3-Chemical Thermodynamics and Energetics

3. CHEMICAL THERMODYNAMICS AND ENERGETICS 1)

Three moles of an ideal gas are expanded isothermally from a volume of 300 cm3 to 2.5 L at 300 K against a pressure of 1.9 atm. Calculate the work done in L atm and joules. Given : V1 = 300 cm3 = 300 × 10–3 L = 0.3 L = 2.5 L V2 T = 300 K P = 1.9 atm To Find : Work done (W) Formula : W = –P∆ ∆V Solution : ∴ W = –P∆ ∆V W = –1.9(2.5 – 0.3) = –1.9 × 2.2 = –4.18 L atm Now, 1 L atm = 101.33 J So –4.18 L atm = –423.56 J ∴ W = –423.56 J

∴

0.36 =

–1.2 × (V2 – 0.5)

∴ ∴

0.36 = 1.2 0.3 = V2 =

–V2 + 0.5 0.2 L

∴

V2

200 cm3

∴

=

–(V2 – 0.5)

3)

Calculate the maximum work when 24 g of oxygen are expanded isothermally and reversibly from a pressure of 1.6 × 105 Pa to 100 kPa at 298 K. Given : Mass = 24 g P1 = 1.6 × 105 Pa P2 = 100 K Pa = 100 × 103 Pa = 105 Pa n

= = = =

24 32 0.75 moles 8.314 J K–1 mol–1 298 K

R T To Find : Maximum mark (Wmax) 2) One mole of an ideal gas is compressed Formula : 3 from 500 cm against a constant pressure 5 of 1.216 × 10 Pa. The work involved in P Wmax = –2.303nRT log10 1 the process is 36.50 J. Calculate the final P2 volume. Solution : Given : V1 = 500 cm3 P Wmax = –2.303nRT log10 1 = 0.5 L P 2 P = 1.216 × 105 Pa = –2.303 × 0.75 × 8.314 × 298 = 1.2 atm W = 36.50 J 1.6 × 10 5 × log 10 = 0.36 L atm 10 5 To Find : = –2.303 × 0.75 × 8.314 × 298 Find Volume (V2) × log10 1.6 Formula : = –4273.39 × log 1.6 W = –P∆ ∆V = –4279.39 × 0.204 Solution : Wmax = –873 J W = –P∆ ∆V Chemical Thermodynamcs and Energetics

.. 2 Three moles of an ideal gas are 5) 2.8 × 10–2 kg of nitrogen is expanded compressed isothermally and reversibly isothermally and reversibly at 300 K to a volume of 2L. The work done is from 15.15 × 105 Nm–2 when the work 0 done is found to be –17.33kJ. Find the 2.983 kJ at 22 C. Calculate the initial volume of the gas. final pressure. Given : Given : n = 3 moles 2.8 × 10 –2 × 10 3 V2 = 2L n = 28 Wmax = 2.983 kJ = 1 mole = 2.983 × 103 J T = 300 k T = 220C = 15.15 × 105 Nm–2 P1 = 295 K Wmax = –17.33 kJ R = 8.314 JK–1 mol–1 = –17.33 × 103 J To Find : To Find : Initial volume Final pressure Formula : Formula : V –2.303nRT log10 2 Wmax = P V1 Wmax = –2.303nRT log10 1 P2 Solution : Solution : V Wmax = –2.303nRT log10 2 P V1 Wmax = –2.303nRT log 1 3 P2 ∴ 2.983 × 10 = –2.303 × 3 × 8.314 ∴ –17.33 × 103 = –2.303 × 8.314 V2 × 295 log10 V1 15.15 × 105 × 300 log P2 3 2.983 × 10 2 3 – ∴ 17.33 × 10 ∴ = log10 V1 16,945.22 15.15 × 105 3 2 = 5.744 × 10 log P2 ∴ –0.176 = log10 V1 4)

Take antilog on both sides. Antilog [–0.176] = 0.6668

=

∴

V1

=

∴ ∴

V1 V1

∴

2 V1 2 V1

∴

∴

15.15 × 105 = log P2

3.017

15.15 × 105 = log P2

Taking antilog on both sides

2 0.6668 = 2.999 L = 3L

Chemical Thermodynamcs and Energetics

17.33 5.744

Antilog [3.017]

∴

15.15 × 105 = P2

1039.92 =

15.15 × 105 P2

.. 3

∴

P2 P2

6)

15.15 × 10 5 = 1039.92 = 1456.8 Nm–2

A sample of gas absorbs 4000 kJ of heat. i) If volume remains constant, what is ∆U ? ii) Suppose that in addition to absorption of heat by the sample, the surroundings does 2000 kJ of work on the sample, what is ∆U ? iii) Suppose that as the original sample absorbs heat, it expands against atmospheric pressure and does 600 kJ of work on its surroundings. What is ∆U ? Given : i) q = 4000 KJ ii) W = 2000 KJ iii) w = 600 KJ To Find : i) ∆U ii) ∆U iii) ∆U Solution : i) At constant volume, ∆H = ∆U So, ∆ U = 4000 kJ ii)

iii)

7)

i) ii)

To Find : W Formula : W = –∆ ∆ nRT Solution : 2SO3(g) i) 2SO2(g) + O2(g) 1 mole of SO2 will 0.5 mole of O2 to give 1 mole of SO3. W = –∆ ∆ n.RT = –RT(n2 – n1) W = –8.314 × 323(1 – 1.5) = –8.314 × 323 × (–0.5) W = 1342.7 J As W is positive, work is done on system. N2O(g) + 2H2O(g) ii) NH 4NO 3(s) 2 moles of NH 4NO 3(s) gives 2 moles of N2O(g) and 4 moles of H2O(g) So, ∆n = (6 – 0) = 6 W = –∆ ∆ n.RT = –6 × 8.314 × 373 = –18606.75 J W = –18.61 kJ As W is negative, work is done by the system.

The enthalpy change for the reaction Surroundings does 2000 kJ of work on 8) H4(g) + H2(g) C 2 H 6(g) is –620 J C 2 sample when 100 mL of enthylene and 100 mL So, ∆ U = 4000 + 2000 of H react at 1 atm pressure. Calculate = 6000 kJ 2 the pressrure volume work and ∆U. Original sample absorbs heat and expands Given : aginst atmospheric pressure. Work done ∆H = –620 J, on surroundings. V = 200ml = 0.2 L ∆ U = 4000 – 600 1 P = 1 atm = 3400 kJ = 100 ml = 0.1 L V2 Calculate the work done in each of the following reactions. State whether work To Find : Pressure – Volume work (W) is done on the system or by the system.. ∆U The oxidation of one mole of SO2 at 500C Formula : 2SO2(g) + O2(g) 2SO3(g) W = –P∆ ∆V Decomposition of 2 moles of NH4NO 3 Solution : 0 at 100 C W = –P∆ ∆V NH 4 NO 3(s) N2O(g) + 2H2O(g) = –1 × (0.1 – 0.2) = –1 × –0.1 Chemical Thermodynamcs and Energetics

.. 4 Calculate the standard enthalpy of formation of C2H6 from the following data : ∆H 2C2H6(g) + 7O2(g) 4CO2(g) + 6H2O(l), ∆H0 W = –3119 kJ mol–1 10.13 ∆fH 0(CO2 ) = –285.8 kJ mol–1. Given : ∴ P∆V ∆H Standard enthalpy of reaction (∆ ∆ H0) = –3119 kJ ∴ ∆U ∆fH 0(CO2 ) = –393.5 kJ/mol ∆fH 0(H 2O) = –285.8 kJ/mol ∴ ∆U To Find : Standard enthalpy of formation of ∆ f H0 ) C2H 6(∆ 9) Calculate standard enthalpy of the Formula : reaction ∆H0 = ∑∆ fH0 (products) Fe2O3(s) + 3CO(g) 2Fe(s) + 3CO2(g) – ∑∆ fH0(reactants) From the following data : Solution : ∆fH 0(Fe2 O3 )= –824.2 kJ mol–1, 0 –1 Standard enthalpy is given by, ∆fH (CO) = –110.5 kJ mol , = ∑∆ fH0 (products) ∆H0 ∆fH 0(CO2 ) = –393.5 kJ mol–1 – ∑∆ fH0(reactants) Given : = [4 moles of CO2 × ∆fH0 of ∆fH 0(Fe2 O3 ) = –824.2 kJ/mol CO2 + 6 moles of H2O ∆fH0 (CO) = –110.5 kJ/mol 0 × ∆fH0 of H2O] – = –393.5 kJ/mol ∆fH (CO2 ) [2 moles of C2H6 × ∆fH0 of C2H6 + 0] To Find : = [4 × (–393.5) + 6 × –285.8] Standard enthalpy of reaction (∆ ∆ H0) – [2 × ∆fH0 of C2H6] Formula : ∴ –3119 = –1574 – 1714.8 – 2 ∆H0 = ∑∆ fH0 (products) 0 × ∆fH0 of C2H6 – ∑∆fH (reactants) ∴ –3119 = –3288.8 – 2 Solution : 0 × ∆fH0 of C2H6 Standard enthalpy of reaction (∆ ∆H ) 0 ∴ 169.8 = –2 × ∆fH0 of C2H6 = ∆fH (products) 0 ∴ –84.9 = ∆fH0 of C2H6 – ∆fH (reactants) 0 ∴ ∆fH of C2H6 = –84.9 kJ/mol = [2 moles of Fe × 0 + 3 moles of CO2 × ∆fH0 of CO2] – 11) How much heat is evolved when 12 g of CO reacts with NO2 according to the [1 mole of Fe2O3 × ∆fH0 of following reaction, Fe2O3 + 3 moles of CO 4CO(g) + 2NO2(g) 4CO2(g) + N2 , × ∆fH0 of CO] ∆ H 0 = –1198 kJ = [0 + (–1180.5)] – [–824.2 + (3 × –110.5)] Given : ∆H0 = –1198 kJ = –1180.5 + 824.2 + 331.5 To Find : = –24.8 kJ Heat evolved W

= = = = = = = = = = = =

0.1L. atm 0.1 × 101.33 J +10.13 J ∆U + P∆V –P∆V –P∆V –10.13 J –620 J ∆H – P∆V –620 – (–10.13) –620 + 10.13 –609.9 J

Chemical Thermodynamcs and Energetics

10)

.. 5 Solution : 13) Calculate ∆H0 of the reaction According to the reaction, +1198 kJ of CH4(g) + O2(g) CH2O(g) + H2O(g) heat is evolved. When 4 moles of CO From the following data : reacts with NO 2. So heat evolved per Bond C–H O=O C=O O–H 0 –1 mole is +299.5 kJ. ∆H /kJ mol 414 499 745 464 Number of moles of CO Given : Bond C–H O=O C=O O–H mass of CO 0 –1 /kJ mol 414 499 745 464 ∆H = Molar mass of CO To Find : 12 ∆H0 = 28 Solution : = 0.43 moles ∆H0 So, heat evolved when 0.43 moles of CO reacts = 0.43 × 299.5 = +128.8 kJ 12)

= ∑∆H 0(reactant bonds) – ∑∆H0(product bonds) = [4∆ ∆H0(C–H) + ∆H0(O–O)] ∆H0 (C=O)] –[2∆ ∆H0(C–H) +∆ + 2(O–H) = [2 ∆H0 (C–H) + ∆H0 (O–O)] – ∆H0 (C=O) – 2 (O–H) = 2 × 414 + 499 – 745 – 2 × 464 = –346 kJ

38.55 kJ of heat is absorbed when 6.0 g of O 2 react with ClF according to the reation, Cl2O(g) + OF2(g) 2ClF(g) + O2(g) What is the standard enthalpy of the reaction ? 14) Calculate C–CI bond enthalpy from the Given : following data : Heat absorbed = 38.55 kJ CH3Cl(g) + Cl2(g) CH2Cl2(g) + HCl(g) = 6.0 g Mass of O2 ∆H0 = –104 kJ To Find : Bond C–H Cl=Cl H=Cl Standard enthalpy of reaction ∆H0/kJ mol–1 414 243 431 Solution : Given : Number of moles of O2 C–H Cl=Cl H=Cl mass of O 2 414 243 431 = Molar mass of O2 To Find : ∆H 0C–Cl 6 = Solution : 32 ∆H0 = ∑∆H 0(reactant bonds) = 0.1875 – ∑∆H0(product bonds) 38.55 kJ heat is absorbed when 0.1875 = [3∆ ∆H0(C–H) + ∆H0 moles react with ClF. (C–Cl) + ∆H ∆ 0(Cl–Cl)] 38.55 0 –[2∆ ∆H (C–H) + 2∆ ∆ H0 ∴ Heat absorbed for 1 mole = 0 0.1875 (C–Cl) +∆ ∆H (H–Cl)] = 205.6 kJ = [∆ ∆H0(C–H) – ∆H 0(C–Cl)] From reaction 2 moles of ClF reacts with + ∆H0(Cl–Cl)– ∆H0(H–Cl) 1 mole of O 2 so standard enthalpy of ∴ –104 = 414 – ∆H0(C–Cl) + 243 reaction is 205.6 kJ. – 431 ∆H0(C–Cl) = 414 + 104 + 243 – 431 = 300 kJ/mol Chemical Thermodynamcs and Energetics

.. 6 15)

Calculate the standard enthalpy of the reaction, C2H 6(g) from the 2C(graphite) + 3H2(g) following ∆H0 values : 7 O 2 2(g) 2CO2(g) + 3H2O(l), ∆H0 = –1560 kJ

i)

C2H6(g) +

ii)

1 H2(g) + O2(g) 2

H 2 O (l),

C(graphite) + O2(g)

∆H0 = –285.8 kJ CO 2(g), ∆H0 = –393.5 kJ

iii)

Given : Equation are, C2H6(g) + 2CO2(g)

1 H2(g) + O2(g) 2

H 2O (l),

C(graphite) + O2(g)

∆H0 = –285.8 kJ ...(ii) CO2(g), ∆H0 = –393.5 kJ ...(iii)

2CO2 ∆H0 = –787 kJ

2C(g) + 3H2(g)

C2H6(g)

∆ H 0 = 1560 + (–857.4) + (–787) = –84.4 kJ

16)

Given the following equation Calculate standard enthalpy of the reaction, 3 O Fe2O3(s) ∆H0 = ? 2 2(g) 2Al(s) + Fe2O3(s) 2 Fe(s) + Al2O(3)(s) ∆H0 = –847.6 kJ

2Fe(g) + i)

ii)

7 O 2 2(g) + 3H2O(l), ∆H ∆ 0 = –1560 kJ ...(i)

2C + 2O2

2Al(s) +

3 O 2 2(g)

Al 2 O (3)(s) ∆H0 = –1670 kJ

Given : Equation are, 2Al(s) + Fe2O3(s)

2Al(s) +

3 O 2 2(g)

2Fe(s) + Al2O(3)(s) ∆H0 = –847.6 kJ ...(i) Al2 O(3)(s) ∆H0 = –1670 kJ ...(ii)

To Find : ∆H0 Solution : Reverse equation (i), so that it becomes 2Fe(s) + Al2O3(s) 2Al(s) + Fe2O3(s) 7 ∆ H0 = 847.6 kJ ...(iii) C2H6(g) + O2(g) 2CO2(g) + 3H2O(l) 2 Now add equation (ii) to equation (iii) ∆H0 = 1560 kJ ...(iv) 2Al(s) + Fe2O3(s) 2Fe(s) + Al2O3(s) Multiply equation (ii) by 3 and (iii) by 2 ∆H0 = 847.6 kJ then add (iv), (ii) and (iii) 3 2Al(s) + O2(g) Al2 O(3)(s) 7 2 2CO2(g) + 3H2O(l) C2H6(g) + O2(g) 2 ∆H0 = –1670 kJ ∆H0 = +1560 kJ 3 3 2Fe(s) + O2(g) Fe2 O 3(s) 3H2(g) + O2(g) 3H 2 O (l), 2 2 ∆ H 0 = 847.6 + (–1670) ∆H0 = –857.4 kJ = 847.6 – 1670 = –822.4 kJ

To Find : ∆H0 Solution : Reverse equation (i)

Chemical Thermodynamcs and Energetics

.. 7 Given the following equation and ∆H0 values at 250C, i) Si(s) + O2(g) SiO 2(s) ∆H0 = –911 kJ ii) 2C(graphite) + O2(g) 2CO(g) ∆H0 = –221 kJ iii) Si(s) + C(graphite) SiC(s) ∆H0 = –65.3 kJ 0 Calculate ∆ H for the following reaction, SiC(s) + 2CO(g), SiO2(s) + 3C(graphite) Given : Equation are, Si(s) + O2(g) SiO2(s) ∆H0 = –911 kJ ...(i) 2CO(g) 2C(graphite) + O2(g) ∆H0 = –221 kJ ...(ii) Si(s) + C(graphite) SiC(s) ∆H0 = –65.3 kJ...(iii) To Find : ∆H0 Solution : Reverse equation (i), so that it becomes Si(s) + O2(g) SiO2(s) ∆H0 = 911 kJ ...(iv) Add equations (iv), (ii), (iii), SiO2(s) Si(s) + O2(g) ∆H0 = 911 kJ 2C(graphite) + O2(g) 2CO(g) ∆H0 = –221 kJ 17)

18) i) ii) iii)

Given the following equation and ∆H0 values at 250C, 2H3BO 3(aq) B2O3(s)+ 3H2O(l), ∆H0 = +14.4 kJ H3BO 3(aq) HBO2(aq) + H2O(l) ∆H0 = –0.02 kJ H2B4O 7(s)

2B2O3(s) + H2O(l) ∆H0 = 17.3 kJ 0 Calculate ∆ H for the following reaction, 4HBO2(aq) H2B4O7(s) + H2O(l) Given : Equation are, 2H3 BO 3(aq) B2O3(s)+ 3H2O(l), ∆H0 = +14.4 kJ ...(i) HBO2(aq) + H2O(l) H3BO 3(aq) ∆H0 = –0.02 kJ ...(ii) H2 B4 O7(s) 2B2O3(s) + H2O(l) ∆H0 = 17.3 kJ...(iii) To Find : ∆H0 Solution : Reverse equation (i) and multiply by (2) 4H3 BO 3(aq) 2B2O3(s)+ 6H2O(l) ∆H0 = 28.8 kJ ...(iv) Multiply equations (ii) by 4 4H3 BO 3(aq) 4HBO2(aq)+ 4H2O(l) ∆H ∆ 0 = (–0.08) kJ ...(v) Now add equation (iv), (v) and (iii) 2B2O3(s)+ 6H2O(l) 4H3BO3 ∆H0 = 28.8 kJ Si(s) + C(graphite) SiC(s) 4H3 BO 3(aq) ∆H ∆ 0 = –65.3 kJ 4HBO2(aq) + 4H2O(l) ∆H0 = (–0.08) kJ SiO2(s) + 3C(graphite) SiC(s) + 2CO(g) H2 B4 O7(s) 2B2O3(s) + H2O(l) ∆H0 = 17.3 kJ ∆ H 0 = 911 + (–211) + (–65.3) H2B4O7(s)+ H2O(l) 4HBO2(aq) = 624.7 kJ ∆ H 0 = –28.8 + (–0.08) + 17.3 = –11.58 kJ

Chemical Thermodynamcs and Energetics

.. 8 19)

Calculate Kp for the reaction, C2H4(g) + H2(g) C2H 6(g), 0 –1 ∆G = –100 kJmol at 250C Given : ∆ G 0 = –100 kJ/mol = –100000J/mol T = 250C = 298 K To Find : Kp Formula : ∆ G 0 = –2.303RT log10 Kp Solution : ∆G0 = –2.303RT log10 Kp ∴ –100000 = –2.303 × 3.314 × 298 × log10 Kp ∴ ∴ ∴ ∴ ∴

100000

= 5705.85 log10 Kp

100000 = log10 5705.85 Take Antilog on Anti [17.5258] = 3.35 × 1017 = Kp =

Calculate ∆G0 for the reaction at 250C CO(g) + 2H2(g) CH3OH (g) 0 –1 ∆G = –24.8 kJmol If PCO = 4 atm, PH = 2 atm, PCH OH = 2 atm 2 3 Given : 0 ∆ G = –24.8kJ/mol T = 250C = 298 K PCO = 4 atm PH = 2 atm 2 PCH OH= 2 atm 3 To Find : ∆G Formula : ∆ G = ∆G0 + 2.303RT log10 Kp Solution : ∆ G = ∆G0 + 2.303RT log10 Kp 21)

PCH3OH

Kp

∴

QP

=

both sides Kp Kp 3.35 × 1017

∴

QP

=

∴

∆G

= –24.8 + [2.303 × 8.314 × 10–3 × 298 × log(0.125)] = –24.8 + [–5.153] = –24.8 – 5.153 = 29.95 kJ

20)

Kp for the reaction, MgCO3(s) MgO(s) + CO2(s) is 9 × 10–10. Calculate ∆G0 for the reaction at 250C. Given : Kp = 9 × 10–10 T = 250C = 298 K To Find : ∆G0 Formula : ∆ G 0 = –2.303RT log10 Kp Solution : ∆ G 0 = –2.303RT log10 Kp ∴ ∆ G 0 = –2.303 × 8.314 × 298 × log9 × 10–10 = 51,613.96 J/mol = 51.61 kJ

2 4 × (2)

22)

2

=

2 16

= 0.125

Calculate ∆ S (tota l) and hence show whether the following reaction is spontaneous at 250C. HgS(s) + O2(g) Hg(l) + SO2(g) 0 ∆H = –238.6 kJ and ∆S0 = +36.7 JK–1 Given : T = 250C = 298 k 0 ∆ H = –238.6 kJ ∆ S0 = +36.7 JK–1 To Find : ∆ S(total) Formula : ∆Ssurr =

Chemical Thermodynamcs and Energetics

PCO × PH2 2

– ∆H0 T

.. 9 Solution :  –238.6kJ  – ∆H 0 =   T  298k  = 0.8006 kJ/K = 800.6 J/K ∆ Ssys = ∆ S0 = +36.7 J/K ∆S(total)= ∆Ssys + ∆Ssurr = 36.7 + 800.6 = 837.3 J/K As ∆S(total) is positive, the reaction is spontaneous at 298 K

= 50 + 32.5 = 82.5 kJ As ∆G is positive, the reactions is nonspontaneous and the positive value ∆ H indicates that the reaction is endothermic.

∆Ssurr =

∴

For the certain reaction, ∆H0 = –224 kJ and ∆S0 = –153 JK–1. At what temperature will it change from spontaneous to nonspontaneous ? Given : ∆ H 0 = –224 kJ ∆ S0 = –153 JK–1 23) Determine whether the reaction ∆H and = –0.153 kJ/K ∆ S values are spontaneous or To Find : nonspontaneous. State whether they are Temperature (T) = ? exothermic or endothermic. Formula : ∆H = –110 kJ and ∆S = + 40JK–1 i) at 400 K ∆H0 ii) ∆H = –50 kJ and ∆S = – 130 JK–1 at 400 K T = ∆S 0 Given : Solution : ∆H = –110 kJ and ∆S = + 40JK–1 at 400 K ∆H0 ∆H = –50 kJ and ∆S = – 130 JK–1 at 400 K T = ∆S 0 To Find : To state whether the reactions are –224 exothermic or endothermic and = 0.153 spontaneous or nonspontaneous. = 1464.05 K Formula : 0 As ∆ H and ∆ S 0 both are negative, ∆ G = ∆H – T∆ ∆S reaction is spontaneous at lower Solution : temperature. So the reaction will be i) ∆ G = ∆H – T∆ ∆S spontaneous below 1464.05 K and ∆ S = +40 J/K = 0.04 kJ/K nonspontaneous above 1464.05 K. The ∆ H = –110 kJ change over between spontaneous and ∴ ∆ G = –110 – (400 × 0.04) nonspontaneous occurs at 1464.05 K = –110 –16 24)

= –126 kJ Because ∆ G is negative, the 25) reactions is spontaneous and the negative value ∆ H indicates that the reaction is exothermic. ii)

∴

∆G ∆H ∆S ∆G

= = = =

∆H – T∆ ∆S +50 kJ –130 J/K = –0.130kJ/k 50 – (250 × –0.130)

Determine whether the following reaction is spontaneous or nonspontaneous under standard condition ? Zn(s) + Cu2+(aq) Zn2+ + Cu(s) ∆H0 = – 219 kJ, ∆S0 = – 21 JK–1 Given : ∆ H 0 = – 219 kJ ∆ S0 = – 21 JK–1 T = 298 K Chemical Thermodynamcs and Energetics

.. 10 To Find : ∆G0 Formula : ∆ G 0 = ∆H0 – T∆ ∆ S0 Solution : ∴ ∆ G 0 = ∆H0 – T∆ ∆ S0 = –219 – (298 × –0.021) = –219 – (–6.258) = –219 + 6.258 = –212.74 kJ As ∆ G 0 is negative the reaction spontaneous.

Formula : W = – P∆ ∆V Solution : W = – P∆ ∆V = –1 × (22.4) = – 22.4 L atm = – 22.4 × 101.33 J = –2269.8 J Ans : W = –2.27 kJ ∆ H = ∆U + P∆ ∆V ∴ –P∆ ∆V = –2.27 P∆ ∆ V = 2.27 ∴ ∆ U = ∆H – P∆ ∆V = –2043 – 2.27 26) Determine whether the following = –2045.27 kJ reaction is spontaneous under standard ∴ ∆U = –20445.27 kJ conditions. 2H 2O 2(l) 2H2O(l) + O2(g) 28) What is the value of ∆ S surr for the ∆H0 = + 196 kJ, ∆S0 = – 126 JK–1 following reaction at 298 K ? Does it have a cross-over temperatuer ? 6CO2(g) + 6H2O(l) C6H 12O 6(s) Given : ∆ H 0 = 196 KJ, ∆ S0 = –126 J/K + 6O2(g), Solution : 0 –1 ∆G = 2879 kJ mol ∆ G 0 = ∆H0 – T∆ ∆ S0 ∆S0 = – 210 JK–1mol–1. = 196 – (298 × –0.126) Does it have a cross-over temperatuer ? = 196 + (37.55) Given : = 233.55 kJ ∆ G 0 = 2879 kJ mol–1 As ∆ G0 is positive the reaction will be ∆ S0 = – 210 JK–1mol–1 nonspontaneous. As ∆H0 is positive and = –0.210 kJ/K/mol. As ∆S0 is negative. ∆G 0 will be always T = 298 K positive regardless of temperature. The To Find : reaction is therefore nonspontaneous at ∆ Ssurr all the temperature. So it will not have Formula : cross-over temperature. ∆ G 0 = ∆H0 – T∆ ∆ S0 27) Oxidation of propane is represented as Solution 0: ∆ G = ∆H0 – T∆ ∆ S0 3CO2(g) + 4H2O(g), C3H8(g) + 5O2(g) 0 2879 = ∆H – (298 × –0.210) ∆H0 = – 2043 kJ. 2879 = ∆H0 – (–62.58) How much pressure volume work is ∆ H 0 = 2816.42 kJ done and what is the value of ∆ U at 2816.42 constant pressure of 1 atm when volume ∆ Ssurr = 298 change i s+22.4 L. Given : = 9.45 kJ/K P = 1 atm V = +22.4 L To Find : Work done (W) Value of ∆U Chemical Thermodynamcs and Energetics

.. 11 29)

Calculate the enthalpy change for the reaction.

H2O (g) 2H(g) + O(g) And hence, calculate the bond enthalpy of O – H bond H2O from the following data : ∆vapH(H 2O) = 44.0 kJ mol–1 ∆fH(H 2 O) = –285.8 kJ mol–1 = 436.0 kJ mol–1 ∆ aH(H 2 ) = 498.0 kJ mol–1 ∆ H(O 2 ) where ∆ a H is the enthalpy of atomization Given : ∆vapH(H2O) = 44.0 kJ mol–1 ∆fH(H 2O) = –285.8 kJ mol–1 = 436.0 kJ mol–1 ∆ aH(H 2 ) ∆H(O 2 ) = 498.0 kJ mol–1 To Find : ∆H(O–H) ∆H Solution : Reaction can be given as follows H 2O (l) H2O(g) ∆H = 44.0 kJ mol ...(i) H2(g) +

1 O 2 2(g)

H2O

H2O(g)

2H(g) + (O)(g)

∆H

∴

= –44 + 285.8 + 436 + 249 = 926.8 kJ ...(i) Bond enthalpy of O–H bond in H2O can be calculated as follows : ∆ H = Σ (∆ ∆H of reactants) – Σ (∆ ∆H of products) = [2∆ ∆H (O–H)] – [O] ∆ H = 926.8 kJ ... from(i) 926.8 = 2 ∆H (O–H)

∴

∆H (O–H) =

926.8 2 = 463.4 kJ/mol

30)

Calculate ∆ S surr when one mole of methanol (CH3OH) is formed from its elements under standard conditions if ∆fH0(CH3OH) = –238.9 kJ mol–1. Given : ∆H 0( CH

3OH )

= –238.9 KJ/mol

To Find : ∆ Ssum Solution :

∆H = –285.8 kJ mol ...(ii) H2(g) 2H (g) ∆H = 436 kJ mol ...(iii) 2(O)(g) O2(g) ∆H = 498 kJ mol ...(iv) Reverse equation (i) and (ii) and add ∴ them to equation (iii) and (iv) H 2O (l) H2O(g) ∆H = –44.0 kJ/mol 31) 1 H 2O (l) H2(g) + O2(g) 2 ∆H = 285.8 kJ mol H2(g) 2H (g) ∆H = 436 kJ mol 1 O 2 2(g)

498 kJ mol . 2

∆H =

∆H0 T ∆ fH 0 = –238.9 kJ mol = –238900 J/mol

∆Ssurr =

 238900  ∆ Ssurr = –  –  298   = 801.7 J/K

Calculate the total heat to melt 180g of ice at 0 0 C, heat it to 100 0 C and then vaporize it at that temperature. ∆fusH(ice) = 6.01 kJ mol–1 at 00C, ∆vapH(H2O) = 40.7 kJ mol–1 at 1000C. Specific heat of water = 4.28 J g–1 K–1.

2 2(O)(g) 2 Chemical Thermodynamcs and Energetics

.. 12 32)

Given : Wice ∆fusH(ice)

= 180 g = 6.01 KJ/ mol

∆ vap H( H O ) = 40.7 KJ/mol 2

sp. heat of H2O = 4.28 Jg–1k–1 To Find : Heat required to melt ice. Solution : –H2O(g) Latent heat 00 C

of fusion

H2O(l) 0

H2O(l) Heating

6.24 g of ethanol are vaporized by supplying 5.89 kJ of heat energy. What is enthalpy of vaporization of ethanol ? Given : Wethanol = 6.249 q = 5.89 KJ To Find : ∆vapH Solution : Number of moles of ethanol

Mass of ethanol = Molar mass of ethanol

00 C

Latent heat of vaporization

100 C

H2O(g)

Part I : H2O(g) H 2O (l) Heat required = Latent heat for 180 g. 1 mol of H2O = 6.01 kJ 1 mol of H2O = 18 g ∴ 180 g = ? moles ∴ 180 g of H2O = 10 moles of H2O ∴ 10 mol of H2O = 60.1 kJ ∴ Heat required = 60.1 kJ ...(i) Part II : H 2O (l) H 2O (l) 00 C 00 C Heat required = mass × Specific heat × ∆T = 180 × 4.18 × 100 = 75240 J = 75.240 kJ ...(ii) Part III : H 2O (l) H2O(g) 1000C 1000C Heat required = Latent heat of vaporization 1 mol of H2O = 40.7 kJ ∴ 1 mol of H2O = 18 g ∴ 180 g of H2O = 10 moles of H2O ∴ Heat required = 407 kJ ...(iii) From (i), (ii) and (iii) Heat required to melt ice = 60.1 + 75.240 + 407 = 542.34 kJ Heat required to melt ice = 542.34 kJ Chemical Thermodynamcs and Energetics

6.24 46 = 0.1356 moles 5.89 kJ heat is requiered to heat 0.1356 moles. Then heat required to vaporize 1 mole will be,

1000C

=

5.89 0.1356 = 43.44 kJ

=

∴

Enthalpy of vaporization of ethanol = 43.44 kJ

Enthalpy of fusion of ice is 6.01 kJ mol–1. The enthalpy of vaporization of water is 45.07 kJ –1 . What is enthalpy of sublimation of ice ? Given : ∆ fusH = 6.01 kJ/mol ∆vapH = 45.07 kJ/mol 33)

To Find : ∆ sub H Formula : ∆ sub H = ∆fusH + ∆vapH Solution : ∆ sub H = ∆ fusH + ∆vapH ∴ ∆ sub H = 6.01 + 45.07 = 51.08 kJ/mol 51.08 kJ