A) \[=\Delta {{E}^{o}}\]
B) \[>\Delta {{E}^{o}}\]
C) \[=0\]
D) \[<\Delta {{E}^{o}}\]
Correct Answer: D
Solution :
\[{{(C{{H}_{3}})}_{2}}=C{{H}_{2}}(g)+6{{O}_{2}}(g)\xrightarrow{{}}4C{{O}_{2}}(g)\] \[\underset{{}}{\mathop{+\,4{{H}_{2}}O(l)}}\,\] \[\Delta {{n}_{g}}=4-7=-3\] (.i.e., negative) We know that, \[\Delta {{\Eta }^{o}}=\Delta \Eta {{E}^{o}}+\Delta {{n}_{g}}RT\] Here, \[\Delta {{n}_{g}}\] is negative \[\therefore \] \[\Delta {{H}^{o}}<\Delta {{E}^{o}}\]
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