A) \[=\Delta E{}^\circ \]
B) \[>\Delta E{}^\circ \]
C) \[=0\]
D) \[<\Delta E{}^\circ \]
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, \[\Delta H=\Delta E+\Delta {{n}_{g}}RT\] \[=\Delta E-\Delta |\Delta {{n}_{g}}|RT\] \[(\because \Delta {{n}_{g}}=-ve)\] \[\therefore \] \[\Delta H<\Delta E\]You need to login to perform this action.
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