The equilibrium constant of the above reaction is \[{{K}_{p}}\]. If pure ammonia is left to dissociate, the partial pressure of ammonia at equilibrium is given by (Assume that \[{{P}_{N{{H}_{3}}}}<<{{P}_{total}}\] at equilibrium)
A) \[\frac{{{3}^{3/2}}K_{P}^{1/2}{{P}^{2}}}{16}\]
B) \[\frac{K_{P}^{1/2}{{P}^{2}}}{16}\]
C) \[\frac{K_{P}^{1/2}{{P}^{2}}}{4}\]
D) \[\frac{{{3}^{3/2}}K_{P}^{1/2}{{P}^{2}}}{4}\]
Correct Answer: A
Solution :
\[\underset{{{P}_{0}}-2x}{\mathop{2N{{H}_{3}}\,(g)}}\,\underset{x}{\mathop{{{N}_{2}}(g)}}\,+\underset{3x}{\mathop{3{{H}_{2}}(g)}}\,,\,\,K=\frac{1}{{{K}_{p}}}\] \[\therefore \,\,K=\frac{1}{{{K}_{p}}}=\frac{x{{(3x)}^{3}}}{{{P}_{NH_{3}^{2}}}}\] \[\Rightarrow \,\,{{P}^{2}}_{N{{H}_{3}}}={{3}^{3}}{{x}^{4}}{{K}_{p}}\] \[\Rightarrow \,\,{{P}_{N{{H}_{3}}}}={{3}^{\frac{3}{2}}}{{x}^{2}}{{K}_{p}}^{\frac{1}{2}}\] \[=\frac{{{3}^{\frac{3}{2}}}.{{P}^{2}}{{K}_{p}}^{\frac{1}{2}}}{16}\]
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