Chemistry: Atoms First 2e · Fundamental Equilibrium Concepts
Shifting Equilibria: Le Châtelier’s Principle
On this page 9 sections
In 30 seconds
Le Châtelier's principle A stressed equilibrium shifts to counteract the stress Full entry → states that when a system at equilibrium is disturbed, it shifts in the direction that counteracts the disturbance and restores equilibrium. The disturbances are changes in concentration, changes in pressure or volume (for gases), and changes in temperature. Adding a reactant pushes the reaction toward products; removing a product pulls it forward; compressing a gas mixture favors the side with fewer gas molecules; heating an exothermic reaction favors the reactants. Temperature is special — unlike concentration or pressure changes, changing the temperature actually changes the value of K itself, because K depends on temperature.
The quantitative companion is the reaction quotient Q: the reaction runs in whichever direction moves Q back toward K. Catalysts and inert gases added at constant volume do not shift the equilibrium — they change approach speed or total pressure, not the reaction's balance.
Why this matters
Le Châtelier's principle is the design rule of industrial chemistry. The Haber–Bosch process makes ammonia — the source of most nitrogen fertilizer — by exploiting all three levers at once: high pressure (fewer gas moles on the product side), moderate temperature (exothermic reaction, with a catalyst to compensate for slower kinetics), and continuous removal of product ammonia. In the body, the bicarbonate buffer CO2 + H2O ⇌ H+ + HCO3- shifts with blood CO2: rapid breathing removes CO2 and changes blood pH. On exams, Le Châtelier questions test whether you can predict the direction of shift — and, critically, distinguish the cases that change K from those that do not.
The college version
Core Concepts
Concentration changes
Adding a reactant (or removing a product) makes Q < K, so the forward reaction runs until Q = K again. Removing a reactant (or adding a product) makes Q > K, so the reverse reaction runs. The value of K is untouched — only the composition changes.
Pressure and volume changes for gases
Changing the volume changes all partial pressures. The direction of the shift is set by Δn, the change in moles of gas from reactants to products. If the product side has fewer gas moles (Δn < 0), compression (higher pressure) favors products; if Δn > 0, compression favors reactants; if Δn = 0, pressure changes have no effect. Inert gas added at constant volume changes total pressure, not the reacting gases' partial pressures — no shift.
Temperature changes — the special case
Temperature changes alter the equilibrium constant itself. For an exothermic reaction (ΔH < 0), heat is effectively a "product": raising the temperature favors the reactants and decreases K; lowering the temperature favors products and increases K. For an endothermic reaction (ΔH > 0), the logic flips: raising the temperature increases K and favors products.
Catalysts and the no-shift cases
A catalyst lowers the activation energy for both directions equally: it speeds the approach but leaves the position and K unchanged. Inert gas at constant volume changes no partial pressures — no shift. Spotting these no-shift cases is a favorite exam trap.
The Haber–Bosch process as a case study
For N2(g) + 3H2(g) ⇌ 2NH3(g), ΔH°= -92 kJ per mole of reaction. High pressure favors ammonia because Δn = 2 - 4 = -2. Low temperature would favor ammonia but make the reaction too slow, so plants compromise near 400–500 °C with an iron catalyst, very high pressure (roughly 150–300 atm), and continuous condensation of product ammonia.
How It Works / Step-by-Step Process
- Identify the stress: concentration, pressure/volume, or temperature.
- If temperature changed, shift K per exo/endo — and stop there (no Q needed).
- If concentration or pressure changed, compute Q and compare with K: Q < K → forward, Q > K → reverse, Q = K → none.
- Check the no-shift cases: catalyst, inert gas at constant volume, or a pressure change with Δn = 0.
Common Confusions
| Common Confusion | Correct Understanding |
|---|---|
| "Adding an inert gas shifts the equilibrium." | Only if the volume is allowed to change. At constant volume, no shift. |
| "A catalyst changes the equilibrium position." | It speeds both directions equally; position and K are unchanged. |
| "Heating always favors products." | Heating favors the endothermic direction — reactants for an exothermic reaction. |
| "Temperature and concentration both just shift the position." | Concentration changes leave K alone; temperature changes K itself. |
| "Higher pressure always shifts toward products." | It shifts toward the side with fewer gas moles; if Δn = 0, no shift. |
| "The shift completely cancels the disturbance." | It only partially relieves it; the new equilibrium is at a different composition. |
Eli explains
The same idea, in plain words
Explain it like I’m 10
Imagine a seesaw with kids on both sides, perfectly balanced. If you add one more kid to the left, the seesaw tips and the right side rises — the system "fights back" by moving in the direction that uses up the new kid. That is Le Châtelier's principle: when you disturb a balance, the reaction shifts to use up what you added or replace what you removed.
Worked examples
At 445 °C, Kc = 50.2 for H2(g) + I2(g) ⇌ 2HI(g). At equilibrium [H2] = [I2] = 0.022 M and [HI] = 0.156 M. Extra H2 is injected, raising [H2] to 0.050 M before any reaction. Compute the new Q:
Q = [HI]2[H2][I2] = (0.156)2(0.050)(0.022) = 0.02430.00110 = 22.1
Q = 22.1 < K = 50.2, so HI forms and consumes some added hydrogen. Solving the equilibrium expression (ICE-table method, next topic) gives [HI] = 0.172 M, [H2] = 0.042 M, [I2] = 0.014 M. Check: (0.172)2/((0.042)(0.014)) = 50.3, essentially K again — the system shifted forward and K never changed.
Nitrogen dioxide dimerizes: 2NO2(g) ⇌ N2O4(g), ΔH°= -57.2 kJ per mole of reaction (exothermic, Δn = -1). At 25 °C, Kp = P(N2O4)/P(NO2)2 = 6.9 atm⁻¹.
Pressure stress. At equilibrium, P(NO2) = 0.20 atm and P(N2O4) = 0.28 atm:
Q = P(N2O4)P(NO2)2 = 0.28(0.20)2 = 0.280.040 = 7.0 ≈ K
Compress the mixture to half its volume: every partial pressure doubles, P(NO2) = 0.40 atm and P(N2O4) = 0.56 atm:
Q = 0.56(0.40)2 = 0.560.16 = 3.5
Q = 3.5 < K = 6.9, so NO2 dimerizes, reducing the number of gas moles — compression favored the side with fewer gas molecules, as Δn = -1 predicts.
Temperature stress. Heating favors decomposition back to brown NO2 and decreases K for the exothermic direction — the brown color deepens. Cooling favors colorless N2O4, and the mixture lightens.
An ammonia plant runs N2(g) + 3H2(g) ⇌ 2NH3(g), ΔH°= -92 kJ, Δn = -2:
- Pressure: roughly 150–300 atm. Δn = -2, so high pressure favors ammonia.
- Temperature: a compromise near 400–500 °C — cooler would favor the exothermic reaction (larger K) but be too slow, so the iron catalyst restores speed at a smaller K.
- Product removal: ammonia is condensed out of the circulating gas. Removing product keeps Q < K, so the forward reaction keeps running and unreacted N2/H2 is recycled.
Pressure and removal push the position, temperature balances yield against rate, and the catalyst handles kinetics — one long application of Le Châtelier's principle.
Key takeaways
- Le Châtelier's principle: a disturbed equilibrium shifts to counteract the disturbance.
- Adding reactant / removing product → forward shift; adding product / removing reactant → reverse shift. K unchanged.
- Pressure changes matter only for gases, and only through Δn: compression favors the side with fewer gas moles.
- Inert gas at constant volume: no shift. Catalyst: faster approach, no shift, K unchanged.
- Temperature changes K itself: heating favors the endothermic direction; cooling favors the exothermic direction.
Check yourself
4 review questions from the chapter. Try each one, then open the answer.
State Le Châtelier's principle in one sentence.
Show answer
When a system at equilibrium is disturbed, it shifts in the direction that counteracts the disturbance and restores equilibrium.
For N2(g) + 3H2(g) ⇌ 2NH3(g) (ΔH°= -92 kJ), predict the effect of (a) increasing pressure, (b) raising temperature, (c) removing NH3, (d) adding a catalyst.
Show answer
(a) Forward — fewer gas moles on the product side; (b) reverse — exothermic reaction, K decreases; (c) forward — product removal makes Q < K; (d) no shift — only a faster approach.
In Example 1, why did Q drop below K when hydrogen was added?
Show answer
Adding H₂ raised the denominator of the mass-action expression, lowering Q below K, so the forward reaction must run to restore Q = K.
In Example 2, compression made Q = 3.5 < Kp = 6.9. Which direction did the reaction shift, and why does that relieve the stress?
Show answer
Forward, toward N₂O₄: dimerization consumes two NO₂ per N₂O₄ formed, reducing the gas moles and partially offsetting the pressure increase.
Study tools & related lessonsKey vocabulary · Related
Key vocabulary
- Le Châtelier's principle
- A stressed equilibrium shifts to counteract the stress
- Δ n
- Change in moles of gas, products minus reactants
- exothermic / endothermic
- Releases / absorbs heat
- reaction quotient (Q)
- Mass-action ratio at current conditions
Sources & references
This lesson was adapted from the open educational references above; their licenses and attributions are preserved. See Copyright & Licensing.
Educational content only. It is not medical, legal or professional advice. Found an error? Tell us.