A) \[=\Delta {{E}^{o}}\]
B) \[>\Delta {{E}^{o}}\]
C) \[=0\]
D) \[<\Delta {{E}^{o}}\]
Correct Answer: D
Solution :
\[{{(C{{H}_{3}})}_{2}}C=C{{H}_{2}}(g)+6{{O}_{2}}(g)\xrightarrow{{}}\]\[4C{{O}_{2}}(g)+4{{H}_{2}}O(l)\] \[\Delta {{n}_{g}}=4-6=-2\] (i.e., negative) We know that, \[\Delta H=\Delta E+\Delta {{n}_{g}}\,RT\] \[=\Delta E0-|\Delta {{n}_{g}}|RT\] ( \[\because \] \[\Delta {{n}_{g}}=-ve\]) \[\Delta H<\Delta E\]
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