A) \[K_{1}^{2}=\frac{1}{{{K}_{2}}}\]
B) \[2{{K}_{1}}=K_{2}^{2}\]
C) \[{{K}_{2}}=\frac{2}{K_{1}^{2}}\]
D) \[K_{2}^{2}=\frac{1}{{{K}_{1}}}\]
Correct Answer: A
Solution :
For the reaction, \[S{{O}_{2}}(g)+\frac{1}{2}{{O}_{2}}(g)S{{O}_{3}}(g)\] Equilibrium constant,\[{{K}_{1}}=\frac{[S{{O}_{3}}]}{[S{{O}_{2}}]\,{{[C{{O}_{2}}]}^{1/2}}}\] ... (i) For the reaction, \[2S{{O}_{3}}(g)2S{{O}_{2}}(g)+{{O}_{2}}(g)\] equilibrium constant, \[{{K}_{2}}=\frac{{{[S{{O}_{2}}]}^{2}}[{{O}_{2}}]}{{{[S{{O}_{3}}]}^{2}}}\] ... (ii) On squaring both sides in Eq (i), we get \[K_{1}^{2}=\frac{{{[S{{O}_{3}}]}^{2}}}{{{[S{{O}_{2}}]}^{2}}[{{O}_{2}}]}\] …. (iii) Eqs. (ii) \[\times \]Eq (iii), we get \[K_{1}^{2}\times {{K}_{2}}=1\] or \[{{K}_{2}}=\frac{1}{K_{1}^{2}}\] or \[K_{1}^{2}=\frac{1}{{{K}_{2}}}\]
You need to login to perform this action.
You will be redirected in
3 sec
