A) \[\operatorname{CO}(g)+\frac{1}{2}{{O}_{2}}(g)C{{O}_{2}}(g)\]
B) \[{{H}_{2}}(g)+{{I}_{2}}(g)2HI\,(g)\]
C) \[{{\operatorname{PCl}}_{5}}(g)\,\,\,\,PC{{l}_{3}}(g)+C{{l}_{2}}(g)\]
D) \[7{{H}_{2}}+2N{{O}_{2}}(g)\,\,\,\,2N{{H}_{3}}(g)+4{{H}_{2}}O\,(g)\]
Correct Answer: C
Solution :
Using the relation\[{{\operatorname{K}}_{p}}={{K}_{c}}.{{(RT)}^{\Delta n}}\], we get \[\frac{{{\operatorname{K}}_{p}}}{{{K}_{c}}}=\,\,{{(RT)}^{\Delta n}}\] Thus \[\frac{{{\operatorname{K}}_{p}}}{{{K}_{c}}}\] will be highest for the reaction having highest value of M. The \[\Delta n\] values for various reactions are [a] \[\Delta n=1-\left( 1+\frac{1}{2} \right)=-\frac{1}{2}\] [b] \[\Delta n=2\,\,-\left( 1+1 \right)=0\] [c] \[\Delta n=\,\,\left( 1\,\,+\,\,1 \right)-1=1\] [d] \[\operatorname{A}\,\,=\,\,(2+4)\,\,-\,\,(7+2)=-\,3\] Thus, maximum value of \[\Delta n= 1\]
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