Question

The equilibrium constant for the given reaction:
$S{O}_3\left(g\right)\rightleftharpoons S{O}_2\left(g\right)+\frac{1}{2}{O}_2\left(g\right);K_c=5\times {10}^{-2}$
The value of $K_c$ for the reaction will be:
$2{SO}_2\left(g\right)+O_2\left(g\right)\rightleftharpoons 2{SO}_3\left(g\right),$

• A. $400$
• B. $2.40\times {10}^{-3}$
• C. $9.8\times {10}^{-2}$
• D. $4.9\times {10}^{-2}$

Answer 1

Answer: (A)

$400$

$S{O}_3\leftrightharpoons S{O}_2+\frac{1}{2}{O}_2;K_C=5\times {10}^{-2}$
$⟹K_C=\frac{\left[S{O}_2\right][O_2{]}^{1/2}}{\left[S{O}_3\right]}=5\times {10}^{-2}$
For, $2S{O}_2+O_2\leftrightharpoons 2S{O}_3;K_C^{\prime }=\frac{[S{O}_3{]}^2}{\left[S{O}_2{]}^2\right[O_2]}$
$⟹K_C^{\prime }=\frac{1}{K_c^2}=\frac{1}{25\times {10}^{-4}}=4\times {10}^2=400$