The equilibrium constant $\left(K_C\right)$ for the following equilibrium $2…
The equilibrium constant $\left(K_C\right)$ for the following equilibrium
$2 \mathrm{SO}_2(g)+\mathrm{O}_2(g) \rightleftharpoons 2 \mathrm{SO}_3(g)$
at $563 \mathrm{~K}$ is 100 . At equilibrium, the number of moles of $\mathrm{SO}_3$ in the 10 litre flask is twice the number of moles of $\mathrm{SO}_2$, then the number of moles of oxygen is
0.4
0.3
0.2
0.1
Solution
Let, number of moles of $\mathrm{SO}_2=x$ moles of $\mathrm{SO}_3=2 x$
$2 \mathrm{SO}_2(g)+\mathrm{O}_2(g) \rightleftharpoons 2 \mathrm{SO}_3(g)$
At equilibrium,
$\begin{aligned}
K_c & =\frac{\left[\mathrm{SO}_3\right]^2}{\left[\mathrm{SO}_2\right]^2\left[\mathrm{O}_2\right]} \\
100 & =\frac{\left[\frac{2 x}{10}\right]^2}{\left[\frac{x}{10}\right]^2\left[\frac{n_{\mathrm{O}_2}}{10}\right]} \\
100 & =\frac{40}{n_{\mathrm{O}_2}} n_{\mathrm{O}_2}=\frac{40}{100}=0.4
\end{aligned}$
Hence, number of moles of oxygen $=0.4$