Question

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

$ S{{O}_{2}}(g)+\frac{1}{2}{{O}_{2}}S{{O}_{3}}(g) $
$ {{K}_{c}}=\frac{1}{4.9\times {{10}^{-2}}} $
$ 2S{{O}_{2}}(g)+{{O}_{2}}(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 $