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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 understandingA sealed mixture has constant concentrations of 0.80 mol dm-3 reactant and 0.15 mol dm-3 product. Does the unequal ratio rule out equilibrium?Think it through, then reveal the answer
No. Equilibrium requires equal forward and reverse rates, not equal concentrations. If the reversible reactions continue at equal rates under unchanged conditions, these unequal concentrations can be equilibrium values.
02

Predict the response to a disturbance

Track the imposed change separately from the subsequent reaction.

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 example

Follow 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.

  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 understandingAn equilibrium 2SO2(g) + O2(g) ⇌ 2SO3(g) has an exothermic forward reaction. Which two changes increase the equilibrium proportion of SO3?Think it through, then reveal the answer
Cooling favours the exothermic forward direction. Compression favours the side with fewer gas moles: 2 rather than 3. Cooling may nevertheless slow production, so this does not alone establish the best industrial conditions.
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 example

Convert 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.

  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 understandingCan initial concentrations be inserted directly into Kc?Think it through, then reveal the answer
Only if those concentrations already describe an equilibrium mixture. Usually they do not: calculate the equilibrium amounts using the stoichiometric changes first. The same algebraic ratio before equilibrium need not equal Kc.
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 example

From 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.

  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.

Worked example

Find 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.

  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 understandingFor A(aq) ⇌ B(aq), Kc = 3.00. The total concentration [A] + [B] remains 0.800 mol dm-3. Find each equilibrium concentration.Think it through, then reveal the answer
[B] = 3[A], so 4[A] = 0.800. Therefore [A] = 0.200 and [B] = 0.600 mol dm-3. The ratio is 3.00 and the sum is 0.800, checking both conditions.
05

Choose a useful production rate and yield

Industrial conditions combine equilibrium, kinetics and cost.

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 example

Read 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.

  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 understandingA proposal says: "More iron increases the equilibrium yield of ammonia." Repair the statement.Think it through, then reveal the answer
A suitable increase in available catalyst can increase the rate of reaching equilibrium, subject to other process limits. It does not change Kc or the equilibrium yield at the same temperature, pressure and feed composition.

Quick revision

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

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 and references

Learning outcomes and sources

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