Question

【题目】已知热化学方程式：2H2(g)+O2(g) 2H2O(l) H= -571.6 kJ·mol-1，2CH3OH(l)+3O2(g) 2CO2(g)+4H2O(l) H= -1452 kJ·mol-1，H+(aq)+OH－(aq) H2O(l) H= -57.3 kJ·mol-1。据此判断下列说法正确的是

A. CH3OH的燃烧热为1452 kJ·mol-1

B. 2H2(g)+O2(g) 2H2O(g) H > -571.6 kJ·mol-1

C. CH3COOH(aq)+NaOH(aq) H2O(l)+CH3COONa(aq) H= -57.3 kJ·mol-1

D. 2CH3OH(l)+O2(g) 2CO2(g)+4H2(g) H= -880.4 kJ·mol-1

Answer 1

【答案】B

【解析】A.根据方程式2CH3OH(l)+3O2(g) =2CO2(g)+4H2O(l) H= -1452 kJ·mol-1可知，CH3OH的燃烧热为×1452 kJ·mol-1=726 kJ·mol-1，故A错误；B.液态水变成水蒸气会吸热，因此2H2(g)+O2(g) =2H2O(g) H > -571.6 kJ·mol-1，故B正确；C. 醋酸为弱酸，电离需要吸热，CH3COOH(aq)+NaOH(aq) =H2O(l)+CH3COONa(aq) H＞-57.3 kJ·mol-1，故C错误；D. ①2H2(g)+O2(g) =2H2O(l) H= -571.6 kJ·mol-1，②2CH3OH(l)+3O2(g) =2CO2(g)+4H2O(l) H= -1452 kJ·mol-1，根据盖斯定律，将②-①×2得：2CH3OH(l)+O2(g) =2CO2(g)+4H2(g) H= -308.8 kJ·mol-1，故D错误；故选B。