Le Châtelier’s Principle: Temperature and Catalysts

Unlike changes in concentration or pressure, a change in temperature changes the value of the equilibrium constant: heating shifts an equilibrium toward the endothermic direction, cooling shifts it toward the exothermic direction, and a catalyst speeds up the approach to equilibrium without shifting it. In brief: the dissociation N₂O₄ ⇌ 2NO₂ is endothermic (ΔH = +57.2 kJ), so heating from 25 °C to 50 °C raises K from 0.0060 to about 0.036 and the mixture turns darker brown.

Temperature is the only common stress that changes K. The direction of the change follows directly from the sign of ΔH, which makes the effect easy to predict. Catalysts, on the other hand, have no effect on the position of equilibrium at all. This article explains how temperature and catalysts act, gives a table of the effects, shows how K changes numerically using the van ‘t Hoff equation and describes how industry balances the rate against the yield.

How does temperature shift an equilibrium?

Treat heat as if it were a reactant or a product:

• For an endothermic reaction (ΔH > 0), heat is a reactant: reactants + heat ⇌ products. Raising the temperature adds heat, so the system shifts to the right, K increases and more products form.
• For an exothermic reaction (ΔH < 0), heat is a product: reactants ⇌ products + heat. Raising the temperature shifts the system to the left, so K decreases.
• Cooling has the opposite effect in each case.

The reason is thermodynamic: the equilibrium constant is related to the standard Gibbs energy change by ΔG° = −RT ln K, and ΔG° = ΔH° − TΔS° depends on temperature. The van ‘t Hoff equation relates the temperature and the constant: ln(K₂/K₁) = −(ΔH°/R)(1/T₂ − 1/T₁).

Effect of temperature on K

How much does K change?

Example 1: an endothermic reaction

For N₂O₄ ⇌ 2NO₂, ΔH° = +57.2 kJ/mol and K_c = 0.0060 at 298 K. Find K_c at 323 K (50 °C).

1. 1/T₂ − 1/T₁ = 1/323.15 − 1/298.15 = 0.0030945 − 0.0033540 = −2.595 × 10⁻⁴ K⁻¹.
2. ln(K₂/K₁) = −(57,200 / 8.314)(−2.595 × 10⁻⁴) = 6880 × 2.595 × 10⁻⁴ = 1.785.
3. K₂/K₁ = e^1.785 = 5.96, so K₂ = 0.0060 × 5.96 = 0.036.

A rise of 25 °C increases K by a factor of about 6, and the gas turns darker brown as more NO₂ forms.

Example 2: an exothermic reaction

The equilibrium constants (K_p, in atm⁻²) for ammonia synthesis, which has ΔH° = −92.2 kJ/mol, decrease steeply with temperature.

What do catalysts do?

A catalyst lowers the activation energy of the forward and of the reverse reactions by the same amount. Both rate constants increase by the same factor, so their ratio k_f/k_r = K stays unchanged. The consequences are:

• the equilibrium is reached faster, but the equilibrium composition is the same;
• K, ΔH and ΔG° do not change;
• the catalyst does not appear in the equilibrium expression;
• a reaction mixture that is already at equilibrium is not disturbed by adding a catalyst.

How does industry choose conditions?

Chemical engineers have to balance the yield (which depends on K) and the rate (which depends on k).

• Haber process: the reaction is exothermic, so a low temperature would give the best yield, but it would be too slow. A compromise of about 400–450 °C with an iron catalyst and a pressure of 150–250 atm gives a conversion of roughly 15–20 % per pass, and unreacted gas is recycled after the ammonia has been condensed.
• Contact process: 2SO₂ + O₂ ⇌ 2SO₃ is exothermic (ΔH ≈ −198 kJ per equation as written). Vanadium(V) oxide is used at about 420–450 °C, with several catalyst beds and cooling between them, so that the overall conversion reaches about 99 %.
• Steam reforming: CH₄ + H₂O ⇌ CO + 3H₂ is endothermic and uses high temperatures, around 800–900 °C, since a high temperature favors the products.

What classroom demonstrations show the effect?

• NO₂/N₂O₄ tubes: a sealed tube is dark brown in hot water and almost colorless in an ice bath.
• Cobalt chloride: [Co(H₂O)₆]²⁺ (pink) + 4Cl⁻ ⇌ [CoCl₄]²⁻ (blue) + 6H₂O is endothermic, so heating turns the solution blue and cooling turns it pink.

Where does the effect come from?

Jacobus van ‘t Hoff derived the quantitative relation between K and temperature in 1884, in the same year in which Le Châtelier stated the qualitative principle. Both pieces of work were part of the early development of chemical thermodynamics. The practical lesson, that an exothermic equilibrium needs a compromise between yield and rate, was central to the development of the Haber–Bosch process by BASF in the early 20th century.

Which mistakes are common?

• Saying that a catalyst shifts the equilibrium: it does not.
• Thinking that K is the same at all temperatures: temperature is the only common factor that changes K.
• Mixing up the sign of ΔH: an exothermic reaction is favored by cooling.
• Assuming that a higher temperature always gives a better yield: only for endothermic reactions.
• Forgetting the effect on the rate: a lower temperature improves the yield of an exothermic reaction, but it is slower.