Calculate 1) ?G for the decomposition of liquid water, H 2 O (l), to hydrogen ga
ID: 1021855 • Letter: C
Question
Calculate
1) ?G for the decomposition of liquid water, H2O (l), to hydrogen gas and oxygen gas at the following 5 temperatures:
2) Equilibrium constant, Kc, for the decomposition of liquid water, H2O (l), to hydrogen gas and oxygen gas at the following 5 temperatures:
1) 25.0 0 C ?G = ? and equilibrium constant = ?
2) 927 0 C ?G = ? and equilibrium constant = ?
3) 1427 0 C ?G = ? and equilibrium constant = ?
4) 16270 C ?G = ? and equilibrium constant = ?
5) 18270 C ?G = ? and equilibrium constant = ?
Recommendations: Tabulate and plot these values
Rationalize very briefly why Kc values are not constant at these temperatures.
A-10 APENDIX B THERMODYNAMIC PROPER AMIC PROPERTIES AT 25 (kJ/mol) TABLE B.1 Inorganic Substances (continued) -1015 -166.9 and State (kJ/mol) Uk 162.2 396.6 -1140 174.4 Cr203(s) 1058 -187.9 -217.4 PbCl2(s) 424.7 237.9 179.4 CoO(s) -625.5 -813.2 -214.2 PbO2(s) PbCOa(s) Pb50,(s) 337.4 64.8 137.2 297.7 166.3 CuCI(s) CuCl2(s) 126.6 -293.3 -587 7 Lithium 175.7 -129.7 -146.0 108.1 -157.3 LiF(s) LiCI(s) LiBr(s) 342.0 -270.3 561.2 7 Cu2S(s) CuSO4(s) 120.9 109.2 597.9 487.5 158.7 LiOH(s) -278.8 202.7 Magnesium 273.3 Mg(s) 112.5 -591.8 147.1 148.6 Hydrogen MgCl2(s) MgO(s) MgCOs(s) 1096 H2(g) 130.6 0 230.0 -1284.9 1170.6 -157.3 237.2 H2O() H20(g) H2020) H202(g) H2O2(a) Manganese -241.8 188.7 Mn(s) -120.4 -105.6 280.7 385.2 520.0 136.3 238.5 191.2 465.1 106.8 -55.2 180.7 116.1 12(g) 31.85 174.8 82 164.4 153.5 -178.6 210.7 206.5 82C Hg0(s) 416.3 FeCl2(s) FeCl3(s) FeO(s) Fe2O3(s) -399.5 118.0 70.3 824.2Explanation / Answer
G = H - TS
G = -2.303 R T log [Kc] { R = 8.314 J/K.mol)
2H2(g) + O2 (g) <=====> 2H2O(l)
from the table H = -285.8 KJ/mol and S = 69.9 J/K.mol
1. G at 25.0 C (25 +273 = 298 K) temperature
G = -285800 J/mol - (298 K x 69.9 J/K.mol)
G = -306630.2 J/mol = -306.63 KJ/mol
for reverse of the previous reaction
2H2O(l) <====> O2 (g) + 2H2 (g)
G = 306630.2 J/mol = 306.63 KJ/mol
G = -2.303 R T log [Kc]
306630.2 J/mol = -2.303 x 8.314 x 298 log[Kc]
Kc = 1.82 x 10-58
2. at 927.0 C = 1200 K
G = -285800 J/mol - (1200 K x 69.9 J/K.mol)
G = -369680 J/mol = -369.680 KJ/mol
2H2O(l) <====> O2 (g) + 2H2 (g)
G = 369680 J/mol = 369.68 KJ/mol
G = -2.303 R T log [Kc]
369680 J/mol = -2.303 x 8.314 x 1200 x log[Kc]
Kc = 8.139 x 10-17
3. at 1427.0 C = 1700.0 K
G = -285800 J/mol - (1700 K x 69.9 J/K.mol)
G = -404630 J/mol = -404.63 KJ/mol
2H2O(l) <====> O2 (g) + 2H2 (g)
G = 404630 J/mol = 404.63 KJ/mol
G = -2.303 R T log [Kc]
404630 J/mol = -2.303 x 8.314 x 1700 x log[Kc]
Kc = 3.7 x 10-13
4.at 1627.0 C = 1900 K
G = -285800 J/mol - (1900 K x 69.9 J/K.mol)
G = -418610 J/mol = -418.61 KJ/mol
2H2O(l) <====> O2 (g) + 2H2 (g)
G = 418610 J/mol = 418.61 KJ/mol
G = -2.303 R T log [Kc]
418610 J/mol = -2.303 x 8.314 x 1900 x log[Kc]
Kc = 3.11 x 10-12
5. at 1827.0 C = 2100 K
G = -285800 J/mol - (2100 K x 69.9 J/K.mol)
G = -432590 J/mol = -432.59 KJ/mol
2H2O(l) <====> O2 (g) + 2H2 (g)
G = 432590 J/mol = 432.59 KJ/mol
G = -2.303 R T log [Kc]
432590 J/mol = -2.303 x 8.314 x 2100 x log[kc]
Kc = 1.7 x 10-11