Question

For a reaction, $2S{O}_{2\left(g\right)}+O_{2\left(g\right)}\rightleftharpoons 2S{O}_{3\left(g\right)},1.5$ moles of $S{O}_2$ and $1$ mole of $O_2$ are taken in a $2$L vessel. At equilibrium the concentration of $S{O}_3$ was found to be $0.35mol{L}^{-1}$. The $K_c$ for the reaction would be ?

• A. $2.35Lmo{l}^{-1}$
• B. $1.4Lmo{l}^{-1}$
• C. $0.6Lmo{l}^{-1}$
• D. $2.95Lmo{l}^{-1}$

Answer 1

The correct option is A $2.35Lmo{l}^{-1}$

For the given reaction,

$2{SO}_2+O_2\rightleftharpoons 2{SO}_3$

Initial moles ; $1.5$ $1$ $0$

At eqm : $1.5-2x$ $1-x$ $2x$ (let us suppose)

Now given that conc. of ${SO}_3$ at eqm $=0.35$ $mol/L$

$\Rightarrow 2x=0.70\Rightarrow x=0.35$ ($\because$ the container is of $2L$)

Now, $K_C=\frac{{\left[{SO}_3\right]}^2}{{\left[{SO}_2\right]}^2\left[O_2\right]}$

$=\frac{{\left(2x\right)}^2}{{(1.5-2x)}^2(1-x)}$

$\Rightarrow K_C=\frac{{\left[\frac{0.7}{2}\right]}^2}{{\left(\frac{1.5-0.7}{2}\right)}^2\times \left(\frac{1-0.35}{2}\right)}$

$=\frac{0.7\times 0.7\times 2}{0.8\times 0.8\times 0.65}=2.35{Lmol}^{-1}$