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Q. If $ {{K}_{1}} $ and $ {{K}_{2}} $ are the equilibrium constants of the equilibria A and B respectively, what is the relationship between the two constants? (i) $ S{{O}_{2}}(g)+\frac{1}{2}{{O}_{2}}(g)\rightleftharpoons S{{O}_{3}}(g);{{K}_{1}} $ (ii) $ 2S{{O}_{3}}(g)\rightleftharpoons 2S{{O}_{2}}(g)+{{O}_{2}}(g);{{K}_{2}} $

EAMCETEAMCET 1998

Solution:

$ S{{O}_{2}}+\frac{1}{2}{{O}_{2}}\overset{{{K}_{1}}}{\mathop{\rightleftharpoons }}\,S{{O}_{3}} $ $ {{K}_{1}}=\frac{[S{{O}_{3}}]}{[S{{O}_{2}}]{{[{{O}_{2}}]}^{1/2}}} $ ?(i) $ 2\,S{{O}_{3}}\overset{{{K}_{2}}}{\mathop{\rightleftharpoons }}\,2S{{O}_{2}}+{{O}_{2}} $ $ {{K}_{2}}=\frac{{{[S{{O}_{2}}]}^{2}}[{{O}_{2}}]}{{{[S{{O}_{3}}]}^{2}}} $ (ii) On squaring the equation (1) and multiply equation (1) and equation (2), we get $ K_{1}^{2}=\frac{[S{{O}_{3}}]}{{{[S{{O}_{2}}]}^{2}}[{{O}_{2}}]} $ $ K_{1}^{2}\times {{K}_{2}}=\frac{{{[S{{O}_{3}}]}^{2}}}{{{[S{{O}_{2}}]}^{2}}[{{O}_{2}}]}\times \frac{{{[S{{O}_{2}}]}^{2}}+[{{O}_{2}}]}{{{[S{{O}_{3}}]}^{2}}} $ $ K_{1}^{2}\times {{K}_{2}}=1 $ $ {{K}_{2}}=\frac{1}{K_{1}^{2}} $ or $ K_{1}^{2}=\frac{1}{{{K}_{2}}} $