# 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.

You're right that at first glance, this is indeed self-contradictory. However, on a closer look, here's how you can realize that it isn't!

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)

I must state that this isn't a rigorous proof but a layman's explanation for why this is happening, and how to understand it in simple terms. I would leave it up to a more experienced physical chemist, to demonstrate the applicability of the rules I've stated in bold above, via proofs or other methods.

Let the subscript (1) represent the initial equilibrium state, the subscript (0) the initial non-equilibrium state, and the subscript (2) the final equilibrium state.

Assuming the values proposed by Gaurang Tandon:

$$K_c=\frac{C_{B_1}C_{C_1}}{C_{A_1}}=\frac{(1)(1)}{(1)}=1$$

For simplicity, we can define the ratio of volumes between (2) and (1) as:

$$\frac{V_2}{V_1}=\Omega$$

The ratio of concentrations of A between (1) and (0) is:

$$\frac{C_{A_1}}{C_{A_o}}=\frac{\require{cancel}\cancel{n_{A_1}}}{\cancel{n_{A_o}}}\frac{V_o}{V_1}=\frac{V_2}{V_1}=\Omega$$

The ratio of concentrations of A between (2) and (0) is:

$$\frac{C_{A_2}}{C_{A_o}}=\frac{n_{A_2}}{n_{A_o}}\frac{\cancel{V_o}}{\cancel{V_2}}=1-\alpha$$

Dividing both concentration ratios:

$$\frac{\frac{C_{A_2}}{\cancel{C_{A_o}}}}{\frac{C_{A_1}}{\cancel{C_{A_o}}}}=\frac{C_{A_2}}{C_{A_1}}=\frac{1-\alpha}{\Omega}$$

Using the following relationship between $$\alpha$$ and $$K_c$$, we can obtain a function of $$\Omega$$ in terms of $$\alpha$$:

$$\frac{\left(1+\frac{1}{3}\alpha\right)^2}{\left(1-\alpha\right)^3}=K_c\;C_{A_o}=K_c\frac{C_{A_1}}{\Omega}=\frac{1}{\Omega}$$

Substituting above, we get a function of the ratio of concentrations of A between (2) and (1) only in terms of $$\alpha$$:

$$\pmb{\boxed{\frac{C_{A_2}}{C_{A_1}}=\left(\frac{1+\frac{1}{3}\alpha}{1-\alpha}\right)^2}}$$

Since $$\alpha$$ is the dissociation fraction, its domain is:

$$0<\alpha<1$$

As $$\alpha$$ approaches 0, $$C_{A_2}$$ approaches $$C_{A_1}$$, and as $$\alpha$$ approaches 1, $$C_{A_2}$$ approaches infinity:

$$\alpha\to0\implies\frac{C_{A_2}}{C_{A_1}}\to1$$

$$\alpha\to1\implies\frac{C_{A_2}}{C_{A_1}}\to\infty$$

So the concentration ratio can only result in the following values:

$$1<\frac{C_{A_2}}{C_{A_1}}<\infty$$

In other words, the concentration ratio necessarily increases even though an equilibrium shift occurs in the forward direction as a result of a decrease in volume.

We can use this formula to evaluate and confirm the specific example given above by Gaurang Tandon:

$$\alpha=\frac{ax}{C_{A_o}}=\frac{3(0.117)}{2}=0.176\implies \frac{C_{A_2}}{C_{A_1}}=1.65$$