A) \[{{H}_{2(g)}}+{{I}_{2(g)}}\]⇌ \[2H{{I}_{(g)}}\]
B) \[{{C}_{(s)}}+{{O}_{2(g)}}\]⇌\[C{{O}_{2(g)}}\]
C) \[{{N}_{2(g)}}+3{{H}_{2(g)}}\]⇌\[2N{{H}_{3(g)}}\]
D) \[HC{{l}_{(aq)}}+NaO{{H}_{(aq)}}\]⇌\[NaC{{l}_{(aq)}}+{{H}_{2}}O\]
Correct Answer: C
Solution :
In this reaction \[\Delta n=2-4=-2\] so \[\Delta H\ne \Delta E\].You need to login to perform this action.
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