1 \(\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_{3}(\mathrm{~g})\)
2 \(\mathrm{I}_{2}(\mathrm{~g})+\mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{HI}(\mathrm{g})\)
3 \(\mathrm{N}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g})\)
4 \(\mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}_{2}(\mathrm{~g})\)
Explanation:
Increase in pressure favours less moles side.
\({\rm{i}}{\rm{.e}}{\rm{.,}}\,\,\Delta {\rm{n}} = - {\rm{ve.}}\)
\((1)\,\,{{\text{N}}_{\text{2}}}{\text{ + 3}}{{\text{H}}_{\text{2}}} \rightleftharpoons {\text{2N}}{{\text{H}}_{\text{3}}}\,;\,\,\Delta {\text{n = }} - 2\)
\({\text{(2)}}\,\,{{\text{I}}_{\text{2}}}{\text{ + }}{{\text{H}}_{\text{2}}} \rightleftharpoons {\text{2HI }}\,{\text{; }}\Delta {\text{n = 0}}\)
\({\text{(3)}}\,\,{{\text{N}}_{\text{2}}}{\text{ + }}{{\text{O}}_{\text{2}}} \rightleftharpoons {\text{2NO}}\,{\text{;}}\Delta {\text{n = 0}}\)
\({\text{(4)}}\,\,{{\text{N}}_{\text{2}}}{{\text{O}}_{\text{4}}} \rightleftharpoons {\text{2N}}{{\text{O}}_{\text{2}}}\,{\text{;}}\,\Delta {\text{n = + 1.}}\)
Hence, option (1) is correct.