2. Chemical Thermodynamics and Energetics - I

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1 . 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] = cal = cal W = J [... 1 lit-atm = J] = 06 J W = ergs = ergs Ans : Work done is 0 lit-atm, cal, 06 J and erg. Given : = Nm = 1m 3 V = 10m 3 Work done W = ext V W = (V = (10 1 = = J = kj Ans : Work done is kj. 3. Given : = 1.0 atm = A = 100cm S = 10 Work done W = ext A S W = J Ans : Work done by the system is 1013 J. 4. Given : = Nm = 10 m 3 V = 10 m 3 q = J E

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

3 6 Mahesh Tutorials Science = AL T = K Ans : Temperature is K. 7. Given : = atm [ ] log.303 = log = log = = Nm = Nm = 15 dm 3 = m 3 V = 45 dm 3 = m 3 n = moles T = 98K W, q, E W = ext V W = ext V E = [( ( ] = = = J... q = E W q = 0 W = 0 [... process is isothermal] q = ( J = J Ans : Work done is J, 8. Given : 1 = 1 = W CO =. 10 kg T = 300K R = J/mol Work done Formulae : =.303nRT log 10 n = n CO = = wt mol.wt = 0.5 = log 10. = = J Ans : Work done is J. 9. Given : Mass of Ar = Kg = m 3 W = J T = 300 K 1 1

4 Mahesh Tutorials Science 7 V =.303 nrt log 10 ( V At. wt. of Ar = 40 No. of moles (n = Mass Mol. wt = = The gas expands isothermally and reversibly, work done is maximum and is given as : =.303 nrt log 10 ( V log 10 ( V = =.303 nrt = V = AL( = 9.87 V = (9.87 = = m 3 V = dm 3 Ans : Final volume of gas is = dm Given : 1 = a = a = a T = 300 K W = J n =.303 nrt log 1 10 The gas expands isothermally and reversibly, works done is maximum and is given as : =.303 nrt log x 10 3 =.303 n (8.314 (300 log ( x 10 3 = n (.303 (8.314 (300 n = n = n = 1 Ans : No. of mole = Given : T = 373 K = 1 atm log( (.303 (8.314 (300 ( 1 R = 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

5 8 Mahesh Tutorials Science 1. Given : q = kcal mol 1 T = 353K = 1 atm R = Jk 1 mol 1 work done, E, H q = 60 J gm 1 = 60 mol.wt = J mol 1 = 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 = = J E = q + V = = J H = q at (constant pressure H = J Ans : q, W, E and H of water is J mol 1, J, J and J respectively. = (V vap V liq V liq is negligible and can be neglected W = V vap For ideal gas W = RT = = J =.934 kj E = q + V (1 st law = cal + (.934 = ( = = kj H = E + V = E + nrt = = kj Ans : Work done, E and H of the solution is.934 kj, kj and 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 = ( 1 = ( (98 = kj Ans : Heat of reaction = kj

6 Mahesh Tutorials Science Given : H = kj mol 1 T = 98 K R = JK 1 mol 1 = 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 = ( 4 ( (98 = (8.314 (0.98 = = 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 : kj R = 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 = = 0.5 At 398 K = [( = ( 1.38 = kj ] = [( = = kj Ans : H at 98 K is kj H at 398 K is kj ] 16. Given : E = Heat of combustion at constant volume = kj mol 1 T = 98K R = 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 O 6CO (g (g +3H

7 30 Mahesh Tutorials Science n = no. of moles of gaseous products no. of moles of gaseous reactants = 6 15 = 1.5 H = [( ] = = kj mol 1 Ans : Heat of combustion at constant pressure is kj mol Given : H = kj T = 98 K R = 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 = ( =.5 E = [( ] = ( = kj/mol Ans : Heat of reaction at constant volume is kj/mol. 18. Given : E = Heat of combustion at constant volume = 84 kj T = 98 K R = 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 = kj Req. equation : C (s + 3H (g + ½O (g C OH, Formulae : = = 0.5 H = 84 + [( ] = = 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/

8 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 = kj C (s + ½O (g 3H (g C OH (l H = 75.9 kj Now, E = H nrt = 75.9 ( 7/ ( (300 = (8.314 (0.3 = = kj Ans : Change in internal energy is kj. 0. Given : T = 300 K K = JK 1 mol 1 Zn (s + HCl (aq ZnCl (aq + H (g H = Joules E n = n products n reactants (only gaseous = 1 0 n = 1 E = H nrt = = E = Joules Ans : Change in internal energy for the reaction is Joules. 1. Given : Hf CO = 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 ] = [ ( ( 41.8] [( ] (... O = 0 = [ ( 483.6] 5.30 = 13.0 kj/mol Ans : Heat of combustion of ethylene is. Given : 13.0 kj/mol Hf CO = 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

9 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 ( ( 86] [ ] (... O = 0 = [ ] (0.8 = = kj Ans : Heat of combustion of cane sugar is 3. Given : kj. H f CH 3 Cl = 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 = [ ( ] = = 79.1 kj. Ans : Enthalpy change is 79.1 kj. 31. Given : i C (s + O (g CO (g H = 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 = kj H = kj + 4H (g + O (g 4H O (b H = kj C 3 H 8(g 5O (g 3CO (g 4H Ans : 3C (s + 4H (g C 3 H 8(g H = kj H = kj 34. Given : i C (s + O (g CO (g H 1 = kj ii H (g + 1 O (g H iii C OH + 3O (g H = 85.6 kj CO + 3H H 3 = kj

10 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 = kj v 3H (g + 3 O (g 3H Add. equation (iv and (v vi H = 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 = 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 = 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 = kj Heat of formation of SO = kj Heat of combustion of CS = kj Heat of formation of CS i C (s + O (g CO (g ii S (s + O (g SO (g H = kj H = kj iii CS (s + 3O (g CO (g + SO (g H = kj

11 34 Mahesh Tutorials Science Multiply equation (ii by iv S (s + 4O (g 4SO (g H = Adding equation (i and (iv v C (s + S (s + 5O (g CO (g + 4SO (g H = 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 = kj iii 1 H(g + 1 Cl (g HCl (aq H = kj iv K (s + 1 O (g + 1 H (g + aq KOH (aq H = 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 = kj 1 + H(g + 1 Cl (g HCl (aq H = 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 = kj... (vi From (ii and (iv H (g + 1 O (g H H = kj + KCl (s + aq KCl (aq H = 18.4 kj H (g + 1 O (g + KCl (s + aq H + KCl (aq H = kj...(vii From (vi and (vii H (g + K (s + 1 Cl (g + 1 O (g KCl (aq + H O H = kj H (g 1 O (g KCl (s H KCl (aq H = kj Ans : K (s + 1 Cl (g KCl (s H = kj H = H 1 + H +...

12 Mahesh Tutorials Science Given : i C H 4(g + 3O (g CO (g + H 39. Given : H = kj ii H (g + 1 O (g H H = 85.8 kj iii C H 6(s + 7 O (g CO (g + 3H H = 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 = kj + H (g + 1 O (g H H = 85.8 kj CO (g + 3H C H 6(s + 7 O (g H = kj Net Rxn : C H 4(g + H (g C H 6(s By Hess s Law H = = 136.9kJ Ans : Heat of Reaction is 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 = = 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 = kj Req Equation : H By Hess s Law H O (g H = H 1 + H

13 36 Mahesh Tutorials Science H (g + 1 O (g H O (g H = 38.6 kj H H (g + 1 O (g H = kj By Hess s Law H H O (g H = = kj Ans : Heat of reaction is kj.