Answers to Problem Sheet 5 1. (a) spontaneous (b) nonspontaneous (c) nonspontaneous (d) spontaneous (e) nonspontaneous 2. (a) Heat will flow from the warmer block of iron to the colder block of iron until the maximum number of microstates of the two blocks is achieved. Suniverse > 0. (b) The thermal energy of the gas is constant, but the thermal energy is dispersed over a larger volume of space. Ssystem > 0. (c) The thermal energy is increased and the number of microstates of the water is increased. Ssystem > 0. (d) After mixing, the gas expands to twice its volume. The thermal energy is dispersed over a larger volume. Ssystem > 0. 3. (a) This is an adiabatic free expansion of an ideal gas. The internal energy does not change. The temperature does not change. The final temperature is 298K. (b) q=0 (adiabatic), w = 0 (free expansion), E=0 (no temperature change), H=0 (no temperature change), S = qrev/T We need to find qrev , therefore, we need to expand the gas using a reversible path. The path is an isothermal expansion following the curve as shown in the PV graph below. wrev = ‐nRT ln(3) = ‐(1)(8.314)(298)ln(3)=‐2722 J. qrev = 2722 J S = 2722/298 =9.13 J/K 4. 10
8.314 273 ln 10
5.2 5. Isothermal condition, E=H=0 q=‐w .
.
43 q = 43 J 0.14 JK
S
6. w = ‐PextV = ‐(1) (50‐24.6) = ‐25.4 Latm = ‐2.57 kJ E = Cv T = 3/2RT = 3/2PV = (3/2)(1)(50‐24.6)=38.1 Latm =3.86 kJ q = E ‐ w = 3.86 ‐ (‐2.57) = 6.43 kJ H = q = E+PV = 3.86+2.57=6.43 kJ 7. (a) Nitrogen is an ideal gas. The heat required is p∆
5
1.98 120
30 (b) The increase of the internal energy is ∆
∆
5 120
21 (c) w = E‐q = 21‐30 = ‐9 kcal (d) If it is a process of constant volume, the required heat is q = ncvT=21 kcal (=E) 8. (a) (i) negative (ii) positive (iii) positive (iv) positive (b) (iv) 9. ∆
∆
. .
Hvap = S  Tb.p.= 88  (96+273) = 3.2x104 Jmole‐1 10. ∆
∆
20.5 . .
∙
11. Hvap = (86 cal g‐1)(84 gmole‐1)=7224 cal mole‐1 ∆
∆
. .
.
20.4 ∙
12. 92 cal g‐1 = 384.9 J g‐1 Hvap =88426=3.75 x 104 J mole‐1 Molar mass = 3.75 x 104/384.9 = 97.4 g mole‐1 13. Estimate is too low since strong H bonding makes the liquid state more ordered. Therefore, requires more energy to vapourize. 1 8.314
14. ∆
10
19.1 15. (a) So = So(AgI (s)) + So(HCl(g)) ‐ So(AgCl(s))‐ So(HI (g)) = (115.5+186.8)‐(96.2+206.5)= ‐0.4 JK‐1 (b) So =2 So(AgCl(s))‐ 2 So(Ag (s)) ‐ So(Cl2 (g)) = 296.2 ‐ (242.6) ‐165.2= ‐58.0 JK‐1 (c) So =2 So(Fe(s)) + 3 So(CaO(s)) ‐ So(Fe2O3(s)) ‐3 So(Ca(s)) = (2 27.8 + 362.3) ‐ (87.4 + 341.6) = 30.3 JK‐1 16. Devise a reversible path S H2O (s, 1 atm, ‐10oC)  H2O (l, 1 atm, ‐10oC) cooling warming S3 S1 S2
o
H2O (s, 1 atm, 0oC)  H
2O (l, 1 atm, 0 C) li
S = S1 + S2 + S3 ∆
8.7 ln
∆
1440
273
∆
18 ln
0.33 5.28 0.67 S = 0.33 + 5.28 ‐ 0.67 = 5.0 cal K‐1 17 . Devise a reversible path S H2O (l, 1 atm, 125oC)  H2O (g, 1 atm, 125oC) cooling warming S1 S3 S2
o
H2O (l, 1 atm, 100oC)  H
2O (g, 1 atm, 100 C) li
S = S1 + S2 + S3 ∆
18 ln
1.17 ∆
9713
373
∆
7.256 ln
0.47
26.0 0.05
398
2.298 10
0.0028
0.52 373
.
398
S = ‐1.17 + 26.0 + 0.52 = 25 cal K‐1 18. Devise a reversible path S o
H2O (l, 3 atm, 125oC)  H2O (g, 1 atm, 100
C) cooling S1 vapourization
S3 S2
o
H2O (l, 3 atm, 100oC)  H
2O (l, 1 atm, 100 C) li
S = S1 + S2 + S3 1 18 ln
∆
1.17 ∆
≅ 0 since water is a liquid ∆
26.0 S = ‐1.17 + 26.0 = 25 cal K‐1 19. (a) Free expansion, no heat absorbed or given off. Ssurroundings = 0 (b) ∆
ln
∆
0.97
0.5 ∙ 1.98 ln
20. (a) (b) Go = Ho ‐TSo 0 = ‐58 x 103 ‐T(‐14) T = 4143 K Go < 0 spontaneous when T < 4143 K .
0.97
373 21. (a) (b) Go = Ho ‐TSo 0 = 25 x 103 ‐T(69.36) T = 360 K Go < 0 spontaneous when T > 360 K 22. Formation of a double bond requires input of energy. H > 0. Formation of gaseous species means S > 0. Reaction is spontaneous at high temperatures. 23. (a) The change in entropy of the water is ∆
1000 ∙ 4.18 ∙ ln =1305 JK‐1 (b) The change in entropy of the heat source is S
1305 JK
Suniverse = Swater +Ssurroundings = 0 24. (a) Calculate Go for the equilibrium. ΔH° = (∑ ΔHf°,products) – (∑ ΔHf°,reactants) = (–285.8) – (–230) = –55.8 kJ mol–1 ΔS° = (∑S°,products) – (∑S°,reactants) = (70.0) – (–10.9) = +80.8 J K–1 mol–1 Go = Ho ‐ TSo = ‐55.8 ‐ 298(0.0808)=‐79.9 kJ mol–1 Go < 0, spontaneous (b) Go =‐RT ln K ln K = Go/RT = ‐(‐79.9 x 103/(8.314298))= 32.2 K = 1.0 x 1014 The reverse is the autoionization equilibrium constant for water at 25oC. (Kw=1.0 x 10‐14) 25. (a) ΔGo = 2ΔGfo(Al(s)) + 3ΔGfo(H2O(l)) ‐ [ΔGfo(Al2O3(s)) + 3ΔGfo(H2(g))] = 870.4 kJ mol−1 (Reaction is not spontaneous at 25oC and 1 atm) (b) ΔGo = 4ΔGfo(NO(g)) + 6ΔGfo(H2O(l)) ‐ [4ΔGfo(NH3(g)) + 5ΔGfo(O2(g))] = ‐1009 kJ mol−1 (Reaction is spontaneous at 25oC and 1 atm) 26. (a) Go = Ho ‐TSo Evaluate So: So = So((NH2)2CO(s)) + So(H2O(l)) ‐ [So(CO2(g)) + 2So(NH3(g))] = 104.6 + 70.0 ‐ 213.6 ‐ (2192.5) = ‐424 JK‐1 Evaluate Ho: Ho = Hf o((NH2)2CO(s)) + Hf o(H2O(l)) ‐ [Hf o(CO2(g)) + 2Hf o(NH3(g))] = ‐333.2 + (‐285.9) + 393.5 +(246.2) = ‐133.2 kJ Go = ‐133.2 ‐ 298(‐0.424) = ‐6.85 kJ Since Go < 0 it is spontaneous under standard condition. Side note: You could find Go by this equation: Go =  nprod Gf o(prod) ‐  nreactGf o(react) = Gfo ((NH2)2CO(s)) + Gfo (H2O(l)) ‐ [Gf o(CO2(g)) + 2Gf o(NH3(g))] From the table,  Gfo (H2O (l)) and Gfo (NH3(g)) are available.  Gfo (H2O(l)) = ‐237.13 kJ/mole  Gfo (NH3(g)) = ‐16.64 kJ/mole Need to calculate: 
Gfo ((NH2)2CO(s)) from: N2(g) + 2H2(g) + C (s) + 1/2 O2 (g)  (NH2)2CO(s) o
S (Joules mole‐1K‐1) 191.61 130.68 5.74 205.14 104.6 So = 104.6 ‐[191.61+2(130.68)+5.74+0.5(205.14)] = ‐456.7 JK‐1 Gfo ((NH2)2CO(s)) = Hfo ((NH2)2CO(s)) ‐ TSo ((NH2)2CO(s)) = ‐333.2 x 103 ‐ (298)(‐456.7) = ‐197.1 kJ 

Gfo (CO2(g)) from: C (s) + O2 (g)  CO2 (g) o
S (Joules mole‐1K‐1) 5.74 205.14 213.6 So = 213.6 ‐ (5.74 + 205.14) = 2.72 JK‐1 Gfo (CO2(g)) = ‐393.5 x 103 ‐ (298)(‐2.72) = ‐394 kJ Gfo ((NH3(g)) can also be calculated from: 1/2N2(g) + 3/2H2(g)  NH3 (g) o
S (Joules mole‐1K‐1) 191.61 130.68 192.5 So = 192.5 ‐ [( /2)(191.61) + ( /2)(130.68) ]= 192.5 ‐ 291.8 = ‐99.3 JK‐1 1
3
Gfo (NH3(g)) = Hfo (NH3(g)) ‐ TSo (NH3(g)) = ‐46. 2 x 103 ‐ (298)(‐99.3) = ‐16.3 kJ Go = Gfo ((NH2)2CO(s)) + Gfo (H2O(l)) ‐ [Gf o(CO2(g)) + 2Gf o(NH3(g))] = ‐197.1 kJ + ‐237.13 ‐ [‐394 kJ + 2(‐16.64)] = ‐7.63 kJ (b) Cross‐over temperature 0 =  Ho ‐TSo T = ‐133.2 / ‐0.424 = 314.2 K When the temperature is greater than 314.2 K, the reaction is nonspontaneous. 27. ΔGo = ΔGfo(Zn2+(aq)) + ΔGfo(Cu(s)) ‐ [ΔGfo(Zn(s)) + ΔGfo(Cu2+(aq))] = ‐147.06 ‐ 65.49 = ‐212.6 kJ ΔGo = ‐212.6 =‐ nFEo Eo = ‐ 212.6 x 103 / ‐2 (96500) = 1.102 V Eo > 0; spontaneous reaction. This is a voltaic cell. 28. Reaction 1: Ag+ + e‐  Ag (s) Eo =0.7991 V ΔGo = ‐ nFEo = ‐ (1) (96500) (0.7991) = ‐ 7.71 x 104 J ΔGo = ‐RT lnK = ‐8.314298ln(1.56 x 10‐10) = 5.60 x 104 J Reaction 1: Ag+ + e‐  Ag (s) Reaction 2: AgCl (s)  Ag+ (aq) + Cl‐ (aq) AgCl (s) + e‐  Ag (s)+ Cl‐ (aq) ΔGo = ‐ 7.71 x 104 + 5.60 x 104 = ‐2.11 x 104 J ΔGo =‐ nFEo Eo = 2.11 x 104 / 196500 = 0.22 V Reaction 2: Ag (s)  Ag+ (aq) + Cl‐ (aq) Ksp = 1.56 x 10‐10 The low potential makes this desirable as a reference electrode. This is the silver chloride electrode that is used in the combination pH electrode. 29. ΔGo = ΔGfo(Pb(s)) + ΔGfo(PbO2(s)) + 2ΔGfo(H2SO4(aq))‐[ 2ΔGfo(PbSO4(s)) + 2ΔGfo(H2O(l))] ΔGo = ‐217.33 + 2(‐744.52) ‐ 2(‐813.14) ‐ 2(‐237.13) = 394.15 kJ ΔGo =‐ nFEo The reaction involves 2 electrons. n=2 Eo = ‐(394.15 x 103)/2(96500) = ‐2.042 V ΔGo > 0 and Eo < 0, the charging reaction is not spontaneous.