Le Chatelier's Principle: Predicting How Equilibria Respond to Change

A stress changes Q or K, and the system restores equilibrium

At equilibrium the reaction quotient Q equals the equilibrium constant K. A disturbance, often called a stress, breaks that equality in one of two ways. Adding or removing a species, or changing the volume of a gas mixture, changes Q while K stays the same. Changing the temperature changes K itself. Either way, the system then reacts until Q equals K again.

If Q is now smaller than K, the reaction runs forward, toward products. If Q is now larger than K, it runs in reverse. Thinking in these terms is more reliable than memorized slogans, because it also tells you when there is no shift at all.

The word partly matters. The system never fully undoes the change. If you add a reactant, its concentration at the new equilibrium is still higher than it was before you added it.

Adding or removing a species

Adding a reactant or removing a product lowers Q below K, so the reaction shifts toward products. Adding a product or removing a reactant raises Q above K, so the reaction shifts toward reactants. Industrial and laboratory processes use this constantly. Distilling off a volatile product, letting a gas escape, or precipitating a product keeps Q below K and drives the reaction further than its K alone would allow.

Adding more of a pure solid or pure liquid does not shift the equilibrium. Those substances do not appear in Q, so Q does not change. Adding more CaCO₃(s) to a sealed vessel at equilibrium with CaO(s) and CO₂(g) leaves the CO₂ pressure unchanged.

Worked example: extra hydrogen in the hydrogen iodide equilibrium

OpenStax section 13.3 gives Kc = 50.0 at 400 °C for H₂(g) + I₂(g) ⇌ 2 HI(g). Suppose an equilibrium mixture at that temperature has [H₂] = [I₂] = 0.0150 mol/L. Then [HI] = √(50.0 × 0.0150 × 0.0150) = 0.106 mol/L.

Now H₂ is injected so that [H₂] jumps to 0.0450 mol/L before anything reacts. Q = (0.106)² / (0.0450 × 0.0150) = 16.7. That is less than 50.0, so the reaction shifts forward and uses up some of the added H₂.

To find how far it shifts, let x mol/L of H₂ react. Then [H₂] = 0.0450 − x, [I₂] = 0.0150 − x and [HI] = 0.106 + 2x. Setting (0.106 + 2x)² / [(0.0450 − x)(0.0150 − x)] = 50.0 and solving the quadratic, keeping unrounded intermediate values, gives x = 0.00728. The physically sensible root is the one smaller than 0.0150, because I₂ cannot fall below zero. The new equilibrium has [H₂] = 0.0377, [I₂] = 0.00772 and [HI] = 0.121 mol/L.

Read the result against the principle. The system consumed about a quarter of the added hydrogen, so [H₂] ended at 0.0377 mol/L, still well above the original 0.0150. Meanwhile about half of the iodine was converted to HI. Adding a cheap reactant to use up an expensive one is a standard way to raise yield.

Squeezing or expanding a gas mixture

Reducing the volume of a gas mixture raises every concentration by the same factor. Whether that changes Q depends on Δn, the moles of gaseous products minus the moles of gaseous reactants. If Δn is negative, compression lowers Q and the reaction shifts toward fewer gas molecules. If Δn is positive, compression raises Q and the reaction shifts toward the reactants, which again is the side with fewer gas molecules. If Δn is zero, as in H₂ + I₂ ⇌ 2 HI, compression does not change Q and there is no shift.

Raising the total pressure by adding an unreactive gas such as argon at constant volume changes nothing. The partial pressures and concentrations of the reacting gases stay the same, so Q stays equal to K. Adding an inert gas at constant pressure is different: the container must expand, which has the same effect as increasing the volume.

Worked example: halving the volume of an NO₂–N₂O₄ mixture

For N₂O₄(g) ⇌ 2 NO₂(g) at 25 °C, OpenStax section 13.2 describes an equilibrium mixture with [N₂O₄] = 0.042 mol/L and [NO₂] = 0.016 mol/L. These give Kc = (0.016)²/0.042 = 6.1 × 10⁻³.

Halve the volume at constant temperature. Both concentrations double, to [N₂O₄] = 0.084 and [NO₂] = 0.032 mol/L. Q = (0.032)²/0.084 = 1.22 × 10⁻², exactly twice Kc, because the NO₂ term is squared. Q > K, so the system shifts toward N₂O₄, the side with one mole of gas instead of two.

Let y mol/L of N₂O₄ form. Solving (0.032 − 2y)²/(0.084 + y) = 6.1 × 10⁻³ gives y = 0.0044, so the new equilibrium has [NO₂] = 0.023 mol/L and [N₂O₄] = 0.088 mol/L, to two significant figures like the data. The brown NO₂ is more concentrated than before compression (0.023 against 0.016 mol/L), so the gas looks darker, but a smaller share of the nitrogen is now in NO₂: 12 percent compared with 16 percent before.

Temperature shifts K itself

Temperature is the only common change that alters the value of K. For an endothermic reaction (ΔH > 0), K increases as the temperature rises, so heating shifts the equilibrium toward products. For an exothermic reaction (ΔH < 0), K decreases as the temperature rises, so heating shifts it toward reactants. A useful way to remember this is to treat heat as a reactant in an endothermic reaction and as a product in an exothermic one.

OpenStax section 13.3 gives ΔH = +57.20 kJ for N₂O₄(g) ⇌ 2 NO₂(g), so warming the mixture makes it browner. The size of the change can be estimated with the van 't Hoff equation, ln(K₂/K₁) = −(ΔH°/R)(1/T₂ − 1/T₁). It follows from ΔG° = −RT ln K and ΔG° = ΔH° − TΔS° if ΔH° and ΔS° stay roughly constant.

Worked example. From 25 °C (298.15 K) to 50 °C (323.15 K), ln(K₂/K₁) = −(57 200 J/mol ÷ 8.314 J mol⁻¹ K⁻¹)(1/323.15 − 1/298.15) = 1.785, so K₂/K₁ = 6.0. Starting from Kp = Kc × RT = (6.1 × 10⁻³)(0.08206 × 298.15) = 0.15 at 25 °C (Δn = 1, pressures in atm), Kp at 50 °C is roughly 0.9. A 25-degree rise increases the equilibrium constant about sixfold.

A catalyst does not shift an equilibrium. It lowers the activation energy of the forward and reverse reactions equally, so equilibrium is reached sooner, but K and the equilibrium composition are unchanged.

Haber–Bosch ammonia: balancing yield against speed

The industrial synthesis of ammonia, N₂(g) + 3 H₂(g) ⇌ 2 NH₃(g), shows how the principle guides real choices. OpenStax section 13.3 gives ΔH = −92.2 kJ. The reaction is exothermic, so lower temperatures give a larger K and more ammonia at equilibrium. Four moles of gas become two, so high pressure also favors ammonia.

The difficulty is speed. At low temperature the reaction is far too slow to be useful. OpenStax describes the compromise: pressures of about 150 to 250 atm, temperatures of 400 to 500 °C with a catalyst to keep the rate practical, and continuous removal of ammonia from the mixture so that Q stays below K. The temperature is chosen for rate, at the cost of a smaller K, and the other conditions recover as much yield as possible.

Le Chatelier predictions that go wrong

When a prediction feels uncertain, write Q, change the quantity in question, and compare the new Q with K. That check settles nearly every case below.

• Saying K changes when a concentration or the volume changes. Only temperature changes K.
• Predicting a shift after compressing a reaction with Δn = 0, such as H₂ + I₂ ⇌ 2 HI.
• Predicting a shift after adding an inert gas at constant volume.
• Predicting a shift after adding more of a pure solid or liquid.
• Claiming that a catalyst increases the equilibrium yield. It only speeds the approach to equilibrium.
• Assuming the system fully cancels the change. An added reactant ends above its original concentration.
• Counting all moles, not just gas moles, when deciding how a pressure change acts.