# Prove that concentrations of all species increase ultimately in a chemical equilibrium when pressure is increased

The statement:

For a chemical equilibrium in a sealed container, the concentration of "all" species always increases when the pressure of the mixture is increased by decreasing the volume of the container at constant temperature (assuming the ideal gas law is followed).

Here’s what I think I understand so far:

• According to Le Chatelier's principle, increasing the pressure (by decreasing the volume) of a gaseous equilibrium will shift the position of the equilibrium towards the side with fewer gas molecules.
• Initially, when the volume decreases, the concentrations of all gaseous species should increase because concentration is inversely proportional to volume.

However, I’m struggling to provide a rigorous mathematical proof showing that, even after the system re-establishes equilibrium , the concentrations of all species end up being higher than they were before the volume change.

I’ve tried attempting many examples in hopes of disproving this statement, but I consistently end up verifying it instead.

I’d greatly appreciate any detailed explanations or rigorous proofs that could clarify this for me.

Thanks a lot for your help!

• It an interesting question. I'm not sure if this is the right approach, since I've not tried it, but I'd start by seeing if I could get this to work: First, get a general expression for how the equilibrium position (the "advancement") of a reaction depends on pressure at constant T (that's the easy part). Then substitute into it a general expression for how the concentrations of the reactants and the products depend on advancement (that's the trickier part, because you'd need a general expresion that worked for all stoichiometries). Commented Aug 6 at 3:16
• As volume decreases so pressure increases over that initially present, it does not mean that the equilibrium will not change its extent, (i.e. $\alpha$ degree dissociated) just that the concentrations are greater than before. The equilibrium constant $K_p$ is unchanged as the temperature is constant, so $\alpha$ must change. Commented Aug 6 at 14:10

As the volume decreases so the pressure increases over that initially present, The equilibrium constant $$K_p$$ is unchanged as the temperature is constant, but $$\alpha$$ the degree of dissociation must change.

\displaystyle \begin{align}&A\;\rightleftharpoons\;2B\\& 1-\alpha,\quad 2\alpha\end{align}

and so at total pressure $$p$$ and partial pressures $$p_a,p_B$$, $$\displaystyle K_p=\frac{p_B^2}{p_A},\quad p_B=\frac{2\alpha}{1+\alpha}p,\quad p_{A}=\frac{1-\alpha}{1+\alpha}p$$

and substituting and solving for $$\alpha$$ gives

$$\displaystyle \alpha =\sqrt{\frac{K_p}{4p+K_p}}$$

In the reaction $$N_2O_4=2NO_2$$, $$K_p=0.14$$, and at $$1$$ atm, $$\alpha = 0.18$$, but at $$10$$ atm $$\alpha =0.06$$. If you substitute these values into the partial pressures which are proportional to concentration, the values are higher at the higher pressures. At $$1$$ atm, $$p_A =0.69, p_B=0.31$$ and at $$10$$ atm, $$p_A =8.9, p_B=1.1$$.

Perhaps you're overthinking the issue.

• Consider a single species, e.g., helium, reasonably close to an ideal gas at STP. If the pressure doubles, and the number of atoms stays the same, then the same number of atoms occupy one-half the volume -- the density has doubled, and the concentration of He atoms/L has doubled.

• Consider a mix of gases that do not react, at STP, e.g., 50% He and 50% Ne. Each behaves independently, and when compressed to half the volume, the concentration of each doubles.

• However, what if the gases do react, and equilibrium changes as pressure increases. Can pressure alone force ideal gases to shift equilibrium so far that even in the reduced volume, where the initial concentration of all species increases, before equilibrium is reached, that one concentration of one of those compressed species is below it's initial level before compression?

It depends what one means by "equilibrium". A mix of 50% $$\ce{H2}$$ and 50% $$\ce{Cl2}$$, (left in the dark for hours or months) might be considered in equilibrium -- or in a metastable state. Add sufficient pressure, and there would be virtually none of the original reactants. After the BANG!, the container would hold $$\ce{HCl}$$ only. If one considers the initial conditions in equilibrium, then it violates that premise, but if merely a metastable state, then not.

I would argue proving this for generalized stoichiometry would be practically impossible since variable stoichiometric coefficients wouldn't allow to explicitly obtain $$\alpha$$ as a function of volume change.

However, we will consider the following reversible reaction:

$$\ce{2A(g)<=>C(g)}$$

The change in moles for this reaction is:

$$\Delta n=1-2=-1$$

The subscript (0) will represent the initial non-equilibrium state (i.e. immediately after decreasing volume, but before reaching the second equilibrium state).

The subscript (1) will represent the initial equilibrium state (i.e. before decreasing volume).

The subscript (2) will represent the final equilibrium state (i.e. after decreasing volume).

Since there is no change in amounts between (1) and (0):

$$C_{A_1}V_1=n_{A_1}=n_{A_0}=C_{A_0}V_2\implies \boxed{C_{A_0}=C_{A_1}\left(\frac{V_1}{V_2}\right)}$$

$$C_{C_1}V_1=n_{C_1}=n_{C_0}=C_{C_0}V_2\implies \boxed{C_{C_0}=C_{C_1}\left(\frac{V_1}{V_2}\right)}$$

The concentrations of both species in (2) are:

$$C_{A_2}=C_{A_0}\left(1-\alpha\right)\implies \boxed{C_{A_2}=C_{A_1}\left(\frac{V_1}{V_2}\right)\left(1-\alpha\right)}$$

$$C_{C_2}=C_{A_0}\left(\theta_{C_0}+0.5\alpha\right)\implies \boxed{C_{C_2}=C_{A_1}\left(\frac{V_1}{V_2}\right)\left(\theta_{C_1}+0.5\alpha\right)}$$

The total concentrations in (1) and (2) are:

$$C_1=C_{A_1}+C_{C_1}\implies \boxed{C_1=C_{A_1}\left(1+\theta_{C_1}\right)}$$

$$C_2=C_{A_2}+C_{C_2}\implies \boxed{C_2=C_{A_1}\left(\frac{V_1}{V_2}\right)\left(1+\theta_{C_1}-0.5\alpha\right)}$$

The relationship between the equilibrium constant in terms of mole fraction in states (1) and (2) is:

$$\frac{K_{x_2}}{K_{x_1}}=\left(\frac{C_1}{C_2}\right)^{\Delta n}\implies \boxed{K_{x_2}=K_{x_1}\frac{\left(\frac{V_1}{V_2}\right)\left(1+\theta_{C_1}-0.5\alpha\right)}{1+\theta_{C_1}}}$$

Now we need a formula that generates the value of $$\alpha$$ only in terms of molar composition in state (1) and volume ratio:

$$\frac{\left(\theta_{C_1}+0.5\alpha\right)\left(1+\theta_{C_1}-0.5\alpha\right)}{\left(1-\alpha\right)^2}=K_{x_2}=K_{x_1}\frac{\left(\frac{V_1}{V_2}\right)\left(1+\theta_{C_1}-0.5\alpha\right)}{1+\theta_{C_1}}$$

Solving for $$\alpha$$:

$$\boxed{\alpha=\frac{\sqrt{16\;\theta_{C_1}K_{x_1}\left(\frac{V_1}{V_2}\right)+\theta_{C_1}+8K_{x_1}\left(\frac{V_1}{V_2}\right)+1}-4\theta_{C_1}\sqrt{\theta_{C_1}+1}-\sqrt{\theta_{C_1}+1}}{\sqrt{16\;\theta_{C_1}K_{x_1}\left(\frac{V_1}{V_2}\right)+\theta_{C_1}+8K_{x_1}\left(\frac{V_1}{V_2}\right)+1}+\sqrt{\theta_{C_1}+1}}}$$

Where:

$$\boxed{K_{x_1}=\frac{X_{C_1}}{X_{A_1}^2}}$$

$$\boxed{\theta_{C_0}=\frac{X_{C_0}}{X_{A_0}}=\frac{n_{C_0}}{n_{A_0}}=\frac{n_{C_1}}{n_{A_1}}=\frac{X_{C_1}}{X_{A_1}}=\theta_{C_1}}$$

Now, we can define the following concentration ratios:

$$\boxed{\frac{C_{A_2}}{C_{A_1}}=\left(\frac{V_1}{V_2}\right)\left(1-\alpha\right)}$$

$$\boxed{\frac{C_{C_2}}{C_{C_1}}=\left(\frac{V_1}{V_2}\right)\frac{\left(\theta_{C_1}+0.5\alpha\right)}{\theta_{C_1}}}$$

By specifying a particular initial equilibrium composition state, $$\alpha$$ becomes a function only of the volume change ratio, which means the concentration ratios also become functions only of the volume change ratio:

$$\begin{array} {|r|r|}\hline \pmb{X_{A_1}} & \pmb{X_{C_1}} & \pmb{K_{x_1}} & \pmb{\theta_{C_1}} &\pmb{\alpha} & \pmb{\frac{C_{A_2}}{C_{A_1}}} & \pmb{\frac{C_{C_2}}{C_{C_1}}} \\ \hline \to 1 & \to 0 & \to 0 &\to 0 & \to 0 &\to \frac{V_1}{V_2} &\to \frac{V_1}{V_2} \\ \hline 0.25 & 0.75 & 12 & 3 & \frac{\sqrt{672\frac{V_1}{V_2}+4}-26}{\sqrt{672\frac{V_1}{V_2}+4}+2} & \frac{V_1}{V_2}\left(1-\frac{\sqrt{672\frac{V_1}{V_2}+4}-26}{\sqrt{672\frac{V_1}{V_2}+4}+2}\right) & \frac{V_1}{V_2}\left(1+\frac{\sqrt{672\frac{V_1}{V_2}+4}-26}{6\sqrt{672\frac{V_1}{V_2}+4}+12}\right)\\ \hline 0.50 & 0.50 & 2 & 1 & \frac{\sqrt{48\frac{V_1}{V_2}+2}-5\sqrt{2}}{\sqrt{48\frac{V_1}{V_2}+2}+\sqrt{2}} & \frac{V_1}{V_2}\left(1-\frac{\sqrt{48\frac{V_1}{V_2}+2}-5\sqrt{2}}{\sqrt{48\frac{V_1}{V_2}+2}+\sqrt{2}}\right) & \frac{V_1}{V_2}\left(1+\frac{\sqrt{48\frac{V_1}{V_2}+2}-5\sqrt{2}}{2\sqrt{48\frac{V_1}{V_2}+2}+2\sqrt{2}}\right) \\ \hline 0.75 & 0.25 & \frac{4}{9} & \frac{1}{3} & \frac{\sqrt{\frac{160}{27}\frac{V_1}{V_2}+\frac{4}{3}}-\frac{7}{3}\sqrt{\frac{4}{3}}}{\sqrt{\frac{160}{27}\frac{V_1}{V_2}+\frac{4}{3}}+\sqrt{\frac{4}{3}}} & \frac{V_1}{V_2}\left(1-\frac{\sqrt{\frac{160}{27}\frac{V_1}{V_2}+\frac{4}{3}}-\frac{7}{3}\sqrt{\frac{4}{3}}}{\sqrt{\frac{160}{27}\frac{V_1}{V_2}+\frac{4}{3}}+\sqrt{\frac{4}{3}}}\right) & \frac{V_1}{V_2}\left(1+\frac{\sqrt{\frac{160}{27}\frac{V_1}{V_2}+\frac{4}{3}}-\frac{7}{3}\sqrt{\frac{4}{3}}}{\frac{2}{3}\sqrt{\frac{160}{27}\frac{V_1}{V_2}+\frac{4}{3}}+\frac{2}{3}\sqrt{\frac{4}{3}}}\right) \\ \hline \to 0 & \to 1 & \to \infty & \to \infty & \to 1 & \to 0 & \to 0 \\ \hline \end{array}$$

It doesn't matter what initial equilibrium composition state we choose, but I will arbitrarily select the first one shown in the table.

The corresponding concentration ratio profiles are:

We can observe that when the volume ratio is 1 (i.e. there is no volume change), both concentration ratios are equal to 1. This is obvious, since the driving force for equilibrium disturbance is a decrease in volume. You can verify the same thing happens when using a different initial equilibrium composition state.

We also need to remember that since we are decreasing volume in this demonstration, the volume ratio $$\frac{V_1}{V_2}$$ must be greater than 1.

For every value of volume ratio greater than 1, it follows that both concentration ratios are also greater than 1:

$$\frac{V_1}{V_2}>1\implies \frac{C_{A_2}}{C_{A_1}}, \frac{C_{C_2}}{C_{C_1}}>1$$

Which ultimately means both species necessarily exhibit an increase in concentration after volume is decreased, and the second equilibrium state is reached.

A similar derivation can be done for other reactions with different stoichiometry, but that can result in significantly more complex equations.