Question

$\Delta U^{\theta }$ of combustion of methane is $–XkJmo{l}^{-1}$. The value of $\Delta H^{\theta }$ is

(i) $=\Delta U^{\theta }$
(ii) $>\Delta U^{\theta }$
(iii) $<\Delta U^{\theta }$
(iv) $=0$

Answer 1

Since $\Delta H^{\theta }=\Delta U^{\theta }+\Delta n_{g}RTand\Delta U^{\theta }=–XkJmo{l}^{-1},\Delta H^{\theta }=(–X)+\Delta n_{g}RT.\Rightarrow \Delta H^{\theta }<\Delta U^{\theta }$
Therefore, alternative (iii) is correct.