已知25℃时，△fHm⊖(PCl 3,g)=-287.0kJ/mol△fHm⊖(PCl 5,g)=-374.9kJ/mol△fGm⊖(PCl 3,g)=-267.8kJ/mol△fGm⊖(PCl 5,g)=-305.0kJ/mol试计算此时反应PCl3 (g)+Cl2 (g) →PCl5 (g) 的焓变为______ kJ/mol（保留一位小数）
举一反三
- 【单选题】已知反应: (1) CO(g) + H2O(g) = CO2(g) + H2(g) , (298K) = - 41.2 kJ/mol (2) CH4(g) + 2 H2O(g)= CO2(g) + 4 H2(g), (298K) = 165.0 kJ/mol 则 CH4(g) + H2O(g) = CO(g) + 3 H2(g) 的 (298K) = 。 A. - 206.2 kJ/mol B. 123.8 kJ/mol C. 206.2 kJ/mol D. - 123.8 kJ/mol
- 已知反应B4C(s)()+4O2(g)()=2B2O3(s)()+CO2(g)的ΔrHmΘ=-2859()kJ·mol-1；而且ΔfHmΘ(B2O3)()=()-1273()kJ·mol-1；ΔfHmΘ(CO2)()=()-393()kJ·mol-1。则ΔfHmΘ(B4C)为()。A.()-()2859()kJ·mol()-()1B.()-()1666()kJ·mol()-()1C.()1666()kJ·mol()-()1D.()-()80.0()kJ·mol()-()1
- 【填空题】已知 CO2(g)的△fHmθ(298.15K)= -394 kJ/mol,CO2(g)= C(石墨)+O2(g)反应的 △rHmθ (298.15K)为 kJ/mol
- 已知H2O (l) = H2 (g) +1/2 O2 (g) DrHmy = 285.8 kJ·mol–1 则反应2 H2 (g) + O2 (g) = 2H2O (l) 的 DrHmy 为 ( ) A: 285.8 kJ·mol–1 B: - 285.8 kJ·mol–1 C: 571.6 kJ·mol–1 D: - 571.6 kJ·mol–1
- 已知H2O(g) = H2(g) +0.5O2(g)，ΔrHm(1) = 241.8 kJ·mol–1；H2(g) = 2H(g)，ΔHm(2) = 436.0 kJ·mol–1；0.5O2(g) = O(g)，ΔHm(3)=247.7 kJ·mol–1则H2O中H–O键的平均键焓(单位：kJ·mol–1)为（） A: 462.8 B: 925.5 C: –462.8 D: 241.8