题目内容

9.工业上用CO2和H2反应合成二甲醚.已知:
CO2(g)+3H2(g)?CH3OH(g)+H2O(g)△H1=-49.1kJ•mol-1
2CH3OH(g)?CH3OCH3 (g)+H2O(g)△H2=-24.5kJ•mol-1
写出CO2(g)和H2(g)转化为CH3OCH3(g)和H2O(g)的热化学方程式2CO2(g)+6H2(g)?CH3OCH3(g)+3H2O(g)△H=-122.7kJ•mol-

分析 由①CO2(g)+3H2(g)?CH3OH(g)+H2O(g)△H1=-49.1kJ•mol-1
②2CH3OH(g)?CH3OCH3 (g)+H2O(g)△H2=-24.5kJ•mol-1
结合盖斯定律可知,①×2+②得到2CO2(g)+6H2(g)?CH3OCH3(g)+3H2O(g),以此来解答.

解答 解:由①CO2(g)+3H2(g)?CH3OH(g)+H2O(g)△H1=-49.1kJ•mol-1
②2CH3OH(g)?CH3OCH3 (g)+H2O(g)△H2=-24.5kJ•mol-1
结合盖斯定律可知,①×2+②得到2CO2(g)+6H2(g)?CH3OCH3(g)+3H2O(g),则△H=(-49.1kJ•mol-1)×2+(-24.5kJ•mol-1)=-122.7 kJ•mol-1,即热化学方程式为2CO2(g)+6H2(g)?CH3OCH3(g)+3H2O(g)△H=-122.7 kJ•mol-1
故答案为:2CO2(g)+6H2(g)?CH3OCH3(g)+3H2O(g)△H=-122.7 kJ•mol-

点评 本题考查热化学方程式的书写,为高频考点,把握物质的量与热量的关系、焓变为解答的关键,侧重分析与应用能力的考查,注意盖斯定律的应用,题目难度不大.

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