查表得到ΔfHmθ、Smθ
· CaCO3（s）=CaO（s）+CO2（g）
ΔfHmθ -1207.6 -634.9 -393.5
Smθ 91.7 38.1 213.8
ΔrHmθ = - 393.5 + (-634.9) - (-1207.6) = 179.2 kJ/mol
ΔrSmθ = 213.8+38.1-91.7 = 160.2 J/(mol·K)
ΔrGmθ = ΔrHmθ - TΔrSmθ = 179.2 - 298.15*160.2/1000 = 131.4 kJ/mol
ΔrGmθ = ΔrHmθ - TΔrSmθ = 179.2 - T *160.2/1000＜0
所以T ＞ 1118.6 K
因此要指控温度大于1118.6 K,即845.5 ℃