C3-Allens Made Chemistry Exercise Solution

Let n2 photons are re-emitted then, (n2 or  Energy converted into K.E. = 0.68 × 10–19J   )    % of energy used in kinetic energy = hc 0.68  10 19 100 = 8.68% Total energy re-emitted out = n2 × emitted 7.83 10 –19 As given 47 4 6 . Energy given to I2 molecule Eabsorbed × 100 = Ere-emitted out hc 6.626  1034  3  108 hc 47 hc =   4500  10 –10 = 4.417× 10–19 J  n1   n2  absorbed 100 emitted Also energy used for breaking up of I2 molecule  n1  47  emitted = 47  5080 240 103 = 3.984 × 10–19 J n2 100 absorbed 100 4530 = 6.023 1023 n1  Energy used in imparting kinetic energy to two I  n2 = 0.527 atoms 4 2 . H2 + Br2 hv 2HBr Br2 hv 2Br = [4.417 – 3.984] × 10–19 J BE = 192 kJ / mole  K.E./iodine atom = [(4.417 – 3.984)/2] × 10–19 = 0.216 × 10–19 J 192 hv 192 1240 48. = 150 = 3.88 × 10–2 Å = 3.88 pm eV/mole = or = (nm) 103 100 93.368  96.368  = 6235 Å 4 9 .  = 6.6 1034 6 1024  3 106 0.2 n 0.2 n  0.01 4 3 .  0.01 mole 1 128 = 1 1  1065 = 3.68 × 10–65 m 3 Na 5 0 . V = 30 × 102 cm/sec 0.2  n  1 127 10 127 100 2×n= = 5000 Å m = 200 g 10 127 12.7 h h n =  6 = mV 500 = m  V 10 2 2 No. of protons 6 1022  3 1022 500 102 = = 6.626 1026 2 P = mV = 30 × × 200 44. 243  1240  1.75 1029 96.368 (nm) 1240  96.368 5 1 . v = 40 m/sec v = 0.01 = 243 = 491.75×10–9 m  4.9 × 10–7m h  x = 4  9.1  1037  99.99  40 4 5 . Energy required to break H–H bond 430.53 103 J/molecule = 7.15 × 10–19 J 0.53 100 1054 = 6.023 1023 = 40  99.99  9.1 1037 hc 0.53 103 100 h Energy of photon used for this purpose = = m.x.x =  40  9.1  99.99 4 6.625 1034  3.0 108 = 7.83 × 10–19 J 5.27 1034 = 253.7 10–9 x = = 1.447 × 10–3 × 100 9.1 1031  40  0.04  1  Energy left after dissociation of bond = 100 (7.83 – 7.15) × 10–19

ATOMIC STRUCTURE EXERCISE # 4[B] 1 . Given that 1 = 486.1 × 10–9 m 4 . Since we obtain 6 emission lines, it means electron = 486.1 × 10–7 cm comes from 4th orbit energy emitted is equal to, 2 = 410.2 × 10–9 m = 410.2 × 10–7 cm less than and more than 2.7 eV. So it can be like 1 1  this : 1  and v  v2  v1   2   (6 4th  2.7eV 1 1  1 1  = RH =  2 2    R  2 2  n12     )  n 2 H 2  E4 – E2 = 2.7 eV, E4– E3 < 2.7 eV, 1 1  E4– E1 > 2.7 eV v = RH  n12   ........(i) (a) n = 2, n 2 2 For line I of Balmer series (E4 – E2)atom = (E4 – E2)H × Z2 2.7 = 2.55 × Z2 = 1.029 1  RH 1  1  1  1  1  2 2   2 2 n12  n 2 = 109678 (b) IP = 13.6 Z2=13.6 × (1.029)2 = 14.4 eV 1 (c) Maximum energy emitted=E4–E1 = (E4 – E1)H × Z2 1 1  1  = 12.75 × (1.029)2 456.1  107 n12  or = 109678  2 2  = 13.5eV  Minimum energy emitted=E4 – E3 =(E4 – E3)H × Z2  n1 = 4 = .66 × (1.029)2 = 0.7eV For line II of Balmer series ; 5 . n 2E = 27.2eV(17 + 10.2)  E3 – E2  17eV 1  RH 1  1  1  1  n 3E=10.2eV(4.25+5.95.2)  1    2 2 n 2  = 109678  22 n 2  17 eV = 1.89 × Z2  Z = 3  2  2 E2 = –3.4 × 9 = –30.6 eV 1 1  1  En – E2 = 27.2 eV 410.2 107  2 2  or = 109678 n 2 En = 27.2 + E2 = –3.4 eV 2  n2 = 6 1 3 .6  3 2 Thus given electronic transition occurs from 6th to n En = – 3.4 = – 2  n2 = 36  n = 6 4th shell. Also by eq. (i) 6 .  = 975 Å (6th4th) E=c = 6.626 10–34  3 108 =2.03×10–18J=12.75eV 975 1010  v= 1  109678 1  1  So electron will excite to 4th energy level and when   42 62  comeback number of emission line will be 6.   = 2.63 × 10–4 cm minimum energy emitted = E4 – E3 = 0.66 eV 2. Eext = 2.18 × 10–19 1  1  × 6.023 × 1023 = (4th   9       6 ) 116.71 kJ/mol H hc 1.9878 10–25 = 1.882 × 10–6 m= D.E. = 116.71 × 2.67 = 311.62 kJ/mol H = = .66 1.6 10–19 2 E 18820 Å n= PV 1 = 0.04 7. (a) kE = qV = 2 × 1.6–19 × 2 × 106 = 6.4 × 10–13J  (b) At distance d = 5 × 10–14 m let K.E. is x J and RT 0.082  300  T.E. = 0.04 × 311.62 + 0.08 × 116.71 = 21.8kJ PE = k q1q2 = 9 109  2 1.6 10–19  47 1.6 10–19 3. E(ev) = 1240 d 5 1014 (n m ) PE = 4.33 × 10–13 J Energy of 1st photon = 1240 = 11.428 eV By energy conservation : 6.4 × 10–13 = x + 4.33 × 10–13 108.5 x = 2.06 × 10–13 J, Energy of 2st photon = 1240 =40.79 eV 30.4 kE = PE En = 52.217– 54.4 = – 2.182 eV (E1 = –54.4 eV) 6.4 × 10–13 9  109  2  47  (1.6  10 –19 )2 = 1 3 .6  4 d –2.182 =  n2  n= 5  d = 3.384 × 10–14 m

8. pE = ke2 , since F =  du =  ke2  = 1.9878 10–25 = 0.3039 × 10–7 m = 303.9 3r3 dr r4 6.54 1018 Å mv2 ke2 mv2 E1=–8.72 × 10–18=–21.79 × 10–19 × Z2  Z = 2 For stable atom F = r so r4  r ...(1) 0.529 1 (ii) r1 = 2 A°=0.2645A°= 2.645 × 10–11m mv2 = ke2 ...(2) 1240 r3 11.(a)  = 12.4 nm, E (ev) = = 100 eV 12.4 kE = 1 mv2  ke2 , PE = ke2 W0 = 25 eV 2 2r3 3r3 kE = E – W0 = 75 eV  V = 75 volt ke2 ke2 ke2 150 T.E = 2r3  3r3 = + 6r 3 ...(3) A° = 2 A° = 1.414 A° (b)  = V nh nh (c) since p = h     dp  h d Form bohr's postulate mvr =    V=  2 2 2mr putting this in equation (2)  nh 2 ke2  n2h2  ke2 d = 2 dp  (1.414 10 –10 )2  6.62 10 –28  2m r3  r3 h 6.626 10–34 m     m   r   2 2r2   4 m d =2 × 10–14 m r  4 2 mke2 n2h2 1 2 . Since electron is in some exited state of He+ so it's energy  13.6 eV so energy need to exitation putting this in equation (3) is also < 13.6 eV & only for hydrogen E3 – E1 < 13.6 eV. So Z =1. Now for He+ this T.E.  ke2  ke2 energy is equal to the energy gap of 2nd and 6th  4 2 m 2ke2 3 6 64 6 m 3k3e6  orbit so initial state is 2 and final state is 6. 6    n2h2   n6h6  He+       13.6 eV    <13.6 eV   E3 – E1 < 13.6 eV  Z=1  n6h6  He+    2nd 6th    E = 384 m 3 6k2e4 2  6  9.(a) (E3 –E2) =68 eV = (E3 – E2)H × Z2 68 = 1.89 × Z2 1 3 . mvr = nh  3.1652 × 10–34 = n 2 z=6 6.626  1034  (b) (kE)1 = – E1 =13.6 × 36 = 489.6 eV   (c) Energy required = –E1 = 489.6 eV  2  3.14  1240  = = 2.53 nm 489.6 1 0 . E1 = IP = –4 R = –4 × 2.18 × 10–18 J = –8.72 × 10–18 J E1 = –2.18 × 10–18 J n=3 4 E2 = c   R 1 – 1  =  8R  = 1 32   9  E = E2 – E1 = 6.54 × 10–18 J 



UNIT # 07 (PART - II) IONIC EQUILIBRIUM EXERCISE # 1 9 . pH = 1 pH = 2 28. pH = pKa + log H C O  [H+] = 0.1 [H+] = 0.01 3 V = 50 V = 50 H2CO3 7 = 7 – log + log H C O  H C O  4 3 3 [H+] of mixture is[H+]= N1V1  N2 V2  50(0.1  0.01) H2CO3 (H2CO3 ) 100 V1  V2 – = 4 100  3 5 [H+]  0.11  0.055 % H C O 80 % 2 30 . 2NaOH + H3PO4  Na3PO3 pH = 1.26 50mL,0.1M 1 1 . pH = 7 [H+] = 10–7 , [OH–] = 10–7 60mL,0.15 new pH after addition of base 0 7.33 pH = 12 [H+] = 10–12 H3PO4  H+ H 2 P O – 4 [OH–] = 10–2 [mass H+ obtain from first ionization of H3PO4] [OH+] concentration increase 105 times. pH = 1 pK a  1 log C = 1 pK a  1 log C 2 2 2 2 K a1  3 104 13. Relative strength = K a2 1.8 105   4 : 1 = 1.5 + 1.17 = 2.67 1 6 . HCOOH + KOH 3 2 . Mg(NO3)2 + 2NaF   MgF2 + 2NaNO3 2.5 5 40mL,0.5 (M) 50mL,0.2M HCOOH + H2O after reaction is forms Buffer solution 0 0 2.5 10 10 2.5 [HCOOH] = [HCOOK] = (MgF2) = 35 MgF2  Mg+2 + 2 F 90 90 Ionic product pH = pKa + log [salt] pH = pKa [acid] 4 1 . AgCNS  Ag+ + CNS pH = 4 – log (1.8) pH = 3.75 (1–1) (+2) 1 AgCl  Ag+ + Cl 1 9 . Let weak acid is HA its sodium salt is NaA (1–2) (+2) 2 Ka = Kw KH = CH2  0.1 × (0.03)2  K sp AgCNS  (1 ) (1  2 ) .........(i) KH Ka =  1 × 10–10  K sp AgCl  (2 ) (1  2 ) ........(ii) 1 0 14 Ka = 9  105 adde equation (i) and (ii) 2 2 . CH3COOH + NaOH  CH3COONa + H2O    (1 + 2)2 =K sp K sp  100 mL, 0.4 M 100 mL, 0.2M AgCl AgCNS 40 20    (1 + 2) = =K sp  K sp AgCl 1.0 1012  1.7 1010 AgCNS After reaction (AgT) = (1 + 2) =1.3 × 10–5 divide eq. ii from i 20 20 [Cl ] 1.7  1010 [CH3COOH] = , [CH3COONa] = [C NS  ] 2 1.0 1012 250 200  1 =  1.7 × 102 [salt] 4 7 . IP > Ksp I.P. = (Ca+2) (F)2 pH = pKa + log [acid] I.P.  (10–2) (10–3)2  10–8 IP > Ksp pH = pKa [H+] = Ka = 1.8 × 10–5 5 3 . pH = pHIn + log In 6 = 5 + log In 2 4 . CH3COONa + HCl  CH3COOH + NaCl HIA HIn 20 In  at equivalence the [CH3COOH] =  0.1  10 200 HIn 1 1 pH = 2 pK a  2 log C 5 4 . At Half way [HIn] = In– pH = 1 [5 log 2 – log101 ] pH = 5.5 + log [salt] 5.5 = pKa + log [salt] 2 [acid] [acid] pH = 1 [6 – log 2]  pH = 3 – log 2 log [salt] =0.75  [salt]  5.62 2 [acid] [acid]

IONIC EQUILIBRIUM EXERCISE # 2 4 . HCl + NaCl pH = 4.2 1 ml, 0.1 99 ml C6H5COOH + NaOH  C6H5COONa + H2O [H+] = 0.1 pH = 3 22  0.001 100 0 02 6 . HF + H2O  F– + H3O+ pH = 7 + 1 + 1 log C pKa 22 Ka × Kb= Kw pKa + pKb= pKw 12 pKa =14 – 10.83 pKa = 3.17 pH = 7 + 2.1 + log Ka = 6.75 × 10–4 2 200 9. h = Kh Kw pH = 9.1 – 1 pH = 8.1 C Ka C h= 2 5 . CH3COOH + NaOH  CH3COONa + H2O 0.1 0.1 10 –14  8.0 h = 2.48 % At 1/3 neutralization h = 1.3 10–9 1 2/3 1/3 1 3 . Ksp = s2 s = K sp = 6.4 10–5 (salt) s = 8 × 10–3 s = 8 mol/m3 pH = pKa + log + (acid) 1 5 . AgNO3 + NH 3  [Ag(NH3)2 ] pH1 = pKa + log 1 / 3 ...(1) x 2/3 ...(2) 1.6 5 × 10–8 (x – 1.6) 0.8 at 2/3 neutralization Ks = [Ag(NH3 )2 ] 2/3 (AgNO 3 )(NH3 ) pH2 = pKa + log 1 / 3 pH2 = 1 – log 2 pH1 – log 0.8 2 108= (5 10–8 )(x – 1.6)2 (x – 1.6)2 = 0.16 1 = log = – 2 log 2 x=2M mole of NH3 = 4 4 1 8 . C6H5COOH + NaOH  C6H5COONa + H2O 27. H SO –  H+ + SO 2 – 4 4 21 1 1 01 (1– )       pH = pKa + log [salt] pH = 4.2 + log 1 10–2 = 2   0.09 pH=1.02 [acid] 1 1 IONIC EQUILIBRIUM EXERCISE # 3 COMPREHENSION BASED QUESTIONS 1 c = [H3PO4] = 0.05 % Comprehension # 1 pH = 2 [pK a1  log c] 1. Suppose volume of H CO – = V mL = 0.05 10 mol L1 (M ) = 5.1 × 10–3 M 3 98 millimoles of H C O – = 5V 3 millimoles of H2CO3 = 20 –log c = 2.3, pK a1  2.12 pH = 2.21 pH = pK + log [H C O  ] [H  ]3 [P O 3  ] a 3 4 [H2CO3 ] 3. [H3PO4 ]  K a1 K a2 K a3 V V = 78 mL 7.40 = 6.11 + log , 4 3log [H+] + log[PO 3–] 4 3 . If CO2 escapes, [H+] decreases, hence pH increases. = log K a1 + log K a2 + log K a3 – log [H PO ] 34 (CO2 ,[H+] pH ) 3 p H – l og [ P O 3 – ] = l og [ H 3P O 4] = pK a1  pK a2  pK a3 Comprehension # 2 4 1. Phosphoric acid with three ionisable hydrogens ions 21 – lo g [ P O 3 –] – 3 = 2.12 + 7.21 + 12.32 4 M is a tribasic acid. H-atoms are attached to O-atoms, 4 log[PO43–] = – 3.65 ( [PO43–]=2.24 × 10 – H-)4 .Zn3(PO4)2  3Zn2+ + 2PO 3– 4 2 . If first step is only taken ( Ksp = [Zn2+]3 [PO43–]2 9.1 × 10–33 = [Zn2+]3 (2.2 × 10–4)2 ) [Zn2+]3 = 1.88 × 10–25 [Zn2+] = 5.73 × 10–9 M

IONIC EQUILIBRIUM EXERCISE # 4[A] 1 . ( i ) H O Ka  H+ + OH– 1 4 . For weak acid 2 10–7 10–7 [H+] = K1C1  K 2C2...  K w = 1.8 10–5  0.02  6.4 10–5 .01  10–14 [H  ][OH – ] 10–7 10 –7 10 –14 =1.8×10–16 K = = = a [H2O] 1000 / 8 55.5 = 100 10–8 ( i i ) K × K = 10–14 [H+] = 10–3 [H ][ACO – ] ab [ACO] = 3.6 × 10–4 K= [ACOH ]2 a 2. K = C2  2 = C1 = 1 same [C H O ] = 6.4 × 10–4 a 1 C2 1 /100 =10 252 3 . Ka = C2 1 5 . HCN is a weak acid so H+ due to it can neglect As comparision to HF 1  Ka1 1.8 10–5 [H+] = KC = 6.7 10–4  0.1 = 67 10–6 2 Ka2 = 6.2 10–10 = 8.18 × 10–3 11 pH = – log [8.18 × 10–3] = 3 – log [8.18] pH = pK – log C 4.(a) pH = 2.087 2 a2 5.(c) 1 6 . H S  2H+ + S–2 (e) 4.50 × 2 = pK –log (0.1) 2 (f) a [H+] = 2 × 10–4, [H S] = 0.1 M K = 10–8 (g) 9 – 1 = pK pK = 8 a 2 a a K = 10–7 × 10–7 [H+] = Ka C = 1.8 10–6 = 3 – log 1.8 = 2.87 K = K× K [H+] = 10–8 + 10–7 = 10–7 [0.1 + 1] 12 K = 10–21 pH = 7 –log 1.1 = 6.95  10–21 = [2  10 –4 ]2 [S –2 ] 10–21 = 4 × 10–8 [S–2] [OH–] = 10–10 + 10–7 = 10–7 [1.001] [0.1] POH = 7 –log 1.001 = 6.99 pH = 7.0004 1 10 –14 = [S–2] 2.5 × 10–15 = [S–2] 4 [H+] = Ka C = 1.8 10–5 10–6 [H+] = 1.8 10–11 = 18 10–12 = 4.24 × 10–6 17.(i) H PO   H+ + H PO – K = 7.225 × 10–3 34 24 1 pH = 6- log 4.24 = 5.37 0.01 M C (1 –) C1 C1 H PO – HPO –2+ H+ K = 6.8 × 10–8 6. pK = 14 - log 2.56 = 13.59  13.6 (ii) 24    4 2 w C1 (1 – 2) C12 [C1] pH = pKw =6.795 HPO –2 PO –3 H+ K = 6.8 × 10–8 (iii) 4   4 + 3 2 [H+] = 10–11.5 12C1(1– 3) C (123) [C1] 1 0 . pH = 11.5 7.225 × 10–3 = C12 = 0.01  12 (1  1 ) 1   [OH–] = 10–2.5 RxN. (i) NH OH Kb  NH + + OH– 4 4 10–2.5 10–2.5 (1 – ) × 0.7225 =  2 1 10–5 = 1.8 × 10–5 1 12 + 0.7225  – 0.7225 = 0 C C = =0.556 M 1 = 0.562 1 1 . C = 10–2, [H+] = 10–3 1.8  [H+] = 0.01 × 0.562 [H+] = 5.6 × 10–3 10 –3 10–3 10 –3 = 1.1 × 10–4 [H PO ]  5.6 × 10–3 RxN. (ii) K= 10–2 – 10 –3  24 a 9  H + CHCl COO– 6.8 × 10–8= [H P O –2 ][H  ] 2 4 1 3 . CHCl COOH + from [i] reaction. 2 – [H P O 4 ] 0.01 2 0.01 – x 0.01+ x x  [HPO –2] = 6.8 × 10–8 M RxN. (iii) 4 x(0.01  x) = 2.55 × 10–2 0.01 – x 4.5 × 10–13 = [P O –3 ][H  ] 4 0.01 x + x2 = 2.55 × 10–4 – 2.55 × 10–2 x x2 + 0.355x – 0.000255 = 0 [H P O –2 ] 4 x = –0.0355  0.04775 =1.1 × 10–2 4.5 10–13  6.8 10 –8 = [PO4–3] 2 5.6 10 –3 CHCl COO– = 6.126 × 10–2 5.464 × 10–18 = [PO4–3] 2

20. NH4Cl  N H + + Cl– 30. CH COO + H O  CH COOH + OH 4 32 3 + NH4OH  N H 4 + OH– 10 –14 [N H  ][O H – ] 0.08 K= 1.8 10 –5 4 b Kb = [NH 4O H ] x2  10 10 –10 x2 = 0.8 10 –10 0.08 1.8 1.8 [NH +] = is due to salt because NH OH ionise in x2 = 0.44 × 10–10 44 x = 0.66 × 10–5 less amount due to common ions effect 1.8 × 10–5 = 0.1  [OH – ] 9 × 10–6 = [OH–] 32. C H N+ + H O  C H NOH + H+ 56 2 55 0.05 CH COO 1 3 pH = [pK – pK – log C] 21. HC H O + NaOH   + HO 2w b 232 2 50ml, 0.2M 50ml, 0.1M 1 2.699 = [14 – pK + 0.6] 10m mol 5m mol 2b O 2.398 = 14.6 – pK  b OH  CH3COO + H2O CH3–C–OH + pK = 14.6 – 5.398 = 9.802 b 10 5 K = 10–9.802 50 5 5 b pH = pK =5 – log 1.8 pH = 4.74 3 8 . pH = pK1  pK2 a 2 2 2 . (NH4)2SO4 pH = 11  7 – 2 log 4.5 = 9 – log 4.5 = 8.54 x 2 Molarity (NH ) SO =  100  40. C H C O O H + O H    C H C O O – + H O (WASB) 42 4 3 3 2  2x  1 Molarity of NH4+=  100  pH = [pK + pK + log C] 2w a  0.1  11 Molarity of NH4OH =  100  =10–3 = [14 + 5 – log1.9 + log ] 2 20 1 = [19 – log1.9 – log 20] 2  2x /100  1 14 – 9.26 = 4.24 + log  0.1 / 100  pH= [19 – log 20 × 1.9] = 8.78 2 pOH= 5.28 [OH] = 10–5.28 0 = log (20x)  1 = 20x [OH] = 10–6 × 100.72 [OH] = 5.24 × 10–6 x = 1/20 mole x = 0.05 mole IONIC EQUILIBRIUM EXERCISE # 4[B] 1 . Q+ + H2O  QOH + H+ kw/k1 adding both ( )  n1  n '1 (x + y)2 = k w  n1'  n2'  = kw (k2n1'  k1n2') V  k1 k2  k1k2 n'1 – x x x+y kw R+ + k1k2V H 2O  ROH + H+ [H+] = (x + y) = (k2n1  k1n2) n2  n 2' pH = –log [H+] V n'2 – y y y+x pH = 1 log  k1k2  (k2n1 V k1n2 )  kw = x(x  y)  x(x  y) ..........(i) 2  kw       k1 n1'  x n1' k w  1.818  10 4 k4 kw = y(x  y)  y(x  y) .........(ii) 2. Na Y+H O  Na HY+NaOH 42 3 k 2 n2'  y n 2' 0.1 Assuming () x << n1' & y << n2' 0.1 – x xx from equation (i) & (ii) ((i)  (ii) ) x2 =1.818× 10–4 5500.55x2 + x – 0.1 = 0 0.1  x x (x + y) = kw n1' & y (x + y) = kw n2' x = 4.17 × 10–3 k1 k2

 kw  1.445 108 5. Cu3(AsO4)2  3 C u +2 + 2 A s O –3 Ksp=8×10–36 k3 4 N a H Y + H 2 O N a H 2 Y + N a O H 3 2 3x 2x+2y x Pb3(AsO4)2  3Pb+2+2AsO4–3 Ksp=4.096×10–36 3y 2y+2x x–y y y+x since y << x x – y ~ x, x + y ~ x 1.445 × 10–8 = y.x  y Let solubility of Cu3(AsO4)2 & Pb3(AsO4)2 is x & y x respectively. (Cu (AsO )  Pb (AsO )  3 42 3 42 x  y )  Na2H2Y+H2O  NaH3Y+NaOH k w  4.7 1012 108 x3 (x + y)2 = 8 × 10–36 .....(i) k2 108 y3 (x + y)2 = 4.096 × 10–36 ....(ii) y (i) x3 8 (ii)  y3  4.096  x = 1.25 y y–z z z+x y – z ~ y, z + x ~ x z.x z = 1.628 × 10–17 putting this in equation (ii) ((ii))  4.7 × 10–12 = y 108 y3 (2.25 y)2 = 4.096 × 10–36 NaH3Y+H2O  H4Y+NaOH k w  9.8  1013 y = 2.3 × 10–8 x = 1.25 y = 2.875 × 10–8 z k1 [Cu+2] = 3x = 8.825 × 10–8 [Pb+2] = 3y = 7.1 × 10–8 z–t t t+x 6. (a) Al(OH)3  Al+3 + 3OH– Ksp z – t ~ z, t+x~x 9.8 × 10–13 = t.x t = 3.82 × 10–27 A l(O H ) –  Al+3 + 4OH– K z 4 fraction ()= t = 3.82 × 10–26 Al(OH)3 + OH–  Al(OH) – K sp  38.46 4K 0.1 1 0 3 3. s=[Zn(OH)2(aq)]+Zn(OH)++Zn+2+Zn(OH)–3+ Zn(OH)4–2 38.46 =  Al(OH  )4  = [OH ]  [OH  ]  k1k2 k1k2k3   [OH ] [OH  ]2 s=k1+ + +k1k4[OH–] + k1k4k5 [OH–]2 [OH–] = 2.6 × 10–5 s=10–6+ 10 13 + 10 17 +10–3[OH–]+10–2 [OH–]2 pOH = 4.585 [OH ] [OH  ]2 pH = 9.415 (a) pH = 5, pOH = 9, [OH–] = 10–9 (b) K = [Al+3] [OH–]3 s = 10–6 + 10–4 + 10 + 10–12 + 10–20 = 10 M sp (b) pH = 9, pOH = 5, [OH–] = 10–5 5 × 10–33 = [1 × 10–3] [OH–]3 [OH–] = 1.7 × 10–10 s=10–6+10–8+10–7+10–8+10–12 = 1.12 × 10–6 M pOH = 9.767 (c) pH = 13, pOH = 1, [OH–] = 10–1 pH = 4.23 s = 10–6 + 10–12 + 10–15+10–4+10–4 = 2 × 10–4 M 7 . HCl        0.09 M 4 . Given : CH3COOH   C1 = 0.1 M, 1 , K a1  10 –5 1 Cl2  AgCl G1°=–109.7kJ/mole–1 Cl2CHCOOH  C2 = 0.09 M, 2 , K a2  ? Ag+ pH = 1, [H+] = 0.1 2 Ag  Ag+ + e– G°2 = 77.2 kJ/mole–1 0.1 = 0.09 + C11 + C22 e– + 1 Cl2  Cl– G3°=–131.2kJ/mole–1 C11 + C22 = 0.01 ........(i) 2 CH3COOH  CH3COO– + H+ so for reaction (   )  C1 AgCl  Ag++Cl–  G ° =–  G1°+ G ° + G ° C1 – C11 C11 0.1 2 3 Cl HCCOOH  Cl HCCOO– + H+ G° = 55.7 kJ/mole 22 G° = – RT ln Ksp C2 55.7 × 10–3 = –8.314 × 298 ln Ksp C2 – C22 C22 0.1 Ksp = 1.723 × 10–10 1.723 × 10–10 = [Ag+] [Cl–] = s × 0.05 K a1  (C11 )(0.1) ~ 1 × 0.1 = 10–5 C1 (1  1 ) s = 3.446 × 10–9 M 1 = 10–4

putting this in equation (i)((i) ) 1 2 . k1 = 7.5 × 10–3, k2 = 6.2 × 10–8 , k3 = 10–12 H3PO4 + NaOH  NaH2PO4 + H2O 10–4 × 0.1 + 0.092 = 0.01 (a) 63 2 = 0.111 3– 3 K a2  (C22 )(0.1) = (0.111)(0.1) = 1.248 × 10–2 (b) pH = pk1 = 2.12 C2 (1  2 ) 1  0.111 H3PO4 + NaOH  NaH2PO4 + H2O 66 8. C = 10  0.935 1000 = 5.5 M – –6 17 100 pH = pk1  pk2 = 4.66 2 Ka = C2  5.5 × 10–6 = 5.5 2 H3PO4 + NaOH  NaH2PO4 + H2O  = 10–3 (c) 4.8 7.2 [OH–] = C = 5.5 × 10–3 pOH = 2.26 , pH = 11.74 – 2.4 4.8 9. ln k w2  H  1  1  NaH2PO4 + NaOH  Na2HPO4 + H2O k w1 R  T1 T2  4.8 2.4   2.4 – 2.4 5.474  10 14 = H  1  1  (d) pH = pk2 = 7.2 ln 1.08  10 14 8.314  298 323  H3PO4 + NaOH  NaH2PO4 + H2O 4 10 H = 51952.6 J = 51.95 kJ/mole –6 4 1 0 . In begining [H+] = KaC NaH2PO4 + NaOH  Na2HPO4 + H2O 46 [H+] = 1.8 105 = 0.004242 pH = 2.372 – 24 On doubling pH, new pH(pH ,  pH) Na2HPO4 + NaOH  Na3PO4 + H2O 42 = 4.744 [H+] = 1.8 × 10–5 2–2 CH3COOH  CH3COO– + H+ 13. pH = pk3 = 12 C (a) For H2CO3  k1 = 4.2 × 10–7, k2 = 4.8 × 10–11 (b) H3PO4k1=7.5× 10–3, k2=6.2× 10–8, k3 = 10–12 C – C   C C (c) Na2CO3 + HCl  NaHCO3 + NaCl (d) 22 (C)2 , C = [H+] = 1.8 × 10–5 ––2 Ka = C(1  ) pH = pk1  pk2 = 8.347 1.8 × 10–5 = (1.8  10 5 )2 2 C  C Na3PO4 + HCl    Na2HPO4 + NaCl C – C = 1.8 × 10–5 0.8 1.6 + NaCl 0.8 C = 3.6 × 10–5 – 0.8 NaH PO 1 = 2.77 × 104 L Na HPO + HCl   24 V= 24 C 0.8 0.8 0.8 11.(a) PV = nRT –– 1 × 0.959 = n × 0.0821 × 298 pH = pk1  pk2 = 4.66 2 n = 0.03919 volume of H2O = 1 mL (per volume of H2O) Na3PO4 + NaH2PO4  2Na2HPO4 55 10 (H2O   = 1 mL (H2O   )) –– n 0.03919 C = V = 103 = 39.19 M pH = pk2  pk3 = 9.6 pkb = 3.39  kb = 4 × 10–4 2 [OH–] = K bC  0.1252M H3PO4 + Na3PO4  Na2HPO4 + NaH2PO4 44 pOH = 0.9023 pH = 13.097 –– 44 (b) M = 0.1252 for NaOH (NaOH   0.1252) pH = pk2 = 7.2

1 4 . BOH + HCl  BCl + H2O 16. Initial ()pOH= pKb =4.744 4 4 Let x mole of NaOH has been added so (NaOH ––4 x )  At end point m moles of BOH = m moles of HCl [ N H + ] = 0.1 + x, [NH3] = 0.1 – x 4 ( BOH m moles) pOH = 5.744 0.16 × V = 4 V = 25 mL Total volume ( ) = 40 + 25 = 65 mL 0.1  x 5.744 = 4.744 + log 4 [BCl] = 0.1  x 65 0.1  x since BCl is SAWB 1 = log 0.1  x 11 0.1  x 0.9 pH = 7 – 2 pkb  log C 2 = 10  x = = 0.0818 moles 0.1  x 11 1 14 5.23 = 7 – 2 pkb  2 log 65 17. Na3PO4 + H2O  Na2HPO4 + NaOH pkb = 4.75 K = Kw = 0.0222 K3 Now on further adding NaOH  NaOH    Na2HPO4 + H2O  NaH2PO4 + NaOH BCl + NaOH  BOH + NaCl 4 1.8 K = Kw = 1.58 × 10–7 K2 2.2 – 1.8 pOH = pk + log 2.2 = 4.837 pH = 9.1628 NaH2PO4 + H2O  H3PO4 + NaOH b 1.8 0.06 K = Kw = 1.4 × 10–12 15.(a) pH = pKa + log 0.05 K3 pH = 3.744 + log 1.2 = 3.823 since equilibrium constant of 2nd & 3rd reaction is very less, [OH–] will mainly come from 1st reaction. (b) On diluting solution 10 times (10   (2nd 3rd    ) [HCOOH] = 0.005, [HCOONa] = 0.006 ,[OH–] 1st  )  HCOOH  H + + HCOO– Na3PO4 + H2O  Na2HPO4 + NaOH 0.1 0.005 0.005 (1 – ) 0.005 0.005 + 0.006 0.1–x x x Ka = 1.8 × 10–4 = (0.005  0.006) (0.005) x2 = 0.0222  45x2 + x – 0.1 = 0 0.005(1  ) 0.1  x 0.0052  0.006 = 1.8 × 10–4 x = 3.73 × 10–2 1 [OH–] = x = 3.73 × 10–2 M  27.772 + 34.33 – 1 = 0  = 0.0285 [H+] = 0.005 = 1.425 × 10–4 pH = 3.846 Na2HPO4 + NaOH  NaH2PO4 + NaOH x (c) On further diluting solution by 10 times (  x–y y y+x 10 )  x – y ~ x, y + x ~ x, so [HCOOH] = 0.0005, [HCOONa] = 0.0006 HCOOH  H+ + HCOO– 1.58 × 10–7 = (y  x) y  y (x  y) 0.0005 + NaOH NaH2PO4 + H2O  H3PO4 z+x 0.0005(1–) 0.0005   0.0005+0.0006 y Ka = 1.8 × 10–4 = (0.0005  0.0006)(0.0005) y–z z 0.0005(1  ) y – z ~ y, z + x ~ x 0.00052  0.0006 = 1.8 × 10–4 z(x  z) zx z  3.73 10–2 = (y  z) y 1.58  107 1 1.4 × 10–12 = = = 2.772 + 4.33 – 1 = 0   = 0.2047 [H+] = 0.0005 = 1.0235 × 10–4 z = 5.93 × 10–18 [H3PO4] = z = 5.93 × 10–18 M pH = 3.9899

1 8 . pH = 8, [H+] = 10–8, [OH–] = 10–6 2 2 . 4 m mole of H+ ion will produce (H+  4 m H C O –  H+ + C O – 2 K = 5 × 10–13 mol ) 3 3 [H+] = 4 10 3 = 0.04 0.0005 0.1 0.0005–y–z 10–8 y P O –3 + H+  HPO4–2 1 4 k3 0.04 0.08 HCO –+H O  H CO +OH– K= Kw =2 × 10–8 0.02 0.02 0.1 32 23 K1 – 0.0005 z 10–6 H P O –2 + H+  H 2P O – 1 0.0005–y–z 4 4 k2 since equilibrium constant for first reaction is very 0.1 0.02 less y << z (     0.08 – 0.02   y<< z) so now they form a buffer solut ion of HPO –2 & H2PO4– 4 (  HPO –2 H2PO4–   4 2 × 10–8 = z(106 ) )  0.0005  z = + 0.08 (k2 = 6.3 × 10–8) 51 z = 0.0005,  z = 9.8 × 10–6 pH pk2 log 0.02 pH = 7.2 + log 4 = 7.8 [H2CO3] = 9.8 × 10–6 M 2 3 . At equivalence point ( )  [HCO3–] = 0.0005 – 9.8 × 10–6 = 4.9 × 10–4 M meq. of HA = meq. of NaOH = 3.612 10 –8  y NaA + HCl   HA + HCl 4.9 104 5 × 10–3 = 3.612 1.806 –2 10–8 1.356 – 1.806 3 [ C O ] = y = 2.45 × M [S] pH = pKa + log [A] 1 9 . Fe+3 + H2O  Fe(OH)+2 + H+ k = 6.5 × 10–3 x 4.92 pKa + 1.356 0.95x 0.05x 0.05x = log 6.5 × 10–3 = (0.05)2 x 1.806 0.95 x = 2.47 pKa = 5.044 [H+] = 0.05x = 0.1235 pH = 0.908 Now NaOH + HA  NaA 22 –– 2 salt [NaA] = 2  0.1 2 0 . pH = pK2 + log acid 20 y pH = 7 1 1 log C = 7 5.044  1 6.7 = 7.2 + log + 2 pKa + + log 0.1 0.005 2 2 2 y = 1.58 × 10–3 mole pH = 9 2 1 . When indicator is half in ionic form pH(  2 4 . In begining let x m mole of BOH are present (     pH) = pKa = 7.2  BOH  x m mole ) pH = 7.2 + log 5 = 7.898 BOH + HCl   BCl + H2O x now with this pH (  pH )  3x x 7.898 = pK + log 4 = pK = 7.2959 44 a1 a1 pOH 1 again when 50% of new indicator is in ionic form = pkb + log 3 (50% ) 14 – 9.24 = pkb – log 3 pH = pKa1 = 7.2959 pkb = 5.237 kb = 5.8 × 10–6

now BCl + NaOH   BOH 26. Zn(OH)2(s)  Zn+2(aq) + 2OH – Ksp x (aq) 6 4 Zn(OH)2(s) + 2OH–  [ Z n ( OH )4] –2 KC (aq) –– x dissolved Zn(OH)2 is present in form of Zn+2 & x [Zn(OH)4–2] so solubility s = [Zn+2] + [Zn(OH)4]–2 = 6,  x = 24 4 (Zn(OH)2Zn+2 [Zn(OH)4–2]  24  [BOH] =  0.48 50 [OH–] = kb  C = 1.668 × 10–3   s= [Zn+2] + [Zn(OH)4]–2 ) pOH = 2.77 s = K sp + K [OH–]2 pH = 11.22 [OH  ]2 C 2 5 . (a) pH at one fourth neutralization (  For min. solubility (  )  pH) x/4 ds 2K sp + 2KC [OH–] = 0 (pH)1 = pka + log 3x / 4 = pka + log d[OH] = 0  [OH  ]3 1 3 [OH–]  K sp 1/4 pH at three fourth neutralization (     =  KC  pH) [OH–] = 9.8 × 10–5 (pH) = pk + log 3x / 4 = pk + log pOH = 4.00869 pH = 9.9913 2a x/4 a 3 2 7 . AgCl(s)  Ag+ + Cl– Ksp pH = (pH) – (pH) = 2 log 3 = 2 1 0.9542 Ag+ + 2NH3  A g ( N H ) + K × K2 2 3 1 (b) 4.45 = pk + log x/3 = pk – log 2 AgCl(s)+2NH3  Ag(NH3)2++Cl–, K = Ksp K1K2 a 2x /3 a pka = 4.751 0.2 xx (c) pH = 2 i.e. 0.2–2x (pH)1 = pka + 1, (pH)2 = pka – 1 K= x2 = K K K = 0.002828 sp 1 2 [S]  10 (0.2  2x) [A] For pk + 1  a x x = 10   x = 10a – 10x = 0.05318 ax 0.2  2x 10a x = 0.009613 x= Solubility () = 9.6 × 10–3 M 11 2 8 . [Cl–] = 0.02 M 10 th i.e. stage Ag(CN) –  Ag ++2C N– K = 4 × 10–19 2 inst 11 [S] 1 0.05 For pka – 1   0.05–x~0.05 x 2x [A] 10 x1 4 × 10–19 x.(2 x )2 = = 0.05 a  x 10 a 4x3 = 4 × 10–19  x = 1.7 × 10–7 x= 0.05 11 i.e. 1 th stage [Ag] [Cl] = 1.7 × 10–7 × 0.02 = 3.4 × 10–9 > Ksp 11 so AgCl will precipitate. ( AgCl   )

2 9 . After mixing with equal volume ( A–2 + H2O  HA– + OH– kw   ) k2 [Ag+] = 0.01 M, HCN = 0.01 M HA– + H O  H A + OH– kw 22 k1 HCN  H+ + CN– Ka Ag+ + CN–  AgCN(s) 1 kw [HA  ][OH  ] [HA–] = k w [A 2 ] K sp k2 = [A 2 ]  k2 [OH  ] HCN + Ag+  H+ + AgCN(s) , kw = [H2A][OH  ]  [H2A] = kw [HA  ] k1 [HA ] k1[OH ] K = K a = 2.25 × 106 [H2A] = k 2 [A 2 ] K sp w k1k 2 [O H  ]2 0.01 0.01 From mass balance ( )  x x 0.01 + kw [A 2 ] k 2 [ A 2 ] k2 [OH ] w since K value is very high almost all of reactant will s = x + convert into product k1k 2 [OH – ]2 (K [H ]x [H  ]2 x s=x+  )  k2 k1k2 0.01 = 2.25 × 106 X = 6.6 × 10–5 s x2 x = 1  [H  ]  [H  ]2 [Ag+] = 6.66 × 10–5 M k2 k1k2 3 0 . M A  M + 2 + A–2 ss ksp = [M+2] [A–2] = s . x = s2 [H  ] [H  ]2 Let solubility is s. ( s)  1  k2 k1k2 But some amount of A–2 will undergo hydrolysis.  [H  ] [H  ]2  k sp 1    Let x is the amount of A–2 left in solution.  k2 k1k2  s= (A –2  x     A–2 ) 

IONIC EQUILIBRIUM EXERCISE # 4[B] 1 .  pH = 1 ; H+ = 10–1 = 0.1 M 6. H2CO3(aq) + H2O()  HCO3–(aq) + H3O+(aq) pH = 2 ; H+ = 10–2 = 0.01 M 0.034–x xx  M1 = 0.1 V1 = 1 M = 0.01 V = ? [H C O  ][H O  ] xx 3 22 K = 3 = From 1 [H2CO3 ] 0.034  x M1V1 = M2V2  4.2 × 10–7  x2  x = 1.195 × 10–4 0.1 × 1 = 0.01 × V 0.034 2 As H2CO3 is a weak acid so the concentration of H2CO3 will remain 0.034 as 0.034 >> x. V2 = 10 litre x = [H+] = [HCO3–] = 1.195 × 10–4  volume of water added = 10 – 1 = 9 litre. Now, HCO3– (aq) + H2O()  CO32–(aq) + H3O+(aq) 2 . H+ = C ;  = [H ] or 10 3 = 10–2 x–y y y C = As HCO3– is again a weak acid (weaker than H2CO3) 0.1 with x >> y. Ka = C 2 = 0.1 × 10–2 × 10–2 = 10–5 3 . Cr(OH)3(s)  Cr3+ (aq.) + 3OH–(aq.) 27S4 = Ksp  K sp 1/4  1.6  10 30 1/4 [C O 2  ][H O  ] y  (x  y) 3 (x  y)   K2 = 3  S =  27  = 27  [H C O  ]     3 4 . pH = 5 means Note : [H3O+] = H+ from first step(x) and from [H+] = 10–5 second step(y) = (x + y) HA  H+ + A–1 [As x >> y so x + y  x and x – y  x] t=0 c 0 0 So, K  y  x = y c 2x teq c(1 – ) c (c)2 [H  ]2  K2 = 4.8 × 10–11 = y = [ C O 2– ]  3 [H  ][A  ] Ka =  So the concentration of [H+]  [ H C O – ] [HA ] c(1  ) c  [H  ] 3 = concentrations obtained from the first step. As But, [H+] << C the first step. As the dissociation will be very low  Ka = (10–5)2 = 10–10 5 . AgBr  Ag+ + Br– in second step so there will beno change in these Ksp = [Ag+] [Br–] concentrations. For precipitation to occur [H+] = [ H C O 3– ] = 1.195 × 10–4 & [ C O 2 – ] Ionic product > Solubility product 3 = 4.8 × 10–11 7. AgBr  Ag+ = Br– [Br–] = K sp  5 10 –13 = 10–11 Ksp = [Ag+] [Br] [Ag ] 0.05 For precipitation to occur Ionic product > Solubility product i.e., precipitation just starts when 10–11 moles of [Br–] = K sp  5  10 13  10 11 KBr is added to 1 AgNO3 solution [Ag  ] 0.05  Number of moles of Br– needed from KBr = 10–11 i.e., precipitation just starts when 10–11 moles of  Mass of KBr = 10–11 × 120 = 1.2 × 10–9 g KBr is added to 1 AgNO3 solution

 Number of moles of Br– needed from KBr = 10–11 1 1 . MX2  M++ + 2X–  Mass of KBr = 10–11 × 120 = 1.2 × 10–9 g s 2s 8. Na2CO3 2Na+ + CO 2– Where s is the solubility of MX2 3 then Ksp = 4s3 ; (2s)2 = 4 × 10–12 = 4s3 ; s = 1 × 104  [M++] = s = 1 [M++] = 10 × 10–4 1× 10–4M 1× 10–4M 1× 10–4M KSP(BaCO3) = [Ba2+] [CO32–] [Ba2+] = 5.1× 10 –9 = 5.1× 10–5 M 12. pH = –log[H+] = log 1 1  10 4 [H ] 9 . In corresponds to choice (c) which is correct answer. BA + H2O  BOH + HA 5.4 = log 1 [H ] Base Acid On solving, [H+] = 3.98 × 10–6 Now pH is given by 1 3 . MX4  M4+ + 4X– 111 S 4S pH = 2 pKw + 2 pKa – 2 pKb Substituting given values, we get KSP = [s] [4s]4 = 256 s5 1  s =  K sp 1/5 pH = (14 + 4.80 – 4.78) = 7.01  2    256  1 0 . Let s = solubility 1 4 . Mg(OH)2  [Mg2+] + 2[OH–] x 2x AgIO3  Ag+ + IO3– ss Ksp = [Mg] [OH]2 = [x] [2x]2 = x.4x2 = 4x3. 1 5 . AB  A+2 + 2B– Ksp = [Ag+] [I O – ] = s × x = s2 3 2 Given Ksp = 1 × 10–8 [A] = 1.0 × 10–5, [B] = [2.0 × 10–5], Ksp = [B]2[A] = [2 × 10–5]2 [1.0 × 10–5] = 4 × 10–15  s = K sp  1  108 = 1.0 × 10–4 mol/lit = 1.0 × 10–4 × 283 g/lit ( Molecular mass of Ag IO3 = 283) 1.0 104  283 100 = gm/100ml 1000 = 2.83 × 10–3 gm/100 ml

REDOX REACTION EXERCISE # 1 1 . No. of equivalent = mole × n-factor 8. xHI + yHNO  NO + I + HO 3 2 2 ( = ×n-)  (2 I   I2  2e)  3  .....(1)   SO –2 + HO  SO –2 + 2H+ + 2e– ...(1) 5 2 3 2 4 (   NO 3  3C  N O )  2  .....(2 ) n-factor for R × N (1) is (2)  50 × .1 × n = 25 × .1 × 2 Adding (1) and (2) 2.5  2  6I– + 2NO –  2NO + 3I n= 3 2 5 6HI + 2HNO3  2NO + 3I2 + 4H2O n=1  x=6, y=2  Final oxidation state will be (3 – 1) = 2 ( (3–1)= 2 ) +7 +3 +4 2 . Meq.( )ofK Cr O = Meq.( )9 . MnO4– + C2O–42+ H+  Mn+2 + 2CO2 + H2O 2 27 2× 1 of ABD n-factor of K Cr O in acidic medium = 6. 5 2 27 (  KC r O n =6) 2MnO4–+5C2O4–2+H+(6) 2Mn+2 + 10CO2 + 8H2O 2 27  (A) 2, 5, 16 6 × 1.68 × 10–3 = x × 3.26 × 10–3  x=3 10. Molarity () = Normality( ) New oxidation state of A–n will be = –n + 3 n factor( ) (A–n    =–n +3 )  (e) 1 M H PO = 1/3 M H PO 34 34 OO OO 11. M= No. of equivalent( )  Volume of sol ( )(in L ) O––S––O––S=O Na–O––S––S––S––S=O +6 +6 +5 +5  Meq. = 50 × 2 = 10 3 . (A) OH OH O ONa M mole = 10  5 H2S2O7 (6, 6) Na2S4O6 (5, 0, 0, 5) 2 M H2C2O4.2H2O = 24 + 16 × 4 + 2 + 2 × 18= 126gm. H2SO5  Mass of H C O .2H O ( )  2 O 22 4 S = 126 × 5 × 10–3 = .63gm + – +4 Na2S2O3 O=S––O––O––H 1 2 . 63 % (w/v) H2C2O4.2H2O O  100 mL contain = 63 gm (B) NaO––S=O O–Na+ (6) 125 mL  = 63 125gm 100 (4, 0) (C) SO > SO > H S > S 8 Mole of HCO 63 125   5  3 22 22 4 = 126 100  8      +6 +4 –2 0 40  x 100 125 (D) H SO > SO > H S > H S O . 24 22 22 8 +6 +4 –2 (6, 6) 125  40  5  Mole of NaOH = 100  40 =  4  4. (NH ) Cr O  N + Cr O + 4H O 42 2 7 2 23 2 Decomposition () R × N H2C2O4 + 2NaOH  Na2C2O4 + 2H2O 5 . NaN3  N3–  N  –1/3 2 × mole of Acid () = Mole of NaOH NH  N  –1 55 22 2  NO  N  +2  5(A) 84 NO  N  +5 5 25  4  6. S O –2 + I  S O –2 + I– And will have mole of NaOH (NaOH  25 2 46  Redox R × N () I2 Re dn I  5   4  S 2 O 2 Oxd.  S 4 O 2   )  3 6 (4, 0) (5, 0, 0, 5)  Sol. is neutral ( ) 

1 3 . CaCO3 + HCl  CaCl2 + H2O + CO2 (224 mL) 1 9 . M eq.()ofKMnO = M eq. () 4 CO mole = 224  102 of C 2O –2 2 22400 4 10 2 1 90  1 = 100 × NC 2 O 2 200  10 3 20 20 4 HCl M= HCl N =   .05 99 M. eq. ()ofH3PO4 = M. eq. () M mole of oxalate =  22 4 of Ba(OH) Wt of oxalate = 9  88 103  22  9 × 10–3=198×10–3 2 4 % C O –2 = .198 100 = 66 % 1.5  v  3  90 .5  2 90 2 .5 2 4 .300 2 0 . M eq. of KMnO4=M eq. of C2O4–2 = M eq. of CaCO3 v = = 20 mL 3 1.5 40 × .25 = M eq. of CaO 1 5 . KMnO n factor in Acidic medium = 5 10 103  Mole of CaO 4 2 (KMnO4   n ) K Cr O n factor in acidic medium = 6 5 103  56 100 2 27 % CaO = (K Cr O  n ) .518 2 27 CaO = 54 % 6 × 0.1 × V = 5 × 0.3 × V 21. 2CrO + 3H SO  Cr (SO ) + 3H O + 7/2O 12 5 24 2 43 2 2 6 V1  V2 1 mole CrO Liberate ()  7/4 mole of O 15 5 2 V2  2 V1 22. +3 +3 5 AsO–43+ 2H+ + 2I–  AsO3 + H2O + I2 2 1 6 . K Cr O have greater n factor as compaire KMnO molar mass ( )NaAsO 2 27 4 34 so same volume of K2Cr2O7 will oxidise more = 23 × 3 + 75 + 15 × 4 amount of Fe+2. molar mass ( )=208 (K2Cr2O7  n KMnO4   K Cr O   F e+2     eq.()of AsO – =  1   1  2 27 4 208  104   )  2  17. Mole of V2O5= 10 16  10 80  10 =.055 equivalent ()of Na AsO = equivalent () 51 2  5 102  182 34 Mole of V+2 = .055 × 2 of I2 = .1098 mole ~ 0.11 = equivalent ()of Na2S2O3 . 4 1  .2  V 104 V 2  V O 2  2e I2  2e 2I 1 L  V = 48.1 mL 104 .2  Mole of I = Mole of V+2 = .11 2 3 . M eq. of KMnO = 25 × .2 = 5 2 4 18. Cl2 + S 2O –2  S O –2 + Cl– + S 3 4 (A) M eq. of FeSO = 25 .2  5 41 50 × .01 × n factor () = 5 × 10–4 × 2 × 103 10 104  103 (C) M eq. of HO = 25 .1  2  2.5  2  5 .5 22 n factor S2O3–2 = = 2 × 10–3 × 103 (D) M eq. of SnCl = 25 .1  2  5 2 n factor = 2      (i) Cl + HO + Na S O  Na SO + S + 2HCl 24. N   N 1 V1  N 2 V2   3 250  750 1  1500  1.5 2 2 22 3 24 V1  V2 1000 1000 Balanced equation Molarity () = 1.5  0.75  3 (ii) Mole of S O –2 = 50 × 10–3 × 10–2 = .0005 24 23 (iii) Equivalent of oxidising agent (  1 )= 5 × 10–4 × 2 = .001  HO  HO + O 22 2 22 5 104 1L H O , 1 Mole H O give O = 11.2 L 50 103 22 22 2 (iv) Molarity of Na SO (  )= 1L HO, 0.75 HO  = 11.2 × 3 24 22 22 = 10–2 = .01 M 4  = 8.4 V O 2

 Volume strength( )= 8.4 V 2 8 . (320 mL, 10V H2O2) + (80 mL, 5NH2O2) Alternative( )V =5.6× N=5.6× 1.5=8.4V (A) (B) S  10  2 5 . M eq. of KMnO4 = .2 × 50 × 5 = 50 NA =  5.6  M eq. of H O = 2 × 25 × .5 = 25  NC  N A VA  N B VB  10  320  5 80 22 VA  VB 5.6 M eq. of KMnO remaining (KMnO   320  80 44 ) = (50 – 25) = 25 Mole of KMnO = 25 103  5 103 = .005 400  1000  4 45 =7 26. 100x  3   2 NC 400  11  10 = (17/7) 1000  24  7 x = 20  2.5  VS  5.6  17 VS  13.6V 8 7 17 2 7 . Ca(HCO3)2 + CaO  2CaCO3 + H2O 17 M = Mole / L 2 × 100 C 14 56 gm 2 gm M= 72 Mol/L C 200  2 C = 17  34 gm / Ltr. 56 x 14 x = .56 gm Concentration () =41.285 gm/Ltr. REDOX REACTION EXERCISE # 2 1 . Fe + 1 5. HNO3 + N H +  N2 + NO2 1 2 O2  FeO 4 0 0.65 Meq of HNO3 = Meq of N H + 2FeO + 4 0.15 1 mole × n-factor = mole × n-factor 1 1 × mole = 1 × 6 2 O 2  Fe O mole of HNO3 = 6 23 1 0.15 6 . Z x  KMnO 4 H  M n2  Z y (1–0.60) 0.30 0.4 0.30 Meq of Zx = Meq of KMnO 4 Mole ratio ( ) FeO = 0.40  4 25 × 0.1 × (y – x) = 25 × 0.04 × 5 Fe2O3 0.30 3 (y – x) = 0.04  5  2 2 . FeSO4 0.1 1 mole of SO 2– than 1 mole Fe2+ Z2+  Z4+ 4 In Fe (SO ) (4 – 2) = 2 2 43 2– Fe3+ 3 mole of S O 4 than = 2 mole 7 . Cl– + KMnO4  Mn2+ + Cl2 Meq of NaCl = Meq of KMnO 1 mole of SO 2– than = 2 mole Fe3+ 43 4 10 5 mole × n-factor =  ratio ()= Fe2  1  3 158 2 Fe3 2 2 mole × 1 = 10  5 158 2 33 3. 2Fe + O  Fe O 2 2 23 10 5 volume of Cl2 = × 22.4 = 3.54 L Let assume n mole of Iron 158 2 Initial n 0 8. I    I– + IO – 3 2 n–x x I2  2I– I2    2IO3–  2  2e–  I2   2I– × 5 I2 + 6H2O   2IO3– x 10e–  5I2   10I– I + 6H O + 12OH– wt. (n – x) × 56 +  2  × 160 = n × 56 × 1.1 22 24x = 5.6 n    2 I O – + 12H O 3 2 I + 12OH–    2 I O – + 6H O + 10e– 2 3 2 x 12OH– + 6I2  10I– + 2 I O – + 6H2O  n  = 0.2323 3 % total Iron = 23.3% ratio of IO   2 1:5 3 10 I

9 . Eq. mass ( ) 1 8 . Half meq of salt (Na2CO3) in neutralize using Hph molecular mass () indicator = n  factor( ) ((Na CO ) Hph  23 )  As +3  2As+5 n-factor 4 1 2 meq of salt = meq of HCl 2 6 24 2 1 S3  3 S total n-factor (n-) = 28 (20 × 0.1 × 2) = 0.05 x .........(i) 2 Eq. mass ( )=m.wt. complete meq of salt (Na2CO3 . NaHCO3 . 2H2O) is neutralise using MeOH indicator 28 ((Na CO . NaHCO . 2H O)    23 32 2 2 5 2 5 6 4 MeOH    )   1 0 . Zn S  HNO3  Zn (NO3 )2  H2SO4  NO2 Meq. of salt = Meq of HCl change in O.N. of Zn (Zn     20 × 0.1 × 3 = 0.05 y ..........(ii) ) eq (ii) – eq (i) Zn = 0 0.05 (y – x) = (6 – 2) S = 6–(–2) = 8 N=5–4=1 4 (y – x) = 4 × 20 (y – x) = 0.05 1 4 . From Hit and trial method (y – x) = 80 mL 3H O + CN–  NO + CO + 6H+ + 7e– ...(i) 1 9 . Meq of I2 = Meq of Hypo solution ( )  2 2 = 20 × 2.5 × 10–3 3CN– + 9H O  3NO + 3CO + 18H+ + 21e– 2 2 Meq of 10ml I–=Meq of I = 20 × 2.5 × 10–3 = 0.05 2 21e– + 7 N O – + 28H+  7NO + 14H2O 3 Meq of 100 mL I– = 0.5 Balance equation ( ) Meq of CaCO = 0.5 3 3CN– + 7NO3– + 10H+  10NO + 3CO2 + 5H2O 1 5 . K2Cr2O7 + Sn2+  Sn4+ + Cr3+ w 1000  0.5 w = 0.06175 123.5 Sn4+ + Fe2+  Fe3+ % = purity ()= 0.06175 100  61.75% meq. of Sn4+ = meq of K2Cr2O7 meq. of Sn4+ = meq of Fe3+ 0.1 20. I + Na S O  I– + S O 2– 2 22 3 46 meq. of Fe3+ = meq of K2Cr2O7 let x mL of I react with Hypo (xmL, I  22 4.9  6 N K2Cr2O7  294  0.1  1 )  millimol () × n -factor = 1 × 10 meq of I2 = meq of Hypo xN = 15 × 0.4 xN = 6 .........(i) meq of H SO used by base (  HSO millimol () = 10 24 24  )=10 × 0.3 × 2 = 6 1 7 . Vol. of O at NTP meq of NaOH used by I (I   NaOH 2 22 VO = 500 1 273  )=(30 – 6) 2 300 (150 – x) N = 24 ..(ii) VO = 455 mL from eq (i) & eq (ii) 2 150  x  4  5x = 150 35 mL of H2O2 gives 455 mL at N.T.P. x N = M × n-factor (35 mL H2O2N.T.P. 455 mL )  x = 30 mL 30 N = 6 455 1  1 mL of HO gives =  13 N= 22 35 5 = 13 mL of O2 at NTP 1 =M×2 hence volume strength of H O = '13 V' 22 5 M = 1  0.1 (H 2O2  ) 10

2 1 . Let a gm H2SO4 and (3.185 – a) g H2C2O4 2 2 . K2Cr2O7 + KI  I2 + Cr3+ Meq of 10 mL mixture = 0.3 meq of 1000 mL mixture = 0.3 × 1000 = 30 I+ Na S O  I + S O 2– meq of H2SO4 + meq of H2C2O4 = 30 2 22 3 2 46 Na2S2O3 + K2Cr2O7  meq of Na S O  meq. of K Cr O 22 3 2 27 a 1000  (3.185  a) 1000  30 ......(i) 1 1 49 45 30 × N = 15 × N= 20 40 In another ex. meq. of I = meq. of Hypo meq of 100 mL mixture = meq of KMnO4 2 = 4 × 0.02 × 5 meq. of I2 =meq. of KI meq of 100 mL mixture = 0.4 meq of KI = meq. of K2Cr2O7 meq of 1000 mL mixture = 4 meq of H C O = 4 1 24 × = meq. of 25 mL K Cr O 22 4 40 2 2 7 24 500 meq. of 500 mL K2Cr2O7 =  40 25 w  6 1000  12 w = 0.588 294 3.185  a 1000  4 .............(ii) % purity ()= 0.588 100  73.5% 45 0.8 REDXO REACTION EXERCISE # 3 COMPREHENSION BASED QUESTIONS 4. eq. of H2SO4 + eq. of SO3 = eq. of NaOH x  2  (1  x)  2 = 54 × 0.4 × 10–3 Comprehension # 1 98 80 1 . H2O + SO3  H2SO4 ; 18 g water combines with 80 g SO3 % of free SO = 1  0.74 100  26% 31 (18 g  80 g SO   )  3  4.5 g of H2O combines with 20 g of SO3 Comprehension # 2  4.5 g H2O, 20 g SO3    1 L of H O (aq) provide 11.2 L of O at STP 22 2  100 g of oleum contains 20 g of SO3 or 20% free 1 . SO3 moles of O2 11.2  0.5 =  100g  20g SO3  20% SO3 22.4 2 . Initial moles of free SO3 present in oleum nH2O2 required 0.5 × 2 = 12  2 moles M H2O2  nH2O2 1M 18 3 Vsolution   SO3   = moles of water that can combines with SO3 2. Strength in percentage mean how many g H2O2 combined with water = 9  1 mole present per 100 mL 18 2 (100 mLg  SO3  H2O2   )   moles of free SO remainsSO    33  = 2  1  1 mole  M  1 and mol. wt. of HO = 34 32 6 22 34 H O present per litre of solution or 3.4 H O 22 22 volume of free SO at STP (STP  SO   33 present per 100 mL of solution. ) = 1  22.4  3.73L  34 H2O2 3.4H2O2 100 mL   6 3 . Na2CO3 + H2SO4  Na2SO4 + H2O + CO2 3. m.eq. of H2O2 = m.eq. of KMnO4 20 × N = 0.05 × 5 × 80  N = 1 moles of CO2 formed = moles of Na2CO3 reacted 5.3 N = volume strength ()of H2O2 = = 0.05 5.6 106  volume strength of H2O2 = 5.6 volume of CO2 formed at 1 atm pressure and 300 K = 0.05 × 24.63 = 1.23 L

4 . m-eq. of Ba(MnO ) = m. eq. of H O Comprehension # 3 42 2 2 n-factor () = 5 × 2 =10  M  33.6  3  1. H3PO2 is a monobasic acid ( ) 11.2  2.  n-factor =1 w 3. n-factor =  3 – 2  × 0.95 = 0.8075 375 × 10 × 1000 = 3 × 125 × 2 ; w = 28.125  0.95  M  eq. wt. = 0.8075 % purity ()= w 100 28.125 100 = 70.31 4. n-factor of VO = 3; Fe2O3= 1 × 2 =2 ; = 40 40   x and y are 2 and 3 REDOX REACTION EXERCISE # 4[A] 4 . KMnO4 + X+n  X O – + Mn+2 1 3 . CaCO3 + 2HCl  CaCl2 + CO2 + H2O 3 0.1 mole 0.25 1.61×10–3mole 2.63×10–3mole – 0.05 Eq. ()of KMnO4 = Eq. of X+n HCl + KOH  KCl + H2O 0.05 2×V 1.61 × 10–3 × 5 = 2.63 × 10–3 × (5 – n) n=2  56 = M  35.5 M = 41 V = 0.05 L = 25 mL 2 2 6 . CuS + Cu2S + KMnO4  Mn+2 + Cu+2 + SO2 1 4 . 2NaOH + NaH2PO4  Na3PO4 + 2H2O 6 85 12 Eq. wt. () ofCuS = M1/6 1×V = 0.1 Mole Eq. wt. () ofCu2S = M2/8 Eq. wt. () ofKM nO4 = M3/5 120 8. N =  53   0.06 V × 1 = 0.1 × 2  250  V = 0.2 lit = 200 ml. n-factor = 2 1 5 . CaCO3 + 2HCl  CaCl2 + H2O + CO2 0.06 x mole 2x M =  0.03 MgCO3 + 2HCl  MgCl2 + H2O + CO2 2 y mole 2y 2x+2y= (50  0.8 – 16  0.25) x+y=0.018...(1) 9 . H2SO4 + 2 NH3  (NH4)2SO4 1000 (30 – 25) Meq. 25 Meq. x × 100 + y × 84 =1.64 ...(2) (30 × 0.2) Meq.  % CaCO3 = x 100 100 = 48.78%  1.64 VNH3 = 25 × 10–3 × 22400 = 537.6 ml   % MCO3 = 51.22%  1 0 . n1 × 56 + n2 × 74 = 4.2 ....(1) 1 6 . M Eq.()of CaCO3 = M Eq. of HCl – n1 × 1 + n2 × 2 = 0.1 ....(2) M Eq. of NaOH % of KOH = n1  56 100 = 35% ....(1) 4.2 w 1000 = 10 × 4 – 4 × 18.75 × 0.2 = 25 (100 / 2) Ca(OH)2 = 65% w = 1.25 gm 1 1 . n × 106 + n × 84 =1 n × 2 + n × 1 = 0.1 × V × 1000 ....(2) % CaCO3 =  1.25  100 = 83.33%  1.5  V = 157.89 ml 1 2 . Eq. of H2SO4 = Eq. of NaOH 1 7 . Na2CO3 + NaHCO3 n × 2 = 0.0267 × 0.4 x g milli mole n = [0.0267 × 0.2] mole of H2SO4 total. x= 4×1 ....(1) [N × 98 – 0.5] = mass of H2O added mole of H2O = mole of SO3 2x + 4 = 10.5 % of SO3 = 20.72 % y = 2.5, x = 4 Na2CO3 = 4 × 106 mg = 0.424 mg NaHCO3 = 0.21 gm

1 8 . Na2CO3 NaOH 26. Sn + K2Cr2O7  SnCl4 + Cr+3 x y m mole 1 0.1 N V ml x + y = 19.5 × 0.995 = 19.4025 ... (1) 1  ... (2)  M / 4  × 1000 = 0.1 × V  V = 337 ml 2x + y = 25.0.995 = 24.875 x = 5.4775 m mole 5.4725 106 2 7 . Meq. of Cu = 1000 [20 × 0.0327] = 32.7 Na2CO3 = = 23.2 gm/lit. 20 25 NaOH = 22.28 gm/lit. w (63.5/1) 19. NaOH + Na2CO3 1000 =32.7  w = 2.07645 gm x y m mole 1 ... (1) % Cu = 2.07645 100  41.53% x + y = × 17.5 = 1.75 5 10 2 8 . Meq. of Fe = Meq. of K2Cr2O7 0.84  x / 100 1000 = x × N y = 0.25 ... (2) x = 1.5, y = 0.25 m mole 56 N = 0.15 NaOH = 1.5  40 gm = 0.06 gm   1000 0.25 106  1000 Na2CO3= = 0.0265 g m   2 9 . H2C2O4.2H2O + KHC2O4.H2O+NaOH product 2 0 . Na2CO3 NaHCO3 x mole y mole 18.9ml, 0.5 N x y Meq. H2C2O4.2H2O+KHC2O4.H2O+KMnO4Mn+2+CO2 x = 2 × 0.2 = 0.4 ... (1) x mol y mol 21.55ml, 0.25N 4 4 y + x = 2.5 × 0.4 ... (2) =1 x×2+y×1= 18.9  0.5 .... (1) y = 0.6 x = 0.4 1000 2 1 . Same as 19. Ce+2 x  y 2 × 1000 = 21.55 × 0.25 ....(2) 2 2 . C e +4 + S n +2  S n +4 +  4 4  × 40.05 + 20 ml % H2C2O4. 2H2O = 14.36% % KH2C2O4.H2O = 81.7% 1M 1M 3 0 . Meq. of Ca(OH)2 = Meq of HCl Meq. of Ce+4 = Meq. Sn+2 w  40.05 × 1 × (4 – n) = 20 × 1 × 2  74 / 2  × 1000 = (50 × 0.5 – 0.3 × 20) (4 – n) = 20 2  1 w 40.05 n=3 % Ca(OH)2 = × 100 = 1.406 50 4 23. + CrSO4  Cr+3 + Se+2 3 1 . Meq. of Na2CO3 = Meq of HCl SeO2 w Meq. of SeO2 = Meq. of CrSO4 106 / 2 × 1000 = 50 × 0.1 – 10 × 0.16 12.53 × 0.05093 × (4 – n) = 25.52 × .1 × 1 w 4–n~4  n=0 % purity = × 100 =90.1% 24. K2C2O4.3H2C2O4.4H2O+MnO4– Mn+2 4 1 + CO2 3 2 . x gm substance ()  V ml, 0.1 M 0.6 x gm NaCl, 0.37 x gm KCl 1   0.6x  0.37x  1000 = 25 × 0.1 – 5.5 × 0.1  5 0 8   58.5 74.5  ×  8 1000 = V × 0.1 × 5 x = 0.1281 gm  V = 31.68 ml 3 3 . 12 = 5.6 × N  N = 2.1428.57 2 5 . H2O2 + KMnO4  Mn+2 + O2 700 × 2.1428 = 1000 × N 1  x /100 N1 = 1.5 = M1 × 2 (34 / 2) × 1000 = x × N M1 = 0.75  gm/lit = 0.75 × 34 = 25.5 Volume strength of final solution (   20  )= 5.6 × 1.5 = 8.4 N = = 0.5882 34

3 4 . 50 × N = 20 × 0.1 4 0 . 100 ml  1.62 mg Ca(HCO3)2 N = 0.04 = M × 2 60 × 103 ml 1.62 × 60 × 103 = 972 mg M = 0.02  gm/ = 0.02 × 34  100 gm/ = 0.68 Ca(OH)2 + Ca(HCO3)2  2 CaCO3 + 2 H2O 35. 5 106 =41.66 ppm w 0.972  100× 103 74  162   1 1   1 0 3  0.972   111 120  w =  162  × 74 = 0.444 gm  100 4 1 . Bleaching powder + Mohr salt excess  product. 36. 1000 106 = 1.734 ppm (+  ) Mohr salt ()+ KMnO4  product (. 3 7 . 0.001 100 106 =100 ppm 42. Meq. of SeO 2 = Meq. of B rO  used 1000 3 3 3 8 . Meq. of H2O2 = Meq. of KMnO4 w  2 1000 = 20  1  5  5  1  2  x 0.632 M 60 25  34 / 2 = 158 / 5 w = 0.084 gm = 84 mg % Purity ()= x 100 = 85% 43. 1  0.552 × 1000 = 100 × 17 × 0.0167 × n 0.4 M 25 3 9 . 5 × x = 5.5 × N 28 n = 6 = No. of electron taken up by oxidant. = 5.6 × N (    )   5.5 5 × x = 5.5 × 0.909 N = 0.909 x=1 REDOX REACTION EXERCISE # 4[B] 1. meq. of Hypo = 5 = meq. of I2 3. Let H2C2O4 . 2H2O  x g in 100 mL (=5=I2 )  (H2C2O4 . 2H2O  100 mL x g ) moles of I2 = 2.5 m moles On reaction with NaOH with phenolphthalein (I2  = 2.5 m moles) (N aOH )  2CuSO4 + 4KI  Cu2I2 + 2K2SO4 + I2 g Eq. of acid in 50 mL (50 mL ) from reaction moles of CuSO4 ( CuSO4 x2  ) = 2.5 × 2 = 5 m moles = Mw of hydrated CuSO4 ( CuSO4 Mw) 2 126 = 159.5 + 18x g Eq. of NaOH (NaOH   )  so () 1.2475 = 5 × 10–3 x = 5. = 1  0.11905 10 159.5  18x x 2 0.11905 1 so =  x = 1.5 g 2 . meq. of Hypo = 100 × 10 = 10 = meq. of I2 2 126 10 so mass of Na C O = 2.5 – 1.5 = 1 g ()  (I2 ) 22 4  meq of ClO3– (ClO3–  )=10 Now, Na2C2O4   in 0.5g m of same mixture( ,  0.5g)  – (ClO3– m moles of ClO 3 moles) = 2 H2C2O4 . 2H2O  0.3 g 6H2O + 6Cl2  10Cl– + 2ClO3 + 12H+ Na2C2O4  0.2 g 2 m moles g Eq. of H2C2O4.2H2O(H2C2O4.2H2O   )  so moles of Cl = 6 m moles 2 0.3 2 ( Cl2  = 6 m moles) = 6e– + 14H+ + C r 2O – 2  2Cr+3 + 7H2O 126 7 (2Cl–  Cl2 + 2e–) 3 ) =0.2 2 14H+ + Cr2O7–2 + 6Cl–  3Cl2 + 2Cr+3 + 7H2O g Eq. of Na2C2O4 (Na2C2O4 134 6 m moles g Eq. of KMnO4 (KMnO4 ) =V10–3 m moles of Cr 2O –2 = 2 m moles 10 7 – 2 10–3 × wt. of C r 2O 7 = 2 × 294 = 0.588 g 0.3  2 0.2  2 V 103 so + = V = 77.46 mL (Cr2O7–2  )  % purity ()= 58.8 % 126 134 10

4 . First HCl will react with KIO3 to from I2 & Cl2 then 6 . 4OH– + 2H2O + SO2  SO42– + 4H2O + 2e– this Cl2 produced will again react with KI to form 4OH– + SO2   S O 2– + 2H2O + 2e– ...(1) 4 I2. (HCl, KIO3  I2 Cl2    Cl2KI    I2 2H2O + H2O2 + 2e–  2H2O + 2OH–   ) H2O2 + 2e–  2OH– ...(2) Eq. () (1) + (2) Let initially x moles of KIO3 were mixed with y 2OH– + H2O2 + SO2  S O 2– + 2H2O ...(3) 4 moles of HCl then( KIO3x   NaOH + HC l  NaCl + H2O H Cl y  ) Meq. 30 × 0.04 0.024 × 22.48 – 10Cl–  I2 2IO 3 + + 5Cl2 1.2 ~ 0.53952 xy 0.66048 × 10–3 – yy – From equation (3) ((3) ) 10 2 Cl2 + 2KI   I2 + 2KCl 2OH– + H2O2 + SO2   S O 2– + 2H2O y 4 0.66048× 10–3 0.33024 × 10–3 2  moles of SO = 0.33024 × 10–3 2 y – (SO2  )  2 wt. ()= 0.33024 × 10–3 × 32=10.5676×10–3 Total moles of I2 formed (I2   )= y y 3y % of S sample(S %)= 10.5676 103 ×100 =1.76% += 0.6 10 2 5 3 y 0.021  24 10 3 7. (Mn2+  Mn3+ + e–) ... (1) so =  y = 0.00042 mole 4e– + 8H+ + M n O –   Mn3+ + 4H2O ... (2) 52 4 so concentration of HCl (HCl  ) equation() (1) × 4 + equation() (2) 0.00042 8H+ + 4Mn2+ + MnO –  4Mn3+ + Mn3+ + 4H O = = 0.0168 M = 0.0168 N 4 2 0.025 or 8H+ + 4Mn2+ + MnO4–  5Mn3+ + 4H2O ...(1) moles of KIO consumed (KIO  )  2e– + 8H+ + Mn3O4  3Mn2+ + 4H2O ...(2) 33 0.00042 from equation (1) milli equivalent of MnO4– =N×V = ((1) MnO4–  )  5 = M × V.f. × V = 31.1 × 11.7 × 5 = 0.72774 volume of KIO consumed (KIO  ) 33 0.00042  5 = = 0.00042 L = 0.42 mL milli equivalent of Mn2+ = milli equivalent of 5 – 5 . As2O3 + 6HCl  2AsCl3 + 3H2O M n O 4 ×4 AsCl3 + 2H2O  HAsO2 + 3H+ + 3Cl– (Mn2+ = MnO4– × 4)  gram equivalent of I2 = gram Eq. of HAsO2 = 0.72774 × 4 = 2.91096 I2  =HAsO2    1 = gram Eq. of AsCl3 from equation (3) milli eq. of Mn3O4 = milli 3 (AsCl3   )  equivalent of Mn2+ = gram Eq. of As2O3 ((3) Mn3O4  =1 Mn2+ (As2O3   )  3 gram equivalent of As2O3 = 2 × 0.04134 × 23.04  )=1 × 2.91096 = 0.97032 × 10–3 = 0.9524 × 10–3 × 2 3 (As2O3  )   equivalent of Mn3O4 = 0.97032 × 10–3 gram equivalent of KMnO4 = 0.9524 × 10–3 × 2 (Mn3O4  )  (KMnO4  )  W = 0.97 × 10–3 229 Let amount of KMnO4 used = w g then W = 222.20 × 10–3 ( KMnO4  =w g ) w  5 = 0.9524 × 10–3 × 2 % of Mn3O4 in the sample (Mn3O4  %) 158.5 w = 0.06 g = 222.20 103 × 100 = 40.77% 0.545

8 . H2S + SO2 1 0 . BaCrO4  0.0549 g xy Cr  0.0549 × 52 × 25 = 0.282 g S–2 2 – = 6) ((n-= 6)) 253 4  S O (n -factor % of Cr (Cr  %) = 0.282 × 100 = 2.82% for H2S (H2S  ) x  6 = 0.534975 × 10 34 10–3 =(20 × 0.0066× 6 – 7.45 × 0.0345) × 10–3 Cr2O7–2  0.282 = 0.002711 mole 52 2 x = 3.031525 × 10–3 g Eq. of MnO4– (MnO4–   ) =15.95 × 10–3 × 0.075 × 25 – 0.002711 × 6 = 0.01364 SO2  SO42– (n -factor = 2) (n- = 2) wt. ()= 0.01364 × 158.5 0.432388 = 5 y 2 for SO (SO  ) 64 wt. of Mn (Mn  ) = 0.01364  55 = 0.15 g 22 5 2y = (25 × 0.396 – 12.44 × 0.0345) × 10–3 % of Mn (Mn %) = 0.15 × 100  1.5% 64 10 2y = 0.56082 × 10–3 1 1 . CH3(CH2)nCOOH + O2  (n + 2)CO2 + H2O 64 y = 17.94624 g a (n + 2)a CO2 + 2NaOH  Na2CO3 (n + 2)a b – b – 2(n + 2)a (n + 2)a concentration of H2S(H2S  )  solution has (  )= 3.031525 × 10–3 = 0.1212mg NaOH  b – 2(n + 2)a = 25 Na2CO3  (n + 2)a On dividing in equal part moles get halfed. concentration of SO2 (SO2  )  () 17.94624 Part - I (  -I) : = 25 = 0.7178 mg SO2/L b  2(n  2)a (n  2)a ... (i) 9 . Let mass of KClO3 (KClO3 ) xg + = 0.05 ... (ii) Let mass of KCl (KCl  )  yg 22 ... (iii) ... (iv) Part - II (  -II) : b  2(n  2)a +2× (n  2)a = 0.08 KClO3  x/122.5 AgCl = 108 + 35.5 22 KCl  y/74.5 = 143.5 (ii) – (i) 6e– + 6H+ + C l O –  Cl– + 3H2O (n  2)a  0.03 3 2 x y 0.1435 (n + 2)a = 0.06 1225 745 143.5  = = 0.001 ... (i) 1.16 and = a for complete oxidation of an oxidizing agent = reacted FeSO4 solution – unreacted FeSO4 60 14n from equation (iii) & (iv) ( (iii)  (iv) ) (   =  1.16 0.06 FeSO  – FeSO ) 60 14n = (n  2) 19.33n + 38.66 = 60 + 14 n 44 5.33n = 21.33  n = 4 = N1V1 – N2V2 = 30 × 0.6 – 37.5 × 0.8 N = 3 milli eq.() from equation (iii) ((iii) ) x 0.003 6a = 0.06 = = 0.0005 a = 0.01 1225 6 from equation (i) ((i) ) put above value in eq. (i) (  (i) b )  – 0.06 + 0.03 = 0.05 2 y ... (ii) b = 0.08 moles of NaOH (NaOH 0.08 ) 745 = 0.0005 2 moisture()=1–(1225+745) × 0.0005 = 0.015g b = 0.16 mass () = 0.16 × 40 = 6.4 g

1 2 . Total m mol of AgCl from 20 mL solution  mass % of NaOH (original) (NaOH   = 0.4305 1000  3 %) = 20  40 100  80 143.5 1000 (20 mL   AgCl  m mol) Now, let us assume that in 20 mL, x m mol of m moles of AgCl from HCl = 0.8  m moles of NaOH has got converted to Na2CO3 AgCl from CaCl = 2.2 ( , 20 mL , x m mol NaOH, Na2CO3 2 )  (HCl  AgCl m mol = 0.8  CaCl2   AgCl m = 2.2)  In 20 mL, m mol of NaOH = 4 – x  1.1 m mole of CaCl2 was consumed for m of Na2CO3 x precipitation of oxalate from 20 mL solution. mol = 2 (20 mL   In 2nd titration, HCl used in titration of NaOH + 1.1 m CaCl2  ) Na2CO3 = 5 × 0.1 – 9 × 0.2 = 3.2 Hence, total m mol of oxalic acid in 250 mL (2nd ,NaOH + Na CO    23 1.1 solution = × 250 = 13.75 HCl)  20 upto phenolphthalein end point, m mol of HCl  (, 250 mL m xx required = 4 – x + = 4 – = 3.2 22 ) m % of oxalic acid (  %) (, HCl m mol) = 13.75 103  90 100 = 82.5  x = 1.6 1.5 13 In presence of methyl orange, the whole NaOH  Total Na CO formed ( Na CO ) = x × 5 and Na2CO3 are neutralized 23 232 (NaOHNa2CO3  5x   )  = =4 2 m mol of NaOH left unreacted (NaO H m mol) = 20 – 4 × 2 = 12  meq of HCl = 16 × 0.25 = 4 = meq of (NaOH + Na2CO3) = meq. of NaOH original  weight of 1.0 g of exposed sample = 1 – (HCl   = 16 × 0.25 = 4 = (NaOH + 8  40 4  (106  18) + = 1.176 g Na2CO3)  =NaOH  1000 1000 )  (1.0g )   Total meq of NaOH in origi nal 1.0 g sample  weight % of Na2CO3 in exposed sample = = 4 × 5 = 20 4 106 1000 1.176 × 100 = 36.05 % (1.0 g NaOH  )  (  Na2CO3 %)

REDOX REACTION EXERCISE # 5[A] 1 . (i) 7 – 2 = 5 3 . In this oxidation number of N is changing (ii) 7 – 6 = 1 4 . +4 + x – 6 = 0  x = 2 (iii) 7 – 4 = 3 5 . x + 4(0) – 2 = +1 (iv) 7 – 3 = 4 x=3 O–Cl+1 6 . Final product will be Cr2O3 in this oxidation state 2 . Ca of Cr is +3 Cl–1 REDOX REACTION EXERCISE # 5[B] 1 . 2 + 2(2 + x – 4) = 0 [ Ba(H2PO2) is neutral 1 7 . TIPS/Formulae : molecule] or 2x – 2 = 0  x = +1 The highest O.S. of an element is equal to the 4 . TIPS/Formulae : number of its valence electrons (i) Write balance chemical equation for given (i) [Fe(CN)6]3–, O.N. of Fe = +3 change. oxidation state. [Co(CN)6]3–, O.N. Of Co = +3 (ii) CrO2Cl2, O.N. of Cr = +6, (ii) Identify most electronegative element in the (Highest O.S. of Cr) reaction and has the oxidation states of –1 (in H2O2) [MnO4]– O.N. of Mn = +7, and –2(in BaSO4). In H2O2, peroxide ion is present. (Highest O.S. of Mn) 5 . TIPS/formulae : (iii) TiO3, O.N. of Ti = +6, Balanced the reaction by ion electron method MnO2 O.N. of Mn = +4 Oxidat ion react ion : C2O4–2  2CO2 +2e–1] × 5 (iv) [Co(CN)6]3–, O.N. of Co = +3 Reduction reaction : MnO3, O.N. of Mn = +6 M nO – + 8H+ + 5e–  Mn2+ + 4H2O] × 2 2 4 . TIPS/formulae : 4 Use molarity equation to find volume of H2SO4 Net reaction : solutions. 2 Mn O – + 16H+ + 5 C O 2 –  4 4 2 2Mn2+ + 10CO2 + 8H2O 1 1 . TIPS/Formulae : CuCO3  H2SO4  CuSO4 + H2O + CO2 (i) In an ion sum of oxidation states of all atoms 63.5 12  48.98 98 g is equal to charge on ion and in a compound 123.5 g sum of oxidation states of all atoms is always For 123.5 gms of Cu(II) carbonate 98 g of H2SO4 are required. For 0.5 gms of Cu(II) carbonate weight zero. Oxidation state of Min in MnO4– = +7 98  0.5 Oxidation state of Cr in C r( C N ) 3 – = +3 of H SO reqd. = g = 0.39676 g H SO 6 24 123.5 24 Oxidation state of Ni in NiF62– = +4 Oxidation state of Cr in CrO2Cl2 = +6 Weight of required H2SO4 = 0.39676 g Weight of solution in grams 1 4 . TIPS/Formulae : (i) Mass of one electron = 9.108 × 10–31 mol. wt.  Molarity  Volume in mL = (ii) 1 mole of electron = 6.023 × 1023 electrons 1000 weight of 1 mole of electron 98  0.5  V 0.39676 = = Mass of one electron × Avogadra number 1000 = 9.108 × 10–31 × 6.023 × 1023 × 1023 kg or V = 0.39676  1000 ml 90  0.5 No. of moles of electrons in 1 kg Volume of H2SO4 solution = 8.097 ml

THERMO CHEMISTRY EXERCISE # 1 9 . Heat evolve ( )=1939.1 12 = 581.73 3 . Hr = [(Hf)TiO2 + 4(Hf)HCl – (Hf)TiCl4 – 2(Hf)H2O] Hr = – 944.7 – (4 × 92.3) + 763.2 + (2 × 241.8) 40 Hr = – 67.1 kJ/mole PV 1 2 . nC2H4  RT 4 . Hr = [3(Hf)CO2 + 4(Hf)H2O – (HC)C3H8] VC 2 H 4 2 VC H 4 1   3.67   3.67 3 3 –2221.6 = 3 × (–394) – 4(285.8) – (HC)C3H8 ( H ) C3 H8 = – 103.6 kJ/mole n C2H 4  1  2  3.67 3.67 C 0.082  3  298 nCH4  3  0.082  298 5 . Hr = [4(Hf)CO2 + 2(Hf)H2O – 2(HC)C2H2] Heat evolve = 2  3.67  (1400) 3  0.082  298 –2601 = – 4(394) – 2(285.8) – 2(HC)C2H2 (HC)C2H2 = 226.7 Heat evolve = 3.67  900 6 . Hr = [2(Hf)NaOH – 2(Hf)H2O] 3  0.082 298 total heat evolve from mixture()  281.9 = 140 + 45 = 185 kJ 2 = (Hf)NaOH + 285.8 11 (Hf)NaOH = – 426.8 Kg 1 3 . 2 H2  2 Cl2  HCl (Hf)HCl = 52 + 24 – 1039 = – 22 kcal THERMOCHEMISTRY EXERCISE # 2 9 6. Hr =  1 (H f )C2H2  2(H f )CO2  1 ( H f ) H 2 O  1 . C3H6 + 2 O2  3CO2 + 3H2O 2 2  3C + 3H2  C3H6 H = 20.6 kJ/mole 11 C + O2  CO2 H = –394 kJ/mole Hr =  2 (–1300) + 2(–390)  × 572 2 1 7. Hr = 234 H2 + 2 O2  H2O H = –285.8 kJ/mole Hr = 2(H f )CO2  3(H f )H2O  (H f )C2H5OH  (H C )C3H6  [3 H CO2  3 H f(H2O )  H f (C3H6 ) ] Hr = 2(393.5)  3(241.8)  277.7 = [3 × (–394) – 3(285.8) – 20.6] (H C )C3H6  2060 kJ / mole Hr = –1234.7 kJ/mole 8 . Applying Hess's law, 2 . HC = [57(–285.8) – 52(393.5) + 7.870] Hr = [2(–414) + 2(86) + 571.6] HC = 34117.4 kJ/mole Hr = –84.4 kJ Applying Hess's law, energy liberated for 1 gm fat(1 9. ) = 34117.4 = 38.4 kJ/mole Hr = [3(110.5) – 28.9 + 2(–285.8) + 3(–74.8)] 887 = –747.5 3 . Hr = [4(90.2) – 6(241.8) + 4(46.1)] heat realeased for 3 gm = 905.6  3 = 39.9 13 4 17 2 N 2  2 Cl2  NCl3 (3 )  Hr = H1  H 2  3 H 3  2 2  4 . Heat lost copper = heat gain by gold 1 2 . Hr = 30 × 0.385(318 – T) = 15 × 0.129 (T – 298)        2  final temperature T = 315.1 K H f 4 H f 4 H 5  H f P4 O1 0 N2O5 HPO3 HNO3 T = 42.1°C H r  2 43.1  4 948.5   4 174.1  2984.0  5 . Applying Hess's law. = – 199.8

1 4 . Hr° = 2 2 . C2H5OH  C2H4 + H2O ... (i) H = 45.54 8a 8a 4 HCH  4 H ClCl  4 H CCl  4 H CCl  C2H5OH CH3CHO + H2 ...(ii)  H = 68.91 aa = 8a + a = 1 1 4  414  4  243  4  331 – 4  4313 a= 9 Hr° = 420 energy involve in (i) reaction 1 5 . For Hg = 0, H = E ((i)   ) = 45.54 × 8 Hg  0, H  E 9 energy involve in (ii) reaction 1 8 . Heat evolve = mCVt = 100 × 4.2 × 10 = 4.2 kJ ((ii)   ) = 68.91 × 1 for 0.1 mole the enthalpy change = 4.2 kJ 9 for 1 mole the enthalpy change = 42 kJ total involve in (i) + (ii) are  48.137 Kg 2 3 . HAuBr4 + 4HCl  HAuCl4 + 4HBrH = 8.8 1 9 . HCl + NaOH  NaCl + H2O enthalpy change = mC dT = 100 × 4.2 × 3 % conversion ()= 0.44 100  5% V 8.8 = 1.26 kJ EXERCISE # 3 enthalpy change for 5 millimole = 1.26 kJ 1.26 enthalpy change for 1 mole  5 10–3  2.52 × 102 kJ THERMOCHEMISTRY COMPREHENSION # 1 COMPREHENSION # 3 1 . Hr = (H f )C2F4  2(H f )HCl  2(H f )CHClF2  1 . (i) H = (v + w + x + y + z) = [–658.3 + 2(–92.3) + 2(485.2)] = 127.5 kJ/mole w (ii) (Hf)K+ = 2 2 . Add eq. (i), (ii) and (iii) y (iii) (H)EA for H = 2 CX4(g)  C(g) + 4X z H = –H1 + 718 + 2D(X – X) (iv) (H)lattice for KH = 2 X=F 2.(i) electron affinity is exothermic ()  H = +679.6 + 718 + 2 × 154.7 (ii) ionization is endothermic ( )   H = 1707 3 . (H)r=2× 90 + 2 × 418 + 436 – 2 × 78 – 2 × 710 Average bond energy of C – F bond (H)r = – 124 kJ/mole = 1707  426.75 4. ( H) = – 124  –62 kJ/mole 4 f KH 2 (C – F )  X = Cl 6 . Meq. of KH = Meq. of HCl H = 106.6 + 718 + 2(246.7) = 1318 0.1 Average bond energy of C – Cl bond = 329.5 Kg 1000 3 . C – Cl bond energy = 329.5 E KH = 25 × 0.1 C – H bond energy = 416.1 C – F bond energy = 426.75 Valency factor ( )of K is 1 hence Order of reactivity C – Cl > C – H > C – F EK = MK MK = 39 EKH = 40 EKH =EK = EH 40 = E + 1 E  39 K K

THERMOCHEMISTRY EXERCISE # 4[A] 1 . 2 C2H6 + O2  2 CO2 + 3 H2O 19. Cs(s) + 1 Cl2(g) –388.6 CsCl(s) 2 mol 2 (H) / mole = – 01560 kJ 81.2 1 (243) 2 = 2 (–345) + 3(286) – (H°f)C2H6 H°f = – 790 – 858 + 15 – 98 kJ Cs (g) Cl x = –88 kJ/mol 375.7 –348.3 Cs + Cl(g) 5 . CuSO4 + 5H2O H CuSO4 . 5H2O 81.2 + 375.7 + 121.5 – 348.3 + x = –388.5 15.9 kcal 2.8 578.4 – 348.3 + 388.5 = –x – x = 966.9 – 348.3 CuSO4(aq.) x = – 618.6 20. 2C(g) + 6H(g) 2839.2 CH 26 Applying Hess's law H + 2.8 = –15.9 2C(g) + 4H(g) 2275.2 CH 24 H = –15.9 – 2.8 H = 18.7 9 . C H COOH(s) + 15/2 O  7CO + 3H O 6C(g) + 6H(g) 5 50 6  CH 66 65 2 22 H = qp = 7 × (–393) + 3 (–286) + 408 (C – C) + 6(C – H) = –2839.2  C – C = 373.98 = – 2751 – 858 + 408 = –3201 (C = C) + 4(C – H) = –2275.2  C = C = 637.72 H = U +  n R T –6(410.87)+3(373.98) + 3(631.72) + RE = –5506 g U = – 3201 –8.3 × 300 × 0.5 = –3201 + 1.247 –5482.3 + RE = 0.5506 RE = –23.68 kJ/mol = –3199.75 11. 1 H2 1 Cl2 92.3 HCl (g) 21. q = 0 U = w 2 2 x –167.44  nRTf  nRTi  n Cv T =P  Pi  H+ (eq.) Cl–(aq.) avg.   Pf –92.30 + x = –167.44 n5  n R Tf n R Ti  2  Pf Pi  x = –75.14 kJ/mol R T  Pavg    13. C H O (s) + O (g)  6CO (g) + 6H2O () 6 12 6 6 2  Tf 300  1.0 g 5/2 (T – 300) = –  2  5  U = m Cv dt 5/2 T – 750 – Tf  60 2 U = –10 kJ × 1.56 = 15.6 kJ for 1 mole = 15.6 × 180 = –2808 kJ 3T = 810 T = 270 K 13 5 R (300) = – 150R 14. PH  P+ 2 H2 U = w=2 × 2 = –1247.1J 3 22 954 = 3 (P – H) H= –150 R + 2 R(–30)= –210R = – 1745.9J P H  P + 2H 2 2 . 2Al(s) + Fe2O3(s)  Al2O3(s) + 2Fe(s) 24 22 0.2mole 1485 = 4 (P–H) + (P–P) 0.1mole 0.1mole 0.2mole 1485 = 4 × 954 + (P–P) 0.254 kg ice melted 3 254 1.436 H = 18 = 20.26 kcal (P–P) = – 1272 +1485 = 213 kJ/mol Heat liberated for 0.1 mole = 20.26 kcal 16. 2C (g) + 6H (g) –676 C2H6(g) Heat liberated for 1 mole = –202.6 kcal 2× 171.8 3(104.1) Hf 2 4 . CH2=O H (CH2O) 2C (s) + 3H2(g) –134 –122 rH=–676+343.6+312.3 = –676 + 655.9 = 20.1 nCO2+nH2O 4 (C – H) = 396 (C–C) + 6(99) = 676 applying Hess law C – H = 99 K (C–C) = 676 –594 = 84 H – 122 = – 134 H = 12 Kcal

THERMOCHEMISTRY EXERCISE # 4[B] 1 . Given 4. Ca(s)C a(g)C a+1(g)Ca +2  CH3 HC = CH CH3 (g)   CaC2(s) HC = CH CH3 C(s)C(g)C2(g)C2–(g)C –2 2(g) CH3 –60=[179+590+1143+718×2–614–315+410+ L.E.] cal kcal ...(i) L.E. = – 2889 kJ/mole H1 = –950 mole = –0.95 mole 5 . O2 consumed by body in 1 hr. = 20 × 60 × 200 (0.2 – 0.1) = 24000 mL. CH3 HC = CH CH3 CH2 = CH – CH – CH 2 3 H2 = cal kcal ...(ii) so volume of O2 at 273K is let V then +1771 = 1.771 V 24000 mole mole = CH2 = CH – CH2 – CH3 + 6O2  4CO2 + 4H2O 273 310 kcal ...(iii) V = 21135.48 mL moles of O2 = 0.9435 H3 = –649.8 mole (ii) + (iii) – (i) C6H12O6 + 6O2  6CO2 + 6H2O HC = CH CH3+ 6O2  4CO2 + 4H2O() H = –2880 kJ/mol CH3 kcal 0.9435 moles of glucose  H = –647.079 ...(iv) mole 6 H2O()  H2O(g) Heat released = 2880  0.9435 = 452.9 kJ 6 kcal ...(v) Heat used for muscular work H4 = 11 mole = 452.9 × 0.25 = 113.22 kJ (iv) + 4 × (v) HC = CH CH3 so distance = 1.132 km CH3 +6CO2 4CO2 + 4H2O(g) 6 . Given : H = –603.079 31 H = 88 ...(i) 2C(s)+ 2 H2(g)+ 2 N2(g)CH3CN(g) [2 BC – C + BC = C + 8 BC – H] H = –84 ...(ii) + 6BO = O–8 BC = O–8BO – H = –603.079 2C(s) +3H2(g)  C2H6(g) H = 717 BC = C = 192.921 kcal/mole H = 946 2 . Given C(s)  C(g) H = 436 n(CH2=CH2)(–CH2–CH2–)n H = –72 N  2N 2(g) (g) i.e., BC = C – 2 BC – C = –72 ...(i) H2(g)  2H(g) 6C(s) + 3H2(g)  C6H6() H = 49 ...(ii) BC – H = 410 C6H6()  C6H6(g) H = 30 from equation (i) R.E. of C6H6 = – 152 (2×717+1.5×436+0.5× 946)–(3× 410+BC – C+BC  N) = 88 1 H = 218 BC – C + BC  N = 1243 ...(iii) 2 H2  H C(s)  C(g) H = 715 from equation (ii) BC–H = 415 for equation (2) (2 × 717 + 3 × 436) – (BC – C + 6 × 410) = –84 (6×715+6× 218)–(3BC–C+3BC=C+6×415 – RE) = 79 BC – C = 366 kJ/mole from equation (iii) BC–C + BC=C = 959 ...(iii) from equation (i) and (iii) BC  N = 877 kJ/mole BC–C = 343.66 BC = C = 615.33 7 . Given : 3 . Given NaCl(s)+aq  N a + + C l – H = –2 kJ/mole 6C(s) + 3H2(g)  C6H6()H = 49 (aq) (aq) C6H6()  C6H6(g) H = 45 Na+ +Cl–  NaCl(s) H = –772 kJ/moles (g) (g) so 6C(s) + 3H2(g)  C6H6(g)H = 94 .........(i) so N a + + C l –(g)+ a q .  Na + + Cl – q.) H = –774 (g) (g) (a 2C(s) + H2(g)  C2H2(g) H = 75 .........(ii) & Na+(g) + aq.  N a + H = –390 (g) (i) – 3 × (ii) 3C2H2(g)  C6H6(g) = 131 so enthalpy of hydration of Cl– = –384 3[BC  C+2BC–H]–[3BC–C+3BC=C + 6BC – H – RE]=–131 similarly enthalpy of hydration of I– = –307 3[BC  C – BC – C– BC = C] + RE = – 131 RE = – 131 + 99 = –32

UNIT # 08 (PART - I) CHEMICAL KINETICS EXERCISE # 1 2 . r = 1 d[H2 ] 16. d[a  x]  K 1 [a  x]  K 2 [a  x] 3 dt dt d[H2 ] = 3 × 2.5 × 10–4 =7.5× 10–4 mol L–1 S–1 2 0 . r = K[X][C] dt [X] K' = [ A ][B ] 3 . d[O2 ]  1 d[SO3 ] r = K'[A] [B] [C] dt 2 dt 2 1 . r = K[NO] [Cl ] 22 d[SO3 ] = 2 × 2.5 × 10–4 = 5 × 10–4 mol L–1 S–1 K' = [NO ]2 dt [NO ]2 4 . 1 d[H2 ]  1 d[NH3 ] r = KK' [NO]2 [Cl ] 3 dt 2 dt 2 d[NH3 ]  2  0.3 104  35  0  dt 3 2 3 . K × 15 = ln  35  9  = 2 × 10–5 = 0.2 × 10–4 43. K × 32 = ln  100  ... (i) 6 . r = k[A]2 [B]  1  ... (ii) r = k[x]2 [y] 100  r' = k[3x]2 [3y] K × t = ln  0.1  r' = 27 r eq. (ii)/(i) 8 . r = 1 d[C] = 1 × 1 = 0.5 mol L–1 S–1 t 3 ln10 2 dt 2  32 2 ln10 t = 48 min 2.303  a0  44. Given t  1 t  a  10. K = log 1 a 2 2  t1 1 1 2 a n 1 = 2.303 log 10 n–1=2 500 1 K n=3 100 t  1 200  1 n 1 100  0.5 45. 1 a n 1  2  = 2.303 log10 = 4.606 × 10–3 sec–1 2 = 2n–1 n–1=1 n=2 500 5 1 . K  AeEa / RT 1 2 . r = K[A] = 5 × 10–5 [1] = 5 × 10–5 M s–1 Ea ln K = ln A – 1 3 . r = K [N O ] [O ] ....(i) 2 22 2 RT K1  [N2O2 ] or [N O ] = K1 [NO ]2 from eq. (i) 2.303 log K = – Ea + 2.303 log A K 1 [N O ]2 22 K 1 RT r = K K1  [NO]2 [O2 ] log K = – Ea + log A 2 K 1 = K [NO]2 [O 2.303RT compair r ] slope Ea = 5000 2 2.303R K = K K1 Ea = 5000 × 8.31 × 2.303 = 95.7KJ k–1 mol–1 2 K 1 5 2 . – Ea = –40000 15. r = 1 d[N2O5 ] = 1 d[NO2 ] = d[O2 ]  K[N2O5 ] R – 4 dt dt 2 dt Ea = 40000 × 2 = 8 × 104 cal K = 2K 53. K  AeEa / RT  AeEa / RT  1 3 2 1 Ae RT K = 4K = 2K 21 a = E + E – E= 60 + 10 – 30 = 40 kJ 1 3 2 K = K = K /2 31

CHEMICAL KINETICS EXERCISE # 2 1 . A (g)  2B (g) 1 1 1  t(n 1)  10–5 mole 100 mole 10. k   C n 1  n 1   0 Kf = 1.5 × 10–3 s–1 C [1 0 0 ]2  Kf 1  2n1  1 K = 10 –5 10 Kb k(n 1)   t=  C o n 1 C on 1  ... (1) 1/2 Kb = 1.5 103 10 –7 104 1  4 n1  1 k(n 1)   Kb = 1.5 × 10–11 L mol–1 s–1 t=  C o n 1 C on 1  ... (2) 3/4 2 . A + B  C + D equ. (2)  equ. (1) rate = k [A]1/2[B]1/2 t3 / 4  4 n1 1  22 (n1)  1 (2n1  1)(2n1  1) t1 / 2 1  (2n1  1) (2n1  1) dx  k (a  x)(a – x)   2 n 1  = = dt  dx  k(a  x) t = t [2n–1 + 1] dt 3/4 1/2 11. Fraction of reactant consume f=   C  1 C0  2.303 log10  a     t= k    a x for a reaction : df  k(1  f) (Remaining amount). dt 2.303  1  t= 2.31 10–3 log10  0.25  1 2 . 2 A + B K  C + D t = 600 sec. rate = k [A] [B]2 t=0C 2C 00 3 . X k1  A + B and Y k2  C + D t = 30 min, C0 7C0 Rate at that time 24 k1  log(2) k1  log 2 k2 log(100 / 4) 2 k1 k2 2 log10 – 2 log 2 Rate = k  C0   7C0 = 4 9 C 3 k2 = 4.06  2   4  0 5 . P(mmHg) 500  32 250 14. A  B K = 1015 e–2000/T CD A K = 1014 e–1000/T At T (K = K ) C AC t (in min.) 235 950 C105 e–2000/T = 1014 e–1000/T 1/2 1 10 e–2000/T = e–1000/T 10 = e1000/T a n1 t1 / 2   1000   2.303  log 10 = 1000/T T = 1000 T = k e loge 10 ( t1 / 2 )1  a2 n 1 235  250 n 1 (t1/ 2 )2  a1 950  500        log10 k2 = 2.303 Ea  T2  T1  22 = (2)n–1 n–1 = 2  n = 3 15. k1 R  T1 T2    6 . t1k1 = t2k2 20 × k1 = 5 × k2 k1 = 4 Now Arrhenius equation R log 2  280  290 k2 = Ea 2.303 10  k2  Ea  T2 – T1  log k4 = 2 .3 0 3  R log 2  280  290  10   k1  2.30  R  T1 .T2  k3 2.303 10  R  0 300  log   =   2 9 10 Ea  20  log10  k4   280 log 2 k4 =1.91 log10(4) = 8.314  2.3  300  320   k3  300 k3   Ea = 55.332 kJ/mole 1 6 . A + 2 B  products 7 . k = 3.0 × 10–4 s–1 ; Ea = 104.4 kJ mol–1 a–x 2a – 2x A = 6.0 × 1014 s–1 k2= Ae–Ea/RT  d[A]  k [A] [B] But when T   e–  1 dt k = A  k = 6.0 × 1014 s–1 Reactant are in their stoichiometric proportion

  d(a  x)  k (a  x) 2(a  x) dt dt 2 8 . = – N, dx 2 k(a  x) 0.693 dt  t1/2= 2k dt For max. no of nuclei dN  0 dt d[A ] = k[A]  = N  N = / 1 7 . 2A +B  products dt 2 9 . Let n is the moles of reagent 'R' when R is reacted when t l/k A = A e–kt with A at time t = 0 0 2 A  B + C A = A e–1 A = A /e 0 0 2 4 . Rate = K [A] [B ] t=0 n 0 0 2 [A ]2 at t n–x 2x 3x K = [A 2 ]      [A] = {K[A2]}1/2 at t  0 2n 3n order = 1 1  5n = n   n = n2 2 2 5 r = K' [A2]1/2 [B2] n + 4x = n   x = n1  n 1 4 100 2 6 . For A K × t = ln ...(1) 2.303  n  k = t log  n – x  1 100 – 94 For B 100 ...(1) K × t = ln  4n2  2 100 – 50  5(n2 –    Eq. (1) and (2) so k = 1 ln t n1 ) 100 3 0 . Overall rate constant =k=k +k +k = 6.93 × 10–3 ln 123 K1 6 K2   K1  4.058 t= 0.693 ln 100 K2 1/2 6.93 10–3 =100 sec; 50 After half-life, P + P + P = 4 atm BCD 2 7 . % B = K1 100 = 76.83% PB = k1  200 K1  K2 PB  PC  PD k1  k 2  k 3 693 %C = 100 – B% = 23.17% P =4× 200 = 1.154 atm B 693 CHEMICAL KINETICS EXERCISE # 4[A] 1.(a) N + 3H  2NH3 3 . 2A + B + C  D + E 2 2  dx  Rate = [NH 3 ] = 2 × 10–4 mol L–1 s–1  dt 1 = K [A] [B]2 [C]0 t 1  dH 2 = 1 dH2 = 1 dNH 3 if increases conc. 2 time dt 3 dt 2 dt  dx  K [2A] [2B]2 [2C]0 =  dx   dt 2 = 1 8  dt 1 dN2 = 1 × 10–4 rate increase by 8 time dt 4. 2 A + B  2 AB  dx   K[A ]2 [B 2 ] (b)  dH2 = – 3  2 10–1 2  dt 1 dt 2 = –3 × 10–4 V if V is decrease to 1 3 2. NO (g)  2 NO (g) + O (g)   dx   [3A ]3 [3B]  dx   27  dx  25 2 22  dt 2  dt 2  dt 1 –d [N O ] / dt = k [N O ] reaction increase by 27 times 25 1 25 d [NO2] / dt = k2[N2O5] 5 . 2H O  2H O + O dO2 = 3.6 M min–1 d [O ] / dt = k [N O ] 22 22 dt 2 3 25  d N2O5  1 d NO2  2 d O2 (a) Rate of formation of H O 2 dt 2 dt dt dH2O = 2 × 3.6 = 7.2 M min–1 dt K = K 2 = 2K 12 3 dH2O2 = 7.2 M min–1 2K = K = 4K dt 12 3

6 . A  B K = 1.2 × 10–2 M s–1 46. (i) A  product (ii) B  product initially = 10 M from the unit of rate constant we find that reaction is of zero order X = kt log  k2  = Ea  10  1  k1  2.303R 310  300    remainng = 10 – 1.2 × 10–2 × 10 × 60 = 10 – 7.2 [Remaining mole = 2.8 M] log 2 × 2.303 × 8.314 × 31 × 300 = Ea Ea = 53.6 kJ/mole If 10 = 1.2 × 10–2 t 1000 =t  0.693  1.2 t = 1000 ddd t = 13.88 min k=  30  60 1.2 A2 28. HO + O Ea 2 HO° H = 72kJ  0.693 min –1 2 Eb k=  B2   T = 500 K 15 Ea = 77 kJ/mole  H = Ea – Eb Ea = 26.8 kJ/mole 72 = 77 – Eb Eb = 77 – 72 log k2 Ea  –10  kB 2.303 310  300  Eb = 5 kJ  33. 2 NO + Br  2NOBr 2 (i) NO + Br  NOBr (East) 22  k2  –26.8 1000  kB  2.303  8.314  30  (ii) NOBr + NO Slow 2NOBr log    3.10 2 Rate is determined by the solwest step  Rate = k [NOBr ] [NO]   2  =  k  But NOBr2 is a visible of Intermediate log k 2 –0.151 B [NOBr2 ] 10 [NO][Br2 ] k=  Rate = k1 [NO]2 [Br ] –1 2 2 sec  3rd order RxN k =  0 .7 06  .69 3  2   4 5 . A  B Ea = 70 kJ/mole t = 40 min 15  60 1/2 k =0.0327 min–1  0.693  k=2.8875× 10–4 2 k =  40 min  47. 2H O    2 H O + O 22 2 2 log k2 Ea  T2 – T1  ml. eq of H O in 10 ml diluted 10 k1 =  T1 T2  22   2.303R = M eq. of KMnO titrated = 25× .025 × 5 = 3.125 14 ml. eq in 100 ml solu. = 31.25 = 70 103  15  but 22400 ml O = 68 gm H O 2.303  8.314 298  313  2 22 k2 = 5.88 × 10–4 × 103 = 100.588 68 k1 1 ml = 22400 k2 = 3.87 20 ml = 68 20 ln 1 ml H O k1 22400 22 k2 log a / a – x  68 20 10 = in 10 ml H O k1 = log10 (a / 3ka) 22400 22 log a / a – x  No. of Meq. in 10 ml of 20%kg ml H O 22 3.87= log(4 / 3) 68 20 10 1000 Initialy = 17 22400 = 35.71 a 10.4835=  a – x  2.30 3 log  35.71  6  31.25   a  =10–.4835 a  k=  –   a – x  ×100=32.84% a x  % decomposition = 100 – 82.84 = 67.16% k = 0.022 hr–1

CHEMICAL KINETICS EXERCISE # 4[B] 1. Initial m mol of cyclobutene (k31 40  k3 = 8.317 .....(iii) = ln m )= 10 6 10 Let after 20 min, x m mol cyclobutene isomerized. k1 = Ae–80,000/RT (20  ,x m mol  k = A eEa1 / RT E a1  Act ivat ion energy in 2  )  presence of 1st catalyst m mol of cyclobutene left = 10 – x and m mol of (1st )  diene formed = x ln k2 = 80,000  E a1 1.3862 k1 RT = ln (mmol=10– x  0.2231 m mol = x) E a1 = 75443.8 J = 75.44 kJ m mol of Br2 required after 20 min = 10 – x + 2x = 10 + x = 16 k 3  Ae Ea2 / RT E a2  Activation energy in (20   Br2 m mol = 10 – x presence of 2nd catalyst + 2x = 10 + x = 16) (2nd  )  10 .........(i) ln k3 = 80,000  E a2 8.317  x = 6  20K = ln = ln k1 RT 0.2231 4 If y m mol of cyclobutene isomerized after 30 min. Ea2 = 70974.9 J = 70.975 J ( 30   ymmol  ka  ka )  4. NO NO + NO 25 23 10 ........(ii) NO2 + NO3 kb  NO2 + O2 + NO 30 K = ln NO + NO kc  2NO 10  y 32 From Eqs. (i) and (ii) ((i) (ii) ) y = 7.47 d[N2O5 ] dt  m mol of Br2 required (Br2   m mol) = ka [N2O5] – k–a [NO2] [NO3] ........(i) = 10 + y = 17.47 d[NO3 ] dt  Vol. of bromine solution required (   = ka [N2O5] – k–a [NO2] [NO3] – kb [NO2]  ) = 17.47 mL [NO3] – kc [NO] [NO3] = 0 ..........(ii) 14000 20000 d[NO] =k [NO ][NO ]–k [NO] [NO ] = 0 .......(iii) RT 2. 102 e RT = 103 e Solving (  ), dt b2 3c 3 T = 313.42 K From equation (iii) ((iii) )  Rate constant ( )K1 = K2 = 0.464 hr–1 [NO] = kb [NO2] kc n0 (A2 ) putting this in equation(ii)((ii))   K1t = ln n0 (A2 )  n1  n1 = 0.37 k [N O ] = [NO ] {k [NO ] + 2k [NO ] a 25 3a 2 a2 K2t = ln n0 (B3 )  n2 = 0.37 [NO3] = ka [N2O5 ] n0 (B3 )  n2 ka [NO2 ]  2kb[NO2 ]  Total moles of gases after 1.0 hr. (1.0   putting this in equation (i)((i) )    ) =1.37 + 1.74 = 2.11 d[N2O5 ] = ka [N2O5] – kak a [N2O5 ] dt ka  2kb P = 5.42 atm 3 . Let rate constant in absence of catalyst is k1 d[N2O5 ] = 2kakb [N2O5 ] dt ka  2kb ( k1)  Let rate constant in presence of first catalyst is k 2 ( k2)  r = 1 d[N2O5 ] = kak b [N2O5 ] Let rate constant in presence of second catalyst is k3 2 dt ka  2kb ( k3)5 . 2 t+  k1 × 1 = ln  k1 = 0.2231 hr–1 .......(i) 2 d       d   =  ()1 ......(ii) dt dt 2 2 k2 × 0.5 = ln 80  k2 = 1.3862 40 order () = –1

6 . Given ( ) : r = k' [complex]a [[]a ] 10. A  B + C t=0 & since ( )  t = 2t t = 20 a 3/4 1/2 so () a=1 min. a–x xx ln 2 ln 2 t= – a a t1/2 = k ' = k[H  ]b On doubling concentration of [H+] ion t1/2 gets half 60(a – x) + 40x – 80x = 5 so b = 1 40a – 80a = – 20 ([H+]  t  on solving (  ) a = 0.5, x = 0.25 1/2  b = 1) so t1/2 = 20 min. 7 . Let at 20°C rate constant(20°C ) =k Average life = 1/k = 1.443 t1/2 = 28.86 min. then at 3°C rate constant (3°C )=k/3 1 1 . At t =  when equilibrium is established ln k = Ea 1  1  Ea = 43.45 kJ (t =   ) k/3 R 276 293  Let at 40°C rate constant is k1 then k = [P]  7  2.33 [A] 3 ln k1 43.45 103 1 1  k = 293  313  k1 k 1 8.314 & = 2.33  k1 = 2.33 k–1 (40°C  k1 ) k1 = 3.125 k [A] = k 2 [A ]0 [1  e (k1 k2 )t ] k1  k2 so time required for juice to get spoil at 40°C = 64 0.725 = 1 [1  2.33 e 3.33 k13600 ] = 20.47 hr. 3.33 3.125 (4 0°C  )  8 . Given : k–1 = 4.16 × 10–5 sec–1 k1 1 , k1 = 1.3 × 10–5 k1 = 2.33 k–1 = 9.7 × 10–5 sec–1 k2  k2 = 9k1 1 3 . CH3OCH3    CH4 + CO + H2 9 [A] = [A]0 e (k1 k2 )t t=0 0.4 k 2 [A ]0 [1  e (k1 k2 )t ] t = 4.5hr. 0.11 0.29 0.29 0.29 k1  k2 [C] = P0 0.4 ln kt = ln   4.78 × 10–3 × 4.5 × 60 = PP [C ] k2 9k1 [A]  k1  k2 [e (k1 k2 )t  1] = 10k1 [e10 k1t  1] = 0.537 P = 0.11 atm 9 . k = 0.16, k = 3.3 × 10–4 at t = 0 P0 = 0.4 1 M0 = 46 so k2 = k1/k = 2.0625 × 10–3 k1 + k2 = 0.0023925 at t = 4.5 hr. P = 0.11 + 0.29 × 3 = 0.98 atm [B]eq. = k1 [A]eq. = k1[A]0 0.11  46  0.29(16  28  2) k2 k1  k2 M = 0.98 = 18.77 given [B] = [B]eq.  k1 [A ]0 r0  P0 M  0.4 18.77 2 2(k1  k2 ) = 0.26 r P M0 0.98 46  & [B] = k1 [A ]0 1  e (k1 k2 )t 1 4 . k1 B k = k1 + k2 k1  k2 so 1  1 – e(k1k2 )t A 0.693 0.693 + 0.693 2  90 (k1 + k2) t = ln2 k2 C t1/ 2 60 t = 289.71 sec. = 4.82 min. t1/2 = 36 min.

1 6 . A + B k1 C, C k3  D 2 0 . For reaction 1 (1  ) k2 r = k1[A] [B] – k2[C] k e =E a / R  1  1  10–3 e 6000  1  1  d[C]  T1 T   509 T  dt = k1 [A] [B] – k2 [C] – k3 [C] = 0 k2 =   2.79 × 1 [C] = k1[A ][B] For reaction 2 2   k2  k3 d[D] k1[A ][B] k2 = 1.52 × 10–4 e1 2250  1  1  = r = k [A] [B] – k  510 T  × dt 1 2 k2  k3 r = k1k3[A ][B] For given condition (   )  k2  k3  1 1  50 1 1  since k2 >> k3  509 T  510 T  e =600 0  122  10–3 10 e–4 k = k1 k3 2.79 × 1.52 × net k2 18.355 = e1 2250 1  1 6000 1  1  510 T 509 T  A1 A3 so Anet = A2 (E ) = E + E – E ln 18.355 = 12.33 – 6250  2.9 a net a1 a3 a2 T 18. 2P(g)    4Q(g) + R(g) + S() T = 670.6 K = 397.6 °C t=0 P P/2 0 P /2 t = 30 min. P0–P 2P 0 2B k1 t= – 2P 21. k2 C [A] = [A] e (k1 k2 )t 0 0 A so P0 – P + 2P + P/2 = 317 – 32.5 i.e. P + 1.5 P = 284.5 .........(i) 2k1 [A ]0 [1  e (k1 k2 )t ] 0 k1  k2 & 2.5 P0 = 617 – 32.5 = 584.5 [B] = so P = 233.8 0 P = 33.8 k 2 [A ]0 [1  e (k1 k2 )t ] k1  k2 k × 30 = ln 233.8  k = 0.0052 [C] = 200 since V & T are constant (V T  )P At t = 75 min 233.8  moles 0.0052 × 75 = ln at t = 0, P = 1 atm so P  P P° – P = 158.23  P = 75.57 PT = 32.5 + P0 + 1.5 P = 347.155 + 32.5 [A]0 = 1 PT = 379.65 mm Hg t = 10 sec. P = 1.4 atm so (ii) 0.0052 × t = ln 8 t = 399.89 min. at t = 10, [A] + [B] + [C] = 1.4 ....(2) t =  P = 1.5 atm...(3) 19. Bn+  B(n+4)+ t=0 a t=10 min. a – x at t = , [B] + [C] = 1.5 v.f. of Bn+ = 2 from equation ((3) ) (3) v.f. of B(n+4)+ = 5 Let normality of reducing agent    =N 2k1  k2  1.5 k1  k2 k1  k2 so 2a = 25 N 2(a –x) + 5x = 32 N 7 2k1  k2  1.5 2a + 3x = 32 N x= N k1  k2 3 k× 10 = ln a a x  ln 12.5N 1 + k1  1.5  .5N – 7 k1  k2 12 N 3 k1 = k2 ... (4) k = 0.02 min–1.

from equation   (2) & (4) 25. Po218 1  Pb214 2  Bi214 84 82 83 2k1 k 2 (1  e (k1 k2 ) t ) Number of nuclei of Pb214 at time t are N k1  k2 k1  k2 2 +   + = 1.4e (k1 k2 )t 1  e (k1 k2 )t  = 1N 0 t = 10 sec & k1 = k2 so e  e1t 2 t (2  1 ) e 20k1  1 – e 20k1  0.5 – 0.5e 20k1  1.4 ( t  Pb214 N2) 0.1 = 0.5 e 20k1 For max. value of N2 (N2     )  0.2 = e 20k1 dN2  0 20k1 = 1.6094 dt 23. k1 = 0.0804 = k2 82Pb208 + 6 2He4 + 4 –10 so t = 1 ln 2 90Th232  (2  1 ) 1 t=0 a 6x when 2 = 0.693 , 1 = 0.693 time t a – x 2.68 , 3.05 given : on putting these values (  ) a – x 5  10 7 = 2.155 × 10–9 mole = 232 3.05 t = 31.87 ln = 4.12 min 2.68 8  10 5 6x =  x = 5.9523 × 10–10 mole 22400 2 6 . Let the mass of sample in a g & initial mass of U238 so a = 2.75 × 10–9 is w g the 0.693 (gU238 k = 1.39 1010 w g ) k × t = ln a U238  Pb206 ax 0.693 2.75  10 9 t=0 w 1.39 1010 = ln 2.155  10 9 = 0.2438 t = 4.89 × 109 year t w-x 206x given w – x = 0.5 a 2.68 0.693 24. k = 8 kt = ln A 0  0.693  4 = ln A 0 206x = 2.425 a  0.93 = 0.0225525 a A8 A 238 100 x = 0.026 a A A0 = 0.707 Total activity is 70.7% of the original activity but so w = 0.526 a only 67.7% found in the thysoid so mass of stable iodide ion had migrated to the thyroid gland is w t = ln w  x (70.7%  67.7%  0.526 a 0.5 a ) 1.52 × 10–10 × t= ln = 67 .7  0 .1 = 0.09575 mg t = 3.33 × 108 year. 70 .7

CHEMICAL KINETICS EXERCISE # 5[A] 1 d[HI] 12. N = N  1 n where n is number of half life periods. 1 . rate of appearance of HI = t 0  2  2 dt  rate of formation of H = d[H2 ] n  total time  24  6 2 dt half life 4 = d[I2 ]  N = 2 0 0  1 6  3.125g dt t  2 rate of formation of I   2 1 3 . Enthalpy of reaction (H) = Ea(f) – Ea(b) hence d[H2 ]   d[I2 ]  1 d[HI] for an endothermic reaction H = +ve hence for dt dt 2 dt H to be negative. or  2d[H2 ]   2d[I2 ]  d[HI] Ea(b) < Ea(f) dt dt dt 1 4 . The molecularity of a reaction is the number of reactant molecules taking part in a single step of the reaction. 2 . Order is the sum of the power of the Note : The reaction involving two different concentrations terms in rate law expression. reactant can never be unimolecular. hence the order of reaction is = 1 + 2 = 3 2.303 1 2.303 4 log  log 15. t1/4 = K 3/4 K 3 3 . For a zero order reaction. rate = k [A]° i.e., rate = k = 2.303 (log 4  log 3)  2.303 (2 log 2  log 3) hence unit of k = M.sec–1 KK For a first order reaction = 2.303 (2  0.301  0.4771)  0.29 KK rate = k [A] k = M.sec–1/ M = sec–1 1 6 . Since the reaction is 2nd order w.r.t. CO. Thus, rate law is given as 4 . Rt = log C – log C 0t r = k [CO]2 It is clear from the equation that if we plot a graph Let initial concentration of CO is a i.e. [CO] = a between log C and time, a straight line with a  r1 = k(a)2 = ka2 when concentration becomes doubled,i.e., t [CO] = 2a slope equal to  k and intercept equal to 2.303  r2 = k(2a)2 = 4ka2  r2 = 4r1 log[A ] will be obtained. So, the rate of reaction becomes 4 times. 0 1 7 . In Arrhenius equation K = Ae–E/RT, E is the energy 6 . It is zero order reaction of activation , which is required by the colliding Note : Adsorption of gas on metal surface is of molecules to react resulting in the formation of zero order. products. 7 . In equation K = Ae–Ea/Rt ; A = Frequency factor 1 8 . (i) NO(g) + Br (g)  NOBr (g) K = velocity constant, 2 2 R = gas constant and Ea = energy of activation (ii) NOBr (g) + NO(g)  2NOBr(g) 8 . r = k[O ][NO]2. When the volume is reduced to 2 2 Rate law equation = K[NOBr2] [NO] 1/2, the conc. will double. But NOBr is intermediate and must not appear  New rate = k[2O2][2NO]2 = 8 k [O2] [NO]2 2 The new rate increases to eight times of its initial. in the rate law equation 9.  Rate2  k[2A ]n [12 B]m  [2]n [1 2]m  2n.2 m  2nm from Ist step KC = [NOBr2 ] Rate1 k[A ]n [B]m [NO][Br2 ] 1 0 . As the concentration of reactant decreases from  [NOBr2] = KC [NO] [Br2] 0.8 to 0.4 in minutes hence the t is 15 inutes.  Rate law equation = k.KC [NO]2 [Br2] 1/2 To fall the concentration from 0.1 to 0.025 we need two half lives i.e., 30 minutes. hence order of reaction is 2 w.r.t. NO. 1 1 . The velocity constant depends on temperature only. It is independent of concentration of reactants.

19. HR = E – E = 180 – 200 = – 20 kJ/mol k[H2S]  k ' [Cl2 ][H2 S] f b H [H ] Rate = k[Cl2] K The nearest correct answer given in choices may be obtained by neglecting sign. hence only, mechanism (i) is consistent with the given rate equation. 0.693 2 4 . For the reaction 2 0 . For a first order reaction t1/2 = K i.e., for a A  Product first order reaction t does not depend up on the given t1/2 = 1 hour 1/2 for a zero order reaction concentration. From the given data, we can say tcompletion = [A0]  initial conc. k rate constant that order of reaction with respect to B = 1 because change in concentration of B does not change half life. Order of reaction with respect to A = 1 because  t1/2 = [A0] rate of reaction doubles when concentration of B 2k is double keeping concentration of A constant.  Order of reaction = 1 + 1 = 2 and units of o r k = [A 0 ]  2  1 mol lit1 hr 1 second order reaction are L mol–1 sec–1. 2t1/2 2 1 2 1 . The rates of reactions for the reaction Further for a zero order reaction dx change in concentration 1 A 2B 2 k=  can be written either as dt time –2 d [A ] with respect to 'A' 0.50  0.25 dt 1= 1d tim e or [B] with respect to 'B'  time = 0.25 hr. 2 dt 2 5 . Since for every 10°C rise in temperature rate from the above, we have doubles for 50°C rise in temp increase in reaction rate = 25 = 32 times. –2 d [A]  1 d [B] dt 2 dt 26. k1  A eEa1 / RT .......(i) 1 dA 1 dB or – = A eEa2 / RT dt 4 dt k2  2 .......(ii) 2 2 . For first order reaction On dividing eq. (i) from eq. (ii) k = 2.303 log 100 k1  A1 (Ea2  Ea1 ) / RT ....(iii) t 100  99 k2 A2 0.693  2.303 log 100 693 t 1 Given Ea2 = 2Ea, 0.693  2.303  2 693 t On substituting this value in eqn. (iii) t = 46.06 minutes k1 = k2A × eEa /RT 1 2 3 . Since the slow step is the rate determining step 2 7 . For a first order reaction hence if we consider option (1) we find 2.303 a 2.303 0.1 k  log  log Rate = k[Cl2] [H2S] Now if we consider option (2) we find t a  x 40 0.025 Rate = k[Cl2] [HS–] ..........(i) = 2.303 log 4  2.303  0.6020  3.47 107 From equation (i) 40 40 [H  ][HS ] or [HS]  k[H2S] R = K(A)1 = 3.47 × 10–2 × 0.01 = 3.47× 10–4 = H2S H k Substituting this value in eqution (i) we find

UNIT # 02 PERIODIC TABLE EXERCISE # 1 7 . 4d35s2 1 5 . The difference of IP4 and IP5 is maximum Block – d Valency = 4 Period – 5 Group – number electrons + (n–1) d electrons = Group = IVA or 14th 2 + 3 = 5 or (VB) Family = Carbon family 8. Z N–3 O–2 F– es– 7 8 9 2 4 . EN of P and H is almost same = 2.1 z/e 0.7 0.8 0.9 3 3 . For A particular atom successive I. P is are always 1 increase. So E1 < E2 < E3 Radius  (z / e) 4 0 . Vanderwall's Radii > Covalent radil Order of radius = N–3 > O–2 > F– PERIODIC TABLE EXERCISE # 2 2 . P–3 S–2 Cl– 1 51. M(g)  M+ M+(g2) Size  negative  (g) 100 ev IP2 Zeff IP1 Order of size = P–3 > S–2 > Cl– H = IP1 + IP2 9 . Al – 1s2, 2s22p6, 3s2 3p1 Al+ – 1s2, 2s22p6, 3s2 IP2 = 250 – 100 = 150 ev Al2+ – 1s2, 2s22p6, 3s1 Al+3 – 1s2, 2s22p6 6 8 . IP1 Be B 1s2, 2s2 1s2, 2s2,2p1 Stability = Al+3 > Al+ > Al+2 Be > B 7 1 . IP1 (more stable) 28. 5B – 1s2, 2s22p1 N Two electrons of 1s and 2s subshells IP1 3 2 . Size  number of shell 83. 1. B C O 3 5 . 2nd pd. NO 2. [He]2s2, 2p1 [He]2s2, 2p2 [He]2s2, 2p3 [He]2s2, 2p4 3rd pd. S Cl (More stable) (Half filled) B<C<O<N EA3rdpd > EA2ndpd M(s)  M(g) (More repulsion of electrons) Cl > S > O > N 1 M(s)  M 2 + 2e– 4 0 . Size  negative charge  positive charge (g) IP  EA 3. M(g)  M  ) 41. X = (g 2 4. M   M 2 + 2e– 2 X – IP – EA = 0 (g (g) He Istpd ) 48. Be N Ne IIndpd 5. M(g)  M 2 + 2e– (g) 1 M(g)  M+ (4)=IP2 M+(g2) Group – Ionization energy  size (g) (3)=IP1 Period – Ionization energy  zeff I.P. – He > Ne > N > Be (5) = IP2 I.P. = IP1 + IP2 (5) = (3) + (4) (5) = I.P – IP1 or (5) – (3)

CHEMICAL BONDING EXERCISE # 1 1 . (i) Vapour pressure of (B) is higher than (A) due to  intra molecular H-bonding present in (B). Cl Cl–  (B) (A) I H-    Cl Cl 5 . Tetracyanoethylene   1 6 .  NC CN (B.P. – L.P.) repulsion at 90° = 8 C=C 2 8 . A+3 B–2 NC CN A2B3 compound 9, 9 29. Valency of x = 2  x+2 Valency of y = 1  dy+2 8. Bond strength  Direction of orbital    x+2 y–1   xy2 P–P > P–S > S–S observer dipole moment 3 0 . Polarisability  Size of anions 1 1 . % Ionic character = Theortical dipole moment 1.03 10–18 esu  cm O = O H/O —O / H +1 = 1.275 10–8 cm – 4.8 10–10 esu ×100=16.83.% O O–1/2 1/2O– 1 3 . Ionic character  EN 3 1 . Bond order 2 1 1.5 1 4 . Due to presence of vacant d-orbital in P atom but 1 not in N atom.  P d-Bond order  Bond length N  Order of Bond length = H2O2 > O3 > O2 CHEMICAL BONDING EXERCISE # 2 4 . According to M.O.T. 38. N   –+ – CO2 3 N NN   1s2 *1s1 H 2 16 outer electrons Total electron – 22 total electron – 22 Bond order = 2 –1 1 = 22 44. MgO BaO L.P. MgO > BaO 7 . Cl– M.P. MgO > BaO I Cl–Be–Cl Ionic character  EN sp, linear, lone pair = 0  4 9 . BF3 has number lone pair and planar Cl dipole moment µ = 0 sp3d, lone pair = 3 NF3 has polar bond and pyramidal µ  O Linear 1 5 0 . Bond length  size of central atom 1 2 . sp3 – 109° 28' 5 2 . Pyrophosphoric acid H4P2O7 Px and Py – 90° H–O–H – 104'.5° OO sp – 180° H—O— P— P—O—H 1 9 . Hydrolysis  Covalent character OO NCl3 > PCl3 > AsCl3 > SbCl3 > BiCl3 HH 21. Ca+2 C 2  2 Tetra basic acid Ionic bond

CHEMICAL BONDING EXERCISE # 4[A] 1. (a) K+ HF2 Cl—Be—Cl F——H---------F– 5. 180° H-bonding sp, hybridisation Cl BeCl2(s), hybridisation is sp3 H–bonding present in KHF2 so it exist but there is no H–bonding in KHCl2 due to less polarity in Cl Cl between HCl and Cl– Be Be Be Be  KHF2  Cl Cl Cl KHCl2 HCl  Cl–   H-  BeCl2  BeCl2 (s) sp hybridisation sp3 hybrisiation  (b) In (CH3)3 N, hybridisation is sp3 and pyramidal 6 . Due to the properly oriented tetrahedral structure  of ice, H+ ions are free and hence move more shape but in (SiH3)3 N, due to present of vacant rapidly in ice than in water where molecular asso- d-orbital in Si atom, lone pair of N shifted in vacant orbital so its hybridisation is sp2 with tirgonal pla- ciations are not so well organized. nar geometry.        (CH3)3N(SiH,3s)3p3NSid-H+H+ d- 7.BCl3 exist in monomer form sp2  (i) Cl— B —Cl (c) Due to presence of vanderwaal's force in between CO2 molecule it is gas but SiO2 is a 3-d network Cl structure so it is solid in nature. sp2, trigonal planar  CO2   exist in dimer form – Al2Cl6 SiO23-d  AlCl3 Cl    Cl Dimer form  Cl  Al Al 2.(a) N2O3ONsp3, tetrahedral NO  Cl C l Cl O P2O3 exist in P4O6 (Ring structure) (ii) BaSO4 (Barium sulphate) BeSO4 H.E > L.E. H.E < L.E.  (iii) In O2 (O=O) due to small size of oxygen strong p- P OO P OP p bonding is possible but in S2 due to larger size OOO of sulphur there is not strong p-p bonding. So bond is weak and break down. P  O2 (O=O)  p-p S2   p-p (b) H–Cl, H–Br, H–I Due to increase in size of I, the difference between electronegativity of H and I is less, so bond length          is more and but bond strength is weak. 8 . (i) 1s (ii) *2py (iii) 2pz (v) 2px HI  (iv) 2s  9 . According to M.O.T. O2 is paramagnetic in nature.      M.O.T. O2 

10.(a) In graphite sp2 hybridisation and due to presence (ii) HH of free electron they are good conductor of elec- N=N tricity in a layer but not so good in between two HH layers. (iii) P4O10 sp2   O (b) In solid states position of ions are fixed.  P O        OO P (iv) POCl3 O P OP O Cl 1 2 . Due to H-bonding in NH3 molecule. It is liquid but OOO Cl Cl there is presence of vanderwaal force in between P HCl molecules instead of H-bonding so it is gas. NH3 H- HCl(v)XeOF4 O  O FF 1 4 . 1.4 Å (vi) C3O2 O=C=C=C=O Xe 1 5 . 84.35 % F  F 16. 21 , 3 F 2 FF (vii) Br F  F 17.(i) S8, Crown structure CHEMICAL BONDING EXERCISE # 4[B] 1.(i) Na[B O (OH) ] O R [B O (OH) ]– OO 33 4 33 4 S HO – O OOO B—O OH SS OB (v) S O O OO 39 HO B—O OH Trimeric metaborate ion cyclic timer of SO3 (vi) (CN)  NC–CN (ii) Na [B O (OH) ] · 8H O (Borax) 2 45 4 2 2 3. H S O 2n 6 OH OO Na2 HO—B O—B–—O HO— S —S- - -S —–—S—OH O B—OH · 8H2O (n–2) O—B–—O OO OH

UNIT # 04 S-BLOCK EXERCISE # 1 1 2 2 . Hydration energy  1 7 . Solubility  size of ions L.E 15. KO2  O  2 2 5 . Reducing agent  negative S & P value. 1s2, *1s2, 2s2, *2p2, 2 p 2x , 2 p 2 = 3 3 . Al4C3 + 12H2O  4Al(OH)3 + 3CH4 y 2 p 2 ,  *2 p 2 ,  * 2 p 1z z y n = 1, Paramagnetic S-BLOCK 2HNO3  Mg(NO3)2 + H2 EXERCISE # 2 5. Mg + 2 9 . CaC2, Al4Cl3 and Be2C are ionic carbides but SiC are covalent. very dilute MNH2 1 — Si—C—Si— 12. M + NH3  + H2 —C—Si—C— 2 —Si—C—Si— Alkali metals Liquid Metal amide 17 . Li + Dry Air  Li2O + Li3N (O2 + N2) Na + Dry Air  Na2O 3 4 . 2BeCl2 + LiAlH4  2BeH2 + LiCl + AlCl3 (X) (O2) 1 8 . NaH + H2O  NaOH + H2 3 8 . 3Mg N2 ,  Mg3N2 H2O Mg(OH)2 + NH3 (Colourless glass) CuSO4 CuSO4·4NH3(Blue (Aq) colour) OR (X) (Y) (Z) (T) H– + H+OH–  OH– + H2 44. 2 Na + Al2O3 High temperaturte 2NaAlO2 CwOa2terin Na2CO3 + Al(OH)3 4 7 . CsBr3 is an ionic compound so exist as Cs Br3– P-BLOCK EXERCISE # 1 1 . H3BO3 SHtreoantegdly B2O3 7 . BCl3 + 3H2O  H3BO3 + 3HCl  1 2 . (SiH3)3 N (trisilyl amine) H3BO3 100°C BO2 160°C H2B4O7 Meta boric acid tetra boric acid pd  4. Al2O3  Al3+ + A lO 3  SiH3  SiH3 SiH3 N SiH3 3 SiH3 SiH3 N Al3+ + 3e–  AlO 3  Al (at cathode) 3 p-d bonding A lO 3   2Al2O3 + 3O2 + 12e– (at anode) 3 1 7 . CO2 + H2O  H2CO3 The overall chemical reaction taking place during Acidic oxide electrolysis H2CO3  H+ + H C O  2Al2O3  4Al + 3O2 3 6 . B2H6 + 3O2  B2O3 + 3H2O + Heat weak acid B2H6 + 6H2O  H3BO3 + 6H2 H C O   H+ + C O 2  2NaH + B2H6 ether  NaBH4 3 3

1 9 . P2O5 + 3H2O  2H3PO4 HNO2 oxidation number = +3 ortho phosphoric It reduces as well as oxidise, act both oxidising and reducing agent. P2O5 + H2O2HPO3 H2O H4P2O7 H2O 2H3PO3 H2SO4 oxidation number = +6 Meta phosphoric Pyrophosphoric Highest O.N., only reduces, act only oxidising agent. acid acid 2 4 . 2Pb(NO3)2  2PbO + 4NO2 + O2 2 5 . 2NO + O2  2NO2 2 2 . 2HNO2  N2O3 + H2O Anhydride Brown fumes 3 1 . HCOOH H2SO4 CO + H2O Removal of H2O from HNO2 is called anhydride. 3 4 . Higher the I.E, higher the acidic strength of 2 3 . HNO3 oxidation number of N is = +5 hypohalus acid (hydroxides) Highest O.N., only reduces, acid only oxidising 3 9 . I2 + 2Na2S2O3  2I2 + Na2S4O6 + 2NaI agent. EXERCISE # 2 P-BLOCK 1 8 . 2NaNO3    2NaNO2 + O2 1 . H3BO3 + 3C2H5OH  B(OC2H5)3 + 3H2O 2Pb(NO3)2  2PbO + 4NO2 + O2 2 . AlCl3 + 3H2O  Al(OH)3 + 3HCl 2Cu(NO3)2  2CuO + 4NO2 + O2 5 . 4B + 3O2  B2O3 2B+N2  2BN NH4NO3  N2O + 2H2O Mixture of oxide and nitride 2 0 . PbS + 4O3  PbSO4 + 4O2 (Black) 6 . Due to higher EN of B it attract lone pair of elec- tron with faster rate. 2 1 . AgCl + 2NH4OH  [Ag(NH3)2]Cl + 2H2O AgCl + 2Na2S2O3  NaCl + Na3[Ag(S2O3)2] 7 . Due to back bonding BF3, BCl3 and BBr3 are exist AgCl + NH3  [Ag(NH3)2]Cl in free form. But BH3 is not. 2 2 . 2KMnO4 + 5H2S H+ K2SO4 + 2MnSO4 + 9 . Na2B4O7 + 7H2O  2Na[B(OH)4] + 2H3BO3 8H2O + 5S + 3H2SO4 Aqueous solution of borax acts as a buffer because it contains weak acid and its salt with strong base. 2 9 . CuSO4 + 2KI  CuI2 + K2SO4 2CuI2  Cu2I2 + I2 1 2 . 2HNO3 – H2O N2O5 Anhydride 3 1 . PI3 + 3H2O  H3PO3 + 3HI OO H2 + I2 Pt  2HI I2 + H2S  2HI + S N—O—N 3 2 . I2 can not dissplace Br2, Cl2, F2 from KBr, KCl, OO KF, because it weakest oxidising agent. 1 3 . (NH2)2CO Molecular mass = 60 3 9 . White or yellow P 470K Black-P Urea mass of nitrogen = 28 (A) (B) 28 % of N = × 100 = 47% 60 570K Red-P CO2-atom

4 0 . Ca3P2 + 6H2O  2PH3 + 3Ca(OH)2 4 1 . Ca + C2  CaC2 N2 Ca(CN)2 59. O 4 6 . P4O10 + 4HNO3  4HPO3 + 2N2O5 P OO 4 9 . B2H6 + 2NH3  B2H6·2NH3 OP O PO O When the addition product is heated at 200°C a O volatile compound borazole or inorganic benzene O is formed. P 3B2H6·2NH3  2B3N3H6 + 12H2 O P4O10 5 7 . CH2 COOH P4O10 C3O2 + 2H2O COOH 150°C (carbon suboxide) HYDROGEN COMPOUND EXERCISE # 1 n2 11. NaCl  Na+ + Cl– 4. rn = 0.529 Å At cathode 2 Na+ + e–  Na for protium, deuterium and tritium the n and 1 z are 1, 1 and 1 respectively. Na + H2O  NaOH + 2 H2 At anode 8 . Laboratory method of formation of H2 gas Cl–  Cl + e– granulan zinc + dil H2SO4  H2 Cl– + Cl  Cl2 BaO2 + 2HCl  BaCl2 + H2O2 1 0 . Be+2NaOH+2H2O Na2B3O2·2H2O + H2 (sodium beryllate) 17. HYDROGEN COMPOUND EXERCISE # 2 1 . Zn + 2NaOH  Na2ZnO2 + H2 1 4 . D2O + CO2  D2CO3 sodium zincate D2O + SO2  D2SO4 2Al + 2NaOH + 2H2O  2NaAlO2 + 3H2 D2O + P2O5  2D3PO4 sodium meta aluminate D2O + N2O5  2DNO3 1 3 . 3Fe + 4D2O   Fe2O4 + 4D2 1 9 . Na2O2 + 2H2O  H2O2 + 2NaOH Magnetic oxide