Question

$2{SO}_2\left(g\right)+O_2\left(g\right)\leftrightharpoons 2{SO}_3\left(g\right)$
Given that the equilirbium constant for the reaction above has a value of $278$ at a particular temperature, what is the value of equilibrium constant for the following reaction at same temperature?
${SO}_3\left(g\right)\rightleftharpoons {SO}_2\left(g\right)+\frac{1}{2}{O}_2\left(g\right)$

• A. $1.3\times {10}^{-5}$
• B. $1.8\times {10}^{-3}$
• C. $3.6\times {10}^{-3}$
• D. $6\times {10}^{-2}$

Answer 1

The correct option is A $1.3\times {10}^{-5}$

The given reaction is :-

$2{SO}_2\left(g\right)+O_2\left(g\right)\rightleftharpoons 2{SO}_3\left(g\right)$

$\Rightarrow K_{eq}=\frac{{\left[{SO}_3\right]}^2}{{\left[{SO}_2\right]}^2\left[O_2\right]}=278\to \left(I\right)$

Now, for ${SO}_3\left(g\right)\rightleftharpoons {SO}_2\left(g\right)+\frac{1}{2}{O}_2\left(g\right)$

$\Rightarrow {K_{eq}}^{{}^{\prime }}=\frac{\left[{SO}_2\right]{\left[O_2\right]}^{1/2}}{\left[{SO}_3\right]}$

$={\left(\frac{1}{K_{eq}}\right)}^2={\left(\frac{1}{278}\right)}^2=1.3\times {10}^{-5}$