Chemistry 2e · Fundamental Equilibrium Concepts

Shifting Equilibria: Le Châtelier’s Principle

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In 30 seconds
Why this matters
The college version
Eli explains
Key takeaway
Check yourself
Study tools
Sources & references

In 30 seconds

A system at equilibrium stays there until something disturbs it. predicts the response: if a is applied to a system at equilibrium, the system shifts in the direction that partially relieves the stress. The shift follows from the reaction quotient Q versus the equilibrium constant K: when a stress makes Q < K, the forward reaction is favored until Q climbs back to K; when Q > K, the reverse reaction is favored.

The common stresses are:

Changes in concentration (adding or removing reactants or products),
Changes in pressure or volume (for gas reactions),
Changes in temperature (the only stress that changes the value of K),
Adding a or an inert gas (which cause no shift, for reasons below).

Why this matters

Real chemistry rarely runs in a sealed flask left alone. Industrial reactors feed reactants and remove products continuously; the body constantly adds and removes the chemicals in its equilibria; environmental systems are pushed by pollution and climate change. Le Châtelier’s principle is the tool for reasoning about all of these:

Industry: The Haber process makes ammonia from nitrogen and hydrogen. Because the reaction is exothermic and reduces gas moles, high pressure and continuous product removal push equilibrium toward ammonia.
Medicine: Hemoglobin binds oxygen reversibly: high oxygen partial pressure in the lungs favors the oxygen-bound form; low partial pressure in the tissues releases it. This is also why carbon monoxide is dangerous — it binds hemoglobin tightly and does not easily leave.
Environment: Dissolving CO₂ in seawater shifts carbonate equilibria that affect shell-forming organisms; adding CO₂ to the atmosphere is a concentration stress on these equilibria.

The college version

Core Concepts

Concentration changes: the Q vs K logic

Consider aA + bB ⇌ cC + dD with K fixed (temperature constant).

Adding a reactant (or removing a product) makes Q < K; the reaction proceeds forward until Q = K.
Adding a product (or removing a reactant) makes Q > K. The reaction proceeds in reverse.
Removing a product — for example, distilling off a gas or precipitating a solid product — pulls the reaction forward. This is the "removal of product drives completion" trick used in synthesis.

Note the shift partially offsets the stress: if you add reactant, some is consumed, but not all. The system moves to a new equilibrium position, not the old one.

Pressure and volume changes: gas reactions only

For gases, changing the container volume changes all partial pressures. Whether the equilibrium shifts depends on Δn, the change in moles of gas:

Decreasing volume (increasing pressure): the system shifts toward the side with fewer moles of gas, because that direction reduces the total pressure.
Increasing volume (decreasing pressure): the system shifts toward the side with more moles of gas.
If Δn = 0 (equal gas moles on both sides), pressure changes cause no shift.

For N₂(g) + 3H₂(g) ⇌ 2NH₃(g), Δn = -2, so compression favors ammonia. The constant K_p is unchanged — the position shifts, the constant does not.

An inert gas (like helium) added at constant volume changes total pressure but not the partial pressures of the reacting gases — no shift. At constant total pressure it forces a volume increase, diluting reactants — effectively a volume stress.

Temperature changes: the stress that changes K

Temperature differs from the other stresses: it changes the value of K itself, because it changes the balance of forward and reverse rate constants. Think of heat as a product (exothermic) or reactant (endothermic), then apply the concentration logic:

Increasing temperature favors the endothermic direction (absorbs the added heat).
Decreasing temperature favors the exothermic direction.

For the exothermic synthesis of ammonia, raising the temperature shifts equilibrium toward reactants, decreasing yield. Industry therefore compromises: moderate temperature for speed (kinetics), high pressure for yield (equilibrium).

Catalysts and the rate of reaching equilibrium

A catalyst lowers the activation energy for both the forward and reverse reactions equally. It does not change Q or K, and it does not change the equilibrium position — it only makes the system reach equilibrium faster. "Faster" is not "farther" — a classic exam trap.

How It Works / Step-by-Step Process

Worked example 1: predicting the direction of shift

Problem. For N₂O₄(g) ⇌ 2NO₂(g) (colorless dinitrogen tetroxide ⇌ brown nitrogen dioxide), predict the effect of (a) adding NO₂, (b) compressing the mixture, (c) heating, given that the forward reaction is endothermic.

Solution. Apply the stress logic step by step.

(a) Adding NO₂ increases the numerator of Q = [NO₂]²[N₂O₄], so Q > K. The reaction shifts in reverse, consuming some added brown gas until Q = K; the color partially fades.
(b) Compressing raises all pressures. Count gas moles: 1 N₂O₄ on the left, 2 NO₂ on the right; Δn = +1. The system shifts toward fewer gas moles — the left, forming colorless N₂O₄ and reducing pressure.
(c) Heating an endothermic reaction adds "reactant heat": N₂O₄ + heat ⇌ 2NO₂. The equilibrium shifts forward, producing more brown NO₂ — a sealed tube turns browner when warmed, paler when cooled.

Worked example 2: the quantitative Q vs K test

Problem. For H₂(g) + I₂(g) ⇌ 2HI(g), K_c = 50.2 at 445 °C. A mixture contains [H₂] = 0.20 M, [I₂] = 0.20 M, [HI] = 1.0 M. Which direction will it proceed?

Solution.

Compute the reaction quotient Q with current concentrations:

$Q = [HI] 2 [H 2][I 2] = (1.0) 2 (0.20)(0.20) = 1.0 0.040 = 25$

Compare to K: Q = 25 < K = 50.2, so the reaction has too little product. It proceeds forward, converting H₂ and I₂ into HI until Q rises to 50.2.

This test works for any concentration stress and reappears in the next topic.

Worked example 3: Haber process reasoning

For N₂(g) + 3H₂(g) ⇌ 2NH₃(g), ΔH < 0 (exothermic), Δn = -2. Maximum-yield reasoning: (1) raise pressure — favors the 2-mole product side; (2) remove ammonia as it forms — pulls the reaction forward; (3) do not overheat — high temperature favors reactants. Temperature is kept moderate as a kinetic compromise.

Common Confusions

Do Not Confuse	With	Difference
Catalyst shifting equilibrium	Catalyst speeding equilibrium	A catalyst lowers activation energy for both directions; position and K are unchanged, only the time to get there.
Temperature changing position only	Temperature changing K	Temperature changes K itself; concentration and pressure stresses change position only, leaving K fixed.
Pressure changes for all reactions	Pressure changes only mattering for gases	Pressure/volume stresses matter only when gases are present and Δn ≠ 0; condensed phases are nearly incompressible.
Inert gas always shifting equilibrium	Inert gas at constant volume	At constant volume an inert gas changes total pressure but not partial pressures of reactants — no shift.
Adding more reactant consuming all of it	Adding more reactant consuming only part	The shift is partial; the system reaches a new equilibrium position, not the original one.

Eli explains

The same idea, in plain words

Explain it like I’m 10

Imagine a seesaw with kids on both sides, balanced in the middle. If you drop a heavy book on one side, the seesaw tips — but the kids shift around to make it as balanced as it can be again. Chemistry does the same thing: if you add more of one chemical, the reaction shifts to use some of it up and get back to balance. Heating an ice-water bath melts ice; cooling it refreezes water — the system pushes back against what you did.

Key takeaways

Le Châtelier’s principle: a stress on an equilibrium shifts the reaction to partially relieve the stress.
Adding reactant or removing product shifts the reaction forward; adding product or removing reactant shifts it in reverse.
For gases: decreasing volume favors the side with fewer gas moles; Δn = 0 means no shift from pressure changes.
Temperature is the only stress that changes K; heating favors the endothermic direction.
Catalysts change the rate of approach to equilibrium, never the equilibrium position.
Inert gas at constant volume: no shift; at constant pressure: behaves like a dilution.
The shift is partial — the system moves to a new equilibrium, not back to the original one.
Q vs K: Q < K → forward; Q > K → reverse; Q = K → at equilibrium.

Check yourself

6 review questions from the chapter. Try each one, then open the answer.

State Le Châtelier’s principle in one sentence.
Show answer
A system at equilibrium, when subjected to a stress, shifts in the direction that partially relieves that stress.
For 2SO₂(g) + O₂(g) ⇌ 2SO₃(g), does increasing the pressure favor products or reactants?
Show answer
Products: 3 moles of gas on the left versus 2 on the right (Δn = -1), so compression favors the fewer-mole side.
Why does raising the temperature lower the yield of an exothermic reaction?
Show answer
Heat is a product of an exothermic reaction; adding heat shifts the equilibrium toward reactants, so less product is present at the new equilibrium.
Does adding a catalyst change the equilibrium constant? What does it change?
Show answer
No — K is unchanged. A catalyst only speeds the approach to equilibrium by lowering the activation energy for both directions.
If Q > K for a reaction at some instant, which direction does the reaction proceed?
Show answer
Reverse: the reaction proceeds toward reactants until Q decreases to K.
For N₂O₄(g) ⇌ 2NO₂(g), what happens to the color when the container volume is increased?
Show answer
Increasing volume lowers pressure; the system shifts toward more gas moles (2 NO₂), so more brown NO₂ forms and the color deepens.

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Study tools & related lessonsKey vocabulary · Related

Key vocabulary

Le Châtelier’s principle: A system at equilibrium shifts to partially counteract an applied stress
stress: A change in concentration, pressure, volume, or temperature applied to an equilibrium
reaction quotient (Q): The equilibrium-constant ratio computed with current concentrations
Δ n: Change in moles of gas: products minus reactants
exothermic / endothermic: Releases heat / absorbs heat
catalyst: Substance that speeds both directions equally without being consumed

Sources & references

openstax.org — Chemistry 2e

This lesson was adapted from the open educational references above; their licenses and attributions are preserved. See Copyright & Licensing.

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