r
Gm 1
CO(g) + 2H2(g) ———→ CH3OH (g) Pθ
ΔrHm 2，ΔrSmrGm 2 Δ
H
m
，ΔrGm 3
3θ
ΔrHm 1=ΔfHm(CH3OH(g)) -ΔfHm(CO(g))=-201.2-(-110.52)=-90.68 kJ·mol–1
ΔrSm 2= Sm(CH3OH (l)) -Sm(CO(g))-2Sm( H2(g))=127.0-2×130.57-197.56 =-331.7 J·K-1·mol-1
–1
ΔrHm 1=ΔrHm 2+ΔvapHm，ΔrHm 2= -90.68-38= -128.68 kJ·mol
-3–1
ΔrGm 2 =ΔrHm 2-TΔrSm 2 =-128.68-298.15×(-331.7)×10= -29.78 kJ·mol
ΔrGm 3 = - RTln Kθ = -RTln(P/ Pθ)
= -8.314×298.15ln(16582/100000) = 4.454 kJ·mol–1
ΔrGm 1 =ΔrGm 2 +ΔrGm 3 =-29.78 + 4.454 = -25.33 kJ·mol–1
ΔrGm 1 = - RTln Kθ，-25.33×103 = -8.314×298.15ln Kθ Kθ= 2.74×104
方法2：
-1 rHm(298.15K)=-201.2-(-110.52)= -90.68 kJ mol,
CH3OH(l)→CH3OH(g)(25℃, 16582Pa)→CH3OH(g)(25℃, 10000Pa)
Sm[ CH3OH(g)]= 127.0+38.0×103/298+Rln(16.582/100)=239.4 J K-1mol-1,
rSm=239.4-(197.56+2×130.57)=-219.3 J K-1mol-1 ,
-14 rGm(298.15K)= rHm-T rSm=-25.3kJ mol, K(298.15K)= 2.7×10
[ -25.3kJ mol-1 2.7×104 ]
3-5 已知 fGm(H2O,l,298.15K) = - 237.19 kJmol-1，25℃时水的饱和蒸气压p* (H2O)=3.167kPa，若H2O(g)
可视为理想气体，求 fGm (H2O,g,298.15K)。?
答： ① ② ③ 1mol H2O(l, 25℃,PΘ) →1mol H2O(l, 25℃, 3.167kPa) →1mol H2O(g, 25℃, 3.167kPa) →1mol H2O(g, 25℃, 100kPa)
①是凝聚相的等温变压过程，由于压力对凝聚相自由能的影响可以忽略不计，可认为ΔGm=0 r
②是等温可逆相变过程ΔrGm=0 ③是理想气体的等温变压过程ΔΔΔ
–1
r
Gm（③） =RTln(P2/P1)= RTln(Pθ/P) = 8.314×
298.15×10-3×ln(100/3.167) = 8.558 kJ·mol–1
ff
Gm(H2O(g),PΘ)- Δ
ff
Gm(H2O(l)，PΘ)=0+0+Δ
r
Gm（③） =8.558，则有
Θ
Gm(H2O(g),P)= ΔΘ
Gm(H2O(l)，P)+Δ
r
Gm（③） =8.558-237.19=-228.63 kJmol
[ - 228.6 kJ mol-1 ]
3-6 已知 fGm(CH3OH,g,298.15K) = -162.51 kJ·mol-1, 25℃时p*(CH3OH)=16.27kPa,若CH3OH(g)可视为理
想气体，求 fGm (CH3OH,l,298.15K)。?
答： 设计途径
① ② ③
Θ
CH3OH(g,298.15K, P)→CH3OH(g,298.15K, 16.27kPa) →CH3OH(l,298.15K, 16.27kPa) →CH3OH(g,298.15K, 16.27kPa)
①理想气体的等温变压过程ΔRTln(16.27/100)=-4498.9J=-4.4989kJ mol②等温等压可逆相变过程ΔrGm=0
③是凝聚相的等温变压过程，ΔG=V(l) (P*-PΘ),由于压力对凝聚相自由能的影响不大，可认为ΔG=V(l) (P*-PΘ)=0
fGm (CH3OH,l)- fGm(CH3OH,g)=RTln(p*/p)+0+0=-4.4989 kJ mol
即 fGm (CH3OH,l)= fGm(CH3OH,g)+RTln(p*/p)+0+0=(-162.51) +(-4.4989)=-167 kJ mol-1
r
Gm
（①）
=RTln(P2/P1)=
-1
-1
[ -167 kJ mol-1 ]
3-7 已知Br2(l)的饱和蒸气压p*(Br2)=28574 Pa，求反应Br2(l) = Br2(g)的 rGm(298.15K)。
答：同理： ① ② ③
Br2(l,PΘ)→Br2(l,P*)→Br2(g,P*)→Br2(g,PΘ)
①凝聚相的等温变压过程，ΔG=V(l) (P*-PΘ),由于压力对凝聚相自由能的影响不大，可认为ΔG=V(l) (P*-PΘ)=0
②等温等压可逆相变过程ΔrGm=0
③理想气体的等温变压过程ΔrGm =RTln(P2/P1)= RTln(100/28.574)=3138J=3.138kJ mol
ΔrGΘm(Br2,298.15,g, PΘ)- ΔrGΘm(Br2, 298.15,l, PΘ)=0+0+ rGm(③)=3.138
[ 3.14 kJ mol-1]
3-8 已知理想气体间的反应CO(g)+H2O(g)==CO2(g)+H2(g)在973.15K时K= 0.71。? (1) 系统中四种气体的分压均为1.50p时，上述反应的自发方向如何?
? (2) p(CO)=10p，p(H2O)=5p，p(CO2)=p(H2)=1.5p时，反应的自发方向又如何??
-1
答： CO(g) + H2O(g) —→ CO2(g) + H2(g)
(1) ΔrGm = RTln(Ja/K)=RTln(1/0.71)>0，反应逆向自发。 (2) Ja = 1.5×1.5/(10×5)=0.045 < 0.71，反应正向自发。
[ (1)逆向自发 (2)正向自发] 3-9
θ
已知反应CO(g)+H2(g)=HCOH(1) rGm(298.15K) = 28.95 kJ mol-1,而298.15K时
p*(HCOH)=199.98kPa,求298.15K时，反应HCHO(g)=CO(g)+H2(g)的K (298.15K)。?
答： Δ
r
Gm
CO(g) + H2(g) —→
Δrm ΔrGm 2
θ
仿照（3-7题）ΔrGm 2 = RTln(P/ P)= 8.314×298.15ln(199.98/100)) = 1.718 kJ·mol–1
ΔrGm 1 =ΔrGm-Δ
θ
Gm 2 =28.95 –1.718 = 27.23 kJ·mol–1 r
HCOH(g) —→ CO(g) + H2(g) ΔrGm = -ΔrGm 1 = -27.23 kJ·mol–1 ΔrGm = - RTln K，-27.23×103= -8.314×298.15ln Kθ Kθ= 5.9×104
[ 6×104 ]
3-10 通常钢瓶中装的氮气含有少量的氧气，在实验中为除去氧气，可将气体通过高温下的铜，使发生下述反应：2Cu(s) + 0.5 O2(g) == Cu2O(s) 已知此反应的 rGm/(J mol-1)= -166732+63.01(T/K)。今若在600℃时反应达到平衡，问经此手续处理后，氮气中剩余氧的浓度为若干??
答：2Cu(s) + 1/2 O2(g) —→ Cu2O(s)
600℃，ΔrGm= -166732+(63.01×873.15)=-111.715 kJ·mol–1