Question

At a certain temperature, the equilibrium constant $K_c$ is $16$ for the reaction,
$S{O}_{2\left(g\right)}+N{O}_{2\left(g\right)}$ $\leftrightharpoons S{O}_{3\left(g\right)}+N{O}_{\left(g\right)}$
If $1.0mol$ each of all the four gases is taken in a one litre container the concentration of $N{O}_2$ at equilibrium would be:

• A. $1.6mol{L}^{-1}$
• B. $0.8mol{L}^{-1}$
• C. $0.4mol{L}^{-1}$
• D. $0.6mol{L}^{-1}$

Answer 1

The correct option is C $0.4mol{L}^{-1}$

The give reaction is :-

${SO}_{2\left(g\right)}+{NO}_{2\left(g\right)}\rightleftharpoons {SO}_{3\left(g\right)}+NO\left(g\right)$

Initial conc : $1$ $1$ $1$ $1$

At eqn : $(1-x)$ $(1-x)$ $(1+x)$ $(1+x)$

Now, $K_C=16$

$\Rightarrow \frac{\left[NO\right]\left[{SO}_3\right]}{\left[{SO}_2\right]\left[{NO}_2\right]}=16\Rightarrow \frac{(1+x)(1+x)}{(1-x)(1-x)}=16$

$\Rightarrow {(1+x)}^2=16{(1-x)}^2$

$\Rightarrow 1+x=4(1-x)$

$\Rightarrow 5x=3$

$\Rightarrow x=3/5$

So, conc. of ${NO}_2=(1-x)=(1-3/5)=2/5=0.4mol/L$