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
• E. $\Delta H=0,\Delta S_{\text{surrounding }}=0,\Delta S_{\text{system }}<0$ and $\Delta G=-$ ve

Answer 1

Answer: (D)

$(i)\because N_{2(g)}+3H_{2(g)}\rightleftharpoons 2N{H}_{3(g)}$

$\Delta H=(-)ve$

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 $\Delta S_{\text{(univ) }}>0$