| 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) \[\frac{{{3}^{3/2}}K_{p}^{1/2}{{P}^{2}}}{4}\]
B) \[\frac{K_{P}^{1/2}{{P}^{2}}}{4}\]
C) \[\frac{3_{{}}^{3/2}K_{P}^{1/2}{{P}^{2}}}{16}\]
D) \[\frac{K_{P}^{1/2}{{P}^{2}}}{16}\]
Correct Answer: C
Solution :
\[2N{{H}_{3}}{{N}_{2}}+3{{H}_{2}};K_{p}^{'}=\frac{1}{{{K}_{p}}}\] \[{{P}_{Total}}=P={{P}_{{{H}_{2}}}}+{{P}_{{{N}_{2}}}}+{{P}_{N{{H}_{3}}}}={{P}_{{{N}_{2}}}}+{{P}_{{{H}_{2}}}}\] \[({{P}_{Total}}>>{{P}_{N{{H}_{3}}}})\] \[{{P}_{{{N}_{2}}}}=\frac{P}{4},{{P}_{{{H}_{2}}}}=\frac{3P}{4}\] \[\frac{1}{{{K}_{P}}}\frac{\left( \frac{P}{4} \right){{\left( \frac{3P}{4} \right)}^{3}}}{{{({{P}_{N{{H}_{3}}}})}^{2}}}=\frac{27{{P}^{4}}}{256P_{N{{H}_{3}}}^{2}}\] \[{{P}_{N{{H}_{3}}}}={{\left( \frac{27{{P}^{4}}{{K}_{P}}}{256} \right)}^{1/2}}=\frac{{{3}^{3/2}}{{P}^{2}}}{16}\sqrt{{{K}_{P}}}\]
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