# $N_2 + 3H_2 \longrightarrow 2NH_3$ This reaction is carried out at constant temperature and pressure. If $\Delta H$ and $\Delta U$ are enthalpy and internal energy change for the reaction, which of the following expression is true?

(a) $\Delta H = 0$
(b) $\Delta H = \Delta U$
(c) $\Delta H < \Delta U$
(d) $\Delta H > \Delta U$

Answer: $\Delta H < \Delta U$
$\Delta n_g$ for this reaction is +ve ($\Delta n_g$ = -4 if $NH_3$ is considered as liquid and $\Delta n_g$ = -2 if $NH_3$ is considered as gas)
Now, $\Delta H = \Delta U + \Delta n_gRT$
$\therefore \Delta H < \Delta U$ for this reaction.