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Question

Chemistry Question on Solutions

Mixing of N2N_2 and H2H_2 form an ideal gas mixture at room temperature in a container. For this process, which of the following statement is true?

A

ΔH=0,ΔSsurrounding =0,ΔSsystem =0\Delta H=0, \Delta S_{\text {surrounding }}=0, \Delta S_{\text {system }}=0 and ΔG=\Delta G=- ve

B

ΔH=0,ΔSsurrounding =0,ΔSsystem >0\Delta H=0, \Delta S_{\text {surrounding }}=0, \Delta S_{\text {system }}>0 and ΔG=\Delta G=- ve

C

ΔH>0,ΔSsurrounding =0,ΔSsystem >0\Delta H>0, \Delta S_{\text {surrounding }}=0, \Delta S_{\text {system }}>0 and ΔG=\Delta G=- ve

D

ΔH<0,ΔSsurrounding >0,ΔSsystem <0\Delta H<0, \Delta S_{\text {surrounding }}>0, \Delta S_{\text {system }}<0 and ΔG=\Delta G=- ve

Answer

ΔH<0,ΔSsurrounding >0,ΔSsystem <0\Delta H<0, \Delta S_{\text {surrounding }}>0, \Delta S_{\text {system }}<0 and ΔG=\Delta G=- ve

Explanation

Solution

(i)N2(g)+3H2(g)2NH3(g)(i) \because N _{2(g)}+3 H _{2(g)} \rightleftharpoons 2 NH _{3(g)}
ΔH=()ve\Delta H=(-) ve

Favourable condition-Low temperature and

and Δn=24=()2\Delta n=2-4=(-) 2 i.e. ()ve(-) ve

Also

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

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

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