Question

Given that the equilibrium constant for the reaction, $2S{O}_{2\left(g\right)}+O_{2\left(g\right)}\rightleftharpoons 2S{O}_{3\left(g\right)}$, has a value of 278 at a particular temperature. What is the value of the equilibrium constant for the following reaction at the same temperature?

$S{O}_{3\left(g\right)}\rightleftharpoons S{O}_{2\left(g\right)}+\frac{1}{2}{O}_{2\left(g\right)}$

• A. $3.6\times {10}^{-3}$
• B. $6.0\times {10}^{-2}$
• C. $1.3\times {10}^{-5}$
• D. $1.8\times {10}^{-3}$

Answer 1

The correct option is B $6.0\times {10}^{-2}$

$2S{O}_2\left(g\right)+O_2\left(g\right)\leftrightharpoons 2S{O}_3\left(g\right)$, $K=278$ ....(i)

By reversing the equation (i), we get

$2S{O}_3\left(g\right)\leftrightharpoons 2S{O}_2\left(g\right)+O_2\left(g\right)$ .....(ii)

Equilibrium constant for this reaction is

$K^{\prime }=\frac{1}{K}=1/278$

By dividing the equation (ii) by 2, we get the desired equation.

$S{O}_3\left(g\right)$ ⇋ $S{O}_2\left(g\right)+1/2O_2\left(g\right)$ .....(iii)

The equilibrium constant for this reaction

$K^{\prime }$ $=(\frac{1}{K}{)}^{1/2}$

$K^{\prime }=5.99\times {10}^{-2}$

So, B option is correct.