Recently, while studying chemical equilibrium, I came across the following statement in my textbook:

An increase in pressure favours the side of an equilibrium reaction that has the smaller number of gas molecules.

Although there is a simple explanation for why this statement holds true using Le Chatelier’s principle, I thought it’d be a fun exercise to derive it mathematically by comparison of the equilibrium constant $K_c$ and the reaction quotient $Q$. (More importantly, writing this blog post provides me an excuse to not learn more chemistry and do math instead, which is always preferable.)

Outline of Approach

Recall that if $Q < K_c$ then the products are favored and conversely if $Q > K_c$ the reactants are favored. Thus, if we can somehow show that, in the event of an increase in pressure, $Q$ is greater than $K_c$ if and only if there are fewer moles of gas on the reactant side (and similar for the case where $Q$ is smaller), then we’ll have proven the desired claim.

Proof

Consider the general reaction

$$ \ce{aA_{(g)} + bB_{(g)} + \dots <=> rR_{(g)} + sS_{(g)} + \dots} $$

where $A$, $B$, $\dots$ are the reactants and $R$, $S$, $\dots$ are the products.

Suppose the system is currently in equilibrium. Then the initial reaction quotient $Q_0$ and the equilibrium constant $K_c$ must necessarily be equal, so

$$ Q_0 = K_c = \frac{[R]^r[S]^s\cdots}{[A]^a[B]^b\cdots}. \tag{1} $$

Our goal now is to relate concentration to the total pressure of the system, say $P_0$. Recall that concentration $c = n/V$ where $n$ denotes the number of moles and the $V$ the volume. Rearranging the ideal gas law $PV = nRT$ to isolate $n/V$ yields

$$ \frac{n}{V} = \frac{P}{RT} = c. \tag{2} $$

Note that we cannot directly substitute this back into Equation 1 yet as each gas has its own partial pressure – that is, the $P$ for each gas is different. To account for this, recall that

$$ p_X = p_{\Sigma} \frac{n_X}{n_{\Sigma}} $$

that is, the partial pressure is the total pressure scaled by the mole fraction. Combining this with Equation 2 yields an expression for the concentration of any gas (say $X$) in the reaction in terms of the total pressure of the system:

$$ c_X = [X] = \frac{p_X}{RT} = \frac{P_0n_{\Sigma}}{RTn_X} \tag{3} $$

Substituting the final expression for concentration into Equation 1 and factoring out the part dependent on pressure, we get

$$ \begin{align*} Q_0 = K_c &= \cfrac{\left(\cfrac{P_0n_{\Sigma}}{RTn_R}\right)^r\left(\cfrac{P_0n_{\Sigma}}{RTn_S}\right)^s\cdots}{\left(\cfrac{P_0n_{\Sigma}}{RTn_A}\right)^a\left(\cfrac{P_0n_{\Sigma}}{RTn_B}\right)^b\cdots} \\ &= \cfrac{{P_0}^{r + s + \dots}}{{P_0}^{a + b + \dots}} \cfrac{\left(\cfrac{n_{\Sigma}}{RTn_R}\right)^r\left(\cfrac{n_{\Sigma}}{RTn_S}\right)^s\cdots}{\left(\cfrac{n_{\Sigma}}{RTn_A}\right)^a\left(\cfrac{n_{\Sigma}}{RTn_B}\right)^b\cdots}. \tag{4} \end{align*} $$

(Aside: Substituting the expression for concentration in terms of partial pressure gives us a rather interesting result; see Appendix 1.)

For brevity write $$ \Delta n = \overbrace{(r + s + \cdots)}^{\text{tot. num of product moles}} - \underbrace{(a + b + \cdots)}_{\text{tot. num of reactant moles}} \tag{5} $$ such that the pressure factor in front is just ${P_0}^{\Delta n}$. Also put

$$ C = \cfrac{\left(\cfrac{n_{\Sigma}}{RTn_R}\right)^r\left(\cfrac{n_{\Sigma}}{RTn_S}\right)^s\cdots}{\left(\cfrac{n_{\Sigma}}{RTn_A}\right)^a\left(\cfrac{n_{\Sigma}}{RTn_B}\right)^b\cdots} $$

so Equation 4 simplifies to just

$$ Q_0 = K_c = C{P_0}^{\Delta n}. \tag{6} $$

Now consider what happens when the pressure is increased to $P_1$. The equilibrium constant $K_c$ does not change since it is only affected by temperature. What about the reaction quotient, $Q_1$? The factor $C$ from Equation 6 remains the same, so the only change is replacing $P_0$ with $P_1$:

$$ Q_1 = C{P_1}^{\Delta n}. $$

Evidently the reaction quotient $Q_1$ is no longer equal to the equilibrium constant, so the equlibrium position will shift to restore equilibrium. To determine whether the forward or reverse reaction will be favored, we just need to compare $Q_1$ to $K_c$. Without loss of generality, consider the case where $Q_1$ is smaller:

$$ \begin{align*} Q_1 &< K_c \\ CP_1^{\Delta n} &< CP_0^{\Delta n} \\ P_1^{\Delta n} &< P_0^{\Delta n}. \end{align*} $$

Since $P_1 > P_0$, this can only be the case if $\Delta n < 0$. But we defined $\Delta n$ as the total moles of gas on the product side subtract the total moles of gas on the reactant side, so

$$ \begin{align*} \Delta n &< 0 \\ \sum n_{\rm{products}} - \sum n_{\rm{reactants}} &< 0 \\ \sum n_{\rm{products}} &< \sum n_{\rm{reactants}}. \end{align*} $$

In other words, $Q_1 < K_c$ if and only if there are fewer moles of gas on the product side. But $Q_1 < K_c$ also implies that the equilibrium position should shift rightward (favoring products.) The analysis for the case in which there are fewer moles of gas on the reactant side is similar – the equilibrium position shifts leftward favoring reactants, which is what we wanted to show. $\square$

Appendix 1: Equilibrium Constant for Partial Pressures

As part of our proof, we derived the following relationship (Equation 3):

$$ c_X = [X] = \frac{p_X}{RT} = \frac{P_0n_{\Sigma}}{RTn_X}. $$

Recall that we originally substituted the final expression,

$$ \frac{P_0n_{\Sigma}}{RTn_X} $$

back into the equilibrium constant expression (leading to Equation 4). However, if instead we substitute the expression

$$ \frac{p_X}{RT} $$

then we get something rather intriguing:

$$ \begin{align*} K_c &= \cfrac{\left(\cfrac{p_R}{RT}\right)^r\left(\cfrac{p_S}{RT}\right)^s\cdots}{\left(\cfrac{p_A}{RT}\right)^a\left(\cfrac{p_B}{RT}\right)^b\cdots} \\ &= \frac{{p_R}^r{p_S}^s\cdots}{{p_A}^a{p_B}^b\cdots} (RT)^{(a + b + \dots) - (r + s + \dots)}. \end{align*} $$

Indeed, observe that at constant temperature, the factor $$ (RT)^{(a + b + \dots) - (r + s + \dots)} $$ is constant. But if the temperature is constant $K_c$ must, too, be constant, so that means the expression

$$ \frac{{p_R}^r{p_S}^s\cdots}{{p_A}^a{p_B}^b\cdots} $$

is a constant as well! This expression may thus be viewed as a kind of equlibrium constant, the difference being that it is calculated using partial pressures instead of concentration. For this reason, it is often referred to as the equilibrium constant using partial pressures, denoted $K_p$. The relationship between $K_c$ and $K_p$ is readily given by isolating $K_p$ in Equation 7:

$$ \begin{align*} K_c &= K_p (RT)^{(a + b + \dots) - (r + s + \dots)} \\ K_p &= K_c (RT)^{(r + s + \dots) - (a + b + \dots)} \\ K_p &= K_c (RT)^{\Delta n} \end{align*} $$

where $\Delta n$ is as defined in Equation 5.