2. Chemical Thermodynamics and Energetics - I

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. Chemical Thermodynamics and Energetics - I 1. Given : Initial Volume ( = 5L dm 3 Final Volume (V = 10L dm 3 ext = 304 cm of Hg Work done W = ext V ext = 304 cm of Hg = 304 atm [... 76cm of Hg = 1 atm] 76 = 4 atm As gas expands against a constant external pressure, work done is given by W = ext V = ext (V = 4 (10 5 = 4 (5 W = 0 lit-atm W = 0 4. cal [... 1 lit-atm = 4. cal] = 484.4 cal = 4.844 10 cal W = 0 101.3 J [... 1 lit-atm = 101.3 J] = 06 J W = 06 10 7 ergs =.06 10 10 ergs Ans : Work done is 0 lit-atm, 4.844 10 cal, 06 J and.06 10 10 erg. Given : =.06 10 Nm = 1m 3 V = 10m 3 Work done W = ext V W = (V =.06 10 (10 1 =.06 10 9 = 183.4 J = 1.834 kj Ans : Work done is 1.834 kj. 3. Given : = 1.0 atm = 1.013 A = 100cm S = 10 Work done W = ext A S W = 1.013 100 10 1013 J Ans : Work done by the system is 1013 J. 4. Given : = 1.03 10 5 Nm = 10 m 3 V = 10 m 3 q = 1673.6 J E

Mahesh Tutorials Science 5 According to 1 st law of thermodynamics, q = E + W i.e. q = E + V E = q (V = 1673.6 1.03 10 5 ( 10 1 10 = 1673.6 1.03 10 5 10 = 1673.6 1030 = 643.6 J Ans : Change in internal energy is = 643.6 J. 5. Given : = 1.01 10 5 Nm = 15 dm 3 = 15 10 3 m 3 V = 45 dm 3 = 45 10 3 m 3 n = moles i W ii q iii E in cal and J W = ext V... Gas expands isothermally, W = ext V = 1.01 10 5 [(45 10 3 (15 10 3 ] = 1.01 10 5 30 10 3 W = 30.3 10 J W = 3030 J E = 3030 cal 4.184 = 74.187 cal = 0 [... the process is isothermal] As q = E W q = 0 W q = + 30.3 10 J = + 74.187 cal Ans : Work done is 3030 J or 74.187 cal Heat absorbed is 74.187 cal 6. Given : Initial volume ( = dm 3 = 10 3 m 3 Final volume (V = 0dm 3 = 0 10 3 m 3 q = 5718.5 J Mass of gas = 3 10 3 Kg T =.303 nrt log 10 ( V Molecular weight O = 3 No. of moles (n = = Mass Mol.wt 3 10 3 = 1 3 10 3... The gas expands isothermally and reversibly, the work done is maximum and is given by : =.303 nrt log 10 ( V 5718.5 = +.303 (1 (8.314 (T log 10( 0 10 3 10 3 T = 5718.5.303 8.314

6 Mahesh Tutorials Science = AL T = 98.46 K Ans : Temperature is 98.46 K. 7. Given : = atm [ ] log.303 = 0.363 + log 8.314 = 0.9198 1.81.4749 log 5718.5 = 3.7573 = 1.01 10 5 Nm =.0 10 5 Nm = 15 dm 3 = 15 10 3 m 3 V = 45 dm 3 = 45 10 3 m 3 n = moles T = 98K W, q, E W = ext V W = ext V E =.0 10 5 [(45 10 3 (15 10 3 ] =.0 10 5 30 10 3 = 60.6 10 = 6.06 10 3 J... q = E W q = 0 W = 0 [... process is isothermal] q = ( 6.06 10 3 J = 6.06 10 3 J Ans : Work done is 6.06 10 3 J, 8. Given : 1 = 1 = W CO =. 10 kg T = 300K R = 8.314 J/mol Work done Formulae : =.303nRT log 10 n = n CO = = wt mol.wt. 10 44 10. 10 4.4 10 3 = 0.5 =.303 0.5 8.314 300 log 10. =.303 0.5 8.314 300 0.3010 = 864.5 J Ans : Work done is 864.5 J. 9. Given : Mass of Ar = 0 10 3 Kg = 11 10 3 m 3 W =.78 10 3 J T = 300 K 1 1

Mahesh Tutorials Science 7 V =.303 nrt log 10 ( V At. wt. of Ar = 40 No. of moles (n = Mass Mol. wt = 0 10 3 = 0.5 40 10 3... The gas expands isothermally and reversibly, work done is maximum and is given as : =.303 nrt log 10 ( V log 10 ( V = =.303 nrt.78 10 3.303 0.5 8.314 300 = 0.9679 V = AL(0.9679 = 9.87 V = (9.87 = 9.87 11 10 3 = 10.157 10 3 m 3 V = 10.157 dm 3 Ans : Final volume of gas is = 10.157 dm 3 10. Given : 1 = 101.35 10 3 a = 1013.5 10 3 a = 101.35 10 4 a T = 300 K W = 5.15 10 3 J n =.303 nrt log 1 10 The gas expands isothermally and reversibly, works done is maximum and is given as : =.303 nrt log 1 10 5.15 x 10 3 =.303 n (8.314 (300 log ( 101.35 103 101.35 10 4 5.15 x 10 3 = n (.303 (8.314 (300 n = n = 0.8966 n = 1 Ans : No. of mole = 1 11. Given : T = 373 K = 1 atm log(10 1 5.15 103 (.303 (8.314 (300 ( 1 R = 8.314 J K 1 mol 1 Mol.wt of water = 18 Latent heat of vaporisation = 60 Jgm 1 q, W, E and H W = V One mole of steam is condensed into water. Energy has left the system as heat q is negative

8 Mahesh Tutorials Science 1. Given : q = 7.350 kcal mol 1 T = 353K = 1 atm R = 8.314 Jk 1 mol 1 work done, E, H q = 60 J gm 1 = 60 mol.wt = 406.80 J mol 1 = 40680 J W = V = (V = (V vap V liq W = V E = q + V H = E + V W = V = (V 10 mol 1 V liq is negligible and can be neglected W = V vap For ideal gas, W = RT W = 8.314 373 = 3101.1 J E = q + V = 40680 + 3101.1 = 37579 J H = q at (constant pressure H = 40680 J Ans : q, W, E and H of water is 406.80 J mol 1, 3101.1 J, 37579 J and 40680 J respectively. = (V vap V liq V liq is negligible and can be neglected W = V vap For ideal gas W = RT = 8.314 353 = 934.84 J =.934 kj E = q + V (1 st law = 7.350 cal + (.934 = (7.350 4.184.934 = 30.750.934 = 7.8184 kj H = E + V = E + nrt = 7.8184 +.934 = 30.75 kj Ans : Work done, E and H of the solution is.934 kj, 7.818 kj and 30.75 kj respectively. 13. Given : SO (g + O (g SO 3(g E = 199. KJ H H = E + ng RT Here, n = n product n reactants (only gaseous = ( + 1 = 1 n = 1 = 199. + ( 1 = 199..477 (8.314 10 3 (98 = 01.637 kj Ans : Heat of reaction = 01.697 kj

Mahesh Tutorials Science 9 14. Given : H = 74.89 kj mol 1 T = 98 K R = 8.314 JK 1 mol 1 = 8.314 10 3 kj K 1 mol 1 E Thermochemical equation for the formation of C 10 H 8, 10C (s + 4H (g C 10 H 8(s n = n products n reactants (only gaseous n = 0 4 n = 4 E = H nrt = 74.89 ( 4 (8.314 10 3 (98 = 74.89 + 4 (8.314 (0.98 = 74.89 + 9.910 = 84.8 kj Ans : Change in internal energy is 84.8 kj. 15. Given : E at 98K = 85.3 kj T = 98K E at 398K = To find : 397.7 kj R = 8.314 J/K/mol H The required thermochemical equation for the combustion of CO (g is CO (g + ½ O (g CO (g n = No. of gaseous products At 98 K no. of gaseous reactants = 1 1 1 + = 0.5 At 398 K = 85.3 + [( 0.5 8.314 = 85.3 + ( 1.38 = 86.538 kj 10 3 98] = 397.7 + [( 0.5 8.314 = 397.7 1.6544 = 399.35 kj Ans : H at 98 K is 86.538 kj H at 398 K is 399.35 kj. 10 3 398] 16. Given : E = Heat of combustion at constant volume = 363.9 kj mol 1 T = 98K R = 8.314 Jk 1 mol 1 H = Heat of combustion at constant pressure. The required thermo chemical equation for the combustion of benzene is C 6 H 6 + 15 O 6CO (g (g +3H

30 Mahesh Tutorials Science n = no. of moles of gaseous products no. of moles of gaseous reactants = 6 15 = 1.5 H = 363.9 + [( 1.5 8.314 10 3 98] = 363.9 3.71635 = 367.6 kj mol 1 Ans : Heat of combustion at constant pressure is 367.6 kj mol 1 17. Given : H = 1560.63 kj T = 98 K R = 8.314 Jk 1 /mol Heat of combustion at constant volume E E = H nrt n = no. of moles of gaseous products no. of moles of gaseous reactants n = (3.5 + 1 =.5 E = 1560.63 [(.5 8.314 10 3 98] = 1560.63 ( 6.1939 = 1554.4 kj/mol Ans : Heat of reaction at constant volume is 1554.4 kj/mol. 18. Given : E = Heat of combustion at constant volume = 84 kj T = 98 K R = 8.314 Jk 1 /mol H = Heat of combustion at constant pressure Required thermochemical equation for the combustion of CO (g is CO (g + ½ O (g CO (g n = no. of moles of gaseous products no. of moles of gaseous reactants 19. Given : i C (s + O (g CO (g H = 393 kj ii H (g + ½O (g H H = 84. kj iii C OH (l + 3O (g CO (g + 3H H = 136.7 kj Req. equation : C (s + 3H (g + ½O (g C OH, Formulae : = 1 1 1 + = 0.5 H = 84 + [( 0.5 8.314 10 3 98] = 84 1.387 = 85.4 kj/mol Ans : Heat of combustion H is 85.4 kj/mol 1. H =? i By Hess s Law H = H 1 + H ii H = E ngrt Here, n = n n 1 = O (3 + ½ = 7/

Mahesh Tutorials Science 31 Now, [Eq.(i] + [Eq.(ii] 3 Eq.(iiii We get, C (s + O (g CO (g H = 786 kj + 3H (g + 3 O (g 3H H = 85.6 kj C OH (l 3O (g CO (g 3H H = 136.7 kj C (s + ½O (g 3H (g C OH (l H = 75.9 kj Now, E = H nrt = 75.9 ( 7/ (8.314 10 3 (300 = 75.9 + 7 (8.314 (0.3 = 8.79 75.9 = 67.17 kj Ans : Change in internal energy is 67.17 kj. 0. Given : T = 300 K K = 8.314 JK 1 mol 1 Zn (s + HCl (aq ZnCl (aq + H (g H = 151460 Joules E n = n products n reactants (only gaseous = 1 0 n = 1 E = H nrt = 151460 1 8.314 300 = 151460 4.94 E = 151435.058 Joules Ans : Change in internal energy for the reaction is 151435.058 Joules. 1. Given : Hf CO = 393.5 kj/mol Hf H O = 41.8 kj/mol Hf C H 4 = 5.30 kj/mol H combustion H = Σ H product Σ Σ H reactant Thermochemical reaction for heat of combustion of ethylene is C H 4(g + O (g CO (g + H O(l H = Σ H product Σ Σ H reactant = [( H of CO (g + H of H O(l] [( H of C H 4(g + H of O (g ] = [ ( 393.5 + ( 41.8] [(5.30 +0] (... O = 0 = [ 787 + ( 483.6] 5.30 = 13.0 kj/mol Ans : Heat of combustion of ethylene is. Given : 13.0 kj/mol Hf CO = 393.6 kj Hf H O = 86 kj Hf C 1 H O 11 = 0.8 kj Heat of combustion H = Σ H products Σ Σ H reactant

3 Mahesh Tutorials Science Thermochemical reaction for heat of combustion of cane sugar is C 1 H O 11(s + 1O (g 1CO + 11H H = Σ H products Σ Σ H reactant = [( H of 1CO + ( H of 11H O] [( H of C 1 H O 11 + ( H 1O ] = [(1 ( 393.6 + 11 ( 86] [ 0.8 + 0] (... O = 0 = [ 473. 3146] (0.8 = 7869. + 0.8 = 5648. kj Ans : Heat of combustion of cane sugar is 3. Given : 5648. kj. H f CH 3 Cl = 84.68 kj H f C H 6 = 90.6 kj H Reaction : C H 6(g + Cl (g CH 3 Cl (g H = Σ H products H Σ H reactant = [( H of CH 3 Cl (g ( H of C H 6(g + H of Cl (g H = [ 84.68 ( 90.6 + 0] = 169.36 + 90.6 = 79.1 kj. Ans : Enthalpy change is 79.1 kj. 31. Given : i C (s + O (g CO (g H = 404.17 kj ii H (g + 1 O (g H H = 87.0 kj iii C 3 H 8(g + 5O (g 3CO (g + 4H Req. Equation : 3C (s + 4H (g C 3 H 8(g H By Hess s is Law H = H 1 + H +... Eq (i 3 + Eq (ii 4 Eq.(ii 3C (s + 3O (g 3CO (g H = 14.17 kj H = 11.51 kj + 4H (g + O (g 4H O (b H = 1148.08 kj C 3 H 8(g 5O (g 3CO (g 4H Ans : 3C (s + 4H (g C 3 H 8(g H = 14.17 kj H = 146.4 kj 34. Given : i C (s + O (g CO (g H 1 = 395.4 kj ii H (g + 1 O (g H iii C OH + 3O (g H = 85.6 kj CO + 3H H 3 = 1366.8 kj

Mahesh Tutorials Science 33 Required : C (s + 3H (g + 1 O (g H f C OH (l Multiply equation (i by and (ii by 3 iv C (s + O (g CO (g H = 790.8 kj v 3H (g + 3 O (g 3H Add. equation (iv and (v vi H = 856.8 kj C (s + 3H (g + 7 O (g CO + 3H Subtract equation (iii from (vi H = 1646 kj vii C (s + 3H (g + 1 O (g C O 4(l = 0 On rearranging we get, H = 80.8 kj C (s + 3H (g + 1 O (g C O 4(l H = 80.8 kj Ans : Heat of formation of ethyl alcohol is 80.8 kj. 35. Given : i CH 4(g + O (g CO (g + H ii C (s + O (g CO (g iii H (g + ½O (g H H = 98.8 kj H = 396.6 kj H = 84.5 kj H f Required : C (s + H (g Multiply equation (iii by CH 4(g iv H + O (g H Adding equation (iv and (ii H = 569 kj v C (s + O + H (g CO + H O Subtract equation (i from (v H = 965.6 kj C (s + H CH (g O 4 (g H = 36.8 kj On rearranging, C (s + H (g CH 4(g H = 36.8 kj Ans : Heat of formation of methane is 36.8 kj. 36. Given : Heat of formation of CO = 394.55 kj Heat of formation of SO = 97.06 kj Heat of combustion of CS = 1071.5 kj Heat of formation of CS i C (s + O (g CO (g ii S (s + O (g SO (g H = 394.55 kj H = 97.06 kj iii CS (s + 3O (g CO (g + SO (g H = 1071.5 kj

34 Mahesh Tutorials Science Multiply equation (ii by iv S (s + 4O (g 4SO (g H = 594.1 Adding equation (i and (iv v C (s + S (s + 5O (g CO (g + 4SO (g H = 988.67 kj Subtracting equation (iii from (v, C (s + S (s + O (g CS (s SO (g On rearranging, C(s + S (s + O (g H = 8.85 kj CS (s + SO (g H = 8.85 kj Ans : Heat of formation of carbon disulphide is 8.85 kj. 37. Given : i KOH (aq + HCl (aq KCl (aq + H O H = 57.3 kj ii H (g + 1 O (g H H = 86.18 kj iii 1 H(g + 1 Cl (g HCl (aq H = 164.43 kj iv K (s + 1 O (g + 1 H (g + aq KOH (aq H = 487.43 kj v KCl (s + aq KCl (aq Req. Equation : H = 18.4 kj K (s + 1 Cl (g KCl H =? By Hess s Law From (iv (iii, (i K (s + 1 O (g + 1 H (g + aq KOH (aq H = 487.43 kj 1 + H(g + 1 Cl (g HCl (aq H = 164.43 kj + KOH (aq + HCl (aq KCl (aq + H O H = 57.3 kj K (s + 1 Cl (g + 1 O (g + H (g KCl (aq + H O H = 709.18 kj... (vi From (ii and (iv H (g + 1 O (g H H = 86.18 kj + KCl (s + aq KCl (aq H = 18.4 kj H (g + 1 O (g + KCl (s + aq H + KCl (aq H = 67.78 kj...(vii From (vi and (vii H (g + K (s + 1 Cl (g + 1 O (g KCl (aq + H O H = 709.18 kj H (g 1 O (g KCl (s H KCl (aq H = +67.78 kj Ans : K (s + 1 Cl (g KCl (s H = 441.4 kj H = H 1 + H +...

Mahesh Tutorials Science 35 38. Given : i C H 4(g + 3O (g CO (g + H 39. Given : H = 1411.3 kj ii H (g + 1 O (g H H = 85.8 kj iii C H 6(s + 7 O (g CO (g + 3H H = 1560. kj Req. Equation : C H 4(g + H (g C H 6(g By Hess s Law H = H 1 + H C H 4(g + 3O (g CO (g + H H = 1411.3 kj + H (g + 1 O (g H H = 85.8 kj CO (g + 3H C H 6(s + 7 O (g H = 1560. kj Net Rxn : C H 4(g + H (g C H 6(s By Hess s Law H = 1411.3 85.8 + 1560. = 136.9kJ Ans : Heat of Reaction is 136.9 kj i H (g + 1 O (g H H = 85 kj ii C (s + H (g + 1 O (g CH 3 CH H = 184 kj iii C (s + O (g CO (g Req. Equation : H = 394 kj CH 3 CH + 5 O (g CO (g + H H =? By Hess s Law H = H 1 + H CH 3 CH C (s + H (g + 1 O (g H = 184 kj + C (s + O (g CO (g + H (g + O (g H Net Rxn : H = 788 kj H = 570 kj CH 3 CH + 5 O (g CO (g +H H = +184 788 750 = 1174 kj Ans : Heat of Reaction is 1174 kj 40. Given : i H (g + 1 O (g H O (g H = 38.6 kj ii H (g + 1 O (g H O (g H = 85.91 kj Req Equation : H By Hess s Law H O (g H = H 1 + H

36 Mahesh Tutorials Science H (g + 1 O (g H O (g H = 38.6 kj H H (g + 1 O (g H = 85.91 kj By Hess s Law H H O (g H = 38.6 85.91 = 47.65 kj Ans : Heat of reaction is 47.65 kj.

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