A) \[{{K}_{2}}K_{3}^{3}/{{K}_{1}}\]
B) \[{{K}_{2}}\,K_{3}^{2}/{{K}_{1}}\]
C) \[K_{2}^{2}\,{{K}_{3}}/{{K}_{1}}\]
D) \[{{K}_{1}}\,{{K}_{2}}/{{K}_{3}}\]
Correct Answer: A
Solution :
The required equation for the oxidation of NH3 by oxygen to give NO is : \[4N{{H}_{3}}+5{{O}_{2}}\underset{{{800}^{o}}C}{\mathop{\xrightarrow{Pt\,(gauze)}}}\,4NO+6{{H}_{2}}O\] For this \[K=\frac{{{[NO]}^{4}}\,{{[{{H}_{2}}O]}^{6}}}{{{[N{{H}_{3}}]}^{4}}\,{{[{{O}_{2}}]}^{5}}}\] For the equation \[(I)\,{{K}_{1}}=\frac{{{[N{{H}_{3}}]}^{2}}}{[{{N}_{2}}]\,{{[{{H}_{2}}]}^{3}}}\] For the equation \[(II)\,{{K}_{2}}=\frac{{{[NO]}^{2}}}{[{{N}_{2}}]\,[{{O}_{2}}]}\] For the equation \[(III)\,{{K}_{3}}=\frac{[{{H}_{2}}O]}{[{{H}_{2}}]\,{{[{{O}_{2}}]}^{1/2}}}\] For getting the K we must do \[K_{1}^{2}=\frac{{{[N{{H}_{3}}]}^{4}}}{{{[{{N}_{2}}]}^{2}}\,{{[{{H}_{2}}]}^{6}}},\,K_{2}^{2}=\frac{{{[NO]}^{4}}}{{{[{{N}_{2}}]}^{2}}\,{{[{{O}_{2}}]}^{2}}}\] \[K_{3}^{6}=\frac{{{[{{H}_{2}}O]}^{6}}}{{{[{{H}_{2}}]}^{6}}\,{{[{{O}_{2}}]}^{\frac{6}{2}=3}}}\] \[K=\frac{K_{2}^{2}\times K_{3}^{6}}{{{[N{{H}_{3}}]}^{4}}{{[{{O}_{2}}]}^{5}}}\] substituting the value we get, \[K=\frac{{{[NO]}^{4}}{{[{{H}_{2}}O]}^{6}}}{{{[N{{H}_{3}}]}^{4}}\,{{[{{O}_{2}}]}^{5}}}\]so the value of K in terms of K1, K2 and K3 is \[K=\frac{{{K}_{2}}K_{3}^{3}}{{{K}_{1}}}\]
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