Question

A system at equilibrium is described by the equation at a fixed temperature T.
$S{O}_{2}C{l}_2\left(g\right)\rightleftharpoons S{O}_2\left(g\right)+C{l}_2\left(g\right)$
What effect will an increases in the total pressure is caused by a decrease in volume have on the equilibrium?

• A. Concentration of $S{O}_{2}C{l}_2\left(g\right)$ increases
• B. Concentration of $S{O}_2\left(g\right)$ increases
• C. Concentration of $C{l}_2\left(g\right)$ increases
• D. Concentration of all gases increases

Answer 1

The correct option is D Concentration of all gases increases

As per Le - Chatelier principle, If pressure is increased equilibrium will shift in that direction where number of moles are less.
So, as per reaction,$S{O}_{2}C{l}_2\left(g\right)\rightleftharpoons S{O}_2\left(g\right)+C{l}_2\left(g\right)$
the equilibrium will shift towards the backward reaction

Concentration of all gases increases and equilibrium shifts toward less no. of moles but new equilibrium concentration of every gas would be higher than earlier.



Theory:

Effect of change in Volume :
According to Boyle's law at constant temperature.
$P\propto \frac{1}{V}$
For a reaction, when the volume V increases the concentration $\left(\frac{n_g}{V}\right)$ of the reactants and the products decrease.

Consider $A\left(g\right)\rightleftharpoons B\left(g\right)$ When, $\Delta n_g=0$, we know that even if pressure changes there is no shift in the equilibrium.

Therefore even if volume changes there is no shift in equilibrium.

For a reaction $\Delta n_g>0$
$A\left(g\right)\rightleftharpoons 2B\left(g\right)$ when $\Delta n_g=+ve$
when volume increases then $\frac{n_g}{V}$ decreases, then the reaction goes where $\frac{n_g}{V}$ increases . Reaction goes forward.

For a reaction when $\Delta n_g<0$ $2A\left(g\right)\rightleftharpoons B\left(g\right)$ when $\Delta n_g=-ve$
Volume $V$ increases then $\frac{n_g}{V}$ decreases, then the reaction goes where $\frac{n_g}{V}$ increases . Reaction goes backward.