A) \[\Delta H<\Delta E\]
B) \[\Delta H=\Delta E\]
C) \[\Delta H=0\]
D) \[\Delta H>\Delta E\]
Correct Answer: D
Solution :
\[{{N}_{2}}{{O}_{4}}(g)\to 2N{{O}_{2}}(g)\] For this reaction \[\Delta ng=2-1=1\] \[\Delta ng\] is positive i.e., there is an increase in the number of gaseous moles then \[\Delta H>\Delta E\]
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