Pressure and chemical equilibrium

Recently, while studying the concept of 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.

While there’s a simple explanation for why this statement holds true using Le Chatelier’s principle, I thought it’d be a fun exercise to convince myself of the statement via a different, more convoluted route: deducing the expected direction of the shift via comparing the equilibrium constant $K_c$ with the reaction quotient $Q$ .

Idea

In general, the reaction quotient $Q$ tends toward the equilibrium constant $K_c$ . Concretely, 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 claim. Let’s try that!

Analysis

Consider the general reaction

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

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

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

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

Our goal is to express the reaction quotient as a function of the total pressure of the system, say $P_0$ . Recall that concentration $c = n/V$ where $n$ is the number of moles and $V$ is the volume. Rearranging the ideal gas law $PV = nRT$ for $n/V$ yields

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

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, we use

p_X = p_{\mathrm{tot}} \frac{n_X}{n_{\mathrm{tot}}}

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 $X$ ) participating in the reaction in terms of the total pressure of the system:

c_X = [X] = \frac{p_X}{RT} = \frac{P_0n_{\mathrm{tot}}}{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_{\mathrm{tot}}}{RTn_R}\right)^r\left(\cfrac{P_0n_{\mathrm{tot}}}{RTn_S}\right)^s\cdots}{\left(\cfrac{P_0n_{\mathrm{tot}}}{RTn_A}\right)^a\left(\cfrac{P_0n_{\mathrm{tot}}}{RTn_B}\right)^b\cdots} \\\\ &= \cfrac{{P_0}^{r + s + \dots}}{{P_0}^{a + b + \dots}} \cfrac{\left(\cfrac{n_{\mathrm{tot}}}{RTn_R}\right)^r\left(\cfrac{n_{\mathrm{tot}}}{RTn_S}\right)^s\cdots}{\left(\cfrac{n_{\mathrm{tot}}}{RTn_A}\right)^a\left(\cfrac{n_{\mathrm{tot}}}{RTn_B}\right)^b\cdots}. \tag{4} \end{align*}

(As an aside: Substituting the expression for concentration in terms of partial pressure instead gives us a rather interesting link to the equilibrium constant using partial pressures.)

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 abbreviate the constant factor as $k$ ,

k = \cfrac{\left(\cfrac{n_{\mathrm{tot}}}{RTn_R}\right)^r\left(\cfrac{n_{\mathrm{tot}}}{RTn_S}\right)^s\cdots}{\left(\cfrac{n_{\mathrm{tot}}}{RTn_A}\right)^a\left(\cfrac{n_{\mathrm{tot}}}{RTn_B}\right)^b\cdots}

so that Equation 4 simplifies to just

Q_0 = K_c = k{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 $k$ from Equation 6 remains the same, so the only change is replacing $P_0$ with $P_1$ :

Q_1 = k{P_1}^{\Delta n}.

Evidently the reaction quotient $Q_1$ is no longer equal to the equilibrium constant, so the equilibrium position will shift to restore equilibrium. To determine whether the forward or reverse reaction will be favored, we need to compare $Q_1$ to $K_c$ . Let us examine the case where $Q_1 < K_c$ :

\begin{align*} Q_1 &< K_c \\\\ kP_1^{\Delta n} &< kP_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 difference between the total moles of gas on the product side and 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.

A connection to the equilibrium constant using partial pressures

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

c_X = [X] = \frac{p_X}{RT} = \frac{P_0n_{\mathrm{tot}}}{RTn_X}.

Recall that we originally substituted the final expression,

\frac{P_0n_{\mathrm{tot}}}{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.

Idea#

Analysis#

A connection to the equilibrium constant using partial pressures#

Idea

Analysis

A connection to the equilibrium constant using partial pressures