A ] 1 – n = [ A 0 ] 1 – n – K (1 – n ) ⋅ t For given graph, 1 – n = –3 ⇒ n = 4 and – K (1 – n ) = tan 45° ⇒ K (4 – 1) = 1 ∴ K = − − 1 3 3 1 M min Now, r d A dt K A rxn = − ⋅ = ⋅ ⋅ 1 3 1 3 4 [ ] [ ] = × × = × − − 1 3 1 3 0 2 16 9 10 4 4 1 ( . ) min M
Solution: 1 0 25 2 8 0 1 2 M M hr t t = × ⎯ → ⎯⎯⎯ / . . ⇒ t 1/2 = 4.0 hr 0.6M M hr t t = ⎯ → ⎯⎯ 1 2 4 0 0 3 / . .
Solution: The successive t 1/2 are double of previous one and hence, order =
Solution: 20. r = K [ A ] n 10 = K (0.8) n (1) 0.625 = K (0.2) n (2) ∴ n = 2
Solution: C 2 H 6 C 2 H 4 + H 2 t = 0 3 bar 0 0 t = ? (3 – x ) bar x bar x bar From question: (3 – x ) + x + x = 5 ⇒ x = 2 From the unit of rate constant, the order of reaction is 2, hence, t K P P x = − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = − − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ × ° 1 1 1 1 0 0015 1 3 1 3 1 10 2 6 2 6 5 C H C H . = 4.44 × 10 –3 hr = 16 seconds
Solution: Let the reaction be first-order. K 1 1 8 100 20 0 201 = ⋅ = ln . K 2 1 18 100 10 0 128 = ⋅ = ln .\n11.46 Chapter 11 HINTS AND EXPLANATIONS As K 1 ≠ 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 = − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = . . K 2 1 18 1 0 1 1 1 0 5 = − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = . . As K 1 = K 2 , order = 2
Solution: Here, t 1/2 is independent from sugar concentration and hence, the order with respect to sugar is
Solution: Now, r = K [Sugar][H + ] n = K ′ ⋅ [Sugar] t K K n 1 2 2 2 / ln ln [ ] = ′ = + H 500 2 10 5 = ⋅ − ln ( ) K n and 50 2 10 6 = ⋅ − ln ( ) K n ∴ n = –1
Solution: r = K [ester] [ H + ] = K ′ ⋅ [ester] ∴ t K K 1 2 2 2 0 693 0 1 0 01 693 / ln ln [ ] . . . = ′ = = × = + H hr
Solution: For n th order reaction ( n ≠ 1) Kt A A n n n = − − − − [ ] [ ] 0 1 1 1 For n = 0.5, Kt A A = − [ ] [ ] / / 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 % / / / ([ ] ) [ ] = = − = 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 % / / / / [ ] [ ] [ ] = = − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ⎟ K ∴ T t 1 2 1 1 1 2 1 0 3 / . = − =
Solution: For A : t t A A B A = ⋅ = ⋅ 1 2 0 0 2 10 2 8 / log log [ ] [ ] log log [ ] [ ] For B : t t B B B A = ⋅ = ⋅ 1 2 0 0 2 20 2 / log log [ ] [ ] log log [ ] [ ] From question, 10 2 8 20 2 0 0 log log [ ] [ ] log log [ ] [ ] ⋅ = ⋅ B A B A ∴ [ ] [ ] B A 0 8 = ⇒ t B A = ⋅ = 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 ⎯ → ⎯ ⎯ → ⎯ ⎯ → ⎯ ⎯ → ⎯ ⎯ → ⎯ ⎯ ⎯ → ⎯ [ ] B 0 8 B B B B B : [ ] [ ] [ ] [ ] 0 20 0 20 0 20 0 2 4 8 ⎯ → ⎯ ⎯ → ⎯ ⎯ → ⎯
Solution: 0 1 0 025 2 40 1 2 . . / min M M t t = ⎯ → ⎯⎯ ⇒ t 1/2 = 20 min Now, r K A t A = = = × [ ] ln [ ] . min . / 2 0 693 20 0 01 1 2 M = 3.465 × 10 –4 M min –1
Solution: + = − = d B dt d A dt K A [ ] [ ] [ ] / 1 3 or, − = ⋅ ∫ ∫ 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 / / / / [ ] ( ) = − ⋅
Solution: N 2 O 5 2NO 2 + 1 2 O 2 t = 0 a mole 0 t = t ( a – x ) mole x V t 2 mole α t = ∞ \u001f 0 a V 2 mole α ∞ K t t a a x t V V V t = ⋅ = ⋅ − = ⋅ − ∞ ∞ 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 ⋅ − = ⋅ − ln . . . ln . . V t ⇒ V t = 7.2 ml
Solution: For zero order reaction, t P 1 2 3 / α NH ° ∴ 315 70 150 1 2 t / = ⇒ t 1/2 = 675 sec
Solution: A n B t = 0 P 0 0 t = t P 0 – x n.x Now, ( P 0 – x ) = P 0 ⋅ e – Kt ⇒ x = P 0 (1 – e – Kt )\n11.47 Chemical Kinetics HINTS AND EXPLANATIONS Now, P total = ( P 0 – x ) + nx = P 0 ⋅ e – Kt + n ⋅ P 0 (1 – e – Kt ) = P 0 [ n + (1 – n ) e – Kt ]
Solution: t K n K n n n n n 1 2 1 1 1 1 2 1 1 2 1 / ( ( ) ( ) [ ] ( ) = − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ − = − − − − − − 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 / ( ) ( ( ) ( ) [ ] ( ) = − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ − = − − − − − − C ) C C 0 0 0 ∴ t t n n n 3 4 1 2 2 1 1 1 1 2 1 2 1 2 / / ( ) ( ) ( ) = − − = + − − −
Solution: r r A A n 2 1 2 1 = ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ [ ] [ ] ⇒ 2 = (4) n ⇒ n = 1 2 Now, t 1/2 a [ A 0 ] 1 – n ⇒ t 1/2 a [ A 0 ] 1/2 100 50 25 16 16 2 t t = = ⎯ → ⎯⎯⎯ ⎯ → ⎯⎯⎯⎯ min / min ∴ Time for 75 % reaction = 16 16 2 27 3 + = . min
Solution: Kt a a x x a = − = − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ln ln 1 1 or, 2.5 × 10 –5 × (100 × 60) = − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ln 1 1 x a ⇒ x a = 0 138 . ∴ Percentage decomposition = × = x a 100 13 8 . %
Solution: ( )( ) / / t t 1 2 1 1 2 2 = ⇒ 0 693 1 1 2 0 . [ ] K K A = ∴ [ ] . . . . . A K K 0 1 2 2 0 693 6 93 10 0 693 0 2 0 5 = = × × = − M
Solution: 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 = ⋅ ⋅ = ( ) log log ( ) log log ( ) ( / / / / ) ) log log 2 4 3 4 ⋅ ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = × − = 3 2 0 6 0 48 0 6 3 10 . . .
Solution: r = K [ A ] n 1 100 0 02 60 0 02 × = . ( . ) K n (1) 1 100 0 04 15 0 04 × = . ( . ) K n (2) ∴ n = 3
Solution: t T A A = ⋅ gen ln ln [ ] [ ] 2 0 ⇒ 60 75 2 0 = ⋅ ln ln [ ] [ ] A A ∴ [ ] [ ] . A A e 0 0 56 =
Solution: ( ) ( ) [ ] [ ] / / t t A A n 1 2 1 1 2 2 0 1 0 2 1 = ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ − ⇒ 37 82 18 95 0 05 0 10 1 . . . . = ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ − n ⇒ n = 2 Now, ( ) . . . / t 1 2 1 37 82 0 15 0 05 = ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ − ⇒ t 1/2 = 12.6 hr
Solution: Now, T N a 0 1000 2 ⋅ = × (1) and T N a x x x t ⋅ = − × + × + × 1000 2 2 1 ( ) (2) ∴ a a x T T T t − = − 0 0 3 2
Solution: B n + B ( n + 4)+ t = 0 a mole 0 t = 10 min ( a – x ) mol x mol Now, 25 1000 2 × = × N a (1) and, 32 5 1000 2 5 . ( ) × = − × + × N a x x (2) Now, K t a a x = ⋅ − = ⋅ − = − 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 – x ) mole t = 0 t = t CHCOOH C Br COOH + H Br →\n11.48 Chapter 11 HINTS AND EXPLANATIONS
Solution: A 2B + C t = 0 a mol 0 0 t = 10 sec ( a – x ) mol 2 x mol x mol Now, r r P P M M A B A B B A = ⋅ ⇒ 1 2 2 4 16 = − ⋅ a x x ⇒ x a = 3 Now, K t a a x a a a = ⋅ − = ⋅ − = − 1 1 10 2 3 0 04 1 ln sec ln . sec
Solution: r = K [ A ] 2 [ B ] = K ′ ⋅ [ A ] 2 as [ B 0 ] >> [ A 0 ] ∴ t K A K B A 1 2 0 0 0 1 1 1 0 5 0 002 2 0 500 / [ ] [ ][ ] . . . min = ′ ⋅ = = × × =
Solution: r = K [ester][H + ] ∴ r r HA HX HA H = = + 1 100 1 0 [ ] . ⇒ [H + ] HA = 0.01 M Now, Ka A HA ( ) [ ][ ] [ ] . . ( . ) HA H = = × − ≈ + − − 0 01 0 01 1 0 01 10 4
Solution: 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 × = × × = − t / . . sec
Solution: − = + dC dt C C α β 1 ⇒ − + ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ⋅ = ⋅ ∫ ∫ 1 0 2 1 2 C dC dt t Co Co β α / / ∴ t C 1 2 0 1 2 2 / ln = + ⋅ ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ α β
Solution: r = K ′ [CH 3 COOH][C 2 H 5 OH] = K ′ ⋅ [CH 3 COOH] 2 ∴ t K 1 2 0 1 / [ = ′ ⋅ CH COOH] 3 ⇒ 50 1 10 0 2 3 = × × − ( ) . K ∴ K = 100 M –2 min –1
Solution: r = K [ A ] x [ B ] y For case-I : r = K [ A ] x [ B ] y = K ′ ⋅ [ B ] y where K ′ = K [ A 0 ] x In equal time interval, the concentrations of B are in G.P. and hence, y = 1 and ′ = ⋅ = − K 1 10 0 01 0 008 0 02 1 ln . . . min For case-II: r = K [ A ] x [ B ] y = K ″ [ A ] x where K ″ = K [ B 0 ] y In equal time interval, the concentration of A are in G.P. and hence, x = 1 and ′′ = ⋅ = − K 1 10 0 02 0 018 0 01 1 ln . . . min Now, r = K [ A ][ B ] ∴ K K A K B = ′ ′′ = = − − [ ] , [ ] . . . . . min 0 0 1 1 0 02 2 0 0 01 1 0 0 01 or or M
Solution: K K C C K C app = ⋅ + ⋅ = + 1 1 1 1 α α lim C K K →∞ = app 1 α From question, K C C K 1 1 1 90 100 ⋅ + ⋅ = × α α or, C C 1 9 10 90 100 1 9 10 5 5 + × = × × ⇒ C = 10 –5 M
Solution: A 2B + C t = 0 1 0 0 t = 12 hr 1 – x 2 x x t = 24 hr 1 – y 2 y y V.P. of solution, P = X 2 ⋅ P o or, 20 180 18 180 18 1 2 24 = + + × / ( ) x ⇒ x = 0.5 ∴ t = 12 hr = t 1/2 Now, t = 24 hr = 2 × t 1/2 ⇒ y = 0.75 Now, V.P. of solution, P X P y = ⋅ ° = + + × 2 10 10 1 2 24 ( ) = 19.2 mm Hg
Solution: [ ] [ ] . . B C K K = = × × = − − 1 2 4 5 1 26 10 3 15 10 4 1 ∴ Percentage of B = × = 4 5 100 80%\n11.49 Chemical Kinetics HINTS AND EXPLANATIONS
Solution: A R t = t a – x x ∴ r = K ( a – x ) ⋅ x For maximum rate, dr dx = 0 ⇒ x a = 2 ⇒ C A = C R
Solution: K dt by y dy t ⋅ = + − ⋅ ∫ ∫ 0 0 2 1 2 1 / / Co Co ⇒ t K b b 1 2 1 1 2 2 / ( ) ln = + ⋅ ⋅ − ⋅ ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ Co Co
Solution: t t B A 50 94 %, %, = or, 1 100 50 1 100 6 2 1 K K ⋅ = ⋅ ln ln ⇒ K K 1 2 4 067 1 = .
Solution: Percentage product by S N 2 mechanism = × × + × × − − − ( . )[ ]( . ) ( . )[ ]( . ) . [ ] 4 8 10 0 01 4 8 10 0 01 2 4 10 100 5 5 6 RX RX RX = = 16 67 . %
Solution: [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 × 0.30 : 0.10 : 0.10 = 6 : 6 : 6 : 1 : 1
Solution: [ A ] + [ B ] + [ C ] = [ A 0 ] when [ A ] = [ B ] = [ C ], [ A ] = [ ] [ ] A A e Kt 0 0 3 = ⋅ − or, 1 3 3 3 = − + ⋅ e t (ln ln ) ⇒ t = 0.5 min
Solution: 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
Solution: After long time, r A = r B ⇒ K 1 [ A ] = K 2 [ B ] ∴ [ ] [ ] A B K K = = 2 1 40
Solution: [ ] [ ] [ ] ( ) [ ] ( ( ) ( ) C A K A K K e A e K K K e K K t K K t = + − ⋅ = + − + − + ⋅ 2 0 1 2 0 2 1 2 1 1 2 1 2 ( ( ) ) K K t 1 2 1 + ⋅ − = − = − ⋅ × × × − 9 10 1 9 10 1 1 1 10 10 1 25 10 3600 1 5 K K e e K t ( ) [ ] . = 0.5112
Solution: At steady state, K 1 [ A ] = K 2 [ B ] ∴ K K A B 2 1 4 3 1 2 5 10 0 2 0 01 5 10 = = × × = × − − − [ ] [ ] . . . min
Solution: Reaction may be considered as A C K 1 ⎯ → ⎯ ∴ [ C ] = [ A 0 ] ( ) 1 1 − − e K t
Solution: A B K 1 2 ⎯ → ⎯ A C K 2 ⎯ → ⎯ t = 0 1 atm 0 1 atm 0 t = 10 min (1 – x – y ) 2 x 1 – x – y y t = ∞ (1 – a – b ) 2 a (1 – a – b ) b ≈ 0 ≈ 0 From question, a + b = 1 and 2 a + b = 1.5 ∴ a = b = 0.5 Now, P P K K a b x y B C = = = 2 2 2 1 2 ⇒ K K x y 1 2 1 = = 0 Now, P x y x y 10 1 2 1 4 min ( ) . = − − + + = ⇒ x = y = 0.4 ∴ P x y A = − − = 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 ⋅ ° = ⋅ = − ln ln . . min ∴ K 1 = K 2 = 0.08 min –1
Solution: r z u N av max * = = ⋅ ⋅ 11 2 2 1 2 πσ = × × × × × × − − − 1 2 4 10 2 10 2 10 8 2 4 1 19 3 2 π ( ( ) ( ) cm) cm s cm = 2.842 × 10 28 cm –3 s –1 = 4.74 × 10 7 mol l –1 s –1
Solution: d K dT E RT a (ln ) = 2 or, 0 2 2 + + = β γ T T E RT a ⇒ E a = ( b T + γ ) R
Solution: K K B C 1 2 40 60 2 3 = = = [ ] [ ] Now, E K E K E K K a a a ( ) overall = ⋅ + ⋅ + 1 1 2 2 1 2 = 32 kcal/mol\n11.50 Chapter 11 HINTS AND EXPLANATIONS
Solution: r r uncat cat = × 1 2 ⇒ K K uncat cat = × 1 2 or, A e A e E RT E RT T a a ⋅ = × ⋅ − − × ( ) ( ) / / uncat cat 0.5 1 2 or, ln . ( ) ( ) 2 20 0 5 − = − − E RT E RT a a uncat uncat ∴ E a (uncat) = 38.58 kcal/mol
Solution: For A B; K 1 = 8 min –1 at T = 300 K ′ = K 1 ? at T = ? ln ′ = − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ K R T 1 8 20 1 300 1 KJ (1) For A C; K 2 = 2 min –1 at T = 300 K ′ = K 2 ? at T = ? ln . ′ = − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ K R T 2 2 28 314 1 300 1 KJ (2) From (1) and (2), ln / / . . ′ ′ = × − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 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 × ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ × T ⇒ T = 379.75 K
Solution: 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 ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ = − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ (1) For reaction 2: ln ( ) ( ) K K E R a 2 310 2 300 2 1 300 1 310 ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ = − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ (2) From (1) ÷ (2) : ln ln ( ) ( ) 2 2 2 310 2 300 K K ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ = or, K K 2 310 2 300 2 ( ) ( ) ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ = ⇒ K 2 (300) = K 2 310 2 2 2 30 2 ( ) = × ln = 0.0327 min –1
Solution: At 27°C, K 1 1 1 21 6 100 25 2 10 8 = ⋅ = − . 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 = − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = × − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = ∴ K K e 2 1 2 7 = = . ⇒ K 2 2 7 2 10 8 2 4 = × = . ln . ln ⇒ ( t 1/2 ) 2 = 4 min ∴ Percentage decomposition in 8.0 min = 75 %
Solution: K A e A e A e A E Rt RT RT a = ⋅ = ⋅ = − − / / . \u001f 0 37
Solution: ln 2 1 280 1 290 = − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ E R a (1) and ln x E R a = − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 1 290 1 300 (2) From (2) ÷ (1), ln ln x 2 280 300 = ⇒ x = 1.91
Solution: First energy barrier is high as Step-I is slow.
Solution: r K A n = ⋅ [ ] ⇒ n r K A = ln( / ) ln [ ] Now, r r A A n 2 1 2 1 = ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ [ ] [ ] ⇒ n r r A A = − − ln ln ln[ ] ln[ ] 2 1 2 1\n11.51 Chemical Kinetics HINTS AND EXPLANATIONS And, t 1/2 a [ A 0 ] 1– n ⇒ ( ) ( ) [ ] [ ] / / t t A A n 1 2 2 1 2 1 0 2 0 1 1 = ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ − ∴ n A A t t = − − − 1 0 2 0 1 1 2 2 1 2 1 ln[ ] ln[ ] ln( )ln( ) / /
Solution: [ A ] = [ A 0 ] (1 – a ) = [ A 0 ] ⋅ e – Kt ⇒ α = 1 – e – Kt
Solution: (a) − = ⋅ d A dt K A n [ ] [ ] f A A A d A A = − = − [ ] [ ] [ ] [ ] [ ] 1 2 1 From question, f d A A = − [ ] [ ] ∴ f A t K A n [ ] [ ] = ⇒ f t K A n = ⋅ − [ ] 1 or, log log ( ) log[ ] f t K n A ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = + − ⋅ 1 (b) [ ] [ ] A A n Kt n n 0 1 1 1 − − − − = ⇒ [ A ] 1 – n = [ A 0 ] 1 – n + ( n – 1) ⋅ 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 / / [ ] [ ] [ ] [ ] ( = − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = − − − − − 2 2 1 1 1 1 2 1 2 ) n n n − − − − = +
Solution: t 1/2 = C ⋅ (C 0 ) 1 – n ⇒ ln t 1/2 = ln C + (1 – n ) ⋅ ln C 0
Solution: K ′ = K ⋅ [H + ] On doubling [H + ], K ′ will double but K will remain unchanged.
Solution: (a) For steady state, 6 93 10 80 0 693 100 6 3 . . [ ] × = × − SO ∴ [SO 3 ] = 1.25 × 10 –5 M (b) n eq SO 3 = n eq NaOH ⇒ 1.25 × 10 –5 × 10 3 × 2 = V NaOH × 1 ∴ V NaOH = 2.5 × 10 –2 L = 25 ml (c) Mole of SO 3 needed = 980 10 98 10 3 4 × = ∴ Air needed = × = × − 10 1 25 10 8 10 4 5 8 . L (d) 1000 days = 10 t 1/2 ∴ [ ] . . SO M 3 5 10 8 1 25 10 2 1 25 10 = × ≈ × − −
Solution: 3A( g ) 2B( g ) + 2C( s ) t = 0 6 atm 0 – t = 20 min (6 – x ) atm 2 3 x atm 0.05 atm t = ∞ ≈ 0 4 atm 0.05 atm But from question, P ∞ = 4.05 atm and hence, (4.05 – 4) = 0.05 atm is the vapour pressure of C( s ). Now, P x x 20 6 2 3 0 05 5 05 = −+ + = ( ) . . ⇒ x = 2 ∴ t = 20 min = t 1/2
Solution: − = ⋅ d dt K θ θ ⇒ Kt = ln θ θ 0 (a) t K = ⋅ = ⋅ = 1 1 0 04 596 298 17 5 0 ln . ln . sec θ θ (b) t = ⋅ = 1 0 04 1192 298 35 . ln sec
Solution: (a) A + B 2C t = 0 2 a a t = t 2 a – x a – x As [A] ≠ [B] throughout, the overall reaction is not fi rst-order. (b) r = K [A] –1 [B] 2 = K ′ ⋅ [B] 2 ⇒ t K B 1 2 0 1 / [ ] = ′ (c) r = K [A] –1 [B] 2 = K ″ [A] –1 (d) As [A] = [B] = stoichiometric ratio, then the mole ratio will remain constant throughout.
Solution: A P K 1 ⎯ → ⎯ ; t K 1 2 1 0 693 / . = B Q K 2 ⎯ → ⎯ ; t K B K 1 2 2 0 2 1 1 / [ ] = = From question, 0 693 1 1 2 . K K = ⇒ K 2 > K 1
Solution: A 4 4A t = 0 a M 0 t = 30 min ( a – x ) M 4 x M As a – x = 4 x ⇒ x a = 5 ∴ Percentage reaction at t = 30 min = × = x a 100 20% Now, 30 2 1 2 = ⋅ − t a a x / log log ⇒ t 1/2 = 90 min\n11.52 Chapter 11 HINTS AND EXPLANATIONS
Solution: (a) Δ r H = ∑ Δ f H Products – ∑ Δ f H Reactants = 2 × (–1263) – [(–2238) + (–285)] = –3 KJ/mol (b) Can not confirm because in aqueous medium, there is no combustion. (d) Concentration in G.P. in equal time interval.
Solution: n = 1 ⇒ t 100% = 1 0 0 K A ⋅ = ln [ ] Infinite n ≠ 1 ⇒ t 100 % = [ ] ( ) ( ) [ ] ( ) A K n A K n n n n n 0 1 1 0 1 0 1 1 − − − − − = − if < 1 = Infi nite if n > 1
Solution: [ ] [ ] B C K K = = 1 2 1 2 ⇒ [ C ] > [ B ] Hence, after long time, the solution will be dextrorotatory. Now, K = K 1 + K 2 = 6.93 × 10 –2 + 13.86 × 10 –2 = 3 × 6.93 × 10 –2 min –1 ∴ t K 1 2 2 2 0 693 3 6 93 10 10 3 / ln . . min = = × × = − A B A C t = 0 2M 0 2M 0 t = t 2 – ( x + y )M x M 2 – ( x + y )M y M From question, x + y = 1.5 and x y = 1 2 ∴ x = 0.5, y = 1.0 Hence, total rotation = 0.5 × 60° + 0.5 × (–72°) + 1.0 × 42° = 36°
Solution: [ B ] : [ C ] : [ D ] = 1 × 3 K : 2 × 2 K : 3 × K = 3 : 4 : 3 A B A 2C A 3D t = 0 1M 0 1M 0 1M 0 t = t 1 – ( x + y + z ) x M 1 – ( x + y + z ) 2 y M 1 – ( x + y + z ) 3 z M t = ∞ 1 – ( a + b + c ) a M 1 – ( a + b + c ) 2 b M 1 – ( 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 ] ≠ 1M
Solution: For S N 1 path : r 1 = (3 × 10 –4 s –1 ) [RX] For S N 2 path : r 2 = (5 × 10 –4 M –1 s –1 ) [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 = ∴ Percentage product by S N 1 = 2 2 3 100 40 + × = %
Solution: − = + d A dt d B dt [ ] [ ] always
Solution: 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 − = + = +
Solution: Increase in temperature will result in greater increase in the rate of reaction A → B than B → C.
Solution: K 1 = K 2 ⇒ − + = − + 14000 5 20000 10 RT RT ⇒ T K = 1200 8 314 . Now, P P e e A B K t K t 2 3 1 2 1 1 1 1 = × × = − − Now, initial pressure P 0 1 1 0 0821 1200 8 314 100 0 237 = + × × = ( ) . . . 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
Solution: 1 1 25 10 6 3 2 K dK dT d K dT T E RT a ⋅ = = × = (ln ) . ∴ E R T a = × = × × = 1 25 10 1 25 10 2 250 10 6 6 4 . . cal/mol
Solution: Δ = − H E E a a f b ⇒ − = − 2 8 E a f ⇒ E a f = 6 kcal/mol Now, the fraction of molecules crossing energy barrier = − e E RT a / and K e H Rt eq = −Δ /\n11.53 Chemical Kinetics HINTS AND EXPLANATIONS
Solution: The overall reaction is first-order.
Solution: K K K 1 2 3 2 4 1 = = ⇒ 2 K 1 = K 2 = 4 K 3
Solution: 2N 2 O 5 4NO 2 + O 2 2 × 108 gm 4 × 46 gm 32 gm 108 gm 92 gm 16 gm Comprehension II
Solution: CO(g) + Cl 2 (g) COCl 2 (g) r d dt K COCl 2 COCl COCl][Cl = + = ⋅ [ ] [ ] 2 5 2 (1) Now, for steady state of COCl, + = d dt [ ] COCl 0 or K 3 [Cl][CO] – K 4 [COCl] – K 5 [COCl][Cl 2 ] = 0 ∴ [ [ [ ] COCl] Cl][CO] Cl = + K K K 3 4 5 2 (2) ∴ For steady state of Cl, d dt [ ] Cl = 0 or 2 K 1 [Cl 2 ] – 2 K 2 [Cl] 2 – K 3 [Cl][CO] + K 4 [COCl] + K 5 [COCl][Cl 2 ] = 0 ∴ [ [ ] / Cl]= Cl K K 1 2 2 1 2 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ (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 ] ( [ ])
Solution: r 4 >> r 5 or K 4 [COCl] >> K 5 [COCl][Cl 2 ] or K 4 >> K 5 [Cl 2 ] ∴ 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 = + ≈ / / / / [ ][Cl ] ( [ ]) [ ] ][Cl ] / / 2 3 2 2 1 2 4 K K
Solution: A overall = A A A A A 1 1 2 3 5 2 1 2 4 / / ⋅ ⋅ ⋅
Solution: E E E E E E a a a a a a overall = + + − − 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 ] – K 2 [Br][H 2 ] + K 3 [H][Br 2 ] + K 4 [H][HBr] – 2 K 5 [Br] 2 = 0 (1) For steady state of H, + = d dt [H] 0 or, K 2 [Br][H 2 ] − K 3 [H][Br 2 ] – K 4 [H][HBr] = 0 (2)
Solution: From (1) and (2), [ [ ] / Br] Br = ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ K K 1 2 5 1 2
Solution: [ [Br][H ] [Br ] [HBr]) [Br ] [H ] / / / H] = + = ⋅ ⋅ ⋅ 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]) +
Solution: + d dt [HBr] = K 2 [Br][H 2 ] + K 3 [H][Br 2 ] – 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])
Solution: 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
Solution: K t t V V t t = ⋅ = ⋅ 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 = ⋅ = − ln . ln . min For t = 20 min, K 2 1 1 20 25 6 10 1 6 10 = ⋅ = − ln . ln . min As K 1 = K 2 , order of reaction = 1
Solution: t K 1 2 2 2 1 6 10 15 / ln log log . min = = ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ =
Solution: Kt a a x x a t = = − = − ln ln ln [H O ] [H O ] 2 2 2 2 0 1 1 or, ln . ln 1 6 10 25 1 1 × = − x a ⇒ x a = 11 16
Solution: Order = 1, but molecularity = 2 (as per reaction). Comprehension V C 8 H 18 O 2 ( g ) → 2CH 3 COCH 3 ( g ) + C 2 H 6 ( g ) t = 0 800 torr 0 0 t = t (800 – x ) torr 2 x torr x torr
Solution: P t t C H O 8 18 2 torr torr = ⎯ → ⎯⎯⎯ = × 800 100 3 1 2 / ∴ t = 3 × 80 = 240 min
Solution: P acetone = 2 x = 1200 ⇒ x = 600 ∴ P t t C H O 8 18 2 torr torr = ⎯ → ⎯⎯⎯ = × 800 200 2 1 2 / ∴ t = 2 × 80 = 160 min
Solution: 800 – x = 700 ⇒ x = 100 ∴ P total = (800 – x ) + 2 x + x = 1000 torr Comprehension VI
Solution: From (1) and (2) data : order w.r.t OH = 1 From (2) and (3) data : order w.r.t H 2 S = 1 ∴ r = K [H 2 S][OH]
Solution: K r = = × × × × − − − − [ . ( . ) ( . ) H S][OH] M s M M 2 1 4 10 2 1 10 1 3 10 6 1 8 8 = 5.1 × 10 9 M –1 s –1
Solution: r = K [H 2 S][OH] = 5.1 × 10 9 × (1.0 × 10 –8 ) × (1.7 × 10 –8 ) = 8.67 × 10 –7 M s –1
Solution: r = 8.67 × 10 –7 × 0.1 = 8.67 × 10 –8 mol s –1 Comprehension VII
Solution: K t P P x x = ⋅ ° 1 ln For t = 100 min, K 1 1 1 100 800 400 2 100 = ⋅ = − ln ln min For t = 200 min, K 2 1 1 200 800 200 2 100 = ⋅ = − ln ln min As K 1 = K 2 , order of reaction = 1
Solution: K = = × − − ln . min 2 100 6 93 10 3 1 ∴ K K rxn = = × − − 2 3 465 10 3 1 . min
Solution: Time for 87.5 % reaction = 3 3 2 6 93 10 1 2 3 × = × × − t / ln . = 300 min
Solution: 2X( g ) 3Y( g ) + 2Z( g ) t = 0 800 0 0 t = t 800 – x 3 2 x x = 700 ∴ P total = 800 + 3 2 x = 950 torr\n11.55 Chemical Kinetics HINTS AND EXPLANATIONS Comprehension VIII
Solution: A + 2B C + D t = 0 a M b M 0 0 t = t ( a – x ) M ( b – 2 x ) M Now, r = K ⋅ C B ⇒ − = − d dt K b x [A] ( ) 2 ⇒ dx dt K b x = − ( ) 2 or, dx b x K dt x t − = ⋅ ∫ ∫ 2 0 0 ⇒ x b e Kt = − − 2 1 2 ( ) ∴ C A = a – x = a b e Kt − − − 2 1 2 ( )
Solution: For C a a a b e A Kt = = − − − 2 2 2 1 2 , ( ) ∴ ( ) ln / t K b b a A 1 2 1 2 = ⋅ −
Solution: For ( ) ( ) , [ ] [ ] / / t t A B a b A B 1 2 1 2 1 2 = = = Comprehension IX
Solution: r n dn dt K n rxn A A = − ⋅ = ⋅ 1 1 ⇒ n A = n A ° ⋅ e – n , kt
Solution: n 1 A n 2 A t = 0 a mole 0 t = t ( a – x )mole n n x 2 1 ⋅ mole = a ⋅ e – n , kt ∴ x = a (1 – e – n , kt ) Now, V V n n a x n n x a 2 1 2 1 = = − + ⋅ 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 = + − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = + − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ⋅ − − ( ) , ∴ V V n n n n e n kt 2 0 2 1 2 1 1 = + − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ − ,
Solution: If n 1 = 1, n 2 = 2, then V 2 = V 0 (2 – e – kt ) Now, [ ] ( ) [ ] A n V n e V e A e e A A kt kt kt kt = = ° ⋅ ⋅ − = ⋅ − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ − − − − 2 0 0 2 2 Comprehension X
Solution: df f K dt f t 1 0 0 − = ⋅ ∫ ∫ ⇒ t f K = − − ln( ) 1 Now, K = − − = − ( ) 3 200 3 200 1 hr ∴ t K 1 2 2 0 693 3 200 46 2 / ln . . = = ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = hr
Solution: t f K = − − ln( ) 1 ⇒ f = 1 – e – Kt = 1 – e –3 t /200 Comprehension XI
Solution: Unit of K = s –1 ⇒ order = 1
Solution: K K B = × = × × = × − − − 10 100 10 100 1 5 10 1 5 10 4 5 1 . . s\n11.56 Chapter 11 HINTS AND EXPLANATIONS Comprehension XII
Solution: [ ] [ ] ( . ) . M . A A e e K t = ⋅ = × = − − × 0 0 04 25 1 1 0 0 368 M
Solution: [ ] [ ] ( ) B K A K K e e K t K t = − − − − 1 0 2 1 1 2 = × − − − × − × 0 04 1 0 0 06 0 04 0 04 25 0 06 25 . ( . . . ( ) . . M) e e = 0.29 M
Solution: [ C ] = [ A 0 ] – [ A ] – [ B ] = 1.0 – 0.368 – 0.29 = 0.342 M
Solution: t K K K K max ln ln . . min = ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ − = ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ − = 2 1 2 1 3 2 0 06 0 04 20
Solution: [ ] [ ] ( . ) max . . . B A K K K K K = ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = × ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ − − − − 0 2 1 0 06 0 06 0 0 2 2 1 1 0 3 2 M 4 4 = 0.3 M
Solution: [ ] [ ] [ ] [ ] A B C A = = = 0 3 Now, t K A A K = ⋅ = = = 1 3 1 1 0 04 27 5 1 0 1 ln [ ] [ ] ln . . . min
Solution: − = + d A dt d C dt [ ] [ ] ⇒ K 1 [ A ] = K 2 [ B ] ⇒ t = 20 min ∴ [ ] [ ] ( . M) e . . A A e K t = ⋅ = ⋅ = − − × 0 0 04 20 1 1 0 0 45M
Solution: ( r C ) max = K 2 [ B ] max = 0.06 × 0.3 = 1.8 × 10 –2 M/ min Comprehension XIII A B t = 0 0.15 M 0 t = 10 (0.15 – x ) M x M = 0.125 M = 0.025 M t = t eq (0.15 – 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 = + ⇒ 10 0 693 min . = + K K f b (2)
Solution: From (1) and (2), K f = 2.31 × 10 –2 min –1
Solution: From (1), K b = 4.62 × 10 –2 min –1
Solution: t 1/2 = 10 min Comprehension XIV
Solution: E K E K E K K a a a ( ) overall = ⋅ + ⋅ + 1 2 1 2 1 1 or, 10 5 12 9 1 2 1 2 . = × + × + K K K K ⇒ K 1 = K 2 or, A e A e E RT E RT a a 1 2 1 2 ⋅ = ⋅ − − / / or, E E RT A A a a 1 2 1 2 − = ln ⇒ ( ) ln 12 9 10 2 2 10 2 10 3 14 14 2 − × × = × × × − T e ∴ T = 750 K
Solution: 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 > .
Solution: 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
Solution: C → D [ : ] . . . E a A B C D 27 5 30 1 4 3 ⎯ → ⎯⎯ ⎯ → ⎯⎯ ⎯ → ⎯
Solution: C → B [ : D A] . . . E a 67 9 16 9 19 9 ⎯ → ⎯⎯ ⎯ → ⎯⎯ ⎯ → ⎯⎯ C B
Solution: C → D [Lowest E a ]
Solution: B → C [Highest E a ]
Solution: D → C [Highest E a ]
Solution: Molecularity can never be fractional.
Solution: t A K n n 100 0 1 1 % [ ] ( ) = − − when n < 1 = Infi nite when n ≥ 1
Solution: For a particular step, rates always increase with increase in temperature.
Solution: Relative increase in rate constant with increase in temperature is higher for the reaction with higher activation energy.
Solution: Δ H = E E a a f b −
Solution: For zero order reaction : t A K t A K 1 2 0 100 0 2 / % [ ] , [ ] = =
Solution: Order is in dependent from stoichiometry of reaction.
Solution: t A K 1 2 0 2 / [ ] =
Solution: (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 = − ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ = − − ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ 1 4 1 2 3 1 1 56 1 1 / ⇒ x a = 7 8\n11.58 Chapter 11 HINTS AND EXPLANATIONS (S) K a a x = − = 2 3 18 30 ⇒ x a = 5 9 (T) K a a x = − = 2 16 28 ⇒ x a = 7 8
Solution: (A) d C dt K B [ ] [ ] = 2 For d C dt [ ] max ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ , [ B ] should be maximum and hence t K K K K K = ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ − = 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 [ ] ( ) [ ] − − = − ⋅ − ⋅ − ⎡ ⎣ ⎢ ⎤ ⎦ − − − − ⎥ ⎥ or, e e K e K e K K t K t K t K t − − − − − = − ⋅ − ⋅ 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 ⋅ − ⋅ = − ⋅ − ⋅ − − − − ∴ 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 ⋅ = − − − − − K K K e K K t 2 1 1 1 1 2 − = − − ( ) ∴ t K K K K K = − ⋅ − 1 2 1 2 1 2 1 ln
Solution: K K K BrO BrO Br − − − = = 3 1 2 3 ∴ K a BrO M s 3 0 06 3 0 02 1 1 − = = = − − . . and K b Br M s − = = × = − − 2 3 0 06 0 04 1 1 . .
Solution: 7 2 10 3600 2 10 15 1 8 2 . ( × = × × − − − M s M) K K = − − 1 200 1 1 M s = 5 ml mol –1 s –1
Solution: − = ⋅ ⋅ dP dt K P P a b NO H 2 1 5 0 25 372 152 . . = ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ a ⇒ a = 2 and 1 60 0 79 289 144 . . = ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ b ⇒ b = 1
Solution: 0 1 0 4 0 20 . . . = x ⇒ x = 0.8 0 1 0 8 0 2 0 05 0 4 . . . . . = × × y ⇒ y = 0.2
Solution: t K = ⋅ + + 1 3 0 3 ln [ ] [ ] Cr Cr = × ⋅ − − − 1 9 10 100 100 80 5 1 s ln = 1.8 × 10 4 sec = 5 hrs
Solution: (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
Solution: Time for certain progress of reaction, t a [ A 0 ] 1 – n 1 10 0 25 10 0 02 0 04 3 3 1 × × = ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ − − − . . . n ⇒ n = 3
Solution: C 4 H 8 2C 2 H 4 t = 0 a mole 0 t = t ( a – x ) mole 2 x mole As, a – x = 2 x ⇒ x a = 3 Now, t t K a a x a a a = ⋅ − = × × − = − − ln ln 1 25 18 10 3 2 5 1 s hrs
Solution: For 2 % reaction, we may assume that rate is almost constant. r = K [ A ] ⇒ 2 100 1 × − [ ] min A = K [ A ] ⇒ K = 0.02 min –1
Solution: H 2 O 2 ( aq ) H 2 O( l ) + 1 2 O 2 ( g ) Δ H = (–287) – (– 187) = –100 KJ/mol Moles of H 2 O 2 reacted per sec = 7.5 × 10 –4 × 0.02 × 2 = 3 × 10 –5 ∴ Heat produced per sec = 3 × 10 –5 × (100 × 10 3 ) = 3 J
Solution: t K a a x = ⋅ − = × ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ⋅ − − 1 1 4 5 3 1536 10 100 40 8 1 ln . . ln s = × × × × = 0 9 3 1536 10 4 5 3 1536 10 2 8 7 . . . . Year Years
Solution: t 1 2 4 0 693 6 93 10 1000 / . . sec = × = − A n B t = 0 a mole 0 t = 1000 sec a 2 mole n a ⋅ 2 mole Now, a n a a 2 2 3 + ⋅ = ⇒ n = 5
Solution: r = K [ester][H + ] x = k 1 [ester] K 1 = K ⋅ [H + ] x 1 0 10 10 10 3 2 . = ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ − − x ⇒ x = 1 and K K y 1 1 3 3 1 0 10 10 1 = = × = = + − − [ ] . H
Solution: t 3/4 = 2 × t 1/2 and hence, a =
Solution: Now, t K K b 1 2 2 2 / ln ln [ ] = = + H 1 0 0 5 0 02 0 01 . . . . = ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ b ⇒ b = 1
Solution: Concentrations are in G.P. and hence, order =
Solution: 18. A 2 B 3 ( aq ) 2A 3+ ( aq ) + 3 B 2– ( aq ) t = 0 a mole 0 0 t = 10 min a – x 2 x 3 x Now, p = CRT = r gh ⇒ total mole a h ∴ a a x + = 4 2 6 ⇒ t = 10 min = t 1/2 Now, at t = t 3/4 = 2 × t 1/2 = 20 min, x a = 3 4 ∴ a a a h + × ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = 4 3 4 2 ⇒ h = 8 mm p = x = r gh = 1 0 1000 0 8 3 2 . ( . gm cm cm s cm) ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ × ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ × = = 800 80 2 dyne cm pascal Now, x y = = 80 20 4
Solution: [ ] A t 4 1 1 = + ⇒ 4 1 1 3 2 8 [ ] [ ] ( ) [ ] A d A dt t A ⋅ = − + = − ∴ − = = = × − − d A dt A [ ] [ ] ( . ) 5 5 5 1 4 0 2 4 8 10 M s
Solution: A 2B t = 0 a mole 0 t = t ( a – x ) mole 2 x mole From mass conservation, a × M 0 = ( a + x ) × M t ∴ x a M M M t t = − ( ) 0 If the reaction is zero order, then K a a x t x t a M M t M t t = − − = = − ⋅ ( ) ( ) 0\n11.60 Chapter 11 HINTS AND EXPLANATIONS For t = 10 min, K a a = − × = ( ) 42 35 10 35 50 For t = 20 min, K a a = − × = ( ) 42 30 20 30 50 As K values are same, the reaction is of zero-order.
Solution: r = K [ A ] n and 2 r = K (4[ A ]) n ⇒ n = 1 2 ∴ t 1/2 a [ A 0 ] 1 – 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.
Solution: P K P K K K e B B A A B C K K K t A B C = ⋅ ° + + ⋅ − − + + ⋅ [ ] ( ) 1 = × × × − = − − − × × − 2 10 13 86 6 93 10 1 2 3 3 6 93 10 100 3 . . [ ] . e atm
Solution: 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 = ⋅ ⋅ = ⋅ − − − − / / ( )/ = = × = − × × 100 1 4 606 10 2 500 3 e . /
Solution: Fraction of molecules having sufficient energy = = = × − − × × − e e E RT a / . / . 83 14 10 8 314 500 9 3 2 10
Solution: ln ln K K t t E R T T a 2 1 1 2 1 2 1 1 = = − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ or, ln 1 3 1 300 1 280 = − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ E R a (1) and ln 16 1 300 1 330 t E R a = − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ (2) From (1) and (2), t = 4 hrs Four Digit Integer Type
Solution: r = K [O 3 ] 2 = 5 × 10 –4 × (2 × 10 –8 ) 2 = 2 × 10 –19 mol l –1 s –1 = 2 × 10 –19 × 6 × 10 23 × 10 –3 × 60 = 7200 molecules ml –1 min –1
Solution: At t = ∞ , 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 ° = C H Br 2 5 0 mm then P unreactive gas = (200 – 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 – x x x t = ∞ 0 P 0 P 0 From question, P 0 + P 0 + (200 – P 0 ) = 390 ⇒ P 0 = 190 and ( P 0 – x ) + x + x + (200 – P 0 ) = 342.5 ⇒ x = 142.5 ∴ Percentage C 2 H 5 Br undecomposed = P x P 0 0 25 − = %
Solution: t t A A = ⋅ gen log log [ ] [ ] 2 0 ⇒ 96 0 30 3 = ⋅ t gen . log ⇒ t gen = 60 hrs
Solution: r dP dt K P P = − = ′ ⋅ ⋅ NO NO O 2 2 and ′ = × × − − K 1 6 10 0 08 600 5 2 2 1 . ( . ) atm s ∴ r = × × × ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ × ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 1 6 10 48 48 190 760 288 760 5 2 . = = − − 1250 760 1250 1 1 atm s mm s
Solution: From the unit of rate constant, the process is zero order. ∴ t K 100 0 % [ ] = + H = × × ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ × = × = − − 3 10 0 05 1000 1 0 10 6 10 1000 7 7 4 . . sec min
Solution: K t A A = = 1 0 ln [ ] [ ] Constant ∴ 1 100 3 1 9 0 0 0 0 ⋅ = ⋅ ln [ ] [ ] / ln [ ] [ ] / A A t A A ⇒ t = 200 min
Solution: K t A A = ⋅ = 1 0 ln [ ] [ ] Constant\n11.61 Chemical Kinetics HINTS AND EXPLANATIONS ∴ 1 20 500 420 1 100 70 ⋅ = ⋅ ln ln t ⇒ t = 40 min
Solution: K t V V V t = ⋅ − = ∞ ∞ 1 ln Constant ∴ 1 40 80 80 40 1 80 80 70 ⋅ − = ⋅ − ln ln t ⇒ t = 120 min
Solution: 2P 4Q + R + S(l) t = 0 P 0 0 0 t = 30 min P 0 – x 2 x x 2 V.P. = 25 t = 60 min P 0 – y 2 y y 2 V.P. = 25 t = ∞ 0 2 P 0 P 0 2 V.P. = 25 From question, 2 2 25 625 0 0 P P + + = ⇒ P 0 = 240 and ( ) P x x x 0 2 2 25 445 − + + + = ⇒ x = 120 Now, 1 30 1 60 0 0 0 0 ⋅ − = ⋅ − ln ln P P x P P y ⇒ y = 180 ∴ P P y y y 60 0 2 2 25 535 = − + + + = ( ) mm
Solution: Initial moles of NH 4 NO 2 = 200 0 02 1000 0 004 × = . . and moles of N 2 O formed = ( ) . . . 785 25 760 49 26 1000 0 0821 300 0 002 − × × = ∴ t req = t 1/2 = 123 min
Solution: For set 1 and 2, r = K ′ [ B ] as [ A 0 ] >> [ B 0 ] and t K K A 1 2 0 2 2 2 / ln ln [ ] = ′ = ⇒ x = 62.5 For set 3 and 4, r = K ″ [ A ] 2 as [ B 0 ] >> [ A 0 ] and t K A K B A 1 2 0 0 0 1 1 / [ ] [ ][ ] = ′′ = ⇒ y = = 625 2 312 5 . ∴ x + y = 62.5 + 312.5 = 375
Solution: t = 43.5 min = 3 t 1/2 Hence, P ether atm = = 4 2 0 5 3 . ⇒ Δ P ether = 3.5 atm ∴ P fi nal = 0.5 + 3.5 × 3 = 11 atm
Solution: Δ = ⋅ t t r r 1 2 1 2 2 / ln ln ⇒ 12 2 0 04 0 03 1 2 = ⋅ ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ t / ln ln . . ∴ t 1/2 = 28 min = 1680 sec
Solution: t t V V V V t = ⋅ − − ∞ ∞ 1 2 0 2 / ln ln ⇒ 120 2 60 20 60 55 1 2 = ⋅ − − t / ln ln ∴ t 1/2 = 40 min ab = 40 Now, [ [ester] HCl] = − ∞ V V V 0 0 ⇒ [ . HCl] 6 0 20 60 20 = − ⇒ [HCl] = 3.0 M ∴ cd = 03
Solution: t 1 2 3 0 693 1 386 10 500 / . . sec = × = − Let the initial moles of A = x , then after 500 sec, A 2B + C x x − 2 2 2 × x x 2 = x 2 = x = x 2 Total moles becomes x x x x 2 2 2 + + ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = . As moles becomes double, volume becomes double and hence, [ ] . . A req M = × = 0 1 2 2 0 025 = 25 millimole per litre
Solution: A 2B + C t = 0 4 a 0 3 a t = t 4 a – x 2 x 3 a + x From question (4 a – x )(40°) + 2 x (10°) + (3 a + x ) (–30°) = 0° ∴ x a = 7 5 Now, t K a a a = ⋅ − = ⋅ = 1 4 4 7 5 1 0 001 20 13 500 ln . ln min\n11.62 Chapter 11 HINTS AND EXPLANATIONS
Solution: Exp (1): r = K ′ [ B ] y as [ A 0 ] >> [ B 0 ] ∵ t t 7 8 1 2 3 / / = × ⇒ y = 1 Exp (2): r = K ″ [ A ] x as [ A 0 ] << [ B 0 ] ∵ t t 7 8 1 2 7 / / = × ⇒ x = 2 Now, for exp (2) and (3), t K A K B A 1 2 0 0 0 1 1 / [ ] [ ][ ] = ′′ = ∴ a = × = 10 2 2 2 5 . and b = 7 × 2.5 = 17.5 And for exp (1) and (4), t K K A 1 2 0 2 2 / ln ln [ ] = ′ = ∴ c = 30 × 2 = 60 and d = 3 × 60 = 180
Solution: Percentage yield = K K K 2 1 2 100 4 8 3 2 4 8 100 60 + × = + × = . . . %
Solution: A + 2B + 3C D t = 0 1.0 M 1.0 M 1.0 M 0 t = t 1 – x 1 – 2 x 1 – 3 x x = 0.9 = 0.8 = 0.7 = 0.1 (given) ∴ r = 2 × 10 –6 × (0.9) 2 – 1 4 10 0 1 0 8 0 7 1 595 10 6 2 6 . ( . ) . . . × × × = × − −
Solution: [ C ] = 0.875 + 0.6 = 1.475 M
Solution: K K K B A eq f b = = [ ] [ ] ⇒ 1 38 300 0 1 0 2 . / . . K b = ⇒ K b = − 1 38 150 1 . min Now, t K K x x x f b e B e B B = + ⋅ − = + ⋅ − × 1 1 1 38 300 2 76 300 0 1 0 1 0 3 25 100 ln . . ln . . . , , = × ⋅ = 300 6 2 4 100 ln ln min
Solution: For completion in 30 min, the rate should be increased by 4 60 30 8 × = times. Assuming temperature coefficient constant, the approximate temperature is 25 10 8 2 55 + × ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = ° C .
Solution: K 1 = K 2 ⇒ A e A e E RT E RT a a 1 2 1 2 ⋅ = ⋅ − − / / ∴ ln A A E E RT a a 2 1 2 1 = − ln ( . . ) . 10 10 171 39 152 30 10 8 3 14 13 3 = − × × T or, T = 1000 K = 727° C
Solution: t K A e E RT a 1 2 2 2 / / ln ln = = ⋅ − or, 1 60 0 7 5 10 13 149 4 10 8 3 3 × = × × − × × . . / . e T ∴ T = 500 K
Solution: K cat = K uncat or, A e A e E RT E RT a a ⋅ = ⋅ − − cat uncat / / 1 2 or, E T E T a a cat uncat 1 2 = ⇒ E E a a uncat uncat − = 20 400 500 ∴ E a uncat kJ/mol = 100
Solution: Greater the specific surface area of adsorbent, greater will be the extent of adsorption.
Solution: Adsorption decreases the surface energy.
Solution: x m K P x m K n P n = ⇒ = + ⋅ . log log log 1 1 From question, log K = 0.3010 = log 2 ⇒ K = 2 And 1 45 1 1 n n = ° = ⇒ = tan ∴ x m P = × = × = 2 2 0 2 0 4 . .
Solution: K A e E RT a = − . / ∴ ln K K E R T T a 2 1 1 2 1 1 = − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ or, ln 10 cal/mol = − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ⇒ = E R E a a 1 600 1 1000 6900
Solution: For a particular combination of adsorbent, adsorbate and temperature, only one value of ‘ n ’ is permissible.
Solution: x m K P n = ⋅ 1 ⇒ 0 2 4 1 . ( ) = × K n (1) 0 5 25 1 . ( ) = × K n (2) 0 8 64 1 . ( ) = × K n (3) From (1), (2) and (3), K n = = 1 10 2 and ∴ x m K n ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = × = required ( ) . 36 0 6 1 Hence, moles of N 2 adsorbed per gm of iron = = 0 6 28 3 140 .
Solution: r r 1 2 < ⇒ A is the catalyst. r r r r r 3 1 1 4 5 < ⇒ = = ⇒ B is the catalyst. C and D are not catalysts. . r r r r r r C 1 7 2 1 2 6 < < ⇒ < < ⇒ D is catalytic poison. is catalytic prom motor.
Solution: A catalyst always involve in the reaction.
Solution: Enzymes are specific.
Solution: A catalyst does not initiate the reaction.
Solution: Catalyst does not alter the equilibrium position.
Solution: Homogeneous catalysis, because the physical states of both reactant and catalyst is aqueous (liquid).
Solution: Activation energy is decreased.
Solution: Catalyst lowers the activation energy.
Solution: Catalysis occurs through chemisorption.
Solution: Viscosity is higher and surface tension is smaller than water.
Solution: K , as it reacts vigorously in water.
Solution: SnCl 4 formed by reaction will adsorb some common Cl – ions.
Solution: True solution or suspension does not show Tyndall eff ect.
Solution: Blood, clay and smoke are negative sol. In strong acidic solution, gelatin adsorbs some H + ions and become positive.
Solution: Informative EXERCISE II (JEE ADVANCED)\n12.24 Chapter 12 HINTS AND EXPLANATIONS
Solution: Volume of metal used = × = − − 1 9 10 19 10 4 5 . cm 3 ∴ N × × × = − − 4 3 10 10 10 7 3 5 3 ( ) cm cm ⇒ N = × 2 39 10 12 . Hence, number of particles per cm 3 = × = × 2 39 10 1000 2 39 10 12 9 . .
Solution: Larger the carbon chain, normally smaller is CMC.
Solution: Sulphide sol have negative charge on colloidal particles
Solution: Adsorption is physisorption.
Solution: Entropy decreased in adsorption.
Solution: Δ H can never be equal to Δ S.
Solution: (a) Chemisorption does not change into physisorption at higher pressure. (b) CO or CO 2 gases leave the surfaces.
Solution: In Lindlar’s catalyst, catalytic poison is used
Solution: All catalytic reaction is multistep reaction
Solution: (a) K K e A e A e e E RT E RT E E RT a a a a cat uncat = = = ′ ′ ′ − 20 . . / / ( )/ ∴ 20 2 2 1000 300 = − = − × ′ E E RT E a a a Kcal ⇒ E a = 14 Kcal/mol (b), (c) Reaction: 2H 2 O 2 (aq) → 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.
Solution: 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.
Solution: Electrophoresis and electro-osmosis are the experimental methods to determine charge on colloidal particles.
Solution: Below CMC, the solution is true solution.
Solution: PO SO Cl 4 3 4 2 − − − > > ⇒ Sol particles are positively charged.
Solution: RCOONa RCOO Na \u001c − + + 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.
Solution: Due to excess Ag+, sol particles will be positively charged.
Solution: (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.
Solution: (a) Basic dye is positively charged and hence, Fe(CN) HPO 6 4 3 2 − − > (c) Slope should not change in Freundlich’s isotherm.
Solution: Tyndall effect is shown by colloids.
Solution: Charge : Mg 2+ (2 unit) > Cl – (1 unit) Hence, better coagulation for negatively charged gold sol.\n12.25 Surface Chemistry HINTS AND EXPLANATIONS
Solution: Polarizability is maximum in Xe.
Solution: CO is polar and hence, more preferential adsorption.
Solution: Adsorption decreases on increasing temperature. Comprehension II
Solution: In case of concentrated KCl, KCl adsorbs on blood charcoal surface, but in case of dilute KCl, blood charcoal dissolves in KCl solution.
Solution: Greater critical temperature, greater the extent of adsorption.
Solution: Adsorption is always exothermic. Comprehension III
Solution: Initial surface area, A 1 2 2 6 2 24 = × = ( ) cm cm Final volume of each cube = cm 3 8 10 12 ∴ Final side length of each cube = × ( ) 8 10 12 1 3 cm 3 / = × − 2 10 4 cm Hence, final surface area of each cube, A 2 = 6 × (2 × 10 −4 cm ) 2 = 24 × 10 −8 cm 2 ∴ Final total surface area Initial surface area = × × − 24 10 10 2 8 12 4 4 10 4 =
Solution: Number of H 2 molecules = × × × × = × 2 0 112 0 0821 546 6 10 3 10 23 21 . . ∴ Specific surface area = × × × − 3 10 0 4 10 5 21 7 2 . ( ) cm gm = × 2 4 10 6 . cm /gm 2 Comprehension IV
Solution: log log log x m K n P ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = + ⋅ 1 Slope = = ⇒ = 1 0 25 4 n n . and for x -intercept, log x m ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = 0 ⇒ log log K n P = − ⋅ 1 = − × − = 1 4 4 1 0 ( ) . ∴ K = 10
Solution: x m = × = 10 16 20 1 4 ( ) / ⇒ x = × = 20 10 200 gm
Solution: 810 1 0 10 1 4 . ( ) / = × P ⇒ P = 3 atm\n12.26 Chapter 12 HINTS AND EXPLANATIONS Comprehension V
Solution: Positively charged due to adsorption of Ag + ions.
Solution: Informative Comprehension VI
Solution: Positively charged due to adsorption of Fe 3+ ions.
Solution: AgI Ag NO fixed la , , - + ↓ ↓ 3 y yer diffused layer
Solution: S 2– ions get adsorbed. Comprehension VII
Solution: Gold number = 0.025 × 1000 = 25
Solution: Colour become less intense due to adsorption.
Solution: Surface particles have higher energy due to unbalanced forces.
Solution: Word 'always' is not suitable because chemisorption increased with increase in temperature.
Solution: Cellulose nitrate sol is lyophilic.
Solution: Scattering is not related to speed of particles.
Solution: Colloidal particles are negatively charged due to adsorption of I - ions and hence, it moves towards anode.
Solution: Peptization occurs due to adsorption of common ion.
Solution: Natural colloids are normally lyophilic.
Solution: Informative\n12.27 Surface Chemistry HINTS AND EXPLANATIONS
Solution: 6 48 0 01 10 162 10 3 3 . ( . ) = × × × − n ⇒ n = 4
Solution: log log log x m K n P = + ⋅ 1 From the given graph, log K = 1.0 ⇒ K = 10 and 1 0 25 n = . ⇒ n = 4 Now, x m K P n = . 1 ⇒ x 1 0 10 8 1 10 3 1 4 . ( . ) / = × × − ⇒ x = 3 gm
Solution: t k A e s e av E RT a = = = × × = − − − × × 1 1 1 1 25 10 4 8 1 16 10 2 400 3 . ( . ) sec / /
Solution: Soap solution of sodium palmitate, gold sol, silicic acid sol, acidic dye, metal sulphide sol, sol of AgCl by excess KCl in AgNO 3 .
Solution: ln P P R T T ads 1 2 1 2 1 1 = − Δ H or, ln H 1 6 32 1 200 1 250 . = − Δ ads R ⇒ ∆ H ads = 6000 cal/mol
Solution: Number of CH 3 COOH molecules adsorbed = × − × × = × 100 0 5 0 49 1000 6 10 6 10 23 20 ( . . ) ∴ Surface area of each molecule = × × = × − 3 10 6 10 5 10 2 20 19 m 2
Solution: Number of N 2 molecules = × × × × × = × − 0 001 2 46 10 0 082 300 6 023 10 6 023 10 3 23 16 . . . . . ∴ Number of active sites per molecule = × × × × = 1000 6 023 10 20 100 6 023 10 2 14 16 . .
Solution: Number of N 2 molecules absorbed = × × × = × − 2 24 10 22 4 6 10 6 10 3 23 19 . . ∴ Specific surface area = × × × = − 6 10 0 15 10 9 19 9 2 . ( )
Solution: Colloid is a heterogeneous system ⇒ min = 2 phases
Solution: ln 20 1 600 1 1000 = − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ E R a ⇒ E a = 9000 cal/mol. Four-digit Integer Type
Solution: Mass of NaCl used = × × = × ( ) 585 1 2 1 100 5 85 1 2 . / . . gm Moles of NaCl used = × = 5 85 1 2 58 5 0 12 . . . . ∴ Coagulation value = × = 0 12 10 200 1000 600 3 . / millimole/litre
Solution: Number of palmitic acid molecules needed = × = × − 480 0 2 10 2 4 10 7 2 17 cm cm 2 . ( ) . Moles of palamitic acid molecules = × × = × − 2 4 10 6 10 4 10 17 23 7 . ∴ Volume of solution needed = × × = × = − − 1 5 12 256 4 10 2 10 20 7 5 3 dm dm mm 3 3 . /\n12.28 Chapter 12 HINTS AND EXPLANATIONS
Solution: The radius of hydrogen molecule = × ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ m d 3 4 1 3 π / = × × × ⎛ ⎝ ⎜ ⎜ ⎜ ⎞ ⎠ ⎟ ⎟ ⎟ = × − 2 6 10 0 16 3 4 2 5 10 23 1 3 8 . . / π π cm Number of hydrogen molecules at the surface per gm Cu = × × = × 224 22400 6 10 25 2 4 10 23 20 / . π π ∴ Specific surface area of Cu = cm /gm m /gm 2 2 2 4 10 2 5 10 150000 15 20 2 8 . ( . ) × × × × = = − π π
Solution: t k av = 1 Now, ln ln k k t t E R T T a 2 1 1 2 1 2 1 1 = = = − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ or, ln . . 0 36 0 72 1 2500 1 2000 = − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ E R a ⇒ E a = 14000 cal/mol
Solution: t t K K A e A e Fe 1 2 1 2 20 10 2 600 8 3 / / / . . ( ) ( ) = = − × × − × Fe charcoal charcoal 1 10 2 600 10 3 1 / × = e
Solution: Number of adsorbate molecules = × × × × = × − 0 10 10 0 25 10 6 10 2 4 10 3 3 23 7 . . . ∴ Eff ective surface area = × = × − 0 06 2 4 10 25 10 17 20 2 . . m
Solution: Moles of gas adsorbed per gm of charcoal = − × × × × ( ) . 700 400 1 52 760 300 6 R Volume of gas adsorbed per gm of charcoal (at 0°C and 1 atm) = × × × × × × 300 1 52 760 300 6 273 1 . R R = 0.091 litre
Solution: Specific surface area of silica gel = × × × × = − 168 22400 6 10 0 16 10 720 23 9 2 . ( ) m /gm 2
Solution: ln P P R T T 1 2 1 1 1 1 = − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ Δ θ H or, ln . . . . 0 4 59 2 16 628 10 8 314 1 200 1 3 = − × − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ T ⇒ T = 400 K
Solution: r K K E S K K K S K K E S K K S = + + + − − 1 2 0 1 2 1 1 2 0 1 1 [ ][ ] [ ] [ ][ ] [ ] \u001b For r max , K S K 1 1 [ ] \u001a − ∴ 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 [ ] [ ][ ] [ ] = + − ⇒ K K S K S − + = 1 1 1 2 [ ] [ ] ∴ K K S 1 1 3 3 6 3 1 1 250 250 10 4000 − − = = = × = [ ] mg dm dm kg dm kg
Solution: e m ratio of cathode rays is independent to the nature of gas.
Solution: e m e m e e m mA A B B ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = × A B ⇒ 2 3 3 2 = × e e A B ⇒ e e A B = 4 9
Solution: eV = 1 2 2 m v ⇒ v e m = × × 2 V = × × × = × 2 1 764 10 200 8 2 10 11 6 . . m/s
Solution: e m e m ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = × ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = mesor -particle α 1 1 1836 208 2 4 17 65 1 .
Solution: eV = = 1 2 2 2 2 mv p m ⇒ p m = 2 eV ∴ p p p e = = 1836 1 42 85 1 .
Solution: υ υ = = × ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ × × ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = × − c s z 3 10 2 10 50 1 2 10 10 6 1 15 cm cm H .
Solution: p E t n hc t = = ⋅ ⋅ = × × × × × × × − − λ 6 10 6 626 10 3 10 1 662 6 10 15 34 8 9 . . = 1.8 × 10 –3 J/s – m 2
Solution: p E t n hc t = = ⋅ ⋅ λ ⇒ 14 100 200 6 626 10 3 10 1 1987 8 7 10 34 8 9 × = × × × × × × − − n . . ∴ n = 4 × 10 19 s –1
Solution: E n hc hc = ⋅ = × × × × = − − λ ( . N ) A 1 75 10 2500 10 84 4 10 J
Solution: E x E abs × = 100 emit ⇒ n hc x n hc 1 1 2 2 100 ⋅ ⋅ = ⋅ λ λ ∴ x n n = × = × = 2 1 1 2 53 100 4530 5080 47 3 λ λ .
Solution: λ = = 1240 5 248 nm
Solution: E = n ⋅ hc υ = 1 × 6.626 × 10 –34 × 3 × 10 8 × 1650763.73 = 3.28 × 10 –19 J/quanta.
Solution: E n h c = λ ⇒ 0 36 6 626 10 3 10 662 6 10 34 8 9 . . . = × × × × × − − n ∴ n = 1.2 × 10 18
Solution: Energy needed for photochemical dissociation = + = + ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = 482 5 1 2 482 5 96 5 1 2 6 2 . . . . . . KJ mol eV eV eV ∴ λ ≈ = 1240 6 2 200 . nm
Solution: λ ≈ ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = 1240 289 5 96 5 413 33 . . . nm
Solution: Energy absorbed per mole of H 2 = × × × × × × × = − − − 6 10 6 6 10 3 10 270 10 10 440 23 34 8 9 3 . KJ ∴ Percentage of absorbed energy corrected into K.E. = 440 429 440 100 − × = 2.5 %
Solution: E = × × = 9 12400 6900 23 372 kcal/mole ∴ Energy conversion efficiency = × = 111 6 372 100 30 . % EXERCISE II (JEE ADVANCE)\n13.45 Atomic Structure HINTS AND EXPLANATIONS
Solution: E n hc = ⋅ λ ⇒ 6 626 6 626 10 3 10 360 10 34 8 9 . . = × × × × × − − n ∴ Mole of photons absorbed = = × × = × − n A ~ . 1 2 10 6 10 2 10 19 23 5 ∴ Quantum efficiency = × × − − 1 10 2 10 0 5 5 5 .
Solution: h h E υ υ 1 0 = + and h h υ υ 2 0 = + ⋅ E K ∴ υ υ υ 0 1 2 1 = − − K K
Solution: 1 2 2 0 mv hc hc max = − λ λ ⇒ v hc m max = − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ 2 0 0 1 2 λ λ λ λ
Solution: h h υ υ = + K.E. 0 ⇒ υ υ = ⋅( ) + 1 0 h K.E.
Solution: c a z λ = − ( ) 1 and c a z 4 1 λ = ′ − ( ) ∴ ′ = + z z z 1
Solution: Number of atoms in the disc = × × = × 1 12 6 10 5 10 23 22 Now, F K q q r = ⋅ 1 2 2 ⇒ 10 9 10 10 5 9 2 2 2 − − = × × q ( ) ⇒ q = − 10 3 10 C ∴ Number of excess electron on negatively charged disc = × = − − 10 3 1 6 10 10 4 8 10 19 9 / . . Hence, Number of excess electron Number of atoms = × = − 10 4 8 5 10 10 9 22 / . 1 14 2 4 .
Solution: r n = r 1 × n 2 ⇒ 21.2 × 10 –11 = 5.3 × 10 –11 × n 2 ⇒ n = 2
Solution: 2 p r n = 26.5 Å ⇒ 2 p × 0.529 × n 2 2 = 26.5 ⇒ n = 4
Solution: r n = 0.529 × n z 2 Å
Solution: r n = r 1 × n 2 ∴ r n – r n – 1 = r 1 × n 2 – r 1 × ( n – 1) 2 = (2 n – 1) ⋅ r 1 Where n is the higher orbit.
Solution: A A r r r r r r 2 1 2 2 2 1 2 1 2 1 2 1 2 2 16 1 = = ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = × ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = ( ) ( ) π π
Solution: r 4 – r 2 = 2.116 Å ⇒ 0.529 × 4 2 z – 0.529 × 2 2 z = 2.116 ∴ z = 3 ⇒ Li 2+ ion
Solution: d = 2 p r × 100 = 2 p × 0 529 2 4 10 2 10 . × × ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ − m × 100 = 3.32 × 10 –8 m
Solution: Circumference = Z p r = 2 p ⋅ r 0 × n 2 1 = 2 p r 0 n 2 and n = 1, 2, 3,.…
Solution: V n = 2.188 × 10 6 z n m/s ⇒ 0.547 × 10 6 = 2.188 × 10 6 × 1 n ∴ n = 4 Now, r n = 0.529 × n z 2 = 0.529 × 4 1 2 = 8.464 Å
Solution: m r ze r ν πε 2 0 2 2 1 4 = ⋅ ⇒ ν πε = ( ) ze m r 2 0 4
Solution: v c ze nh c = ( ) ⋅ 2 4 2 0 π πε
Solution: Solution of Q.36
Solution: F ze r = ⋅ = × × × × × = × − − − 1 4 9 10 2 1 6 10 4 10 2 88 10 0 2 2 9 19 2 10 2 9 πε ( . ) ( ) . N
Solution: T n z n z , . sec = × − 1 5 10 16 3 2 T T 2 3 3 2 3 2 2 2 3 3 , , / / H Li e + 2+ = ⇒ T x 2 2 3 , sec H e + =\n13.46 Chapter 13 HINTS AND EXPLANATIONS
Solution: r r n n 1 2 1 2 2 2 = ⇒ r r n n 4 2 1 2 2 = ⇒ n n 1 2 1 2 = ∴ T T n n 1 1 3 2 3 1 8 2 = =
Solution: T n a n 3 and n a r n ⇒ T n a r n 3 2 /
Solution: N T n z = = × × = × − − − 10 10 1 5 10 2 1 8 33 10 8 8 16 3 2 6 sec , . .
Solution: λ ν = = ⋅ = × × × × = × − − c c T n z , . . 3 10 1 5 10 1 1 4 5 10 8 16 3 2 8 m
Solution: K.E. J V = = ⋅ ( )⋅ = ⋅ 1 2 1 2 2 2 mv mvr r r ν
Solution: K.E. = = ⋅ ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = 1 2 1 2 2 8 2 2 2 2 2 2 mv m nh mr n h mr π π = ⋅ ⋅ = ⋅ n h m a n h ma n 2 2 2 0 2 4 2 2 0 2 2 8 8 π π ( )
Solution: Δ = − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = × − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = E n n ( ) . . I.E. eV 1 1 14 4 1 1 1 4 13 5 1 2 2 2 2 2
Solution: Reduced mass effect: ′ = ⋅ + ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ r r m m e n 1 On increasing the nuclear mass, radius decreases.
Solution: Reduced mass effect: (I.E.) ′ = ( )⋅ + ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ I.E. 1 1 m m e n On increasing the nuclear mass, ionisation energy increases.
Solution: u m m m m m m m m m m p p p p p e = + = ⋅ + = = × 1 2 1 2 2 1836 2 ∴ r pm = = 0 529 918 0 058 . . Å
Solution: n = 2 but z = 3 – 2 = 1 ∴ E 2 2 2 13 6 1 2 3 4 = − × = − . . eV and I.E. = 3.4 eV
Solution: K.E. of emitted electron = 0.5 × 13.6 eV Now, K.E. mV = 1 2 2 or, 6 8 1 6 10 1 2 9 1 10 19 31 2 . . . × × = × × × − − v ⇒ v = 1.55 × 10 6 m/s
Solution: Δ = = = × − E 1240 589 6 2 1 19 . . eV 3.37 10 J = 48.5 kcal/mol
Solution: λ = = × 1240 0 0141 8 8 10 4 . . nm = 8.8 × 10 –5 m = 88 nm
Solution: Minimum is 1 (4 → 1 transition in both atoms) and maximum is 4 (4 → 3 → 2 → 1 in one atom and any other transition in other atom).
Solution: Number of available orbits is only
Solution: Hence, maximum number of spectral lines = 4 6 2 C = .
Solution: 12.1 10.2 1.9 At least two atoms are needed for these three transitions.
Solution: 1 1 1 2 1 2 1 4 2 1 2 2 2 2 2 2 λ = − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = × − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ Rz n n R ⇒ l = 1223 Å ∴ UV region.
Solution: n n ( ) − = 1 2 15 ⇒ n = 6 Now, for shortest wavelength, required transition is 6 →
Solution: ∴ 1 1 1 1 1 6 35 36 2 2 2 λ = × − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = R R ⇒ λ = 36 35 R
Solution: 1 1 1 2 1 4 4 2 2 2 2 2 λ = × − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = ⋅ − R n R n n ( ) ∴ λ = − = ⋅ − 4 4 4 2 2 2 2 n R n K n n ( ) ⇒ K R = 4
Solution: λα 1 2 z ⇒ λ λ λ H He Li : : : : : : + + = = 2 1 1 1 2 1 3 36 9 4 2 2 2
Solution: 1 1 1 1 1 2 2 2 λ = × × − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ R n ⇒ n R R = − λ λ 1\n13.47 Atomic Structure HINTS AND EXPLANATIONS
Solution: Required transition is 4 → 2 1 1 1 2 1 4 3 16 2 2 2 λ = × − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = R R ⇒ λ = 16 3 R
Solution: λα 1 2 z ⇒ λ λ Na H 10 1 10 2 2 + = ⇒ λ Na 10 12 16 + = . Å
Solution: Excited state is n = 6 [3 → 2, 4 → 2, 5 → 2, 6 → 2] ∴ Number of spectral lines in 1 R region =
Solution: 66. Required transition is 3 →
Solution: Modified Rydberg constant is given by, ′ = × = R R R me h c 2 2 4 2 4 0 2 3 as π πε ( ) Now, 1 1 1 2 1 3 2 5 36 2 2 2 λ = ′ × − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = ⋅ R R ∴ λ = 18 5 R
Solution: υ = ⋅ − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = − R n R n n 1 2 1 4 4 2 2 2 2 ( )
Solution: Δ = × − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = E 1312 1 2 1 3 182 22 2 2 . KJ
Solution: All are visible radiations. Next line is from transition 7 →
Solution: 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 . . × = × × − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ − n ⇒ n 1 =
Solution: Hence, the series is Brackett series.
Solution: 1 2 1 2 1 3 1 2 2 2 λ = × − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ R ⇒ λ 1 9 5 = R 1 2 1 1 1 2 2 2 2 2 λ = × − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ R ⇒ λ 2 1 5 = R From question, l 1 – l 2 = 132 nm or, 9 5 1 3 132 10 9 R R − = × − m ⇒ R = 1.11 × 10 9 m –1
Solution: ∞ n + 1 n 1 λ 2 λ υ = × = × × + − ∞ ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 2 725 10 1 09 10 1 1 1 1 6 7 2 2 2 . . ( ) n ∴ n = 3 Now, 1 1 09 10 2 1 3 1 4 7 2 2 2 λ req = × × − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ . ⇒ l req = 471.8 nm
Solution: 1240 108 5 1 2 1 5 2 2 . = − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ B.E. ⇒ B.E. = 54.4 eV
Solution: K.E. of electron = 13 6 2 1 1 1 2 13 6 27 2 2 2 2 . . . × − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ − = eV Now, 27.2 × 1.6 × 10 –19 = 1 2 9 1 10 31 2 × × × − . v ∴ v ≈ 3.1 × 10 6 m/s
Solution: X E n Y E n = = + 2 2 3 and ( ) ∴ X Y n = + 1 3
Solution: 1 1 1 1 1 3 2 2 2 λ = × × − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ R ⇒ λ = 9 8 R
Solution: n r λ π = 2 ⇒ 2 3 3 6 2 π π × × = x x
Solution: λ = h m 2 E ⇒ E m α λ 1 ( ) for same
Solution: λ = h m 2 E ⇒ λ λ α p = × × = 4 2 1 1 2 2 1
Solution: p mv mv v = = = 1 2 1 2 2 const ⇒ l = Constant
Solution: λ min . . = × × = × − − 1 24 10 5 10 2 48 10 6 4 11 m
Solution: m h c h c R = ⋅ = × × − ∞ ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = × − λ 2 1 2 1 2 4 10 2 2 2 35 . kg\n13.48 Chapter 13 HINTS AND EXPLANATIONS
Solution: λ = h m 2 E ⇒ E E 2 1 1 2 2 2 100 99 1 02 = ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ≈ λ λ . ∴ E 2 is about 2 % greater than E 1 .
Solution: Δ = ⋅ Δ = ⋅ = v h m x h m h mv v min 4 4 4 π π π
Solution: λ = h m 2 E ⇒ Δ = − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ E h m 2 2 2 1 2 2 1 1 λ λ = × × × × − × ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ − − − − ( . ) . ( ) ( ) 6 626 10 2 9 1 10 1 50 10 1 100 10 34 2 31 9 2 9 2 = 7.24 × 10 –23 J = 4.5 × 10 –4 eV
Solution: λ = = = × × × × × × × × − − − 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 × 10 –12 m
Solution: λ = ⋅ 3 32 . n z Å ⇒ 3 32 3 32 2 . . = × n ⇒ n = 2 Energy of photon liberated in 2 → 1 transition, Δ = × × − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = E 13 6 2 1 1 1 2 40 8 2 2 2 . . eV ∴ K.E. of emitted electron from H-atom = 40.8 – 13.6 = 27.2 eV Hence, its de Broglie wavelength is given by, λ = = 150 27 2 2 348 . . Å
Solution: K.E. of electrons = 12400 3000 12400 4000 1 03 − = . eV ∴ λ = = 150 1 03 12 05 . . Å
Solution: Δ ⋅ Δ ≥ x λ λ π 2 4 and λ = = 150 6 5Å ∴ Δ = ⋅ Δ = × × × ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = × − − − λ λ π π π min ( ) . 2 10 2 9 11 4 5 10 4 1 10 6 25 10 x m
Solution: Δ E = 2.55 eV = 13 6 1 1 1 2 1 2 2 2 . × − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ n n eV ∴ n 1 = 2 and n 2 = 4 Now, Δ = × − × = λ 3 32 4 1 3 32 2 1 6 64 . . . Å
Solution: Orbital angular momentum = + ⋅ l l h ( ) 1 2 π
Solution: Electron of 1 s level can never emit photon.
Solution: Maximum permissible value of l = ( n – 1)
Solution: m = –1 ⇒ l ≥ 1 ⇒ can not be s -orbital.
Solution: Energy 2 s < 2 p < 3 s < 3 p < 4 s < 3 d
Solution: m = –3, –2, –1, 0, +1, +2, +3
Solution: Number of radial nodes = n – l – 1 = 3 – 2 – 1 = 0
Solution: Probability of fi nding electron at the nucleus = 0
Solution: Mg( z = 12) 1 s 2 2 s 2 2 p 6 3 s 2
Solution: 2(1 s ) + 2(2 s ) + 2(2 p ) + 1(3 s ) = 7
Solution: L l l h h = + + ⋅ = ⋅ ( ) 1 2 5 π π ⇒ l = 4 Number of orbitals = 2 l + 1 = 9
Solution: r a z mp = = 0 26 45 . pm
Solution: For n – l – 1 = 0, r n a z mp = 2 0 and for all orbitals, r n a z l l n av = + − + ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ 2 0 2 1 1 2 1 1 ( )
Solution: S 1 = 3 s ; S 2 = 3 d ; S 3 = 4 s ; S 4 = 3 p For single electron system 3 s = 3 p = 3 d < 4 s
Solution: l l h h ( ) + ⋅ = ⋅ 1 2 3 π π ⇒ l = 3 ⇒ n ≥ 4\n13.49 Atomic Structure HINTS AND EXPLANATIONS
Solution: (a) (K.E.) Initial = (P.E.) at distance of closest approach or 4.0 MeV = K. q q r 1 2 or, 4 × 10 6 × 1.6 × 10 –19 = 9 × 10 9 × ( . ) ( . ) 2 1 6 10 50 1 6 10 19 19 × × × × × − − r ∴ Distance of closest approach, r = 3.6 × 10 –14 m (b) P.E. = K ⋅ q q r 1 2 = 9 × 10 9 × ( . ) ( . ) 2 1 6 10 50 1 6 10 9 10 19 19 14 × × × × × × − − − = 10 25 × (1.6 × 10 –19 ) 2 J = 10 1 6 10 1 6 10 25 19 2 19 × × × − − ( . ) . e V = 1.6 MeV (c) P.E. = K q q r ⋅ = × × × × × × × × × × − − − 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 − × ) K q q r ⋅ = × × × × × × × × × × − − − 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 − × ) = 3.2 MeV ∴ K.E. of a -particle at this distance = 4.0 – 3.2 = 0.8 MeV
Solution: 1 1 1 1 1028 10 1 09 10 1 1 1 1 2 1 2 2 2 10 7 2 2 2 λ = − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ⇒ × = × × − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ − RZ n n n . ∴ n = 3 1 3 2 3 2 1028 Å 1 Induced radiations λ 1 = 1028 Å λ λ λ 2 1 2 2 2 2 2 1 1 1 3 1 2 1 3 6579 2 = − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ⇒ = . Å λ λ λ 3 1 1 1 3 1 1 1 2 1218 4 1 2 2 2 2 3 = − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ⇒ = . Å
Solution: Per atom only one photon is emitted out and hence, the concerned transition is 2 →
Solution: Δ E Z Z = − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = 13 6 1 1 1 2 10 2 2 2 2 2 . . eV
Solution: (a) r 1 : r 2 : r 3 = 1 2 : 2 2 : 3 2 = 1 : 4 : 9 (c) λ = = × × = × = − c v 3 10 6 10 5 10 500 8 14 7 m nm (d) λ λ α = ⇒ h mE m 2 1 ∴ λ λ λ H : : : : : : H e cn 4 1 1 1 4 1 16 4 2 1 = =
Solution: h n = ϕ + (K.E.) max For A: 4.25 = ϕ A + T A and λ A A 2 = h mT For B: 4.20 = ϕ B + T B and λ B B = h mT 2 As T B = T A – 1.50 and λ B = 2 λ A ϕ A = 2.25 eV; ϕ B = 3.70 eV; T A = 2.0 eV; T B = 0.5 eV
Solution: Theoretical\n13.50 Chapter 13 HINTS AND EXPLANATIONS
Solution: Na (11) 1 s 2 2 s 2 2 p 6 3 s 1
Solution: Total nodes = n – 1
Solution: 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 × = × − . J ∴ (K.E.) max of photoelectron = 1.25 × 10 –18 – 4.5 × 10 –19 = 8.0 × 10 –19 J = 5 eV ∴ λ min = 150 5 30 = Å
Solution: i min = 4 × 10 18 × 1.6 × 10 –19 = 0.64 A
Solution: i i max min . . = × × × × × × = − − 9 10 1 6 10 4 10 1 6 10 9 4 18 19 18 19 Comprehension II
Solution: F du dr K r MV r V K mr = − = = ⇒ = 4 4 5 2 4 2 (1) From Bohr’s quantization, V n h m r 2 2 2 2 2 2 4 = π (2) ∴ 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 = ⇒ = = π π π .
Solution: V nh mr nh m nh mK n h m mK = = = 2 2 4 8 2 2 2 π π π π .
Solution: E = K.E. + P.E. = 1 2 2 4 mv K r + − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = ⋅ − = ⋅ ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 1 2 4 4 4 4 4 m K mr K r K nh mK π ∴ E n h m K = 4 4 4 2 256 π Comprehension III Δ E z n n = − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 13 6 1 1 2 2 2 1 2 . eV 10.2 + 17.0 = 13.6 z 2 1 2 1 2 2 − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ n (1) 4.25 + 5.95 = 13.6 z 2 1 3 1 2 2 − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ n (2)
Solution: Δ E = × − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = 13 6 3 1 1 1 7 119 9 2 2 2 . . eV\n13.51 Atomic Structure HINTS AND EXPLANATIONS Comprehension IV
Solution: After excitation, n =
Solution: Hence, initial excited state is n =
Solution: 1 1 1 1 1654 10 1 09 10 1 2 1 3 2 1 2 2 2 10 7 2 2 2 λ = − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ⇒ × = × × − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ − RZ n n z . ∴ z = 2 Þ He + ion
Solution: Δ E z n n = − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = × − ∞ ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = 13 6 1 1 13 6 2 1 3 1 6 04 2 1 2 2 2 2 2 2 . . . eV Comprehension V
Solution: Final excited state, after absorption of 2.7 eV, is
Solution: 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
Solution: 4 2.7 eV 2.7 eV Less than 2.7 eV More than 2.7 eV 3 2 1
Solution: Δ E I E n n I E = − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ⇒ = − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ . . . . . 1 1 2 7 1 2 1 4 1 2 2 2 2 2 ∴ I . E . = 14.4 eV
Solution: Δ E I E n n min . . . = − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = × − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 1 1 14 4 1 3 1 4 1 2 2 2 2 2 = 0.7 eV Comprehension VI
Solution: r n h mze n z = = × ( ) . 4 4 0 529 0 2 2 2 2 2 πε π Å (for H-like atom) For this system, r = × = × = − 0 529 1 207 2 56 10 0 256 3 . . . Å pm
Solution: I . E . = 2 4 13 6 2 2 4 0 2 2 2 2 2 π πε mz e n h z n ( ) . = × eV ( for H-like atom) For this system, I . E . = 13.6 × 207 = 2835.9 eV
Solution: Rydberg constant for this system = 1.09 × 10 7 × 207 m –1 ∴ 1 1 09 10 207 1 1 1 2 5 91 10 7 2 2 10 λ λ = × × − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ⇒ = × − ( . ) . m Comprehension VII
Solution: n n n ( ) − = ⇒ = 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 λ = − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ⇒ × = × × − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ − RZ n n x . ∴ x = 978.6
Solution: For max l , transition : n = 4 to n = 3 ∴ 1 1 09 10 1 1 3 1 4 1 887 10 7 2 2 2 6 λ λ max max . . = × × − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ⇒ = × − m\n13.52 Chapter 13 HINTS AND EXPLANATIONS
Solution: For max n , transition : n = 4 to n = 1 ∴ Hz ν λ max . . = = × × = × − c 3 10 978 6 10 3 066 10 8 10 15
Solution: 1R radiations involve transition : n = 4 to n = 3 only.
Solution: Visible radiation involve transitions : n = 4 to n = 2 (489.3 nm) and n = 3 to n = 2 (660.5 nm). Comprehension VIII
Solution: 5 4 3 2 1 1 2 3 4 ⎯ → ⎯ ⎯ → ⎯ ⎯ → ⎯ ⎯ → ⎯
Solution: 5 3 2 1 ⎯ → ⎯ ⎯ → ⎯ ⎯ → ⎯ and 5 ⎯ → ⎯ ⎯ → ⎯ ⎯ → ⎯ 4 3 1
Solution: 6 + 1 (any possibility after than Q.27)
Solution: 5 5 1 2 10 ( ) − =
Solution: 4 5 3 2 1 1 1 1 1 3 4 2 5 6 2 Comprehension IX Δ E Z n n eV Z n = − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 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 . . ( ) = − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ Z n n (2)
Solution: n = 2 ⇒ 2 n = 4
Solution: E 1 2 2 13 6 4 1 217 6 = − × = − . . eV
Solution: Δ E min . . = × − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = 13 6 4 1 3 1 4 10 58 2 2 2 eV Comprehension X
Solution: 1 1 1 2 1 4 1 4 16 16 2 1 2 2 2 2 2 2 2 2 λ = − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = × − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = − R Z n n R n R n n ( ) ∴ λ = − = − ⇒ = = 4 16 16 4 366 97 2 2 2 2 n R n cn n C R ( ) . nm
Solution: For series limit, n = ∞ ∴ l = C = 366.97 nm
Solution: For n = 5, l = 1019.36 nm For n = ∞ , l = 366.97 nm Comprehension XI
Solution: E E S H 1 13 6 3 2 2 25 2 2 = − × = × . .
Solution: S 2 = 3 p ⇒ l = 1\n13.53 Atomic Structure HINTS AND EXPLANATIONS Comprehension XII
Solution: n = 2 l = 1 j = 3 2 1 2 or m = = − − + − + − 3 2 1 2 1 2 3 2 , , , for j = 3 2 = − + 1 2 1 2 , for j = 1 2
Solution: n = 3 l = 0 ⇒ j = 1 2 ⇒ m = − + 1 2 1 2 , = 2 ⇒ j = 3 2 ⇒ m = − − + + 3 2 1 2 1 2 3 2 , , , = ⇒ = − − − + + + 5 2 5 2 3 2 1 2 1 2 3 2 5 2 m , , , , , Comprehension XIII
Solution: Radial nodes = n – l – 1 = 3 Angular nodes = 1
Solution: n = 5, l = 1 ⇒ Orbital = 5 p
Solution: For radial nodes, s = 1, 2, 6 = 2 5 0 zr a ∴ r a Z max = 15 0
Solution: Orbital angular momentum = + ⋅ l l h ( ) 1 2 π
Solution: Spin quantum number is independent from wave function.
Solution: Be is the reactive element.
Solution: 2 p ⇒ 2 p x + 2 p y + 2 p z ⇒ Total 3 angular nodes.
Solution: dz 2 has two conical nodes.
Solution: 3 p x and 3 p y diff ers in angular function.
Solution: 4s energy level is lower than 3 d .
Solution: Radial nodes = n – l – 1 Angular nodes = l
Solution: (A) V K P E K E n n = = − = − . . . . mV mV 2 1 2 2 2 (B) ε n n r ∝ − ( ) 1 (C) Lowest energy level is 1 s . (D) r z n ∝ 1
Solution: 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
Solution: (A) r n z ∝ 2 (B) V z n ∝ (C) F mv r z n = ∝ 2 3 4 (D) f v r z n = ∝ 2 2 3 π\n13.54 Chapter 13 HINTS AND EXPLANATIONS
Solution: (A) 3 radial nodes ⇒ 4 s , 5 p , 6 d but graph does not start from origin and hence, only 4 s . (B) 3 radial nodes ⇒ 4 s , 5 p , 6 d (C) Only s- orbital (D) l ≥ 1
Solution: 10. (A) V V 6 4 4 6 2 3 = = v ∝ ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 1 n (B) λ λ 3 2 1 1 1 1 2 1 4 1 2 2 2 2 = − ∞ ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ − ∞ ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = (C) λ λ c p = − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = 1 3 1 6 1 1 1 3 3 3 2 2 2 2 2 (D) Δ Δ E E n H e + = = 1 2 1 4 2 2
Solution: ε = = 1240 300 4 13 . eV For photoelectric effect, e ≥ f ⇒ N 0 = 4
Solution: Frequency of reduction ∝ z n 2 3 ∴ T T 3 2 2 3 8 2 2 7 1 3 4 8 10 1 2 1 28 10 1 6 = × × × × = − − . .
Solution: 1 1 1 2 1 2 2 2 λ = − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ R Z n n ⇒ 1 108 5 10 1 30 4 10 1 09 10 2 1 1 1 7 7 7 2 2 2 . . . × + × = × × × − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ − − n ∴ n = 5
Solution: λ 3 – λ 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 − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ − − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = . nm ⇒ z = 3
Solution: Final excited state = 5th orbit As only the wavelengths are longer than absorbed radiation initial excited state = 3rd orbit
Solution: Δ E = 12.75 = 13.6 × 1 2 4 2 1 4 2 n m n n − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ⇒ =
Solution: Δ Δ x h m V min . . = = × × × × = × − − − − 4 6 626 10 4 10 3 313 10 5 10 34 6 3 26 π π π m
Solution: m h x v min . . = ⋅ ⋅ = × × × × = − − − 4 6 626 10 4 10 5 27 10 1 34 11 24 π π Δ Δ kg
Solution: 2 p r = nl ⇒ l = 4 2 nm = 2 nm
Solution: For radial node, y 23 = 0 ⇒ r 0 = 2 a 0 Four-digit Integer Type
Solution: Initial K. E. = P. E. at distance of closest approach or, p q q r 2 0 1 0 2 1 4 m = ⋅ πε . 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 × × × × = × × × × × × × − − − − z 5 5 10 13 × − ∴ 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
Solution: t = = × × × = Distance Speed sec 2 12600 10 3 10 0 084 3 8 .
Solution: c a λ = − (z ) 6 c a 180 27 1 = − ( ) c a z z 144 1 30 = − ⇒ = ( )
Solution: nh n = ms ∙ Δ T or, n × 6.626 × 10 –34 × 2.45 × 10 10 = 245 × 4.2 × (99.5 – 19.5) ∴ Number of photons = 5.04 × 10 27 ∴ Moles of photon = 5 04 10 6 10 8400 27 23 . × × =
Solution: Δ E = Δ E 1 + Δ E 2 = − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ × + × − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ × = 1310 1 1 1 3 45 100 1310 1 1 1 2 40 100 917 2 2 2 2 kJ
Solution: λ = = 1240 13 6 91 17 . . nm
Solution: Moles of H 2 = PV RT x = × × = 1 1 0 08 300 . Δ E x x = × + − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ × = 436 1312 1 1 1 2 2 100 16 2 2 . kJ
Solution: 2 1 ? 360 nm 120 nm ∞ 1 1 120 1 360 90 λ λ = + ⇒ = nm
Solution: λ = ⇒ = ⇒ = 150 2 5 150 1 2 V V 24 ¯ V V .
Solution: λ λα = = ⋅ ⇒ ⋅ h h m KT m T 2 2 3 2 1 mE ∴ λ λ H N e e = × × = 20 1000 4 200 5
Solution: Nuclear forces are same in between any two nucleon and it is attractive at 1 fm but repulsive forces are also there between protons
Solution: Informative (B.E./nucleon is maximum for Fe)
Solution: For lighter nuclei, n p > 1 may make the nucleus unstable
Solution: Number of n and p , both is even in 30 Zn 64 .
Solution: r A N ∝ 1 3 / ⇒ r r 1 2 1 2 = × ⇒ ( ) ( ) / / A 1 1 3 1 3 1 2 56 = × ⇒ A 1 = 7
Solution: For 1 H 1 , n p = = 0 1 0
Solution: For 1 H 3 , 0 693 32 365 24 730 90 0 693 64 90 . . × × × × ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = × × ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ N w N A A (Some isotopes having n p ratio greater that, 1 H 3 are also know, like 2 He 8 )
Solution: Informative Radioactivity
Solution: Experimental reason behind considering α -particle as He- nucleus.
Solution: Experimental fact
Solution: Experimental reason behind considering b-emission as nuclear charge.
Solution: Isotope formation
Solution: Reason of g -emission
Solution: For an increase in mass, large amount of energy is needed and hence, it is non-spontaneous.
Solution: b N a N N c − × = − × + × = α α β α 4 2 1 and ∴ N b N c a b d α β α = − = − + × − 4 2 4 and ( )
Solution: n p n p F F ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ < ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 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.
Solution: 11 23 10 23 Na Ne → + F
Solution: n p n p ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ > ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ Na Na 24 23 Hence, Na 24 should undergo β -decay.
Solution: c N 14 14 → − β
Solution: Δ m m m u = −= − = Au Hg 198 198 197 968 197 966 0 002 . . . ∴ Q – value = 0.002 × 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 –particle = 1.863 – 1.063 = 0.8MeV. HINTS AND EXPLANATIONS EXERCISE (JEE ADVANCED)\n14.16 Chapter 14 HINTS AND EXPLANATIONS Rate Law
Solution: Rate is independent from all external factors.
Solution: r N = = × × × × × ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = × λ 0 693 28 3 15 10 1 90 6 10 5 24 10 7 23 12 . . . dpspg
Solution: r N A 1 0 693 10 10 = × × . ( ); r N A 2 0 693 5 1 = × × . ( ) r N A 3 0 693 2 5 = × × . ( ); r N A 4 0 693 1 2 = × × . ( )
Solution: t ½ is independent from amount.
Solution: N N n = × ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 0 1 2
Solution: w w o n = × ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 1 2 ⇒ 3 1 2 12 3 g w o = × ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ⇒ w o = 48gm
Solution: Moles of He formed = × × 4 5 10 6 10 23 23 . = 0.75 = Moles of decayed ∴ t t = × = 2 2 0 1 2 / hrs
Solution: 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 = . / ∴ Number of atoms decayed = − ( ) . R R T 1 2 0 693
Solution: Rate should decrease 1 64 1 2 6 = times and hence, t t = × = 6 12 1 2 / hrs
Solution: P w w w Q w w : : 10 20 40 20 20 1 1 2 0 day 0 day 0 day ⎯ → ⎯⎯ ⎯ → ⎯⎯ ⎯ → ⎯⎯ (As fi nal mass ratio is 1 : 4) Hence, Q is non-radioactive.
Solution: t ½ = 30 min Now, r = λ N ⇒ 28 0 7 30 1200 = × ⇒ = . N N
Solution: r r o = × × = = 3 10 3 10 1 8 1 2 8 8 3 ⇒ t t = × = × = 3 3 12 26 36 78 1 2 / . . yrs
Solution: w w o n = × ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 1 2 ⇒ 10 1 2 3 6 mg = ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ w o / ⇒ w o = 14 14 . mg
Solution: r N = = × × × × × × × × × λ 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 × ⎛ ⎝ ⎜ ⎜ ⎜ ⎜ ⎞ ⎠ ⎟ ⎟ ⎟ ⎟ = . dps
Solution: t t r r t o = ⋅ ⇒ > ⋅ 1 2 1 2 2 2 2 5 / / log log log log . ∴ t 1/2 = 5.25 days
Solution: Let the sample contains x gm Pu 239 . Now, r N N Pu Pu = + ( ) ( ) λ λ 239 240 or 6 10 0 693 2 4 10 365 24 3600 239 6 022 10 0 693 9 4 23 × = × × × × × × × ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ + . . . . x 7 7 17 10 365 24 3600 1 240 6 022 10 3 23 . . × × × × × − × × ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ x ⇒ x = 0.3112 Hence, mass percent of Pu 239 = × = x 1 100 31 12 . %\n14.17 Nuclear Chemistry HINTS AND EXPLANATIONS
Solution: U Pb 238 206 ⎯ → ⎯ Initial a 0 Present a – x x From question, ( ) . a x x − × × = 238 206 1 0 1 ⇒ x a = 238 2298 Now, age of ore, t t a a x = ⋅ − 1 2 2 / log log = × ⋅ − = × 4 5 10 0 3 1 1 238 2298 7 2 10 9 8 . . log . years
Solution: Th Pb He 232 208 4 6 ⎯ → ⎯ + Initial a mole 0 Present ( a – x ) mole 6 x mole = × = × − − 4 64 10 232 2 10 7 9 . = × ⇒ × − − 6 72 10 22400 5 10 5 10 . ∴ a = 2.5 × 10 –9 Now, age of sample, t t a a x = ⋅ − 1 2 2 / log log = × × × − × = × − − − 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
Solution: 224 is an integer multiple of 4 and hence, Ra 224 belongs to 4n series, which is thorium series.
Solution: r r Th Ra = ⇒ N t N t Th Th Th Ra ( ) ( ) / / 1 2 1 2 = ⇒ N N Th Ra = 80000 1600
Solution: (a) λ AC Yr 227 0 693 22 3 15 10 2 1 = = × − − . . (b) l for the formation of Th 229 = × × = × − − 2 100 3 15 10 6 3 10 2 4 . . Yr (c) l for the formation of Fr 223 = × × = × − − − 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 = ⇒ = × × ≠
Solution: The net rate of formation of radioisotope, + = − dn dt R N λ . After very long time, steady state will be achieved, at which + = dn dt
Solution: Hence, N R = λ .
Solution: Pb Bi hr hr 212 8 212 1 1 2 1 2 t t / / = = ⎯ → ⎯⎯⎯ ⎯ → ⎯⎯⎯ Time for maximum nuclei and hence, maximum activity of Bi212, t max ln = − ⋅ 1 2 1 2 1 λ λ λ λ = = ⋅ = = 1 2 1 2 8 1 1 1 8 3 429 205 7 ln ln ln . . min hr
Solution: λ λ λ α β overall = + ⇒ = + 0 693 0 693 20 0 693 60 1 2 . . . / t ∴ t 1/2 = 15 min ∴ For 87.5 % decay, t t = × = 3 4 5 1 2 / min
Solution: Average energy released = × + × + = 0 05 40 0 15 80 0 05 0 15 70 . . . . MeV\n14.18 Chapter 14 HINTS AND EXPLANATIONS Nuclear Reactions
Solution: 92 235 0 1 54 139 38 94 0 1 3 U n Xe Sr n + ⎯ → ⎯ + +
Solution: 25 55 0 1 25 56 Mn n Mn + ⎯ → ⎯ + γ
Solution: 4 9 1 1 5 10 Be H B (proton) + ⎯ → ⎯ + γ
Solution: 13 27 2 4 15 30 0 1 Al He P n particle + ⎯ → ⎯ + − ( ) α
Solution: 13 27 2 4 14 30 1 1 Al He Si H + ⎯ → ⎯ + = ( ) X 13 27 2 4 15 30 0 1 15 30 14 30 1 1 Al He P n P Si e + ⎯ → ⎯ + = ⎯ → ⎯ + = + ( ) ( ) Y Z
Solution: Q -value is distributed between β -particle and anti- neutrino.
Solution: 92 235 90 23 88 227 89 227 93 235 U Th Ra Ac Np − − − − ⎯ → ⎯ ⎯ → ⎯ ⎯ → ⎯ ⎯ → ⎯ α α β β α 9 91 231 Pa − 89 AC 235 is not possible.
Solution: Activity is independent from all external factors.
Solution: Half-life of a radio isotope is its characteristic property, independent from all factors.
Solution: Actinum series: 92 235 82 207 U Pb →
Solution: 96 242 2 4 97 293 0 1 2 Cm He Bk Incorrect + ⎯ → ⎯ + 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) + ⎯ → ⎯ + + ⎯ → ⎯ + ( ) 9 9 28 1 2 15 29 0 1 Si H P n (Correct) + ⎯ → ⎯ +
Solution: Neutron is projectile and proton is emitted particles.
Solution: 13 27 2 4 15 30 0 1 Al He P n + ⎯ → ⎯ + 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 + ⎯ → ⎯ + ⎯ → ⎯ + + ⎯ → ⎯ γ 4 4 1 1 H +
Solution: 4 9 4 8 0 1 Be Be n + ⎯ → ⎯ + γ 4 9 1 1 4 8 1 2 Be H Be H + ⎯ → ⎯ +
Solution: Δ m u = × + × − = ( . . ) . 8 1 0072 8 1 0086 16 0 1264 ∴ B.E. per nucleon = × = 0 1264 931 5 16 7 36 . . . MeV
Solution: 8 16 2 4 4 O He ⎯ → ⎯ Δ m u = − × = − 15 9944 4 4 0026 0 016 . . . ∴ Energy required in separation = × = 0.016 MeV 931 5 14 904 . .
Solution: 10 20 6 12 2 4 2 Ne C He ⎯ → ⎯ + Energy required = × − × + × × (20 8 03 12 7 68 2 4 7 07 . ) ( . . ) = 11.88 MeV\n14.19 Nuclear Chemistry HINTS AND EXPLANATIONS Comprehension II
Solution: SC 50 50 ⎯ → ⎯ + + τ β ν i Q − = − × = value (49.9516 49.94479 MeV ) . . 931 5 6 34 ∴ K.E. of MeV ν = − = 6 34 0 80 5 54 . . .
Solution: λ = = × × × − × × × = × − − LC E Δ 6 626 10 3 10 4 795 4 611 10 1 6 10 6 75 1 39 8 6 19 . ( . . ) . . 0 0 12 −
Solution: Th Ra Ra 228 224 224 − − ⎯ → ⎯ ⎯ → ⎯ α γ Q-value = 228 028726 224 020196 4 0026 4 0026 931 6 . . . . . ( ) ( ) ⎡ ⎣ ⎤ ⎦ − + + × − 2 217 10 3 5 307 × − = . MeV ∴ × = K.E. of -particle = MeV α 224 228 5 307 5 214 . . Comprehension III
Solution: Overall rate is the rate of slowest step and hence, T required = 270 days
Solution: At transient equilibrium, N N A B B A A = − ⋅ λ λ λ or, N N N N A B Th Ra Ra Th Th = = − = − × × = λ λ λ ln . ln . ln . 2 3 64 2 1 913 365 2 1 913 365 190 0 8 1 .
Solution: At secular equilibrium, N N B A B A = ⋅ λ λ or, N N N N A B Ra Rn Rn Ra = = = × × = λ λ ln ln . . 2 55 2 3 65 24 3600 5733 8 Comprehension IV
Solution: U Pb 238 206 ⎯ → ⎯ Initial a mole 0 Present ( a – x ) mole x mole = 59 5 238 . = × 12 875 206 80 100 . ∴ a = 0.30 Now, t a a x = ⋅ − = × ⋅ − 1 1 1 52 10 0 3 0 25 10 λ ln . ln . . = 1.33 × 10 9 Yrs
Solution: K Ar 40 40 ⎯ → ⎯ Initial a mole 0 Present ( a – x ) mole x mole = 1 = 10.3 ∴ a = 11.3 Now, t t a a x = ⋅ − = × ⋅ 1 2 9 2 1 25 10 0 3 11 3 1 / log log . . log . = 4.375 × 10 9 years
Solution: U Pb 238 206 ⎯ → ⎯ K Ar 40 40 ⎯ → ⎯ Initial mole 0 b mole 0 Present ( a – x ) mole x mole ( b – y ) mole y mole = × − 0 86 10 238 3 . = × − 0 15 10 206 3 . = × − 10 10 40 3 = × − 1 6 10 40 3 . t t a a x t b b y U K = ⋅ − = ⋅ − ( ) log log ( ) log log / / 1 2 1 2 238 40 2 2 ∴ w = 1.7 mg\n14.20 Chapter 14 HINTS AND EXPLANATIONS Comprehension V
Solution: Given in paragraph
Solution: For radioactive tracing, time should be comparable to t 1/2 .
Solution: T c c T c c 1 1 2 2 1 1 = ⋅ = ⋅ λ λ ln ln and As c c T T T T c c 1 2 1 2 1 2 1 2 1 > > − = ⋅ , ln and λ Comprehension VI
Solution: Let the volume of blood be V ml. t T N N N av 1 0 0 1 2 = = ⋅ − ln ⇒ 5 = ⋅ 15 2 1260 15 60 log log / V ∴ V = 4000
Solution: r 0 1260 60 4000 18 9 = × = . dpm per ml Now, r r r r 0 5 5 10 = ⇒ 18 9 15 15 10 . = r ⇒ r 10 = 11.9 dpm per ml Comprehension VII
Solution: Isotopes are B and E, C and F, D and G.
Solution: Mass number of H = 230 – 4 × 4 = 214
Solution: Z A – 4 × 2 + 3 × 1 = 88 ⇒ Z A = 93 Comprehension VIII
Solution: Net rate of formation + = − dn dt N α λ or, dN N dt N N t α λ − = ∫ ∫ 0 0 ⇒ N N e t = − − ⎡ ⎣ ⎤ ⎦ − 1 0 λ α α λ λ ( ).
Solution: t t N = = = 1 2 0 2 2 / ln λ α λ and ∴ N = 1.5 N 0
Solution: t N N → ∞ = = , then α λ 2 0 Comprehension IX
Solution: λ λ λ = + 1 2 ⇒ ln ln ln / 2 2 24 2 8 1 2 t = + ⇒ t 1/2 = 6 hours.
Solution: 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 . . μ ci
Solution: n p ratio does not increases continuously.
Solution: Binding energy increases but the binding energy per nucleon first increases and then decreases.
Solution: Stable\n14.21 Nuclear Chemistry HINTS AND EXPLANATIONS
Solution: All heavy nuclei should not produce 82 Pb 206 .
Solution: β -decay occurs to decrease n p ratio.
Solution: Same mass of U 238 and U 238 F 6 have diff erent numbers of U 238 nuclei.
Solution: t T av 1 2 0 693 1 / . / / = λ λ = 0.693 = Same for all
Solution: Mesons have mass 200 to 300 times mass of electrons.
Solution: 13 Ae 30 have high n p ratio than its stable nucleus 13 Ae 27 .
Solution: 53 I 127 is stable and hence, I 333 is beta emitter and I 121 is positron emitter.
Solution: (a) 92 U 235 82 Pb 207 N N α β = − = = − − × = 235 207 4 7 82 92 2 7 4 ( ) (b) 92 U 238 82 Pb 206 N N α β = − = = − − × = 238 206 4 8 82 92 2 8 6 ( ) (c) 94 Pu 241 83 Bi 209 N N α β = − = = − − × = 241 209 4 8 83 94 2 8 5 ( ) (d) 90 Th 232 82 Pb 208 N N α β = − = = − − × = 232 208 4 6 82 90 2 6 4 ( )
Solution: Number of half-lifes 28 1 2 81 10 . . = ∴ Mass of Sr 90 remained = × = × − 2 048 10 2 2 10 10 6 . gm gm
Solution: r N = = × × × × × × × ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = × − λ 0 7 14 24 3600 86 4 10 164 0 164 100 6 10 3 10 3 23 . . . 1 11 dps
Solution: Z m Z m A B ⎯ → ⎯ + − − 6 12 2 4 3 He t = 0 1 mole 0 t = 20 days 1 3 4 − 3 3 4 × mole = 1 4 mole ∴ V = × 9 4 22.4 = 9 × 5.6 L at 0°C and 1 atom
Solution: 1 2 0 335 2 mV eV = .\n14.22 Chapter 14 HINTS AND EXPLANATIONS ∴ V = × × × × = − − 2 0 335 1 6 10 1 675 10 8000 19 27 . . . m/s Hence, time for travelling 80 km, t d v = = × = 80 10 8000 10 3 sec Now, t t N N = ⋅ 1 2 0 2 / ln ln or, 10 700 2 100 100 = ⋅ − ln x ⇒ x = 0.99 ≈ 1
Solution: t t N N t N N U = ⋅ ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = ⋅ ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 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 × ⋅ = × N x N x ∴ log . N x 0 2 5 = ∴ Age of earth, t N x = × ⋅ = × 7 2 10 2 6 10 8 0 9 . log log years Four-digit Integer Type
Solution: t t r r = ⋅ 1 2 0 2 / log log or, 6 93 6 93 2 5 10 0 15 . . log log = ⋅ × r ⇒ r 0 = 5.35 × 10 15 dpm Now, r N 0 0 = λ . or, 5.35 × = × × × × 10 0 693 69 3 60 6 10 15 23 . . ( ) n ∴ n = 5.35 × 20 –5
Solution: Initial number of H 3 atoms = 0 93 10 18 6 10 2 8 10 3 23 18 . × × × × × × − − = 4.8 × 10 2 Number of half lives = = 36 9 12 3 3 . . ∴ Final number of H 3 atoms
Solution: t T N N N av 1 0 0 1 2 = = ⋅ − ln and 3 6 8 3 1 0 0 1 t T N N N av = = ⋅ − ln ∴ N N T av 1 0 0 116 16 = = . and sec
Solution: Let the mass of water present in body = w gm. Now, 9 × 10 9 = 2.25 × 10 5 × w ⇒ w = 4 × 10 4 gm = 40 kg ∴ Mass per cent of water in body = × = 40 80 100 50%
Solution: λ λ β − = × 32 100 overall ∴ t t 1 2 1 2 100 32 100 32 12 8 40 / / . ( ) = × ( ) = × = − β overall hr
Solution: λ = + 1 1620 1 405 ⇒ λ = − 1 324 1 Yr ∴ t t required = × = × × 2 2 0 693 324 1 2 / . = 449.064 years
Solution: Sr Y 90 90 ⎯ → ⎯ ⎯ → ⎯ other format For radioactive equilibrium, ( .N) ( .N) Sr Y 90 90 λ λ = or, 0 693 32 365 24 730 90 0 693 64 90 . . × × × × ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = × × ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ N w N A A ∴ w = 0.1667 gm
Solution: Energy released = 2 × 120 × 8.1 – 240 × 7.2 = 216 MeV\n14.23 Nuclear Chemistry HINTS AND EXPLANATIONS
Solution: Δ m = 2 × 2.0021 – 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 × × × × × × = × × × − . . ∴ n = 960
Solution: 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 = × = 10 2 10 100 2 2 mg