ChemistryCore20 min read

Reversible Reactions & Equilibrium

Reactions that run both ways, and how to push them where you want

This topic appears in:

Practise the full question bank:

01

Some reactions run both ways

Many reactions are reversible: the products can react to reform the reactants. We write them with a double arrow, ⇌. In an open container the products often escape and the reaction just goes forward, but in a closed system — nothing in or out — something more interesting happens.

The forward reaction starts fast and slows as reactants are used up; the backward reaction starts at zero and speeds up as products build. Eventually the two rates become equal — dynamic equilibrium. The concentrations stop changing, but both reactions are still going, just at the same rate in each direction. "Dynamic" is the key word: it is balance, not stillness.

02

Le Chatelier's principle

The whole topic is one rule: if you change the conditions of a system at equilibrium, it shifts to oppose the change. That is Le Chatelier's principle, and it predicts which way the balance moves.

Add more of a reactant and the system shifts forward to use it up. Remove a product and it shifts forward to replace it. The system always moves in the direction that partly cancels whatever you did.

03

Temperature and pressure

Temperature: raising it favours the endothermic direction (the one that absorbs the extra heat); lowering it favours the exothermic direction. So heating a reaction whose forward direction is exothermic pushes the equilibrium back, giving less product.

Pressure (for gases): raising it shifts the equilibrium toward the side with fewer gas molecules, easing the squeeze. If both sides have the same number of gas molecules, pressure has no effect. A catalyst changes neither position — it only makes equilibrium arrive faster.

A catalyst does not shift equilibrium

A catalyst speeds up the forward and backward reactions equally, so it reaches equilibrium sooner but does not change how much product you end up with. Only concentration, temperature and pressure move the position.

04

The Haber process

The classic industrial equilibrium is the Haber process for ammonia: N₂ + 3H₂ ⇌ 2NH₃, forward exothermic. The conditions are a compromise. High pressure (~200 atm) favours ammonia (4 gas molecules → 2). Low temperature would favour it too (exothermic), but too low and the reaction is far too slow — so a moderate ~450 °C is used with an iron catalyst, trading some yield for a workable rate. Unreacted gases are recycled. The ammonia goes mostly into fertilisers.

05

The Contact process

Sulfuric acid — the most-produced industrial chemical — is made by the Contact process, and its key step is another equilibrium governed by exactly the same rules:

Sulfur is first burned to sulfur dioxide (S + O₂ → SO₂). The crucial reversible step then oxidises it further: 2SO₂ + O₂ ⇌ 2SO₃, forward exothermic. By Le Chatelier, a low temperature and high pressure would favour SO₃ — but in practice a moderate ~450 °C, a vanadium(V) oxide (V₂O₅) catalyst, and only a modest ~1–2 atm already give about a 96% conversion, so extremes are not worth the cost. Finally the SO₃ is absorbed into concentrated sulfuric acid to form oleum, then diluted to give more H₂SO₄.

Haber (ammonia)Contact (sulfuric acid)
Key equilibriumN₂ + 3H₂ ⇌ 2NH₃2SO₂ + O₂ ⇌ 2SO₃
Catalystironvanadium(V) oxide, V₂O₅
Temperature~450 °C~450 °C
Pressure~200 atm~1–2 atm
Both forward reactionsexothermicexothermic

Same principle, different numbers

Both processes run an exothermic gas equilibrium, so both would prefer low temperature (yield) but use a moderate ~450 °C (rate) with a catalyst. The Contact process needs little extra pressure because its conversion is already very high — a neat reminder that industrial conditions balance yield, rate and cost.

06

A worked example

Worked example

For N₂ + 3H₂ ⇌ 2NH₃ (forward exothermic), which way does the equilibrium shift if you (a) increase the pressure, (b) increase the temperature?

  1. (a) Left side has 4 gas molecules, right has 2. Higher pressure favours the side with fewer molecules.The system shifts to reduce the pressure you applied.
  2. So higher pressure shifts it forward → more ammonia.Fewer gas molecules on the product side.
  3. (b) The forward reaction is exothermic, so raising temperature favours the endothermic (backward) direction.The system absorbs the added heat by going the endothermic way.
  4. So higher temperature shifts it back → less ammonia (but faster).This is why the Haber process compromises on a moderate temperature.

(a) Higher pressure → more ammonia; (b) higher temperature → less ammonia (though the reaction is faster) — hence the compromise conditions.

Practice questions

4 questions · 13 marks · full working on every one

Try each one on paper first, then open the working. The marks are shown where they are actually awarded, because that is where they are actually lost.

Short questions

3 · 7 marks

Two marks each, in the style of the short-question section of the paper. Answer in two or three lines.

SQ1[2 marks]
Explain what is meant by dynamic equilibrium.
Model answer

In a closed system, the forward and backward reactions occur at equal rates, so the concentrations of reactants and products stay constant while both reactions continue.

Examiner tip. "Dynamic" = both reactions still happen; "equilibrium" = concentrations no longer change.

SQ2[2 marks]
State Le Chatelier's principle.
Model answer

If the conditions of a system at equilibrium are changed, the equilibrium shifts in the direction that opposes (partly cancels) the change.

Examiner tip. The system always moves to reduce whatever change you impose.

SQ3[3 marks]
For the exothermic reaction N₂ + 3H₂ ⇌ 2NH₃, state and explain the effect of increasing temperature on the yield of ammonia.
Model answer

Increasing temperature shifts the equilibrium in the endothermic (backward) direction, so the yield of ammonia decreases (though the reaction is faster).

Examiner tip. Raising temperature favours the endothermic direction — here that means less product.

Long questions

1 · 6 marks

Theory and numerical together, as they appear in the long-question section.

LQ1[6 marks]
The Haber and Contact processes are both important industrial equilibria.
  1. For the Haber process (N₂ + 3H₂ ⇌ 2NH₃), explain why high pressure increases the yield of ammonia. [3]
  2. Name the catalyst used in the Contact process and give the equation for its key equilibrium step. [3]
Mark scheme
  1. There are 4 gas molecules on the left and 2 on the right[1]
  2. Higher pressure shifts the equilibrium to the side with fewer gas molecules[1]
  3. so it shifts right, increasing the yield of ammonia[1]
  4. The Contact process uses a vanadium(V) oxide (V₂O₅) catalyst[1]
  5. 2SO₂ + O₂ ⇌ 2SO₃[1]
  6. balanced equation with correct formulae[1]

High pressure shifts to fewer gas molecules (right) → more NH₃; Contact process uses V₂O₅ for 2SO₂ + O₂ ⇌ 2SO₃.

Examiner tip. Pressure favours the side with fewer gas moles; a catalyst speeds equilibrium but does not change the yield.