Chemical equilibrium · Lesson 3

Push an equilibrium and it pushes back

Add a chemical, take one away, squeeze a gas or dilute a solution, and an equilibrium responds in a predictable way. Le Chatelier’s principle predicts the response; collision theory explains it.

At the summit of Mount Everest the air pressure is about a third of the pressure at sea level. Every breath delivers about a third as much oxygen, and a climber who arrived there straight from the coast would lose consciousness within minutes. Yet mountaineers have stood on the summit without bottled oxygen. They manage it by giving their blood weeks to adjust.

Oxygen travels around your body attached to haemoglobin in red blood cells. In simplified form, Hb(aq) + O₂(aq) ⇌ HbO₂(aq). In the lungs the oxygen concentration is high, so the equilibrium lies to the right and haemoglobin loads up with oxygen. In working muscles the oxygen concentration is low, the equilibrium shifts to the left, and the oxygen is released where it is needed.

At altitude the concentration of oxygen in the lungs falls, the equilibrium shifts left, and less oxyhaemoglobin forms. Over several weeks the body responds by making more red blood cells, so the concentration of haemoglobin rises. That pushes the equilibrium back to the right and restores much of the lost oxygen transport. Two changes in concentration, two shifts in an equilibrium: this lesson explains how to predict them and, more importantly, why they happen.

Le Chatelier’s principle

If a system at equilibrium is disturbed by a change in conditions, the system adjusts to partially counteract the change, and a new equilibrium is established.

The principle is a quick way to predict the direction of a shift. It is not an explanation. The system has no intentions; molecules do not “want” to oppose anything. The real reason a system shifts is that the change makes one reaction faster than the other for a while. In your answers, use the principle to predict and collision theory to explain.

Changing a concentration

Take N₂O₄(g) ⇌ 2NO₂(g) at equilibrium and inject extra NO₂. At that instant the concentration of NO₂ jumps, so NO₂ molecules collide with each other more often and the reverse rate increases. The forward rate depends only on [N₂O₄], which has not changed yet. Now the reverse rate is greater than the forward rate, so there is a net reverse reaction: NO₂ is used up faster than it forms and N₂O₄ builds up. As that happens the reverse rate falls and the forward rate rises until they are equal again.

Concentrations. Only the added species jumps. Everything after the jump is gradual.

Rates. The reverse rate jumps; the forward rate changes smoothly. They meet again at a new, higher rate.

Notice where the system ends up. The new [NO₂] is lower than just after the injection but still higher than before it. The change has been partially counteracted, never completely undone. Removing a species works the same way in reverse: removing NO₂ lowers the reverse rate, so there is a net forward reaction that replaces some, but not all, of what was removed.

In solution, a species can be removed by precipitating it or by reacting it away. Adding silver ions to the blood-red equilibrium Fe³⁺(aq) + SCN⁻(aq) ⇌ FeSCN²⁺(aq) precipitates silver thiocyanate, lowering [SCN⁻]. The forward rate falls, there is a net reverse reaction, and the red colour fades.

Changing the volume of a gas mixture

Halving the volume of a gas mixture doubles the concentration of every gas at the same instant. Whether the system shifts depends on how many gas particles each reaction needs to bring together.

For N₂O₄(g) ⇌ 2NO₂(g), the forward rate depends on [N₂O₄], so it doubles. The reverse reaction needs two NO₂ molecules to collide, so its rate goes up about four times. The reverse rate is now greater, so the system shifts to the left, the side with fewer gas particles (1 mol of gas against 2 mol). Le Chatelier’s principle predicts the same thing: the system partly counteracts the increase in pressure by reducing the number of gas particles.

Both concentrations double instantly, then NO₂ falls and N₂O₄ rises. Both end higher than before the change.

Both rates jump, but the reverse rate jumps further.

Decreasing the volume (increasing the pressure) shifts the equilibrium towards the side with fewer moles of gas.
Increasing the volume shifts it towards the side with more moles of gas.
If both sides have the same number of moles of gas, as in H₂(g) + I₂(g) ⇌ 2HI(g), both rates increase by the same factor and there is no shift. The purple colour of iodine still darkens, but only because [I₂] has doubled.
Only gases count. Ignore solids and liquids when you count moles.

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Le Chatelier piston

Choose a change, predict the shift, then watch the rates and concentrations respond. The gauge compares the reaction quotient Q with K.

Diluting a solution

Adding water to an aqueous equilibrium lowers the concentration of every dissolved species, the solution equivalent of increasing the volume of a gas. The system shifts towards the side with more dissolved particles. For Fe³⁺(aq) + SCN⁻(aq) ⇌ FeSCN²⁺(aq), the forward reaction needs two dilute ions to find each other, so its rate falls more than the reverse rate, which involves only one species. There is a net reverse reaction, and the red colour fades by more than dilution alone would cause. Liquid water itself does not appear in K, so it is only its diluting effect that matters.

Changes that do not shift an equilibrium

A catalyst lowers the activation energy of both directions, so it increases the forward and reverse rates by the same factor. The rates stay equal, so there is no shift. The system simply reaches equilibrium sooner.
An inert gas added at constant volume, such as argon, increases the total pressure but does not change the concentration of any reacting gas. The rates are unchanged, so there is no shift.
Adding a pure solid or liquid does not change its concentration, so there is no shift.

A catalyst: both rates jump by the same factor and stay equal. The concentrations do not change at all.

Equilibria you can see

Equilibrium	Colours	Useful disturbances
Fe³⁺(aq) + SCN⁻(aq) ⇌ FeSCN²⁺(aq)	pale yellow, colourless ⇌ blood red	add Fe³⁺ or SCN⁻: redder. Add Ag⁺ (removes SCN⁻): paler
Cr₂O₇²⁻(aq) + H₂O(l) ⇌ 2CrO₄²⁻(aq) + 2H⁺(aq)	orange ⇌ yellow	add acid: more orange. Add OH⁻ (removes H⁺): more yellow
Co(H₂O)₆²⁺(aq) + 4Cl⁻(aq) ⇌ CoCl₄²⁻(aq) + 6H₂O(l)	pink ⇌ blue	add concentrated HCl: bluer. Dilute with water: pinker
N₂O₄(g) ⇌ 2NO₂(g)	colourless ⇌ brown	compress: darker at once, then partly fades

A full-mark answer to “predict and explain the effect of …”

State what the change does at that instant: which concentration changes, so which rate (forward or reverse) increases or decreases.
State that this rate is now greater (or smaller) than the other, so there is a net forward or net reverse reaction.
Say what happens to the amounts or concentrations (and any colour) as the system moves to the new equilibrium, where the rates are equal again.
If asked, compare the new equilibrium with the original: the change is only partially counteracted.

Graph clues examiners love

One concentration jumps, the others do not: a species was added or removed.
All gas concentrations jump by the same factor: the volume changed.
No concentration jumps, but both rates jump equally: a catalyst.
No concentration jumps, both rates jump unequally: a temperature change (Lesson 4).

Now put it to workFive worked examples, twenty practice questions and ten harder problems on this lesson.

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