Question:
$\Delta U^{\theta}$ of combustion of methane is $-X \mathrm{~kJ} \mathrm{~mol}^{-1}$. The value of $\Delta H^{\theta}$ is
(i) $=\Delta U^{\theta}$
(ii) $>\Delta U^{\theta}$
(iii) $<\Delta U^{\theta}$
(iv) $=0$
Solution:
Since $\Delta H^{\theta}=\Delta U^{\theta}+\Delta n_{g} R T$ and $\Delta U^{\theta}=-X \mathrm{~kJ} \mathrm{~mol}^{-1}$,
$\Delta H^{\theta}=(-X)+\Delta n_{g} R T$
$\Rightarrow \Delta H^{\theta}<\Delta U^{\theta}$
Therefore, alternative (iii) is correct.
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