A) \[PC{{l}_{5}}(g)\rightleftharpoons PC{{l}_{3}}(g)+C{{l}_{2}}(g)\]
B) \[2NO(g)+{{O}_{2}}(g)\rightleftharpoons 2N{{O}_{2}}(g)\]
C) \[C(s)+{{H}_{2}}O(g)\rightleftharpoons CO(g)+{{H}_{2}}(g)\]
D) \[2HI(g)\rightleftharpoons {{H}_{2}}(g)+{{I}_{2}}(g)\]
Correct Answer: B
Solution :
\[\underset{2\,mol}{\mathop{2NO}}\,+\underset{1\,mol}{\mathop{O{{}_{2}}}}\,\rightleftharpoons \underset{2\,mol}{\mathop{2N{{O}_{2}}(g)}}\,\] If \[\Delta n=0,\]no effect on equilibrium due to pressure. \[\Delta n>0,\]the increase in pressure favours backward reaction, \[\Delta n<0,\]the increase in pressure favours forward reaction for [a] \[\Delta n={{n}_{p}}-{{n}_{r}}=2-1=1,\] [b] \[\Delta n=2-3=-1\] [c] \[\Delta n=2-1=1\] [d] \[\Delta n=2-2=0\] Hence, in case of [b] An < 0, the increase in pressure will favour the forward reaction.
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