Question

Mixing of $N_2$ and $H_2$ form an ideal gas mixture at room temperature in a container. For this process, which of the following statement is true?

• A. $\Delta H=0, \Delta S_{\text {surrounding }}=0, \Delta S_{\text {system }}=0$ and $\Delta G=-$ ve
• B. $\Delta H=0, \Delta S_{\text {surrounding }}=0, \Delta S_{\text {system }}>0$ and $\Delta G=-$ ve
• C. $\Delta H>0, \Delta S_{\text {surrounding }}=0, \Delta S_{\text {system }}>0$ and $\Delta G=-$ ve
• D. $\Delta H<0, \Delta S_{\text {surrounding }}>0, \Delta S_{\text {system }}<0$ and $\Delta G=-$ ve

Answer 1

The Correct Option is D

$(i) \because N _{2(g)}+3 H _{2(g)} \rightleftharpoons 2 NH _{3(g)}$
$\Delta H=(-) ve$

Favourable condition-Low temperature and

and $\Delta n=2-4=(-) 2$ i.e. $(-) ve$

Also

(ii) $\because \Delta S_{\text {(univ.) }}=\Delta S_{\text {(surrounding ) }}+\Delta S_{\text {(system) }}$

and for $\Delta G$ to be $(-)$ ve, i.e. (spontaneous) at room temperature,

$\Delta H<0$
$\Delta S_{(\text {surrounding })}>0$
$\Delta S_{\text {(system) }}<0$ and $\Delta S_{\text {(univ) }}>0$