A) \[=\Delta {{E}^{o}}\]
B) \[>\Delta {{E}^{o}}\]
C) \[=0\]
D) \[<\Delta {{E}^{o}}\]
Correct Answer: D
Solution :
\[{{(C{{H}_{3}})}_{2}}C=C{{H}_{2}}(g)+6{{O}_{2}}(g)\to 4C{{O}_{2}}(g)+4{{H}_{2}}O(l)\] \[\Delta ng=4-6=-2\] (i.e., negative) we know that \[\Delta H=\Delta E+\Delta ngRT\] \[=\Delta E-(\Delta ng)RT\] (\[\because \,\Delta ng=-ve)\] \[\therefore \,\,\Delta H<\Delta E\]You need to login to perform this action.
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