A) \[2.5\times {{10}^{-3}}\]
B) \[2.5\times {{10}^{3}}\]
C) \[1.0\times {{10}^{-5}}\]
D) \[5\times {{10}^{3}}\]
E) \[5\times {{10}^{-3}}\]
Correct Answer: C
Solution :
Given, \[2PQ{{P}_{2}}+{{Q}_{2}};\] \[{{K}_{1}}=\frac{[{{P}_{2}}][{{Q}_{2}}]}{{{[PQ]}^{2}}}=2.5\times {{10}^{5}}\] \[PQ+\frac{1}{2}{{R}_{2}}PQR;\] \[{{K}_{2}}=\frac{[PQR]}{[PQ]{{[{{R}_{2}}]}^{1/2}}}=5\times {{10}^{-3}}\] Required equilibrium is \[\frac{1}{2}{{P}_{2}}+\frac{1}{2}{{Q}_{2}}+\frac{1}{2}{{R}_{2}}PQR\] \[{{K}_{3}}=\frac{[PQR]}{{{[{{P}_{2}}]}^{1/2}}{{[{{Q}_{2}}]}^{1/2}}{{[{{R}_{2}}]}^{1/2}}}\] On multiplying and dividing by\[[PQ],\] \[{{K}_{3}}=\frac{[PQR]}{{{[{{P}_{2}}]}^{1/2}}{{[{{Q}_{2}}]}^{1/2}}{{[{{R}_{2}}]}^{1/2}}[PQ]}\] Or \[{{K}_{3}}={{K}_{2}}\times \sqrt{\frac{1}{{{K}_{1}}}}\] \[=5\times {{10}^{-3}}\times \sqrt{\frac{1}{2.5\times {{10}^{5}}}}\] \[=\frac{5\times {{10}^{-3}}}{0.5\times {{10}^{3}}}=1\times {{10}^{-5}}\]You need to login to perform this action.
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