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8873 / 2027
Chemical Equilibria

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Chemical Equilibria

Explain a changing equilibrium, calculate its composition and connect the Haber process to rate and yield.

A-Level 8873, revised syllabus (2026-2027)

01

Equilibrium is equal rates, not equal amounts

A closed system can look unchanged while molecules continue reacting.

A reversible reaction can proceed in both directions under suitable conditions. For H2(g) + I2(g) ⇌ 2HI(g), hydrogen and iodine form hydrogen iodide, while hydrogen iodide can decompose. Initially, a mixture containing only H2 and I2 has a forward reaction but no reverse reaction: no HI is present yet.

As products accumulate, the reverse reaction becomes possible and its rate increases. Meanwhile, reactants are used up and the forward rate decreases. At dynamic equilibrium, the forward and reverse rates are equal. In a closed system at constant temperature, each concentration then stays constant because each species is made and removed at the same rate. Both reactions continue.

Two different graph signatures of equilibrium

The forward and reverse rates converge to the same nonzero value. Below, reactant and product concentrations become constant at different values. These are qualitative sketches, not the same vertical quantity.

Equal rates explain constant concentrations. The two equilibrium concentrations do not need to be equal.

Check your understanding

A sealed mixture keeps constant concentrations of 0.80 mol dm-3 reactant and 0.15 mol dm-3 product. Do the unequal values rule out equilibrium?

  1. No. Equilibrium means equal forward and reverse rates, not equal concentrations.
  2. Yes. At equilibrium the reactant and product concentrations become equal.
  3. No. Equilibrium needs equal concentrations, so this mixture is still reacting forwards.
Show the answer

No. Equilibrium means equal forward and reverse rates, not equal concentrations.

If both reactions continue at equal rates, the concentrations stay constant at whatever values Kc allows. These unequal values can be equilibrium values.

02

Predict the response to a disturbance

Track the imposed change separately from the subsequent reaction.

Make a guess. You are not marked.

For 2SO2(g) + O2(g) <=> 2SO3(g), the forward reaction is exothermic. The mixture is heated at constant pressure. What happens to the amount of SO3 at the new equilibrium?

  1. It decreases, as heating favours the endothermic reverse reaction
  2. It stays the same, as heating changes the rate but not the position
  3. It increases, as heating speeds up the forward reaction more
Show the answer

It decreases, as heating favours the endothermic reverse reaction

Heating favours the endothermic direction, which absorbs the added energy. Here that is the reverse reaction, so some SO3 breaks down.

Le Chatelier's Principle: when a system at equilibrium is disturbed, its equilibrium position changes in a direction that tends to oppose the disturbance. The response partly offsets the imposed change; it does not promise to restore every original concentration.

Apply the principle to ammonia formation: an exothermic forward reaction
ChangeImmediate disturbanceSubsequent net reaction
Add H2 at fixed volume and temperature[H2] rises immediately.Forward reaction consumes some added H2; more NH3 forms.
Remove NH3[NH3] falls immediately.Forward reaction replaces some removed product.
Compress the gas mixture at constant temperatureAll gas concentrations rise; total pressure rises.Forward reaction reduces gas amount: 4 mol gas on the left become 2 mol on the right.
Increase temperatureHeat is supplied.The endothermic reverse direction is favoured; the equilibrium NH3 yield decreases.
Add a catalystBoth reaction directions gain a faster pathway.No change in equilibrium composition at the same temperature; equilibrium is reached faster.

For a pressure question, state how pressure changes. Compression changes the reacting gas concentrations. For H2(g) + I2(g) ⇌ 2HI(g), equal gas amounts occur on each side, so compression produces no equilibrium shift in this ideal-gas treatment. Adding an inert gas at fixed volume does not change the concentrations of the reacting gases, even though total pressure rises. Do not use total pressure alone to infer a shift.

Treat heat as a bookkeeping aid for the temperature response: an exothermic forward reaction releases heat, so heating favours the reverse direction. Heat is not a species to insert into a Kc expression. A temperature increase can make the forward reaction faster while making the equilibrium proportion of product smaller; rate and final composition are different questions.

Worked exampleFollow a concentration jump

For A(aq) ⇌ B(aq), a little concentrated A solution is added to an equilibrium mixture, with negligible volume change. Describe [A] and [B] immediately and as equilibrium is re-established.

Try it first. Then show the worked stepsHide the worked steps
  1. [A] jumps upwards because A was added. [B] is unchanged at the instant of addition.
  2. The forward rate increases, so net conversion of A to B occurs. [A] falls from its immediate peak; [B] rises.
  3. At the new equilibrium the rates are equal again. Both concentrations can differ from their original values, while their equilibrium ratio is fixed by the same temperature.
Answer

Distinguish the instantaneous physical addition from the later chemical readjustment. Do not sketch both concentrations jumping when only A was added.

Check your understanding

For 2SO2(g) + O2(g) <=> 2SO3(g), the forward reaction is exothermic. Which two changes increase the equilibrium proportion of SO3?

  1. Raising the temperature and raising the pressure
  2. Lowering the temperature and adding a catalyst
  3. Lowering the temperature and raising the pressure
Show the answer

Lowering the temperature and raising the pressure

Cooling favours the exothermic forward direction. Higher pressure favours the side with fewer gas moles, 2 rather than 3. Cooling also slows the reaction, so it is not automatically the best industrial choice.

03

Kc describes an equilibrium ratio

Use equilibrium concentrations with powers from the balanced equation.

For aA + bB ⇌ cC + dD, the concentration equilibrium constant is Kc = [C]c[D]d / ([A]a[B]b). The square brackets mean equilibrium concentration in mol dm-3. Coefficients become powers; they are not multipliers outside the brackets.

Construct the expression before inserting numbers
ReactionKc expressionUnits using mol dm-3
H2(g) + I2(g) ⇌ 2HI(g)[HI]2/([H2][I2])No units; the concentration powers cancel.
N2(g) + 3H2(g) ⇌ 2NH3(g)[NH3]2/([N2][H2]3)dm6 mol-2
A(aq) ⇌ B(aq)[B]/[A]No units.

For a specified reaction equation, only temperature changes Kc. Changing concentrations or volume changes the current ratio, so the reaction adjusts until the equilibrium ratio again equals Kc. A catalyst changes neither Kc nor the final equilibrium composition. For an exothermic forward reaction, raising temperature lowers Kc; for an endothermic forward reaction, raising temperature increases it.

A large Kc indicates a product-favoured equilibrium for that written equation; it does not tell you that the reaction is fast. Keep the equation fixed when comparing constants: reversing an equation swaps numerator and denominator, giving 1/Kc. In expressions involving a pure solid or pure liquid, its constant contribution is incorporated into Kc; do not insert the solid's mass as a concentration. The examples here focus on gases and solutes.

Worked exampleConvert amounts before calculating Kc

At a fixed temperature, a 2.00 dm3 vessel contains 0.200 mol H2, 0.200 mol I2 and 0.600 mol HI at equilibrium. Calculate Kc for H2 + I2 ⇌ 2HI.

Try it first. Then show the worked stepsHide the worked steps
  1. [H2] = [I2] = 0.200/2.00 = 0.100 mol dm-3; [HI] = 0.600/2.00 = 0.300 mol dm-3.
  2. Kc = (0.300)2/(0.100 × 0.100) = 9.00.
  3. Units cancel because the numerator and denominator each contain concentration squared. The vessel volume cancels in this particular expression; that is not true for every reaction.
Answer

Kc = 9.00, with no units for this equation.

Check your understanding

Can initial concentrations be put straight into the Kc expression?

  1. Only if the starting mixture is already at equilibrium; usually you must work out the changes first.
  2. Yes. Kc is a constant, so any set of concentrations for the mixture gives the same value.
  3. No. Kc uses the concentrations measured at the very start of the reaction.
Show the answer

Only if the starting mixture is already at equilibrium; usually you must work out the changes first.

Kc uses equilibrium concentrations. Use the reaction stoichiometry to find how much reacted, then calculate the equilibrium amounts before substituting.

04

Use the equation to link all the changes

An amount table stops incompatible equilibrium quantities entering the expression.

Equilibrium calculation
  1. Write the balanced equation

    Choose either amounts or concentrations for the change table, and keep them consistent.

  2. Define one change

    Every other change follows the coefficients, with opposite signs for consumption and formation.

  3. Insert equilibrium concentrations

    Divide amounts by the final volume, then substitute into the correct Kc expression.

  4. Check the result

    All concentrations must be non-negative, the elemental inventory must be conserved, and substitution must recover Kc.

Worked exampleFrom an initial mixture to a constant

A 2.00 dm3 vessel initially contains 0.500 mol H2 and 0.500 mol I2, with no HI. At equilibrium it contains 0.600 mol HI. Calculate Kc.

Try it first. Then show the worked stepsHide the worked steps
  1. Forming 0.600 mol HI consumes 0.300 mol H2 and 0.300 mol I2, from the 1:1:2 coefficient ratio.
  2. Equilibrium amounts are 0.200 mol H2, 0.200 mol I2 and 0.600 mol HI.
  3. Divide by 2.00 dm3: concentrations are 0.100, 0.100 and 0.300 mol dm-3.
  4. Kc = 0.3002/(0.100 × 0.100) = 9.00.
Answer

Kc = 9.00. Using 0.500 mol as an equilibrium reactant amount would ignore the reactant consumed.

Try it step by step

Work it out in steps: Amount of reactant used up, then Equilibrium amount of reactant, then Kc.

  1. Problem 1. A 2.00 dm3 vessel starts with 0.600 mol PCl5. At equilibrium it holds 0.200 mol Cl2. PCl5 <=> PCl3 + Cl2. Find Kc in mol dm-3 to 3 significant figures.
  2. Problem 2. A 1.00 dm3 vessel starts with 0.300 mol H2 and 0.300 mol I2. At equilibrium it holds 0.400 mol HI. H2 + I2 <=> 2HI. Find Kc, which has no unit here, to 3 significant figures.
  3. Problem 3. A 1.00 dm3 vessel starts with 0.400 mol SO2 and 0.200 mol O2. At equilibrium it holds 0.300 mol SO3. 2SO2 + O2 <=> 2SO3. Track SO2 as the reactant. Give each answer to 3 significant figures, with Kc in dm3 mol-1.
Show the answer

Problem 1.

Amount of reactant used up: Forming 0.200 mol Cl2 uses 0.200 mol PCl5 and forms 0.200 mol PCl3.

Equilibrium amount of reactant: Equilibrium PCl5 = 0.600 - 0.200 = 0.400 mol. Divide by 2.00 dm3: [PCl5] = 0.200, [PCl3] = [Cl2] = 0.100 mol dm-3.

Kc: Kc = (0.100 x 0.100) / 0.200 = 0.0500 mol dm-3.

Problem 2.

Amount of reactant used up: Forming 0.400 mol HI uses 0.200 mol H2 and 0.200 mol I2.

Equilibrium amount of reactant: Equilibrium H2 = I2 = 0.100 mol, so [H2] = [I2] = 0.100 and [HI] = 0.400 mol dm-3.

Kc: Kc = 0.4002 / (0.100 x 0.100) = 16.0, with no units.

Problem 3.

Amount of reactant used up: Forming 0.300 mol SO3 uses 0.300 mol SO2 (ratio 2 : 2) and 0.150 mol O2.

Equilibrium amount of reactant: Equilibrium SO2 = 0.400 - 0.300 = 0.100 mol, and O2 = 0.200 - 0.150 = 0.0500 mol.

Kc: Kc = 0.3002 / (0.1002 x 0.0500) = 180 dm3 mol-1.

Worked exampleFind an unknown equilibrium without a quadratic

For H2 + I2 ⇌ 2HI, Kc = 4.00. Initially [H2] = [I2] = 0.300 mol dm-3 and [HI] = 0. Let x mol dm-3 of each reactant be consumed. Find the equilibrium concentrations.

Try it first. Then show the worked stepsHide the worked steps
  1. Equilibrium concentrations are 0.300 - x, 0.300 - x and 2x mol dm-3.
  2. 4.00 = (2x)2/(0.300 - x)2. Concentrations are positive, so take the positive square root: 2.00 = 2x/(0.300 - x).
  3. 0.600 - 2x = 2x, giving x = 0.150 mol dm-3.
  4. Check: 0.3002/(0.150 × 0.150) = 4.00; all amounts are physically possible.
Answer

[H2] = [I2] = 0.150 mol dm-3; [HI] = 0.300 mol dm-3. H1 equilibrium calculations do not require solving quadratic equations.

Check your understanding

For A(aq) <=> B(aq), Kc = 3.00, and [A] + [B] = 0.800 mol dm-3. What are the equilibrium concentrations?

  1. [A] = 0.200 and [B] = 0.600 mol dm-3
  2. [A] = 0.600 and [B] = 0.200 mol dm-3
  3. [A] = 0.267 and [B] = 0.533 mol dm-3
Show the answer

[A] = 0.200 and [B] = 0.600 mol dm-3

Kc = [B]/[A] = 3.00, so [B] = 3[A]. Then 4[A] = 0.800, so [A] = 0.200 and [B] = 0.600. Check: 0.600/0.200 = 3.00 and the sum is 0.800.

05

Choose a useful production rate and yield

Industrial conditions combine equilibrium, kinetics and cost.

Make a guess. You are not marked.

Ammonia forms exothermically: N2 + 3H2 <=> 2NH3. What does a lower temperature do to the equilibrium yield and the rate?

  1. Yield rises, but the rate falls
  2. Both the yield and the rate rise together
  3. Both the yield and the rate fall together
Show the answer

Yield rises, but the rate falls

The exothermic forward direction is favoured at lower temperature, but fewer collisions have energy above Ea, so equilibrium is reached more slowly.

The Haber process makes ammonia: N2(g) + 3H2(g) ⇌ 2NH3(g), with an exothermic forward reaction. A representative school description uses an iron catalyst, about 450 °C and a pressure of about 200 atm. Actual plants vary; the reasoning behind the conditions matters more than treating one pair of numbers as universal.

Connect each choice to its benefit and limitation
ChoiceChemical reasonPractical trade-off
High pressureCompression favours 2 mol of product gas over 4 mol of reactant gas; it also increases gas concentrations.Compression consumes energy and stronger pressure-resistant equipment costs more. Indefinitely high pressure is not economical.
Moderately high temperatureA higher temperature increases reaction rate, but decreases Kc for the exothermic forward reaction.A compromise gives an acceptable rate while retaining a useful equilibrium yield. Very low temperature gives a favourable equilibrium but a slow approach to it.
Iron catalystLowers activation barriers and accelerates approach to equilibrium.Allows an economically useful rate; it does not increase the equilibrium yield at fixed temperature.
Cool the outlet and separate ammoniaAmmonia is condensed and removed from the gas stream.Separation allows product collection and leaves unreacted nitrogen and hydrogen for recycling.
Recycle unreacted gasesGases not converted on one pass return to the reactor.Raises the overall conversion of fresh feed, although conversion in one pass remains limited by the operating equilibrium.
Follow the gas through the plant
  1. Feed and compress

    Supply purified nitrogen and hydrogen in an approximately 1:3 mole ratio.

  2. React over iron

    Use high pressure and a moderately high temperature; the outlet still contains reactants as well as ammonia.

  3. Cool and separate

    Condense ammonia for collection.

  4. Recycle

    Return unreacted nitrogen and hydrogen to the reaction loop.

Worked exampleRead an industrial data comparison

Two runs use the same catalyst and pressure. At 400 °C the equilibrium NH3 proportion is higher than at 500 °C, but the initial production rate is lower. Explain both observations.

Try it first. Then show the worked stepsHide the worked steps
  1. Ammonia formation is exothermic. The lower temperature favours the forward equilibrium and gives a larger Kc.
  2. At the lower temperature, a smaller fraction of collisions has enough energy to react, so the rate constant is smaller.
  3. The manufacturer must compare the amount made per unit time, equipment and energy costs, not maximise the equilibrium percentage in isolation.
Answer

Higher equilibrium yield and faster reaction are separate advantages that can favour different temperatures.

Check your understanding

A proposal says, "More iron catalyst increases the equilibrium yield of ammonia." Which correction is right?

  1. More catalyst raises the yield, because it speeds up the forward reaction more.
  2. More catalyst helps equilibrium be reached sooner, but the yield at equilibrium is unchanged.
  3. More catalyst lowers the yield, because iron reacts with some of the ammonia formed.
Show the answer

More catalyst helps equilibrium be reached sooner, but the yield at equilibrium is unchanged.

A catalyst speeds up the forward and reverse reactions equally. It does not change Kc or the equilibrium yield at the same temperature, pressure and feed.

Exam-style question

Paper 2 Section B style

This question draws on Chemical Equilibria and Data and Experimental Skills.

Ethanoic acid and ethanol react reversibly to form ethyl ethanoate and water.

CH3COOH(l) + C2H5OH(l) ⇌ CH3COOC2H5(l) + H2O(l)

1.00 mol of ethanoic acid and 1.00 mol of ethanol were sealed in a tube with a few drops of acid catalyst and kept at 60 °C until equilibrium was reached. The mixture then contained 0.333 mol of ethanoic acid. Ignore any water in the catalyst.

Ammonia is made by the Haber process. N2(g) + 3H2(g) ⇌ 2NH3(g), ΔH = -92 kJ mol-1. Table 1 shows the percentage of ammonia in the equilibrium mixture under different conditions.

Table 1. Percentage of ammonia at equilibrium
pressure / atm300 °C400 °C500 °C
100522510
200633618
400744830
  1. (a) State what is meant by dynamic equilibrium. [2]
  2. (b) Write the expression for Kc for the esterification. [1]
  3. (c) Calculate Kc at 60 °C. [3]
  4. (d) Explain why Kc for this reaction has no units, and why the volume of the mixture is not needed. [1]
  5. (e) A student repeats the experiment with 1.00 mol of ethanoic acid and 3.00 mol of ethanol. She claims that 0.90 mol of ethyl ethanoate forms. Use your value of Kc to test her claim, and explain the effect of the extra ethanol. [3]
  6. (f) Use Table 1 to describe and explain the effect of temperature on the percentage of ammonia. [4]
  7. (g) Use Table 1 to explain the effect of pressure on the percentage of ammonia. [2]
  8. (h) A student says: "An ammonia plant should run at 300 °C and 400 atm, because these conditions give the highest yield." Evaluate this claim. [3]
  9. (i) State the effect of the iron catalyst on the percentage of ammonia at equilibrium. Explain your answer. [1]
 [20]

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Quick revision

Revisit the essentials, then return to an explanation when you need it.

Exam definitions: learn these in mark-scheme words
TermDefinition to write
Dynamic equilibriumIn a closed system, the rate of the forward reaction equals the rate of the backward reaction, and the concentrations of reactants and products remain constant (they are not necessarily equal).
Le Chatelier's principleIf a change is made to a system in dynamic equilibrium, the position of equilibrium shifts to reduce the effect of that change.
Equilibrium constant, KcThe ratio of the equilibrium concentrations of the products to those of the reactants, each raised to the power of its coefficient in the balanced equation. It changes only with temperature.
Keep the three questions separate
QuestionReasoning
Which way does equilibrium shift?Use the imposed concentration, gas-volume or temperature change and the balanced equation.
Does Kc change?Only a temperature change changes Kc for the same written reaction.
How fast is equilibrium reached?Use kinetics; a catalyst changes the approach, not the equilibrium composition.

Kc: equilibrium concentrations, products over reactants, coefficient powers. Convert amounts using the final volume. Link changes by stoichiometry and reject negative concentrations.

Haber: high pressure favours fewer gas moles; lower temperature favours the exothermic equilibrium but slows reaction. Iron gives a faster approach; cooling separates ammonia and reactants are recycled.

Scope

Learning outcomes

8. Chemical Equilibria (8873, 2026 revision; examinations 2026 and 2027). Use the outcome map to find the explanation for a particular syllabus requirement.

See the learning outcome map

  1. 8(a) Explain reversible reactions and dynamic equilibrium.

    • Both directions occur; equal forward and reverse rates give constant concentrations in a closed system.

    Equilibrium is equal rates, not equal amounts

  2. 8(b) Apply Le Chatelier's Principle qualitatively.

    • State the principle; deduce concentration, pressure and temperature responses from supplied information.

    Predict the response to a disturbanceChoose a useful production rate and yield

  3. 8(c) Identify changes that affect the equilibrium constant.

    • Concentration, pressure and catalysts do not change Kc at fixed temperature; temperature does.

    Predict the response to a disturbanceKc describes an equilibrium ratioChoose a useful production rate and yield

  4. 8(d) Deduce concentration equilibrium expressions.

    • Correct species, coefficient powers and concentration units.

    Kc describes an equilibrium ratio

  5. 8(e) Calculate concentration equilibrium constants.

    • Use equilibrium data and stoichiometry where needed.

    Kc describes an equilibrium ratioUse the equation to link all the changes

  6. 8(f) Calculate equilibrium quantities.

    • Use supplied data; no solving of quadratic equations is required.

    Use the equation to link all the changes

  7. 8(g) Explain the Haber process conditions.

    • Industrial compromise: temperature, pressure, iron catalyst, separation and recycling.

    Choose a useful production rate and yield