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已知:2CO(g)+O2(g)=2CO2(g)ΔH="-566"kJ·mol-1N2(g)+O2(g)=2NO(g)ΔH="+180"kJ·mol-1则2CO(g)+2NO(g)=N2(g)+2CO2(g)的ΔH是A.-386kJ·mol-1B.+386kJ·mol-1C.-D.+746kJ·mol-1
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已知:2CO(g) + O 2 (g) = 2CO 2 (g) Δ H ="-566" kJ·mol -1 N 2 (g) + O 2 (g) =2NO(g) Δ H ="+180" kJ·mol -1 则2CO(g) +2NO(g) = N 2 (g)+2CO 2 (g)的Δ H 是
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已知:2CO(g) + O 2 (g) = 2CO 2 (g) Δ H ="-566" kJ·mol -1 N 2 (g) + O 2 (g) =2NO(g) Δ H ="+180" kJ·mol -1 则2CO(g) +2NO(g) = N 2 (g)+2CO 2 (g)的Δ H 是
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已知:2CO(g) + O 2 (g) = 2CO 2 (g) Δ H ="-566" kJ·mol -1
N 2 (g) + O 2 (g) =2NO(g) Δ H ="+180" kJ·mol -1
则2CO(g) +2NO(g) = N 2 (g)+2CO 2 (g)的Δ H 是
N 2 (g) + O 2 (g) =2NO(g) Δ H ="+180" kJ·mol -1
则2CO(g) +2NO(g) = N 2 (g)+2CO 2 (g)的Δ H 是
A.-386kJ·mol -1 | B.+386kJ·mol -1 | C.- | D.+746kJ·mol -1 |
已知:2CO(g) + O 2 (g) = 2CO 2 (g) Δ H ="-566" kJ·mol -1
N 2 (g) + O 2 (g) =2NO(g) Δ H ="+180" kJ·mol -1
则2CO(g) +2NO(g) = N 2 (g)+2CO 2 (g)的Δ H 是
N 2 (g) + O 2 (g) =2NO(g) Δ H ="+180" kJ·mol -1
则2CO(g) +2NO(g) = N 2 (g)+2CO 2 (g)的Δ H 是
A.-386kJ·mol -1 | B.+386kJ·mol -1 | C.- | D.+746kJ·mol -1 |
已知:2CO(g) + O 2 (g) = 2CO 2 (g) Δ H ="-566" kJ·mol -1
N 2 (g) + O 2 (g) =2NO(g) Δ H ="+180" kJ·mol -1
则2CO(g) +2NO(g) = N 2 (g)+2CO 2 (g)的Δ H 是
N 2 (g) + O 2 (g) =2NO(g) Δ H ="+180" kJ·mol -1
则2CO(g) +2NO(g) = N 2 (g)+2CO 2 (g)的Δ H 是
A.-386kJ·mol -1 | B.+386kJ·mol -1 | C.- | D.+746kJ·mol -1 |
已知:2CO(g) + O 2 (g) = 2CO 2 (g) Δ H ="-566" kJ·mol -1
N 2 (g) + O 2 (g) =2NO(g) Δ H ="+180" kJ·mol -1
则2CO(g) +2NO(g) = N 2 (g)+2CO 2 (g)的Δ H 是
2 2 H -1N 2 (g) + O 2 (g) =2NO(g) Δ H ="+180" kJ·mol -1
则2CO(g) +2NO(g) = N 2 (g)+2CO 2 (g)的Δ H 是
A.-386kJ·mol -1 | B.+386kJ·mol -1 | C.- | D.+746kJ·mol -1 |
2 2 H -1
2 2 H
A.-386kJ·mol -1 | B.+386kJ·mol -1 | C.- | D.+746kJ·mol -1 |
▼优质解答
答案和解析
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C
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C
C
C 考查盖斯定律的应用,根据已知反应可知,①-②即得到2CO(g) +2NO(g) = N 2 (g)+2CO 2 (g),所以反应热Δ H =―566 kJ·mol -1 ―180 kJ·mol -1 =-746kJ·mol -1 ,答案选C。【题型】选择题 |
考查盖斯定律的应用,根据已知反应可知,①-②即得到2CO(g) +2NO(g) = N 2 (g)+2CO 2 (g),所以反应热Δ H =―566 kJ·mol -1 ―180 kJ·mol -1 =-746kJ·mol -1 ,答案选C。【题型】选择题
考查盖斯定律的应用,根据已知反应可知,①-②即得到2CO(g) +2NO(g) = N 2 (g)+2CO 2 (g),所以反应热Δ H =―566 kJ·mol -1 ―180 kJ·mol -1 =-746kJ·mol -1 ,答案选C。【题型】选择题
考查盖斯定律的应用,根据已知反应可知,①-②即得到2CO(g) +2NO(g) = N 2 (g)+2CO 2 (g),所以反应热Δ H =―566 kJ·mol -1 ―180 kJ·mol -1 =-746kJ·mol -1 ,答案选C。【题型】选择题
考查盖斯定律的应用,根据已知反应可知,①-②即得到2CO(g) +2NO(g) = N 2 (g)+2CO 2 (g),所以反应热Δ H =―566 kJ·mol -1 ―180 kJ·mol -1 =-746kJ·mol -1 ,答案选C。【题型】选择题
考查盖斯定律的应用,根据已知反应可知,①-②即得到2CO(g) +2NO(g) = N 2 2 (g)+2CO 2 2 (g),所以反应热Δ H H =―566 kJ·mol -1 -1 ―180 kJ·mol -1 -1 =-746kJ·mol -1 -1 ,答案选C。【题型】选择题
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