Topic 9 · page 12 of 15

Le Chatelier — concentration and pressure

If you change the conditions of a system at equilibrium, the position shifts to oppose the change. Adding a substance shifts the position away from it, and raising the pressure shifts it towards the side with fewer moles of gas.

13 min read9.10

Read first: Reversible reactions and dynamic equilibrium

Le Chatelier's principle 9.10

A system at equilibrium has a position: the proportions of reactants and products. If you change a condition, the position changes until a new equilibrium is reached. Le Chatelier's principle lets you predict the direction without any calculation.

The specification asks you to predict the effect and to justify it, for a homogeneous system. Homogeneous means that every species is in the same phase, for example all gases or all in aqueous solution.

Use this order for a justification, every time:

  1. Name the change made.
  2. State that the position shifts to oppose it.
  3. Say which way (to the left or to the right, towards reactants or towards products).
  4. Say what results: more or less of a named substance.

Concentration 9.10

Take the Haber process:

NX2(g)+3 HX2(g)⇌2 NHX3(g) \ce{N2(g) + 3H2(g) <=> 2NH3(g)}
Change The system opposes by Shift Result
Add N₂ or H₂ using up some of the added reactant right more NH₃
Remove NH₃ making more NH₃ right more NH₃
Add NH₃ using up some of the added NH₃ left more N₂ and H₂

The second row is used in industry. The ammonia is cooled until it condenses to a liquid and is removed, so the equilibrium keeps shifting to the right to make more.

Colours give a check. Dichromate and chromate ions are a classic example:

2 CrOX4X2−(aq)+2 HX+(aq)⇌CrX2OX7X2−(aq)+HX2O(l) \ce{2CrO4^2-(aq) + 2H+(aq) <=> Cr2O7^2-(aq) + H2O(l)}

Chromate is yellow and dichromate is orange. Adding acid (more H⁺) shifts the position to the right, so the solution turns orange. Adding alkali removes H⁺, so the position shifts to the left and the solution turns yellow.

Pressure 9.10

Pressure only matters when gases are involved. A change in pressure shifts the position towards the side that opposes it:

So you count the moles of gas on each side of the equation, and only the gases.

Particle drawing of the Haber equation. Left side: one nitrogen molecule drawn as two joined circles and three hydrogen molecules drawn as pairs of small circles, a total of four moles of gas. Right side: two ammonia molecules, each a nitrogen circle with three hydrogen circles, a total of two moles of gas. Text says that higher pressure favours the side with fewer gas molecules, here the right.
Four moles of gas on the left and two on the right. A higher pressure shifts the position to the right.

Worked examplePredicting the effect of pressure

For each reaction, predict the effect of increasing the pressure.

NX2OX4(g)⇌2 NOX2(g) \ce{N2O4(g) <=> 2NO2(g)}
  1. Left: 1 mol of gas. Right: 2 mol of gas.
  2. A higher pressure favours the side with fewer moles, the left.
  3. The position shifts to the left, so there is more N₂O₄ and the mixture is paler.
HX2(g)+IX2(g)⇌2 HI(g) \ce{H2(g) + I2(g) <=> 2HI(g)}
  1. Left: 2 mol of gas. Right: 2 mol of gas.
  2. Equal numbers of moles, so changing the pressure has no effect on the position.

Pressure changes the position only when the numbers of moles of gas differ. For any gaseous reaction, a higher pressure also raises the rate, because the molecules are closer together (page 5). Do not mix up the two effects: the rate rises whichever way the position shifts.

Check yourself

  1. State Le Chatelier's principle.
  2. For N₂(g) + 3H₂(g) ⇌ 2NH₃(g), state the shift for (a) adding more N₂, (b) removing NH₃ as it forms, and (c) increasing the pressure.
  3. Predict the effect on the position of equilibrium of increasing the pressure for each reaction: (a) CO(g) + 2H₂(g) ⇌ CH₃OH(g); (b) N₂O₄(g) ⇌ 2NO₂(g); (c) CH₄(g) + H₂O(g) ⇌ CO(g) + 3H₂(g); (d) H₂(g) + I₂(g) ⇌ 2HI(g).
  4. Justify the effect of increasing the pressure on C₂H₄(g) + H₂O(g) ⇌ C₂H₅OH(g).
  5. Dichromate ions react: Cr₂O₇²⁻ + H₂O ⇌ 2CrO₄²⁻ + 2H⁺. Dilute acid is added to a yellow solution at equilibrium. State and explain the colour change.
  6. A student says: "Higher pressure speeds up the forward reaction, so the equilibrium always shifts to the products." Explain what is wrong.

Answers cover these until you have tried the questions

  1. If a system at equilibrium is subjected to a change, the position of equilibrium shifts to oppose the change.
  2. (a) Adding N₂: the system uses up some of the extra N₂, so it shifts right. (b) Removing NH₃: the system makes more NH₃, so it shifts right. (c) Higher pressure: 4 mol of gas on the left, 2 on the right, so it shifts right.
  3. (a) 3 mol of gas on the left, 1 on the right: shifts right, towards CH₃OH. (b) 1 mol on the left, 2 on the right: shifts left, towards N₂O₄. (c) 2 mol on the left, 4 on the right: shifts left. (d) 2 mol on each side: no effect.
  4. The pressure is increased, so the equilibrium shifts to oppose the change by reducing the pressure. There are 2 mol of gas on the left and 1 mol on the right, so the position shifts to the right, and more ethanol is formed.
  5. The added H⁺ makes the position shift to oppose the increase in H⁺, which is to the left. This makes more Cr₂O₇²⁻, so the solution turns orange. (If alkali were added to an orange solution, H⁺ would be removed, the position would shift to the right, and the solution would turn yellow.)
  6. Higher pressure does speed up the forward reaction, but it also speeds up the backward reaction, and the position is decided by the number of moles of gas on each side. It shifts towards the side with fewer moles of gas, which is the products only if the products have fewer moles. If the moles are equal, there is no shift at all.