已知25℃时液态水的△G=-237.19kJ/mol,

这个要设计过程计算.
H2O (l) 25℃ p1=pΘ ——→ H2O (g) 25℃ p3=pΘ
↓1 3 ↑
H2O (l) 25℃ p2=3168 Pa —2—→ H2O (g) 25℃ p2=3168 Pa
定温可逆膨胀,△G1 = Vm(p2-p1) = 18×10^(-6) * (101325-3168) = 1.767 Pa
可逆相变 ,△G2 = 0
定温可逆压缩,△G3 = nRT ln p3/p2 = 1*8.314*298*ln(101325/3168) = 8585.36 J
所以全过程△G = △G1 + △G2 +△G3 = 8.587 kJ
1/2 H2 (g) + O2(g) = H2O (l) △rGmΘ1 = △fGmΘ(H2O,l) = -237.19 kJ/mol
H2O (l) = H2O(g),△rGmΘ = 8.587 J/mol
故,二者相加,1/2 H2 (g) + O2(g) = H2O (g) △rGmΘ2 = △fGmΘ(H2O,s)
△fGmΘ(H2O,s) = △fGmΘ(H2O,l) + △rGmΘ = -228.603 kJ/mol
选 D