Question

【题目】在25℃、101kPa下，0.1 mol甲醇燃烧生成CO2和液态水时放热72.58 kJ，下列热化学方程式正确的是

A. CH3OH(l)＋3/2O2(g)===CO2(g)＋2H2O(l) ΔH＝＋725.8 kJ/mol

B. 2CH3OH(l)＋3O2(g)===2CO2(g)＋4H2O(g) ΔH＝－1451.6 kJ/mol

C. CH3OH(l)＋3/2O2(g)===CO2(g)＋2H2O(l) ΔH＝－725.8 kJ/mol

D. 2CH3OH(g)＋3O2(g)===2CO2(g)＋4H2O(l) ΔH＝－1451.6 kJ

Answer 1

【答案】C

【解析】

试题由1gCH3OH燃烧放热22.7kJ，则2molCH3OH燃烧放热22.7×32×2=1452.8kJ，则该燃烧反应的热化学方程式为2CH3OH(g)+3O2(g)=2CO2(g)+4H2O(l)△H=-1452.8kJ/mol，CH3OH(g)+O2(g)=CO2(g)+2H2O(l)△H=-726.4kJ/mol，故选A。