1. Tardigrade
  2. Question
  3. Chemistry
  4. Consider the following reactions in which all the reactants and the products are in gaseous state. 2PQ leftharpoons P2+Q2; K1=2.5× 105 PQ+(1/2)R2 leftharpoons PQR; K2=5× 10-3 The value of K3 for the equilibrium (1/2)P2+(1/2)Q2+(1/2)R2 leftharpoons PQR, is

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Q. Consider the following reactions in which all the reactants and the products are in gaseous state.
$ 2PQ \rightleftharpoons {{P}_{2}}+{{Q}_{2}}; $ $ {{K}_{1}}=2.5\times {{10}^{5}} $
$ PQ+\frac{1}{2}{{R}_{2}} \rightleftharpoons 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}} \rightleftharpoons PQR, $ is

KEAMKEAM 2010Equilibrium

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}} $