Question

The equilibrium constant for the given reaction: $S{O}_3\left(g\right)\rightleftharpoons S{O}_2\left(g\right)+\frac{1}{2}{O}_2\left(g\right);K_c=4.9\times {10}^{-2}mo{l}^{1/2}litr{e}^{-1/2}$.
What is the value of $K_c$ for the reaction: $2S{O}_2\left(g\right)+O_2\left(g\right)\rightleftharpoons 2S{O}_3\left(g\right)$?

• A. $416litremo{l}^{-1}$
• B. $2.40\times {10}^{-3}litremo{l}^{-1}$
• C. $9.8\times {10}^{-2}litremol{e}^{-1}$
• D. $4.9\times {10}^{-2}litremo{l}^{-1}$

Answer 1

The correct option is A $416litremo{l}^{-1}$

For the reaction $S{O}_3\left(g\right)\rightleftharpoons S{O}_2\left(g\right)+\frac{1}{2}{O}_2\left(g\right),K_c=4.9\times {10}^{-2}$

Multiply theequation by $2$ and reverse it to obtain the reaction: $2S{O}_2\left(g\right)+O_2\left(g\right)\rightleftharpoons 2S{O}_3\left(g\right)$.

Hence, the equilibrium constant is $K^{\prime }=\frac{1}{K^2}=\frac{1}{(4.9\times {10}^{-2}{)}^2}=416$.

Hence, the correct option is $\text{A}$