Chemistry - Chemical Equilibrium Question with Solution | TestHub

$$2{\mathrm{SO}}_2(g)+{\mathrm{O}}_2(g)\rightleftharpoons 2{\mathrm{SO}}_3(g)$$ Equilibrium constant for this reaction, $$\begin{array}{c}K=\frac{{\left[{\mathrm{SO}}_3\right]}^2}{{\left[{\mathrm{SO}}_2\right]}^2\left[{\mathrm{O}}_2\right]}\\ {\mathrm{SO}}_3(g)\rightleftharpoons {\mathrm{SO}}_2(g)+\frac{1}{2}{\mathrm{O}}_2(g)\end{array}$$ 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 $${\left(K^{\prime }\right)}^2=\frac{{\left[{\mathrm{SO}}_2\right]}^2\left[{\mathrm{O}}_2\right]}{{\left[{\mathrm{SO}}_3\right]}^2}$$ $\begin{array}{rl}& =\frac{1}{K}\\ & =\frac{1}{278}\\ \therefore K^{\prime }& =\sqrt{\frac{1}{278}}\\ & =\sqrt{0.003597}\\ & =5.99\times 10^{-2}\\ & \approx 6\times 10^{-2}\end{array}$