Question

The volume of the reaction vessel containing an equilibrium mixture is increased in the following reaction:
$S{O}_{2}C{l}_2\left(g\right)\rightleftharpoons S{O}_2\left(g\right)+C{l}_2\left(g\right)$
When equilibrium is re-established:

• A. The amount of $C{l}_2$ remains unchanged.
• B. The amount of $C{l}_2$ increases.
• C. The amount of $S{O}_{2}C{l}_2$ increases.
• D. The amount of $S{O}_2$ decreases.

Answer 1

The correct option is B The amount of $C{l}_2$ increases.

Checking $\Delta n_g$ for the given reaction:

$S{O}_{2}C{l}_2\left(g\right)\rightleftharpoons S{O}_2\left(g\right)+C{l}_2\left(g\right)$

$\Delta n_g=2-1=1(i.e.\Delta n_g>0)$

For $\Delta n_g>0$ when volume is increased, $\frac{n_g}{V}\downarrow$ the reaction goes where $\frac{n_g}{V}\uparrow$ i.e. the reaction goes where $n_g\uparrow$ .
It increases in the forward direction.
So, the reaction goes in forward direction.
Hence, when equilibrium is re-established, the amount of products formed increases .
$\Rightarrow$ Amount of $C{l}_2$ increases.