题目内容

7.CO2可用于合成二甲醚(CH3OCH3),有关反应的热化学方程式如下:
CO2(g)+3H2(g)═CH3OH(g)+H2O(g)△H=-49.0kJ•mol-1
2CH3OH(g)═CH3OCH3(g)+H2O(g)△H=-23.5kJ•mol-1
则CO2与H2反应合成二甲醚的热化学方程式正确的是(  )
A.2CO2(g)+6H2(g)═CH3OCH3(g)+3H2O(g)△H=-121.5kJ•mol-1
B.2CO2(g)+3H2(g)═$\frac{1}{2}$CH3OCH3(g)+$\frac{3}{2}$H2O(g)△H=-25.5kJ•mol-1
C.2CO2(g)+6H2(g)═CH3OCH3(g)+3H2O(g)△H=+121.5kJ•mol-1
D.2CO2(g)+6H2(g)═CH3OCH3(g)+3H2O(g)△H=-72.5kJ•mol-1

分析 由①CO2(g)+3H2(g)═CH3OH(g)+H2O(g)△H=-49.0kJ•mol-1
②2CH3OH(g)═CH3OCH3(g)+H2O(g)△H=-23.5kJ•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)△H=-49.0kJ•mol-1
②2CH3OH(g)═CH3OCH3(g)+H2O(g)△H=-23.5kJ•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=(-49.0kJ•mol-1)×2+(-23.5kJ•mol-1)=-121.5kJ•mol-1
即CO2与H2反应合成二甲醚的热化学方程式为2CO2(g)+6H2(g)═CH3OCH3(g)+3H2O(g)△H=-121.5kJ•mol-1
故选A.

点评 本题考查热化学方程式,为高频考点,把握已知反应与目标反应的关系为解答的关键,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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