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How does a system at equilibrium respond to a change in concentration, pressure or temperature?

Topic 7.9 Introduction to Le Chatelier's Principle: predict the direction a system at equilibrium shifts in response to a change in concentration, volume or pressure, or temperature, using Le Chatelier's principle.

A focused answer to AP Chemistry Topic 7.9, covering Le Chatelier's principle and how an equilibrium shifts in response to changes in concentration, volume or pressure, and temperature, including the effect on K of temperature, with full worked examples.

Generated by Claude Opus 4.811 min answerUpdated 2026-06-04

Reviewed by: AI editorial process; not yet individually human-reviewed

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Quick answer

Le Chatelier's principle says that when a system at equilibrium is disturbed, it shifts in the direction that partially counteracts the disturbance. Adding a reactant or product shifts the equilibrium away from the added species (to consume it); removing one shifts toward it (to replace it). Decreasing the volume (raising the pressure) shifts toward the side with fewer moles of gas; increasing the volume shifts toward more moles of gas. Adding heat shifts an endothermic reaction toward products and an exothermic reaction toward reactants. Of all these, only a change in temperature changes the value of K: concentration, volume and pressure changes shift the position of equilibrium but leave K unchanged. A catalyst changes neither K nor the position; it only speeds the approach to equilibrium.

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What this topic is asking
Le Chatelier's principle
Concentration changes
Volume and pressure changes
Temperature changes
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What this topic is asking

The College Board (Topic 7.9) wants you to predict the direction a system at equilibrium shifts when you change the concentration, the volume or pressure, or the temperature, using Le Chatelier's principle. The crucial subtlety is that only temperature changes $K$ ; the other changes shift the position without changing $K$ .

Le Chatelier's principle

The principle is a qualitative summary of how $Q$ moves back toward $K$ (Topic 7.10 makes this quantitative). When a stress moves the system away from equilibrium, the reaction proceeds in whichever direction restores the balance, opposing the disturbance.

Concentration changes

This is the most intuitive case: pour in more reactant and the forward reaction runs to use it up, making more product. Take product away and the forward reaction runs to replace it. The system always moves to oppose the change, and once it settles, $K$ has the same value as before.

Volume and pressure changes

For gas-phase equilibria, decreasing the volume raises the total pressure, and the equilibrium shifts toward the side with fewer moles of gas to reduce the pressure; increasing the volume shifts toward more moles of gas. Count the moles of gas on each side from the coefficients. If both sides have equal moles of gas, a volume change causes no shift. Adding an inert gas at constant volume changes the total pressure but not the partial pressures, so it causes no shift.

Temperature changes

Temperature is the special case that changes K. Treat heat as a reactant (endothermic) or a product (exothermic). Adding heat to an endothermic reaction (heat is a reactant) shifts toward products and increases K; adding heat to an exothermic reaction (heat is a product) shifts toward reactants and decreases K. Cooling does the reverse. Because the temperature change alters the rate constants of the forward and reverse reactions unequally, it is the only disturbance that moves the value of $K$ .

Predicting shifts for a gas equilibrium

For $2\text{SO}_2(g) + \text{O}_2(g) \rightleftharpoons 2\text{SO}_3(g)$ , $\Delta H < 0$ , predict the shift for each change.

step 1 Add more $\text{O}_2$

Adding a reactant shifts the equilibrium right (toward $\text{SO}_3$ ) to consume the added oxygen.

step 2 Decrease the volume

Reactants have $2 + 1 = 3$ mol of gas; products have $2$ mol. Decreasing the volume shifts toward fewer moles, so right (toward products).

step 3 Increase the temperature

The reaction is exothermic, so heat is a product; adding heat shifts left (toward reactants) and decreases $K$ .

step 4 Add a catalyst

A catalyst speeds both directions equally, so there is no shift; equilibrium is just reached faster.

Try this

Q1. For $\text{A}(g) \rightleftharpoons 2\text{B}(g)$ , predict the shift when the volume is increased. [2 points]

Cue. Products have more moles of gas (2 vs 1), so increasing the volume shifts toward products (B).

Q2. State which type of change alters the value of $K$ . [1 point]

Cue. Only a change in temperature.

Exam-style practice questions

Practice questions written in the style of College Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AP 2023 (style)4 marksSection II (long FRQ, part). For

\text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g)

\Delta H = -92\ \text{kJ}

. Predict and justify the direction of shift when (a) more

\text{N}_2

is added, (b) the volume of the container is decreased, (c) the temperature is increased, and (d) a catalyst is added.

Show worked answer →

A 4-point conceptual FRQ on Le Chatelier's principle.

(a) Add $\text{N}_2$ (1 point): adding a reactant shifts the equilibrium to the right (toward products) to consume the added $\text{N}_2$ .
(b) Decrease volume (1 point): decreasing the volume raises the pressure, so the equilibrium shifts toward the side with fewer moles of gas; reactants have $1 + 3 = 4$ mol and products have $2$ mol, so it shifts right (toward products).
(c) Increase temperature (1 point): the forward reaction is exothermic, so heat is a product; adding heat shifts the equilibrium left (toward reactants), and $K$ decreases.
(d) Catalyst (1 point): a catalyst speeds both directions equally, so it does not shift the equilibrium; it only reaches equilibrium faster.

Markers reward the rightward shift on adding reactant, the rightward shift on decreasing volume (fewer-moles side), the leftward shift and decreased $K$ on heating, and no shift for a catalyst.

AP 2021 (style)1 marksSection I (multiple choice). For an endothermic reaction at equilibrium, increasing the temperature will (A) shift the equilibrium toward reactants and decrease

K

(B) shift the equilibrium toward products and increase

K

K

. Justify your choice.

Show worked answer →

A 1-point conceptual MCQ. The answer is (B).

For an endothermic reaction, heat is a reactant; adding heat shifts the equilibrium toward products and increases $K$ . Only a temperature change alters $K$ . The trap is forgetting that endothermic and exothermic reactions respond oppositely to heating.

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Sources & how we know this

AP Chemistry Course and Exam Description — College Board (2020)