Question

A mixture of ${SO}_3,{SO}_2$ and $O_2$ gases is maintained in a $10L$ flask at a temperature at which the equilibrium constant for the reaction is $100$ and the reaction is, $2S{O}_2\left(g\right)+O_2\left(g\right)\rightleftharpoons 2S{O}_3\left(g\right)$.
(i) If the number of moles of ${SO}_2$ and ${SO}_3$ in the flask are equal, how many moles of $O_2$ are present?
(ii) If the number of moles of ${SO}_3$ in the flask is twice the number of moles of ${SO}_2$, how many moles of oxygen are present?

Answer 1

(i) $2S{O}_2\left(g\right)+O_2\left(g\right)\rightleftharpoons 2S{O}_3\left(g\right)$

At equilibrium, let the number of moles of each of ${SO}_2$ and ${SO}_3$ be $n_1$ and of oxygen $n_2$, i.e.,
$\left[S{O}_2\right]=\frac{n_1}{10}$, $\left[O_2\right]=\frac{n_2}{10}$, $\left[S{O}_3\right]=\frac{n_1}{10}$

Applying law of mass action,
$K_c=\frac{{\left[S{O}_3\right]}^2}{{\left[S{O}_2\right]}^2\left[O_2\right]}=100$

$n_2=0.1mole$

$\therefore$ Moles of Oxygen $=0.1mole$

(ii) Let the number of moles of ${SO}_2$ be $n_1$

So, number of moles of ${SO}_3=2n_1$

Let the number of moles of oxygen be $n_2$

$K_c=\frac{{(2\times \frac{n_1}{10})}^2}{{\left(\frac{n_1}{10}\right)}^2\left(\frac{n_2}{10}\right)}or100=\frac{40}{n_2}$

$n_2=0.4mole$

$\therefore$ Moles of Oxygen $=0.4mole$