# Increase in equilibrium concentration of the reactant via decrease in container volume

Consider the hypothetical reaction $\ce{3A(g) -> B(g) + C(g)}$; how to increase the equilibrium concentration of $\ce{A}$, given that the reaction is endothermic?

1. increasing the volume of the container
2. decreasing the temperature
3. decreasing the volume of the container
4. adding $\ce{B(g)}$ at equilibrium

Answer given - 2, 3, and 4. I have trouble with the option 3.

I know according to the Le Chatelier's principle, equilibrium will shift forward on reducing the container volume, but the solution says it still manages to have an increase in the overall equilibrium molar concentration of $\ce{A(g)}$; and that's what I'm not understanding.

The catch here is that when the reaction shifts forward - to counter-balance the decrease in volume - the reduction in the moles of $\ce{A}$ is much lesser than the decrease in volume, hence, the overall ratio, $\frac{\text{moles}}{\text{volume}}$ actually increases!
You can demonstrate this via your own reaction. Assume $K_\mathrm{c}=\pu{1M-1}$, and all equilibrium concentrations to be $\pu{1M}$. When volume is halved, the concentrations double, so $Q=\pu{0.5M-1}$. The reaction thus shifts forward as $Q<K$.
Let the new concentrations be $\pu{2M}-3x, \pu{2M}+x, \pu{2M}+x$ for $\ce{A,B,C}$ respectively. Solving for $x$, you'll get $x\approx0.117$. Hence, the new equilibrium concentrations are $\pu{1.64M}, \pu{2.117M}, \pu{2.117M}$. In fact, not only the reactant, but all species show an increase in their concentration!
Moreover, the same result is obtained in the reverse direction. If the container volume was increased instead, all species would show a decrease in their volume. (the overall ratio $\frac{\text{moles}}{\text{volume}}$ decreases, as the moles increase far less proportionately as compared to the volume)