题目内容
已知下列热化学方程式:
(1)CH3COOH(l)+2O2(g) == 2CO2(g)+2H2O(l) ΔH1=-870.3 kJ·mol-1
(2)C(s)+ O2(g) == CO2(g) △H2=-393.5 kJ?mol-1
(3)H2(g) +O2(g) == H2O(l) △H3=-285.8kJ·mol-1
则反应2C(s)+2H2(g) +O2(g) == CH3COOH(l)的△H为( )
(1)CH3COOH(l)+2O2(g) == 2CO2(g)+2H2O(l) ΔH1=-870.3 kJ·mol-1
(2)C(s)+ O2(g) == CO2(g) △H2=-393.5 kJ?mol-1
(3)H2(g) +O2(g) == H2O(l) △H3=-285.8kJ·mol-1
则反应2C(s)+2H2(g) +O2(g) == CH3COOH(l)的△H为( )
A.-488.3 kJ·mol-1 | B.-244.15 kJ·mol-1 | C.+488.3 kJ·mol-1 | D.+244.15 kJ·mol-1 |
A
考查盖斯定律的应用。根据已知的反应可知,(2)×2+(3)×2-(1)即得到2C(s)+2H2(g) +O2(g) == CH3COOH(l),所以其反应热△H=-393.5 kJ?mol-1×2-285.8kJ·mol-1×2+870.3 kJ·mol-1=-488.3 kJ·mol-1,答案选A。
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