A) \[{{H}_{2}}(g)+C{{l}_{2}}(g)\rightleftharpoons 2HCl(g)\]
B) \[{{H}_{2}}(g)+{{I}_{2}}(g)\rightleftharpoons 2HI(g)\]
C) \[2NOCl(g)\rightleftharpoons 2NO(g)+C{{l}_{2}}(g)\]
D) \[{{N}_{2}}(g)+3{{H}_{2}}(g)\rightleftharpoons 2N{{H}_{3}}(g)\]
Correct Answer: A
Solution :
We know that\[{{K}_{p}}={{K}_{c}}{{(RT)}^{\Delta n}}\] \[{{K}_{p}}\]and \[{{K}_{c}}\]are equal when \[\Delta n(g)=0\] Hence \[{{K}_{p}}={{K}_{c}}\]in case of \[{{H}_{2}}(g)+C{{l}_{2}}(g)\xrightarrow{{}}2HCl(g)\]
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