Le Chatelier's principle

Result

Le Chatelier's principle

T-045Home CU-203Threads equilibrium

Statement

How an equilibrium responds to a disturbance.

Why it matters

law-of-mass-action established the equilibrium constant \(K\) as a fixed ratio at a given temperature; Le Chatelier's principle addresses the practical question that follows immediately — what happens when a system already sitting at that balance is disturbed, by adding reagent, changing volume, or changing temperature. reaction-quotient supplies the exact mechanism behind the qualitative rule given here, and the same reasoning underlies solubility-product's common-ion effect. Beyond the classroom, this is directly how industrial reaction conditions (most famously the Haber process) are chosen to favour yield.

Hypotheses

The system starts at true equilibrium (\(Q=K\)) before the disturbance is applied.Le Chatelier's principle describes a response to a disturbance from equilibrium, not the approach to equilibrium from an arbitrary starting mixture (that more general case is reaction-quotient's own subject). A single variable — concentration, pressure/volume, or temperature — is changed at a time, holding the others momentarily fixed.Simultaneous changes to multiple variables can have compounding or opposing effects that are not simply the sum of each individual prediction; the clean, single-variable predictions developed here are the standard teaching case. Temperature changes are treated as fundamentally different from concentration or pressure changes.Only a temperature change actually moves the value of \(K\) itself (via the van't Hoff relation); concentration and pressure changes instead move \(Q\) away from a fixed \(K\). Reasoning about a temperature disturbance by simple "the system opposes the change" intuition, without reference to the sign of \(\Delta H^\circ\), is not reliable.

Proof

\text{Add reactant} \Rightarrow Q

Increasing a reactant's concentration enlarges the denominator of the reaction-quotient expression, dropping \(Q\) below \(K\); by reaction-quotient's own \(QA

\text{Decrease volume (increase pressure)} \Rightarrow \text{shift toward the side with fewer moles of gas}

Reducing volume raises every gas-phase concentration simultaneously; whether \(Q\) rises above or falls below \(K\) depends on \(\Delta n_{\text{gas}}\) (moles of gaseous product minus reactant), and the reaction shifts toward whichever side has fewer gas moles, since that side's concentration terms are raised to smaller total powers and change proportionally less. A

\frac{d\ln K}{dT} = \frac{\Delta H^\circ}{RT^2}

The van't Hoff relation shows \(K\) itself, not merely \(Q\), moves with temperature; for an exothermic reaction (\(\Delta H^\circ<0\)) increasing \(T\) decreases \(K\), shifting equilibrium toward reactants, and the reverse for an endothermic reaction. B

\text{Catalyst: lowers } E_a \text{ equally for forward and reverse} \Rightarrow k_f/k_r \text{ (hence } K\text{) unchanged}

A catalyst accelerates both the forward and reverse elementary steps by the same factor (law-of-mass-action's rate-constant-ratio definition of \(K\)), so it changes only the time taken to reach equilibrium, never the equilibrium position itself. A

Result

\text{Disturbance}\Rightarrow Q\neq K \Rightarrow \text{net reaction restores } Q=K\ \big(\text{with } K \text{ itself shifting only for temperature}\big)

Reading. Every apparent "opposing" response to a disturbance is really the direct, mechanical consequence of \(Q\) being pushed away from \(K\), then relaxing back — except for temperature, which moves \(K\) itself.

Scope. Concentration and pressure/volume predictions are reliable for any system starting at genuine equilibrium; temperature predictions require the sign of \(\Delta H^\circ\); catalysts never shift equilibrium position under any circumstance.

Corollaries & converses

The Haber process's moderate-temperature, high-pressure operating conditions are a direct industrial application of Steps 2 and 3 balanced against each other for the exothermic, gas-mole-reducing synthesis of ammonia.
solubility-product's common-ion effect (adding an ion already present in a saturated solution reduces solubility) is exactly this same \(Q\)-restoring logic applied to a dissolution equilibrium.
Converse: observing which direction an equilibrium shifts under a known disturbance lets you infer the sign of \(\Delta n_{\text{gas}}\) or \(\Delta H^\circ\) without needing \(K\)'s numeric value.

Fails without

Predict a shift when an inert gas is added at constant volume: since no reacting species' concentration or partial pressure actually changes, \(Q\) never moves away from \(K\) at all, and no shift occurs — a case where the naive "add something, expect an opposing shift" intuition gives a flatly wrong prediction.
Reason about a temperature disturbance using the same "oppose the change" logic as a concentration disturbance, without checking the sign of \(\Delta H^\circ\) (Hypotheses): since temperature uniquely moves \(K\) itself rather than \(Q\), this shortcut can give the wrong shift direction entirely, especially for reactions where intuition about "heat as a reactant/product" is not carefully applied.

Common errors

Believing a catalyst shifts equilibrium position rather than merely speeding the approach to the same, unchanged \(K\) (Step 4).
Assuming an inert gas added at constant volume shifts equilibrium; since it changes no reacting species' concentration or partial pressure, it has no effect at all — only reducing volume (genuinely raising concentrations) matters.
Treating "the system opposes the applied change" as itself a fundamental law, rather than as a mnemonic summary of the more precise \(Q\)-versus-\(K\) mechanism (Discussion).
Applying the concentration/pressure "restoring" intuition to temperature changes without recognising that temperature uniquely moves \(K\) itself.

Discussion

Henry Louis Le Chatelier proposed this principle in 1884, well before the reaction-quotient formalism that fully explains it was available in its modern form.

Taken as a bare "the system opposes the applied change," Le Chatelier's principle is a heuristic, not a rigorous law; the precise statement is always the comparison of \(Q\) against \(K\) (with \(K\) itself possibly moving, for temperature). This is why loose application of the heuristic occasionally seems to give an ambiguous or even wrong prediction — for a reaction with \(\Delta n_{\text{gas}}=0\), for instance, pressure changes affect no side of the equilibrium preferentially and produce no shift at all, despite "opposing the change" intuition suggesting some response should occur.

Common misconception: that increasing pressure always shifts a gas-phase equilibrium toward the side with fewer moles, regardless of stoichiometry. This fails whenever \(\Delta n_{\text{gas}}=0\) (equal gas moles on both sides), where a pressure or volume change has no effect on the equilibrium position at all.

Worked examples

\text{N}_2(g)+3\text{H}_2(g)\rightleftharpoons 2\text{NH}_3(g),\quad \Delta H^\circ<0,\quad \Delta n_{\text{gas}}=2-4=-2

Increasing pressure (reducing volume) shifts this equilibrium toward the side with fewer gas moles, i.e. toward \(\text{NH}_3\) (Step 2); increasing temperature shifts it away from the exothermic forward direction, i.e. toward \(\text{N}_2\) and \(\text{H}_2\) (Step 3) — the two effects work in opposite directions for yield, exactly the tension industrial Haber-process conditions must balance. A

\text{Add extra N}_2 \text{ at constant volume, to the system above at equilibrium}

Adding \(\text{N}_2\) drops \(Q\) below \(K\) (Step 1's mechanism), so the reaction shifts net forward, consuming some of the added \(\text{N}_2\) along with additional \(\text{H}_2\) and producing more \(\text{NH}_3\), until \(Q=K\) is restored. A

\text{High pressure favours NH}_3;\quad \text{high temperature disfavours NH}_3\ (\text{but speeds approach to equilibrium})

Reading. The Haber process's real operating conditions (moderate temperature, high pressure, plus a catalyst to compensate for the rate penalty of not using an even lower temperature) are a direct compromise between the equilibrium-yield and reaction-rate consequences derived above.

Scope. The identical reasoning — apply Steps 1–4 to the specific stoichiometry and thermochemistry at hand — predicts the shift direction for any gas-phase equilibrium under any single disturbance.

Problems

Predict the effect of increasing container volume on the equilibrium \(\text{H}_2(g)+\text{I}_2(g)\rightleftharpoons 2\text{HI}(g)\).
Solution
\(\Delta n_{\text{gas}}=2-2=0\); since gas moles are equal on both sides, changing volume changes all concentrations by the same factor and leaves \(Q\) unchanged relative to \(K\) — no shift occurs (Common errors, final bullet's underlying case).
Explain, using the Q-versus-K mechanism, why continuously removing a product as it forms drives a reaction toward completion, even for a reaction with an unfavourably small \(K\).
Solution
Removing product keeps the numerator of \(Q\) small, so \(Q\) is continually held below \(K\); by Step 1's logic, the reaction is repeatedly driven net forward, and since product is never allowed to accumulate enough to let \(Q\) catch up to \(K\), the reaction can in principle proceed essentially to completion regardless of how small \(K\) itself is.
For \(\text{CaCO}_3(s)\rightleftharpoons \text{CaO}(s)+\text{CO}_2(g)\), predict the effect of increasing container volume.
Solution
Only \(\text{CO}_2(g)\) contributes to the equilibrium expression (solids are omitted, law-of-mass-action's Common errors). Increasing volume lowers \([\text{CO}_2]\), dropping \(Q\) below \(K\); the reaction shifts net forward, decomposing more \(\text{CaCO}_3\) to restore \([\text{CO}_2]\) — consistent with the general rule that increasing volume favours the side with more moles of gas (here, the product side has one mole of gas versus zero on the reactant side).