A) \[{{K}_{p}}=K{{}^{2}}\]
B) \[{{K}_{p}}=\sqrt{K{{}_{p}}}\]
C) \[{{K}_{p}}=2K{{}_{p}}\]
D) \[{{K}_{p}}=K{{}_{p}}\]
Correct Answer: B
Solution :
\[\frac{1}{2}{{N}_{2}}(g)+\frac{3}{2}{{H}_{2}}(g)N{{H}_{3}}(g);\,\,{{K}_{p}}\] \[{{K}_{p}}=\frac{{{P}_{N{{H}_{3}}}}}{{{({{P}_{{{N}_{2}}}})}^{1/2}}{{({{p}_{{{H}_{2}}}})}^{3/2}}}\] ... (i) \[{{N}_{2}}(g)+3{{H}_{2}}(g)2N{{H}_{3}}(g)\] \[K{{}_{p}}=\frac{{{({{p}_{N{{H}_{3}}}})}^{2}}}{{{p}_{{{N}_{2}}}}.\,\,{{({{p}_{{{H}_{2}}}})}^{3}}}\] ... (ii) From Eq. (i) and (ii) \[\frac{K_{p}^{2}}{K{{}_{p}}}=1\] or \[K_{p}^{2}=K{{}_{p}}\] \[\therefore \] \[{{K}_{p}}=\sqrt{K{{}_{p}}}\]
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