Lesson 4 · 30 min

Le Chatelier’s principle: how equilibria respond

Add a reagent, compress, heat — disturb real colour-changing equilibria (cobalt chloride, iron thiocyanate, NO₂), watch them shift, and see why they only partly undo the change.

In this lesson

Use Le Chatelier’s principle to predict the effect of concentration, pressure (volume) and temperature changes
Separate the two kinds of disturbance: those that change Q (concentration, volume) and the one that changes K (temperature)
Calculate how K changes with temperature using the van ’t Hoff equation
Explain why an equilibrium only partly undoes a disturbance, and why catalysts and inert gas at constant volume do not shift it

Dissolve cobalt chloride crystals in concentrated hydrochloric acid and you get a violet solution. Pour some into a test tube and dip it in boiling water: within seconds it turns deep blue. Move it to ice water and it goes back to pink. The solution holds both pink $\ce{[Co(H2O)6]^2+}$ and blue $\ce{[CoCl4]^2-}$ , and the temperature decides which wins. In 1884 Le Chatelier summed up the pattern: when a system at equilibrium is disturbed, the equilibrium shifts in the direction that counteracts the disturbance.

Predict first

The forward reaction $\ce{[Co(H2O)6]^2+} + \ce{4Cl-} \mathrel{\ce{<=>}} \ce{[CoCl4]^2-} + \ce{6H2O}$ is endothermic. What happens when the violet solution goes from room temperature into 80 °C water?

Pick an answer and lock it in before you look. A wrong guess teaches more than no guess.

It turns blue It turns pink No change It turns colourless

Making the principle precise with Q and K

Le Chatelier’s principle is a handy rule of thumb, but what really does the work is last lesson’s Q and K:

Concentration: adding a reactant (or removing a product) makes Q < K and the reaction runs forward; adding a product makes Q > K and it runs back. K is unchanged.
Volume (pressure): compressing raises every gas concentration by the same factor. If $\Delta n \neq 0$ this changes Q, and the equilibrium moves towards the side with fewer gas molecules. Reactions with $\Delta n = 0$ (such as $\ce{H2} + \ce{I2} \mathrel{\ce{<=>}} \ce{2HI}$ ) are unaffected.
Inert gas at constant volume: no partial pressure changes, Q is unchanged and nothing shifts (at constant pressure it dilutes the mixture and does shift it).
Temperature: the only thing that changes K. Heating raises K for an endothermic reaction and lowers it for an exothermic one.
Catalyst: speeds up forward and reverse equally; K is unchanged and equilibrium is simply reached sooner.

\ln\frac{K_2}{K_1} = -\frac{\Delta_r H^\circ}{R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right)

The van ’t Hoff equation (taking ΔH° as constant over the range). For an endothermic reaction (ΔH° > 0), heating raises K.

$K_1, K_2$: equilibrium constants at T₁ and T₂
$\Delta_r H^\circ$: standard reaction enthalpy, J/mol
$R$: 8.314 J mol⁻¹ K⁻¹

Worked example

How much NO₂ in boiling water?

For $\ce{N2O4} \mathrel{\ce{<=>}} \ce{2NO2}$ , $K_p = 0.148$ at 25 °C and $\Delta_r H^\circ = +57.2$ kJ/mol. Find $K_p$ at 100 °C.

Real systems · colours computed from absorption spectra

Le Chatelier: disturb it and watch it re-equilibrate

$\ce{N2O4(g)} \mathrel{\ce{<=>}} \ce{2NO2(g)}$

Endothermic: heating makes more of the brown NO₂.

NO₂ partial pressure 0.297 bar

Temperature: 25 °C
K_c: 0.00599
Q / K: 1
$[\ce{N2O4}]$: 0.024 M
$[\ce{NO2}]$: 0.012 M

Q = K: back at equilibrium

Concentrations over time (vertical lines = disturbances)

N₂O₄
NO₂

In the NO₂ syringe, press “Compress to half the volume”: the colour first darkens abruptly (the concentration doubles), then fades as the equilibrium shifts towards N₂O₄ — but never back to where it started. Try argon at constant volume: nothing happens. Or switch to iron thiocyanate and add Fe³⁺ or SCN⁻ to deepen the red.

Compressing a 1.00 bar NO₂/N₂O₄ equilibrium mixture to half its volume at 25 °C (Kp = 0.148, ideal gases)
Moment	$p_{\ce{NO2}}$ / bar	$p_{\ce{N2O4}}$ / bar	Q
Before (equilibrium)	0.318	0.682	0.148
Just after compressing	0.636	1.364	0.297
After re-equilibrating	0.464	1.450	0.148

Source: Computed with the exact ICE solver from NBS thermodynamic data (Kp = 0.148 at 25 °C).

Concept check

For the exothermic reaction $\ce{N2(g)} + \ce{3H2(g)} \mathrel{\ce{<=>}} \ce{2NH3(g)}$ , which change increases the equilibrium mole fraction of $\ce{NH3}$ ?

Raising the temperature Compressing the gas into a smaller volume Adding argon at constant volume Adding an iron catalyst

Try the calculation

The cobalt chloride equilibrium has $\Delta_r H^\circ \approx +62$ kJ/mol. By what factor does K change from 25 °C to 80 °C?

Key takeaways

A disturbed equilibrium shifts to counteract the disturbance — but only partly.
Concentration and volume changes alter Q; temperature alters K; catalysts and inert gas at constant volume shift nothing.
Compression favours the side with fewer gas molecules; reactions with Δn = 0 do not respond to pressure.
Van ’t Hoff: heating raises K for endothermic reactions and lowers it for exothermic ones.

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Data and references

H. Le Chatelier, “Sur un énoncé général des lois des équilibres chimiques”, Comptes Rendus 99, 786–789 (1884).
T. W. Swaddle, L. Fabes, “Octahedral–tetrahedral equilibria in aqueous cobalt(II) solutions at high temperatures”, Can. J. Chem. 58, 1418–1426 (1980).
D. D. Wagman et al., J. Phys. Chem. Ref. Data 11, Suppl. 2 (1982) (N₂O₄ ⇌ 2NO₂).
K. de Berg, M. Maeder, S. Clifford, Inorg. Chim. Acta 466 (2017); H. S. Frank, R. L. Oswalt, J. Am. Chem. Soc. 69, 1321 (1947) (FeSCN²⁺).
A. C. Vandaele et al., J. Quant. Spectrosc. Radiat. Transfer 76, 373–391 (2003) (NO₂ absorption).