A) \[{{N}_{2}}{{O}_{4}}(g)\] ⇌ \[2N{{O}_{2}}(g)\]
B) \[2S{{O}_{2}}(g)+{{O}_{2}}(g)\] ⇌ \[2S{{O}_{3}}(g)\]
C) \[{{H}_{2}}(g)+C{{l}_{2}}(g)\] ⇌ \[2HCl(g)\]
D) \[{{H}_{2}}(g)+\frac{1}{2}{{O}_{2}}(g)\] ⇌ \[{{H}_{2}}O(l)\]
Correct Answer: C
Solution :
\[\Delta n=0\] for this reaction so, \[\Delta E=\Delta H\].
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