Question

The equilibrium constant for the reaction
$ S{{O}_{3}}(g) \rightleftharpoons S{{O}_{2}}(g)+\frac{1}{2}{{O}_{2}}(g) $
is $ {{K}_{c}}=4.9\times {{10}^{-2}} $ . The value of $ {{K}_{c}} $ or the reaction
$ 2S{{O}_{2}}(g)+{{O}_{2}} (g) \rightleftharpoons 2S{{O}_{3}}(g) $ will be

• A. $ 416 $
• B. $ 2.40\times {{10}^{-3}} $
• C. $ 9.8\times {{10}^{-2}} $
• D. $ 4.9\times {{10}^{-2}} $

Answer 1

Answer: (A)

Equilibrium constant for the reaction,
$ S{{O}_{2}}(g)+\frac{1}{2}{{O}_{2}}(g) \rightleftharpoons S{{O}_{3}}(g) $
$ {{K}_{c}}=\frac{1}{4.9\times {{10}^{-2}}} $
and for $ 2S{{O}_{2}}(g)+{{O}_{2}} \rightleftharpoons (g)2S{{O}_{3}}(g) $
$ {{K}_{c}}={{\left( \frac{1}{4.9\times {{10}^{-2}}} \right)}^{2}} $
$ =\frac{{{10}^{4}}}{{{(4.9)}^{2}}}=416.49 $