A) \[{{K}_{1}}={{K}_{2}}\]
B) \[{{K}_{1}}=\frac{1}{{{K}_{2}}}\]
C) \[{{K}_{2}}=K_{1}^{2}\]
D) \[K_{1}^{2}=\frac{1}{{{K}_{2}}}\]
Correct Answer: D
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 \[K_{1}^{2}=\frac{1}{{{K}_{2}}}\]
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