Given that the equilibrium constant for the reaction, $2 \mathrm{SO}_2(g)+\mathrm{O}_2(g) \rightleftharpoons…
Given that the equilibrium constant for the reaction,
$2 \mathrm{SO}_2(g)+\mathrm{O}_2(g) \rightleftharpoons 2 \mathrm{SO}_3(g)$
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?
$\mathrm{SO}_3(g) \rightleftharpoons \mathrm{SO}_2(g)+\frac{1}{2} \mathrm{O}_2(g)$
$1.8 \times 10^{-3}$
$3.6 \times 10^{-3}$
$6.0 \times 10^{-2}$
$1.3 \times 10^{-5}$
Solution
$2 \mathrm{SO}_2(g)+\mathrm{O}_2(g) \rightleftharpoons 2 \mathrm{SO}_3(g)$
Equilibrium constant for this reaction,
$\begin{gathered}
K=\frac{\left[\mathrm{SO}_3\right]^2}{\left[\mathrm{SO}_2\right]^2\left[\mathrm{O}_2\right]} \\
\mathrm{SO}_3(g) \rightleftharpoons \mathrm{SO}_2(g)+\frac{1}{2} \mathrm{O}_2(g)
\end{gathered}$
Equilibrium constant for this reaction,
$K^{\prime}=\frac{\left[\mathrm{SO}_2\right]\left[\mathrm{O}_2\right]^{\frac{1}{2}}}{\left[\mathrm{SO}_3\right]}$
On squaring Eq. (ii) both sides, we have
$\left(K^{\prime}\right)^2=\frac{\left[\mathrm{SO}_2\right]^2\left[\mathrm{O}_2\right]}{\left[\mathrm{SO}_3\right]^2}$
$\begin{aligned} & =\frac{1}{K} \\ & =\frac{1}{278} \\ \therefore \quad K^{\prime} & =\sqrt{\frac{1}{278}} \\ & =\sqrt{0.003597} \\ & =5.99 \times 10^{-2} \\ & \approx 6 \times 10^{-2}\end{aligned}$