Q. $\mathrm{\Delta}E{^\circ}$of combustion of isobutylene is $–x\ kJ\ mol^{- 1}$. The value of $\mathrm{\Delta}H{^\circ}$ is
A) $= \mathrm{\Delta}E{^\circ}$
B) $> \mathrm{\Delta}E{^\circ}$
C) = 0
D) $< \mathrm{\Delta}E{^\circ}$
Explanation
(d)
$\left( CH_{3} \right)_{2}C = CH_{2}(g) + 6O_{2}(g) \rightarrow 4CO_{2}(g) + 4H_{2}O(l)$
$\mathrm{\Delta}ng = 4 - 7 = - 3$($ie$, negative)
We know that $\mathrm{\Delta}H = \mathrm{\Delta}E + {\mathrm{\Delta}n}_{g}RT$
$= \mathrm{\Delta}E - (\mathrm{\Delta}m)RT\ \ \ (\because\ \mathrm{\Delta}n_{g} = - ve)$
$\therefore\ \mathrm{\Delta}H < \mathrm{\Delta}E$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.