Temperature Dependence of Equilibrium Constants

Introduction

An equilibrium constant can change strongly with temperature, even when no material is added. The van 't Hoff equation links its slope to standard reaction enthalpy. Exothermic and endothermic reactions therefore shift in opposite directions when heated, provided the reaction and standard-state convention are held fixed. The derivation also shows why a simple two-temperature formula is only an approximation when reaction enthalpy varies appreciably.

Core explanation

For a reaction at standard states, Δ rG° = −RT ln K. The Gibbs–Helmholtz relation at constant pressure is [∂(G/T)/∂T] p = −H/T² for a fixed composition; the reaction form is d(Δ rG°/T)/dT = −Δ rH°/T², with standard-state details treated consistently. Since Δ rG°/T = −R ln K, differentiation yields −R dln K/dT = −Δ rH°/T². Hence dln K/dT = Δ rH°/(RT²).

If Δ rH° is positive, K increases with T; heating favours products in the equilibrium sense for the reaction as written. If Δ rH° is negative, K decreases with T. This statement concerns K, not an instantaneous change in reaction rate. Heating often speeds both forward and reverse reactions, while the equilibrium position changes according to the thermodynamic enthalpy relationship. The sign depends on the reaction's written direction; reverse the equation and both Δ rH° and the trend of its K reverse accordingly.

If Δ rH° can be approximated as constant from T₁ to T₂, integrate: ln(K₂/K₁) = −(Δ rH°/R)(1/T₂ − 1/T₁) = (Δ rH°/R)(1/T₁ − 1/T₂). Temperatures must be in kelvin, and enthalpy units must match R. A plot of ln K against 1/T then has approximate slope −Δ rH°/R. For a broad interval, heat-capacity differences cause Δ rH° to vary with T; integrate the temperature-dependent enthalpy or use more detailed thermodynamic data.

The expression is not a rule that pressure changes K at fixed temperature. Pressure can change reaction quotient Q and equilibrium composition for gases, but the thermodynamic K for the written reaction at a specified T and standard convention remains a function of T. Nonidealities alter activities used to describe the mixture and may be significant at high pressure.

Step-by-step reasoning

Start with Δ rG° = −RT ln K, divide by T and apply Gibbs–Helmholtz. Check the sign of Δ rH° to predict the qualitative temperature trend. If a numerical two-temperature estimate is requested, decide whether constant Δ rH° is justified, convert both temperatures to kelvin and enthalpy to J mol⁻¹, then compute ln(K₂/K₁) and exponentiate.

Visual explanation

Draw ln K on the vertical axis and 1/T on the horizontal axis. An endothermic reaction has a negative slope −Δ rH°/R, so moving toward smaller 1/T (higher T) raises ln K. An exothermic reaction has a positive slope, so the same move lowers ln K. Label the plot as approximately straight only when Δ rH° is nearly constant.

Real-world analogy

A balance between two arrangements changes as conditions change; a thermometer can tilt which arrangement is favoured. Enthalpy tells which direction heating pushes the equilibrium constant. The analogy does not imply heat is literally a reactant or that kinetics and equilibrium are the same phenomenon.

Real-world example

For an exothermic gas reaction, increasing temperature commonly reduces its equilibrium product preference even if a catalyst makes the system reach equilibrium faster. Industrial process design must weigh this thermodynamic effect against kinetic improvements. The van 't Hoff equation quantifies the equilibrium trend when Δ rH° and K data are known.

Why?

K is linked to standard Gibbs energy, whose temperature dependence reflects enthalpy and entropy. Gibbs–Helmholtz isolates the enthalpy contribution to the derivative of G/T. Substituting −R ln K transforms that thermodynamic derivative directly into the temperature slope of an observable equilibrium constant.

Common misconception

The integrated formula is not exact over every temperature span. It assumes Δ rH° remains approximately constant. Another error is to switch T₁ and T₂ inside the reciprocal-temperature difference while keeping the same sign. Also, a negative enthalpy does not mean K must be less than one; it determines the slope of K with T, not its absolute value at one temperature.

Worked example

Suppose K₁ = 2.0 at T₁ = 300 K and Δ rH° = +20.0 kJ mol⁻¹ is approximately constant up to T₂ = 330 K. Then ln(K₂/K₁) = (20000/8.314)(1/300 − 1/330) ≈ 0.729. Therefore K₂/K₁ ≈ e^0.729 ≈ 2.07, giving K₂ ≈ 4.14. The reaction is endothermic, so the increase is directionally sensible.

Quick check

1. What happens to K on heating an exothermic reaction, assuming its standard reaction enthalpy stays negative? Answer: dln K/dT = Δ rH°/(RT²) is negative, so K decreases as temperature increases for the reaction as written.

Exam focus

Write the differential van 't Hoff equation before the integrated approximation. Use kelvin and consistent J mol⁻¹ units. Check direction qualitatively from the enthalpy sign before calculating. Distinguish changes in K from changes in Q or rate, and name the constant-enthalpy assumption.

Advanced insight

Kirchhoff's relation dΔ rH°/dT = Δ rC p° allows Δ rH°(T) to be updated when heat-capacity data are available. Inserting that function into dln K/dT and integrating gives a curved ln K versus 1/T relation. Such curvature can reveal that the simple straight-line van 't Hoff approximation is inadequate.

Summary

Combining Δ rG° = −RT ln K with Gibbs–Helmholtz gives dln K/dT = Δ rH°/(RT²). Endothermic reaction K rises on heating; exothermic reaction K falls. If Δ rH° is approximately constant, ln(K₂/K₁) = (Δ rH°/R)(1/T₁ − 1/T₂). The relation concerns equilibrium, not reaction speed.

Practice questions

1. A plot of ln K against 1/T has positive slope. What sign is Δ rH° for the reaction as written? Answer: The slope is −Δ rH°/R. A positive slope means Δ rH° is negative, so the reaction is exothermic in the stated direction. 2. Why may K at 500 K not be reliably estimated from a 300 K value using one constant Δ rH°? Answer: Reaction heat capacity can make Δ rH° change across a wide temperature interval. The integrated constant-enthalpy equation then oversimplifies the temperature dependence. 3. At fixed T, pressure changes the equilibrium composition of an ideal gas reaction. Has thermodynamic K changed? Answer: No. K is fixed by T and the chosen standard-state convention. Pressure changes partial-pressure activities and Q, causing composition to adjust until Q again equals K.