Chapter 10
10.1
ANS From an environmental perspective, the recycling process involves the input of
energy and often other resources that may have an environmental cost.
10.13
ANS
(a) This process is not spontaneous at high temperatures.
(b) This process is not spontaneous at very high temperatures.
(c) This process is always spontaneous.
(d) This process may not be spontaneous at high temperatures.
Detail ANS
(a) Because gas molecules have higher entropy than liquids and solids, the entropy
change of the reaction will be negative if the numbers of gas molecules in the
products are less than that in the reactants.
(b), (c), (d) The change of entropy is negative.
(the molar number of gas molecules in products) – (the molar number of gas
molecules in reactants) < 0
(e) The change of entropy is positive.
(the molar number of gas molecules in products) – (the molar number of gas
molecules in reactants) > 0
Also, it is incorrect if one just considers the entropy change of the system.
10.47 ANS
(a) Fe(s) + 2 HCl(g) Æ FeCl2(s) + H2(g)
∆So = 117.9 + 130.6 – (27.30 + 2×186.8) = -152.4 J/K
(b) 3 NO2(g) + H2O(l) Æ 2HNO3(l) + NO(g)
∆So = 2×155.6 +210.7 – (3×240.0 + 69.91) = -268.0 J/K
(c) 2K(s) + 2 Cl2(g) Æ 2 KCl(s)
∆So = 2×82.60 + (2×223.0 + 2×63.60) = -185.0 J/K
(d) Cl2(g) + 2 NO(g) Æ 2 NOCl(g)
∆So = 2× 264 – (2×210.7 + 223.0) = -116 J/K
(e) SiCl4(s) Æ Si(s) + 2Cl2(g)
∆So = 2×223.0 + 18.80 – (330.6) = -134.2 J/K
These results are consistence with 10.28.
10.49 ANS
∆So = -114.6 J/K. This negative value for entropy change arises from the fact that
water molecules solvate ions in solution. In other words, solvation creates ordering
structure in the solution and decreases the entropy.
10.64 ANS
NO(g) + NO2(g) Æ N2O3(g)
(a)
∆Ho = 83.72 – (33.2 + 90.25) = -39.7 kJ
∆So = 321.8 – (240.0 + 210.7) = -128.9 J/K
∆Go =∆Ho – T ∆So = -39.7- 298×(-128.9) = - 1.29 kJ.
or∆Go = ∆Go(N2O3(g)) – (∆Go(NO(g) +∆Go(NO2(g))) = 139.46 – (51.30 + 86.57) =
-1.59 kJ.
It is a “Spontaneous reaction”.
(b) The reaction will be more spontaneous in low temperature than that in high
temperature.
(c) At 273 K, the∆G273K =∆Ho – T ∆So = -39.7- 273×(-128.9) = - 4.51 kJ <∆Go
For example, the ∆Go of (c) is 3×∆Go(Fe(s)) + 4×∆Go(CO(g)) – (∆Go(Fe3O4(s) +
4×∆Go(CO (g)) = (0×3 + -(394.4×4)) – (-1015 + (-137.2×4)) = -13.8 KJ
10.83 ANS
Ga(l) Æ Ga(g)
The Gibbs free energy is 0 at boiling temperature.
0 = ∆H – T∆S = (271.96 – 5.578) – T(169.03 – 59.25)×(10-3)
266.38 = T(109.78)x(10-3), Tbp = 2426.5 K = 2153.5 oC.
Detail ANS
(b) the ∆Go of CO(g) is -169.4 kJ/mol, calculated through (∆Ho of CH3OH(l) – T×(∆So
of CO(g)) = -110.5 - 298.15×197.674 = -169.4 kJ/mol.
(c) ∆S of O2(g) is 205 J/K, calculated through (+162 J/K,∆So of reaction) = (186.2 ,
∆So of CH4(g)) + 0.5 x (∆So of O2) – (126.8, ∆So of CH3OH(l)).
(d) The ΔGo of reaction #1 is -29.26 kJ. It is a spontaneous reaction.
ΔG > 0 non-spontaneous, ΔG ~ 0 became a spontaneous reaction -128.2 kJ x – T×
(-0.332 kJ/K), then T = 385 K.
(e) The reaction #2 is spontaneous at T > 1012 K. Hence, the reaction would be
carried out in two steps.
The ΔGo of reaction #2 is spontaneous when (-74.8 + 238.7 ) - T×(-162)×(10-3) <0 ,
then T > 1012 kJ.
10.95 ANS
(a) MnO is manganese(II) oxide. MnO2 is manganese(IV) oxide. Mn2O3 is
manganese(III) oxide.
(b)
(1) 6 MnO(g) + O2(g) Æ 2 Mn3O4(g)
(2) 3 MnO2(g) Æ O2(g) + Mn3O4(g)
(2) 6 Mn2O3(g) Æ O2(g) +4 Mn3O4(g)
(c)
(1) ΔGo = 2×(-1283) - 6×(-362.9) = -2566 + 2177 = -388.6 kJ.
(2) ΔGo = (-1283) - 3×(-465.2) = 112.60 kJ.
(3) ΔGo = 4×(-1283) - 6×(-881.2) = - 5132 + 5287.2 = 155.2 kJ.
Hence, MnO2 and Mn2O3 are more stable than Mn3O4 at RT.
Considering 2 MnO2(g) Æ 1/2 O2(g) + Mn2O3(g)
ΔGo = (-881.2) - 2×(-465.2) = +49.2 kJ
Therefore, MnO2 is the most stable compound among these three manganese oxide
compounds at RT.