Question

【題目】已知熱化學(xué)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。據(jù)此判斷下列說(shuō)法正確的是

A. CH3OH的燃燒熱為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.根據(jù)方程式2CH3OH(l)+3O2(g) =2CO2(g)+4H2O(l) H= -1452 kJ·mol-1可知，CH3OH的燃燒熱為×1452 kJ·mol-1=726 kJ·mol-1，故A錯(cuò)誤；B.液態(tài)水變成水蒸氣會(huì)吸熱，因此2H2(g)+O2(g) =2H2O(g) H > -571.6 kJ·mol-1，故B正確；C. 醋酸為弱酸，電離需要吸熱，CH3COOH(aq)+NaOH(aq) =H2O(l)+CH3COONa(aq) H＞-57.3 kJ·mol-1，故C錯(cuò)誤；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，根據(jù)蓋斯定律，將②-①×2得：2CH3OH(l)+O2(g) =2CO2(g)+4H2(g) H= -308.8 kJ·mol-1，故D錯(cuò)誤；故選B。