Consider the following reactions in which all the reactants and the products are in gaseous state |
\[2PQ{{P}_{2}}+{{Q}_{2}};{{K}_{1}}=2.5\times {{10}^{5}}\] |
\[PQ+1/2{{R}_{2}}PQR;{{K}_{2}}=5\times {{10}^{-3}}\] |
The value of \[{{K}_{3}}\] for the equilibrium \[\frac{1}{2}{{P}_{2}}+\frac{1}{2}{{Q}_{2}}+\frac{1}{2}{{R}_{2}}PQR,\] is |
A) \[2.5\times {{10}^{-3}}\]
B) \[2.5\times {{10}^{3}}\]
C) \[1.0\times {{10}^{-5}}\]
D) \[5\times {{10}^{3}}\]
Correct Answer: C
Solution :
Given, \[2PQ{{P}_{2}}+{{Q}_{2}};\] |
\[{{K}_{1}}=\frac{[{{P}_{2}}][{{Q}_{2}}]}{{{[PQ]}^{2}}}=2.5\times {{10}^{5}}\] |
\[\operatorname{PQ}+\frac{1}{2}{{R}_{2}}PQR;\] |
\[{{\operatorname{K}}_{2}}=\frac{[PQR]}{[PQ]{{[{{R}_{2}}]}^{1/2}}}=5\times 1{{0}^{-3}}\] |
Required equilibrium is: \[1/2{{P}_{2}}+1/2{{Q}_{2}}+1/2{{R}_{2}}PQR;\] |
\[{{K}_{3}}=\frac{[PQR]}{{{[{{P}_{2}}]}^{1/2}}{{[{{Q}_{2}}]}^{1/2}}{{[{{R}_{2}}]}^{1/2}}}\] |
or \[{{K}_{3}}={{K}_{2}}\times \sqrt{\frac{1}{{{K}_{1}}}}\] |
\[=5\times {{10}^{-3}}\times \sqrt{\frac{1}{2.5\times {{10}^{5}}}}=\frac{5\times {{10}^{-3}}}{0.5\times {{10}^{3}}}=1\times {{10}^{-5}}\] |
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