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 (ie, negative) We know that, \[\Delta {{H}^{o}}=\Delta {{E}^{o}}+\Delta {{n}_{g}}RT\] \[=\Delta {{E}^{o}}-\left| \Delta {{n}_{g}} \right|RT\] \[(\because \,\,\Delta {{n}_{g}}=-ve)\] \[\therefore \] \[\Delta {{H}^{o}}<\Delta {{E}^{o}}\]
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