Factors affecting equilibria: Le Chatelier’s principle

What changes an equilibrium?

The rules below assume ideal gases or dilute ideal solutions, a fixed reaction and standard states, and constant temperature except when temperature is deliberately changed. Selective concentration changes hold the other activities fixed; gas compression/expansion is isothermal. For coupled changes, calculate Q rather than applying a slogan.

In plain words: an equilibrium mixture behaves like a stubborn balance. Push it one way and it pushes back. Add a reactant and some of it gets used up; squeeze a gas mixture and it reacts in the direction that makes fewer molecules; heat it and it absorbs heat by running the endothermic way. Concentration, pressure and inert-gas changes only move the mixture until Q again equals the same K; a temperature change gives a new K.

NCERT states the principle this way (page 185): a change in any of the factors that determine the equilibrium conditions of a system will cause the system to change in such a manner as to reduce or to counteract the effect of the change. It applies to all physical and chemical equilibria. The aim in industry is the largest yield at mild temperature and pressure; NCERT’s example is the Haber process (NCERT gives annual world production of ammonia as about a hundred million tonnes, mainly for fertilisers; page 185).

Effect of concentration change

When the concentration of any of the reactants or products in a reaction at equilibrium is changed, the composition of the equilibrium mixture changes so as to minimize the effect of concentration changes

NCERT Chemistry Part I (Class 11), Section 6.8.1, page 185

Take H₂(g) + I₂(g) ⇌ 2HI(g). Adding H₂ makes the reaction quotient Q_c = [HI]²/[H₂][I₂] smaller than K_c, so more H₂ and I₂ react to form HI. At the new equilibrium [H₂] is less than just after the addition but more than in the original mixture (NCERT Fig. 6.8, page 185).

Removing a product does the same: it keeps Q_c below K_c, so the reaction keeps going forward (page 186). In the manufacture of ammonia, NH₃ is liquefied and removed; in making quicklime, CaO, from CaCO₃, CO₂ is constantly removed from the kiln, which promotes decomposition when the kiln is hot enough and supplies the required heat.

An experiment (page 186). Two drops of 0.002 M potassium thiocyanate added to 1 mL of 0.2 M iron(III) nitrate give a red colour, from the equilibrium

Fe³⁺(aq) + SCN⁻(aq) ⇌ [Fe(SCN)]²⁺(aq) (yellow, colourless, deep red)

• oxalic acid removes Fe³⁺ as [Fe(C₂O₄)₃]³⁻, so [Fe(SCN)]²⁺ dissociates and the red fades;
• aqueous HgCl₂ removes SCN⁻ as [Hg(SCN)₄]²⁻, and the red fades;
• adding potassium thiocyanate shifts the equilibrium to the right, and the red deepens.

Effect of pressure change

Changing the pressure by changing the volume affects a gas reaction only if the moles of gaseous reactants and products differ (NCERT Section 6.8.2, pages 186–187). For these gas-dominated equilibria the pressure dependence of pure solid/liquid activities is neglected. Pressure can affect condensed-phase equilibria when reaction volume changes matter.

For the methanation reaction, CO(g) + 3H₂(g) ⇌ CH₄(g) + H₂O(g), 4 mol of gas become 2 mol. If the equilibrium mixture is compressed to half its volume at constant temperature, every concentration doubles:

NCERT’s printed expression on page 187 leaves out the factors of 2; with them, Q_c = K_c/4.

Since Q_c < K_c, the reaction goes forward, the direction with fewer moles of gas, which lowers the pressure. In general, halving the volume multiplies Q_c by 2^Δn_g, where Δn_g = (moles of gaseous products) − (moles of gaseous reactants); here Δn_g = 2 − 4 = −2. For C(s) + CO₂(g) ⇌ 2CO(g), a rise in pressure sends the reaction backward, because the forward reaction makes more gas (page 187).

Effect of adding an inert gas

If an inert gas such as argon is added at constant volume, the equilibrium is undisturbed: the partial pressures and concentrations of the reacting gases do not change (NCERT Section 6.8.3, page 187). This no-shift result assumes fixed temperature and ideal gases; at constant pressure the result differs, as follows.

Beyond the NCERT text (reasoned): if the inert gas is added at constant pressure, the volume must increase, so the partial pressures of the reacting gases fall, just as if the mixture had been expanded. The equilibrium then shifts towards the side with more moles of gas; there is no shift only when Δn_g = 0.

Effect of temperature change

Changes of concentration, pressure or volume move the mixture because Q_c no longer equals K_c. A change of temperature is different: it changes the equilibrium constant (NCERT Section 6.8.4, page 187). For the thermodynamic constant K (the pressure-standard constant for ideal gases), the van ’t Hoff rule uses Δ_rH^⊖:

• for an exothermic reaction (ΔH negative), K decreases as the temperature rises;
• for an endothermic reaction (ΔH positive), K increases as the temperature rises.

For an ideal gas concentration constant, K_p = K_c(RT)^Δn_g implies d ln K_c/dT = (Δ_rH^⊖ − Δn_gRT)/(RT^²). Thus the enthalpy sign rule must not be applied blindly to K_c when Δn_g ≠ 0; the familiar strongly exothermic/endothermic examples here retain their stated trend.

For N₂(g) + 3H₂(g) ⇌ 2NH₃(g), ΔH = −92.38 kJ mol⁻¹: raising the temperature shifts the equilibrium to the left, so a low temperature favours a high yield of ammonia, but at very low temperatures the reaction is too slow, so a catalyst is used. NCERT Table 6.5 (page 178) shows the effect: K_p = 6.8 × 10⁵ at 298 K, 41 at 400 K and 3.6 × 10⁻² at 500 K. The link between K and temperature comes from Δ_rG^⊖ = Δ_rH^⊖ − TΔ_rS^⊖ = −RT ln K (see Gibbs energy and equilibrium).

An experiment (pages 187–188). Brown NO₂ dimerises to colourless N₂O₄:

2NO₂(g) ⇌ N₂O₄(g); ΔH = −57.2 kJ mol⁻¹

Two sealed 5 mL tubes of NO₂ of equal colour are first kept in water at room temperature for 8–10 minutes; then one is put in a freezing mixture and the other in hot water at 363 K. In the cold tube the exothermic forward reaction is favoured and the brown colour fades; in the hot tube the reverse reaction is favoured and the brown deepens (NCERT Fig. 6.9).

A second example is endothermic (page 188): [Co(H₂O)₆]²⁺(aq) + 4Cl⁻(aq) ⇌ [CoCl₄]²⁻(aq) + 6H₂O(l), pink to blue. NCERT reports the mixture blue at room temperature, turning pink when cooled in a freezing mixture (page 188); the blue needs a high chloride concentration (reasoned).

NCERT slip (page 188): NCERT prints the pink ion as [Co(H₂O)₆]³⁺; it is the cobalt(II) ion, [Co(H₂O)₆]²⁺, as the charges in the equation require. In an exam, write 2+.

Effect of a catalyst

A catalyst increases the rate of the reaction by making available a new, low-energy pathway. It speeds up the forward and reverse reactions that pass through the same transition state, and it lowers the activation energy for both by exactly the same amount, so it does not affect the equilibrium composition; it appears neither in the balanced equation nor in the equilibrium constant expression (NCERT Section 6.8.5, page 188). Strictly, the deeper reason is that a catalyst changes neither the reactants’ nor the products’ Gibbs energies, so K stays the same; real catalytic mechanisms can have several intermediates and barriers (reasoned). In an exam, give NCERT’s reason: the same lowering of both activation energies.

• Ammonia. The reaction is highly exothermic and reduces the moles of gas. K falls as the temperature rises, but at low temperatures the reaction is too slow. Fritz Haber found an iron catalyst that gives a satisfactory rate at temperatures where the yield is still reasonable; high pressure improves the yield. NCERT gives the optimum conditions with a catalyst as around 500 °C and 200 atm (page 188); a real plant’s optimum also depends on its design (reasoned).
• Sulphuric acid (contact process). For 2SO₂(g) + O₂(g) ⇌ 2SO₃(g), NCERT gives K_c = 1.7 × 10²⁶ (no temperature is stated; it is of the same order as a room-temperature estimate, since NCERT Table 6.5’s 298 K K_p, 4.0 × 10²⁴, gives the same order, about 1 × 10²⁶). The reaction is very slow, so platinum or V₂O₅ is used as catalyst.

If K is exceedingly small, a catalyst is of little help: it cannot make more product than the equilibrium allows.

Solved examples

Common mistakes to avoid

Quick revision summary

• Le Chatelier: a disturbed equilibrium shifts to counteract the change.
• Add a reactant or remove a product: forward. Add a product or remove a reactant: backward.
• Compress a gas mixture: shift to fewer moles of gas; no effect if Δn_g = 0.
• Inert gas at constant volume: no effect; at constant pressure: shift to more moles of gas (none if Δn_g = 0).
• Temperature: K falls with rising T for exothermic reactions, rises for endothermic ones.
• Catalyst: equilibrium reached faster; K and the composition unchanged.

Questions students ask

What are some examples of Le Chatelier’s principle in daily life?

Oxygen binding in blood can be understood through a local binding-equilibrium model: higher oxygen partial pressure favours oxygenated haemoglobin, while lower pressure favours release. Circulating blood is an open, regulated system, not a single closed equilibrium flask. Clinicians use oxygen in treating carbon monoxide poisoning; displacement of CO from haemoglobin illustrates part of the chemistry, but CO also disrupts cellular oxygen use (CDC, Clinical Guidance for Carbon Monoxide Poisoning, 2026). In tooth decay, acids remove OH⁻ and protonate PO₄³⁻, promoting dissolution of enamel in the simplified equilibrium Ca₅(PO₄)₃OH(s) ⇌ 5Ca²⁺ + 3PO₄³⁻ + OH⁻.

What are the limitations of Le Chatelier’s principle?

Le Chatelier’s principle is a qualitative summary of the equilibrium law: it gives the direction of a shift but not its size, it applies only to a system already at equilibrium, and it can mislead when one change alters several concentrations at once, where comparing Q with K is the reliable test. A known case is adding N₂ to the ammonia equilibrium at constant total pressure and temperature: the extra gas also dilutes H₂ and NH₃, and for a small addition when the initial mole fraction of N₂ is above 0.5 the net reaction goes backward, the opposite of the simple prediction. It also says nothing about how fast the new equilibrium is reached.

What is the van ’t Hoff equation for the effect of temperature on K?

For the thermodynamic equilibrium constant, the van ’t Hoff equation gives how K changes with temperature: log(K₂/K₁) = [Δ_rH^⊖/(2.303R)](1/T₁ − 1/T₂), taking Δ_rH^⊖ and Δ_rS^⊖ as constant; it follows from Δ_rG^⊖ = Δ_rH^⊖ − TΔ_rS^⊖ = −RT ln K written at two temperatures. For an endothermic reaction with Δ_rH^⊖ = +50.0 kJ mol^⁻¹ and K = 2.0 at 300 K, 50 000/19.15 = 2611 and 1/300 − 1/350 = 4.76 × 10^⁻⁴, whose product is log(K₂/K₁) ≈ 1.24, so K ≈ 35 at 350 K. The equation is beyond NCERT.

Who gave Le Chatelier’s principle?

Le Chatelier’s principle was stated in 1884 by the French chemist Henri Louis Le Chatelier (1850–1936), who generalised van ’t Hoff’s rule of the same year for the effect of temperature on equilibrium into a rule for any change of conditions. The German physicist Karl Ferdinand Braun reached the same idea independently in 1887, so it is also called the Le Chatelier–Braun principle. Le Chatelier also worked on cements, metals and the measurement of high temperatures.

Practice questions

Try each question first, then open the answer to check your working.

1. Q1. An equilibrium mixture of Fe³⁺(aq) + SCN⁻(aq) ⇌ [Fe(SCN)]²⁺(aq) is diluted with an equal volume of water. Using Q_c, say which way it shifts.

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   Every concentration halves, so Q_c = (½[Fe(SCN)²⁺])/{(½[Fe³⁺])(½[SCN⁻])} = 2K_c. Since Q_c > K_c, some [Fe(SCN)]²⁺ dissociates: the shift is to the left, towards more particles in solution.

2. Q2. For which of these does compressing the mixture have no effect on the equilibrium amounts: (a) N₂(g) + O₂(g) ⇌ 2NO(g); (b) PCl₅(g) ⇌ PCl₃(g) + Cl₂(g); (c) CaCO₃(s) ⇌ CaO(s) + CO₂(g)?

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   Only (a), where Δn_g = 0. In (b) compression shifts the equilibrium backward (fewer moles of gas); in (c) it also shifts backward, since the only gas is a product and p_CO2 must return to its fixed value.

3. Q3. NCERT’s blue [CoCl₄]²⁻ mixture turns pink when cooled. Show from this observation alone that the forward reaction must be endothermic, and predict the colour and the change in K on heating.

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   Cooling favours the exothermic direction; here cooling makes the pink [Co(H₂O)₆]²⁺, so the reverse reaction is exothermic and the forward reaction is endothermic. On heating, K for the forward reaction increases and the mixture turns bluer.