Question

The equilibrium constant for the reaction $S{O}_3\left(g\right)\rightleftharpoons S{O}_2\left(g\right)+\frac{1}{2}{O}_2\left(g\right)$ is $K_c=4.9\times {10}^{-2}$. The value of $K_c$ for the reaction $2S{O}_2\left(g\right)+O_2\left(g\right)\rightleftharpoons 2S{O}_3\left(g\right)$ 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)

$416$

Let us consider the reverse reaction ,

equilibrium constant $=\frac{1}{K_c^2}$

New equilibrium constant $=\frac{1}{(4.9\times {10}^{-2}{)}^2}$

$=416$