7.4 Temperature and pressure effects on equilibrium

Temperature and Pressure Effects on Equilibrium

Temperature and pressure are the two main external variables that shift the position of a chemical equilibrium. Understanding how and why they do so connects Le Chatelier's qualitative reasoning to the quantitative thermodynamic framework built around the van 't Hoff equation. This section covers the qualitative predictions first, then develops the math you'll need to calculate new equilibrium constants at different temperatures.

Temperature Effects on Equilibrium

Le Chatelier's Principle and Temperature Changes

Le Chatelier's principle says that when a system at equilibrium experiences a stress, it shifts to partially counteract that stress and reach a new equilibrium.

Think of temperature as adding or removing energy. Heat is effectively a "reactant" in endothermic reactions and a "product" in exothermic ones. That framing makes the predictions straightforward:

• Endothermic reactions ($\Delta H > 0$): Increasing temperature supplies the "reactant" (heat), so equilibrium shifts toward products. Decreasing temperature shifts it toward reactants.
• Exothermic reactions ($\Delta H < 0$): Increasing temperature adds to the "product" side (heat), so equilibrium shifts toward reactants. Decreasing temperature shifts it toward products.

A critical distinction from pressure effects: temperature changes actually alter the value of the equilibrium constant $K$. For endothermic reactions, $K$ increases with temperature. For exothermic reactions, $K$ decreases with temperature. The size of the shift depends on both the magnitude of the temperature change and the magnitude of $\Delta H$.

Examples of Temperature Effects on Equilibrium

• $\text{N}_2(g) + \text{O}_2(g) \rightleftharpoons 2\text{NO}(g)$ (endothermic): Raising the temperature increases $K$ and favors NO production. This is why NO forms in high-temperature combustion engines but is negligible at room temperature.
• $2\text{SO}_2(g) + \text{O}_2(g) \rightleftharpoons 2\text{SO}_3(g)$ (exothermic): Lowering the temperature increases $K$ and favors $\text{SO}_3$. Industrially, this creates a trade-off: low temperatures favor product but slow the reaction kinetically.
• $\text{CaCO}_3(s) \rightleftharpoons \text{CaO}(s) + \text{CO}_2(g)$ (endothermic, $\Delta H° = +178 \text{ kJ/mol}$): Heating drives decomposition to the right, which is why lime kilns operate at high temperatures.
• $\text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g)$ (exothermic): Cooling shifts equilibrium toward $\text{NH}_3$. The Haber process uses moderate temperatures as a compromise between thermodynamic favorability and acceptable reaction rates.

Pressure Effects on Equilibrium

Le Chatelier's Principle and Pressure Changes in Gaseous Reactions

Pressure effects apply to reactions involving gases, and the key quantity is the change in total moles of gas between reactants and products.

When you increase the pressure on a gaseous system (equivalently, decrease the volume), the system shifts toward the side with fewer moles of gas to reduce the total pressure. When you decrease the pressure (increase the volume), it shifts toward the side with more moles of gas.

Two things to keep in mind:

• If the total moles of gas are equal on both sides, pressure changes have no effect on the equilibrium position.
• Unlike temperature changes, pressure changes at constant temperature do not change the value of $K_p$. They change the partial pressures, which causes the reaction quotient $Q$ to differ from $K$, and the system then shifts to restore $Q = K$.

The magnitude of the shift depends on how large the difference in moles of gas is and how large the pressure change is.

Examples of Pressure Effects on Equilibrium

• $\text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g)$: 4 moles of gas on the left, 2 on the right. Increasing pressure shifts equilibrium toward $\text{NH}_3$. This is why the Haber process uses pressures of ~200 atm.
• $\text{N}_2\text{O}_4(g) \rightleftharpoons 2\text{NO}_2(g)$: 1 mole on the left, 2 on the right. Decreasing pressure favors dissociation into $\text{NO}_2$.
• $\text{CO}(g) + \text{H}_2\text{O}(g) \rightleftharpoons \text{CO}_2(g) + \text{H}_2(g)$: 2 moles of gas on each side. Pressure changes have no effect on the equilibrium position.
• $\text{H}_2(g) + \text{I}_2(g) \rightleftharpoons 2\text{HI}(g)$: 2 moles of gas on each side. Despite appearances, pressure changes do not shift this equilibrium because the mole count is equal on both sides.

Enthalpy Change and Equilibrium Constant

Relationship between Enthalpy Change and Temperature Dependence of K

The qualitative Le Chatelier predictions above are made quantitative by the van 't Hoff equation:

$\frac{d \ln K}{dT} = \frac{\Delta H°}{RT^2}$

where $\Delta H°$ is the standard enthalpy change, $R$ is the gas constant (8.314 J mol$^{-1}$ K$^{-1}$), and $T$ is the absolute temperature in Kelvin.

Reading this equation directly:

• When $\Delta H° > 0$ (endothermic), $\frac{d \ln K}{dT} > 0$, so $K$ increases with temperature.
• When $\Delta H° < 0$ (exothermic), $\frac{d \ln K}{dT} < 0$, so $K$ decreases with temperature.
• The larger $|\Delta H°|$ is, the more sensitive $K$ is to temperature changes. A reaction with $\Delta H° = +180$ kJ/mol will show a much more dramatic change in $K$ per degree than one with $\Delta H° = +20$ kJ/mol.

This equation assumes $\Delta H°$ is approximately constant over the temperature range, which is reasonable for small-to-moderate temperature intervals. For large ranges, you'd need to account for the temperature dependence of $\Delta H°$ via heat capacities ($\Delta C_p$).

Examples of Enthalpy Change and Equilibrium Constant Relationship

Notice that the two reactions with the largest $|\Delta H°|$ values (NO formation and $\text{SO}_3$ formation) show the most dramatic temperature sensitivity in $K$.

Equilibrium Constant vs Temperature

Using the van 't Hoff Equation to Calculate Changes in Equilibrium Constant

Integrating the van 't Hoff equation (assuming constant $\Delta H°$) gives the form you'll use for calculations:

$\ln\left(\frac{K_2}{K_1}\right) = -\frac{\Delta H°}{R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right)$

Here's how to use it step by step:

1. Identify your knowns: $K_1$ at temperature $T_1$, the target temperature $T_2$, and $\Delta H°$ for the reaction.

2. Convert units consistently. Use $R = 8.314$ J mol$^{-1}$ K$^{-1}$, so convert $\Delta H°$ to J/mol. Temperatures must be in Kelvin.

3. Calculate the right-hand side: Compute $\frac{1}{T_2} - \frac{1}{T_1}$, then multiply by $-\frac{\Delta H°}{R}$.

4. Solve for $K_2$: Take the exponential of both sides: $K_2 = K_1 \cdot \exp\left[-\frac{\Delta H°}{R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right)\right]$.

5. Sanity check your answer: If the reaction is endothermic and $T_2 > T_1$, $K_2$ should be larger than $K_1$. If exothermic and $T_2 > T_1$, $K_2$ should be smaller.

A common mistake is forgetting to convert $\Delta H°$ from kJ to J, which throws the answer off by a factor of $e^{1000}$. Always check your units.

Examples of Calculating Equilibrium Constant Changes with Temperature

Example 1 (Endothermic): For $\text{N}_2(g) + \text{O}_2(g) \rightleftharpoons 2\text{NO}(g)$, $\Delta H° = +180.7$ kJ/mol, $K_1 = 1.2 \times 10^{-15}$ at $T_1 = 298$ K. Find $K_2$ at $T_2 = 500$ K.

$\ln\left(\frac{K_2}{K_1}\right) = -\frac{180700}{8.314}\left(\frac{1}{500} - \frac{1}{298}\right) = -21740 \times (-0.001356) = 29.5$

$K_2 = 1.2 \times 10^{-15} \times e^{29.5} \approx 1.2 \times 10^{-15} \times 6.5 \times 10^{12} \approx 7.8 \times 10^{-3}$

As expected, $K$ increases dramatically for this strongly endothermic reaction.

Example 2 (Exothermic): For $\text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g)$, $\Delta H° = -92.2$ kJ/mol, $K_1 = 1.2 \times 10^{3}$ at $T_1 = 500$ K. Find $K_2$ at $T_2 = 400$ K.

Since $T_2 < T_1$ and the reaction is exothermic, you should expect $K_2 > K_1$. Plugging in:

$\ln\left(\frac{K_2}{K_1}\right) = -\frac{-92200}{8.314}\left(\frac{1}{400} - \frac{1}{500}\right) = 11090 \times (0.0005) = 5.55$

$K_2 = 1.2 \times 10^{3} \times e^{5.55} \approx 1.2 \times 10^{3} \times 257 \approx 3.1 \times 10^{5}$

Example 3 (Endothermic): For $\text{CaCO}_3(s) \rightleftharpoons \text{CaO}(s) + \text{CO}_2(g)$, $\Delta H° = +178$ kJ/mol, $K_1 = 2.5 \times 10^{-3}$ at 800 K. At 1000 K, the van 't Hoff equation gives $K_2 \approx 0.18$, a roughly 70-fold increase.

Example 4 (Exothermic): For $2\text{SO}_2(g) + \text{O}_2(g) \rightleftharpoons 2\text{SO}_3(g)$, $\Delta H° = -198$ kJ/mol, $K_1 = 4.5 \times 10^{4}$ at 600 K. Cooling to 500 K gives $K_2 \approx 1.8 \times 10^{6}$, consistent with the exothermic nature favoring products at lower temperatures.