Skip to main content
Northern IrelandChemistrySyllabus dot point

How do reversible reactions reach equilibrium, and how do conditions shift it?

Dynamic equilibrium and reversible reactions, Le Chatelier's principle and the effects of concentration, pressure and temperature, the role of a catalyst, the equilibrium constant Kc, and applications to industrial processes such as the Haber process.

A CCEA A-Level Chemistry answer on chemical equilibrium, covering dynamic equilibrium and reversible reactions, Le Chatelier's principle and the effects of concentration, pressure and temperature, the role of catalysts, the equilibrium constant Kc, and the compromise conditions used in industrial processes such as the Haber process.

Generated by Claude Opus 4.89 min answer

Reviewed by: AI editorial process; not yet individually human-reviewed

Have a quick question? Jump to the Q&A page

Jump to a section
  1. What this dot point is asking
  2. Dynamic equilibrium
  3. Le Chatelier's principle
  4. The equilibrium constant
  5. The Haber process
  6. Examples in context
  7. Try this

What this dot point is asking

CCEA wants you to describe dynamic equilibrium in reversible reactions, apply Le Chatelier's principle to predict the effect of changing concentration, pressure and temperature, explain the role of a catalyst, write and use the equilibrium constant KcK_c, and apply these ideas to industrial processes such as the Haber process.

Dynamic equilibrium

Le Chatelier's principle

A catalyst speeds up both the forward and backward reactions equally, so it lets equilibrium be reached faster but does not change its position.

Working through each change

For the exothermic reaction N2+3H22NH3\text{N}_2 + 3\text{H}_2 \rightleftharpoons 2\text{NH}_3, raising the ammonia concentration shifts the equilibrium back towards the reactants to remove the added ammonia, while removing ammonia (as in the real Haber plant) keeps pulling the equilibrium forward. Raising the pressure shifts the position towards the right because there are four gas moles on the left and only two on the right, so the system reduces the pressure by favouring the side with fewer molecules. Raising the temperature shifts an exothermic equilibrium backwards, because the system absorbs the added heat by favouring the endothermic (reverse) direction, which lowers the yield. Each prediction follows the same logic: identify the change, then identify the direction that opposes it.

The equilibrium constant

For the reaction aA+bBcC+dDa\text{A} + b\text{B} \rightleftharpoons c\text{C} + d\text{D}, the equilibrium constant is:

Kc=[C]c[D]d[A]a[B]bK_c = \frac{[\text{C}]^c[\text{D}]^d}{[\text{A}]^a[\text{B}]^b}

KcK_c is constant at a fixed temperature. Only temperature changes its value; changing concentration or pressure shifts the position of equilibrium but leaves KcK_c unchanged. A large KcK_c (much greater than 1) means the products are favoured at equilibrium, while a small KcK_c (much less than 1) means the reactants are favoured. The units of KcK_c depend on the equation: they are found by writing mol dm3\text{mol dm}^{-3} for each concentration term and cancelling. For a reaction with equal numbers of moles on each side, such as H2+I22HI\text{H}_2 + \text{I}_2 \rightleftharpoons 2\text{HI}, the units cancel completely and KcK_c has no units.

The Haber process

The overall yield of the Haber process is improved not by pushing the equilibrium harder but by removing ammonia as it forms (by cooling and liquefying it) and recycling the unreacted nitrogen and hydrogen. Removing a product continually shifts the equilibrium forward, so the gases pass through the reactor many times and the cumulative conversion is high even though a single pass converts only about 15%15\%.

Examples in context

Example 1. The Contact process for sulfuric acid. The key equilibrium 2SO2(g)+O2(g)2SO3(g)2\text{SO}_2(g) + \text{O}_2(g) \rightleftharpoons 2\text{SO}_3(g) is exothermic and runs with fewer gas moles on the product side. The same compromise reasoning as the Haber process applies: a moderate temperature of about 450 C450\ \text{C} balances yield against rate, a vanadium(V) oxide catalyst speeds the reaction, and only a slightly raised pressure is needed because the yield is already over 95%95\% at modest pressure, so the cost of high-pressure plant is not justified. This is a standard CCEA application that contrasts with the Haber process on the pressure decision.

Example 2. Carbon dioxide in the blood. The equilibrium CO2+H2OH2CO3H++HCO3\text{CO}_2 + \text{H}_2\text{O} \rightleftharpoons \text{H}_2\text{CO}_3 \rightleftharpoons \text{H}^+ + \text{HCO}_3^- keeps blood pH stable. When exercise produces extra carbon dioxide, Le Chatelier's principle predicts the equilibrium shifts right, raising the hydrogen ion concentration; the body compensates by breathing out carbon dioxide, which shifts the equilibrium back. This biological buffer is a clear real-world demonstration that adding or removing a species shifts an equilibrium in a predictable direction.

Try this

Q1. Define dynamic equilibrium. [2 marks]

  • Cue. Forward and backward reactions occur at equal rates in a closed system, so concentrations stay constant.

Q2. For an exothermic forward reaction, state the effect of increasing temperature on the yield of product. [1 mark]

  • Cue. The yield decreases, because equilibrium shifts in the endothermic (backward) direction.

Exam-style practice questions

Practice questions written in the style of CCEA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

CCEA 20205 marksFor the equilibrium H2(g)+I2(g)2HI(g)\text{H}_2(\text{g}) + \text{I}_2(\text{g}) \rightleftharpoons 2\text{HI}(\text{g}), a sealed flask of volume 1.00 dm31.00\ \text{dm}^3 was found at equilibrium to contain 0.20 mol0.20\ \text{mol} H2\text{H}_2, 0.20 mol0.20\ \text{mol} I2\text{I}_2 and 1.60 mol1.60\ \text{mol} HI\text{HI}. Calculate the value of KcK_c and state its units.
Show worked answer →

A direct KcK_c calculation; markers want the correct expression, substitution and units.

Write the expression: Kc=[HI]2[H2][I2]K_c = \dfrac{[\text{HI}]^2}{[\text{H}_2][\text{I}_2]}.

Because the volume is 1.00 dm31.00\ \text{dm}^3, the concentrations equal the moles: [HI]=1.60[\text{HI}] = 1.60, [H2]=0.20[\text{H}_2] = 0.20, [I2]=0.20 mol dm3[\text{I}_2] = 0.20\ \text{mol dm}^{-3}.

Substitute: Kc=(1.60)20.20×0.20=2.560.040=64K_c = \dfrac{(1.60)^2}{0.20 \times 0.20} = \dfrac{2.56}{0.040} = 64.

Units cancel here (two concentration terms on top, two on the bottom), so KcK_c has no units for this reaction.

Markers reward the correct expression, the substitution, the value of 64, and the correct statement that there are no units. A frequent error is to forget to square the HI\text{HI} concentration.

CCEA 20184 marksThe reaction 2SO2(g)+O2(g)2SO3(g)2\text{SO}_2(\text{g}) + \text{O}_2(\text{g}) \rightleftharpoons 2\text{SO}_3(\text{g}) is exothermic and is used in the Contact process. Explain, in terms of yield and rate, why a moderate temperature and a catalyst are used rather than a very low temperature.
Show worked answer →

This tests the compromise reasoning behind an industrial equilibrium.

Because the forward reaction is exothermic, a low temperature would shift the equilibrium towards the products and give a higher yield of sulfur trioxide. However, at a low temperature the rate of reaction would be far too slow, because few molecules have the activation energy, so the process would be uneconomic.

A moderate temperature (about 450 C450\ \text{C}) is therefore chosen as a compromise: the yield is still acceptable and the rate is fast enough for continuous production.

A vanadium(V) oxide catalyst is added to speed up the reaction further without affecting the position of equilibrium, so a good rate is reached at the lower, yield-friendly temperature.

Markers reward the yield argument, the rate argument, the idea of compromise, and the correct role of the catalyst (rate only, not yield).

Related dot points

Sources & how we know this