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9476 / 2027
Chemical Equilibria

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Chemical Equilibria

Explain a changing equilibrium through competing rates, then calculate composition and justify industrial choices.

A-Level 9476 (2026-2027)

01

Equal rates give constant composition, not equal amounts

A closed system can keep reacting in both directions with no net change.

A reversible reaction can proceed in both directions under appropriate conditions. In a closed system initially containing reactants, the forward reaction forms products. As products accumulate, the reverse reaction becomes possible and its rate increases. Dynamic equilibrium is reached when the forward and reverse rates are equal, so macroscopic properties such as concentration and colour remain constant.

Follow A <=> B from a reactant-rich starting mixture
  1. Initially

    Forward rate is large relative to reverse rate; A decreases and B increases.

  2. As composition changes

    The forward and reverse rates approach one another. Their dependence on concentration comes from their mechanisms.

  3. At dynamic equilibrium

    Forward rate = reverse rate, both usually nonzero. Amounts remain constant but need not be equal.

A closed system prevents material loss; it need not be thermally insulated. Constant temperature is needed when discussing one fixed equilibrium constant. The same equilibrium composition can be approached from either side if the overall amounts, temperature and volume or pressure constraints are the same.

Check your understanding

A sealed equilibrium mixture keeps the same colour for an hour. Does that prove the reactions have stopped?

  1. Yes. A constant colour shows that no molecules are reacting any more.
  2. No. A constant colour shows the forward reaction has finished, but the reverse still runs.
  3. No. The concentrations are constant because the forward and reverse rates are equal.
Show the answer

No. The concentrations are constant because the forward and reverse rates are equal.

At dynamic equilibrium both reactions continue at the same rate, so nothing visible changes. Constant colour does not mean equal concentrations either.

02

A disturbed equilibrium responds against the imposed change

Name the actual change first, then the direction that partially opposes it.

Make a guess. You are not marked.

A sealed syringe holds N2O4(g) <=> 2NO2(g) at equilibrium. The plunger is pushed in at constant temperature. Which way does the position of equilibrium move?

  1. Towards N2O4, the side with fewer gas molecules
  2. Towards NO2, as more molecules can absorb pressure
  3. It does not move, because K stays constant
Show the answer

Towards N2O4, the side with fewer gas molecules

The total pressure rises, and the system partly opposes it by favouring the side with fewer gas molecules: 1 mol of N2O4 against 2 mol of NO2.

Le Chatelier's Principle: when a system at equilibrium is disturbed, its position changes in the direction that tends to oppose that disturbance. The response usually offsets only part of the imposed change. Adding a reactant does not mean that all of the added material disappears.

Apply the principle to N2(g) + 3H2(g) ⇌ 2NH3(g), an exothermic forward reaction
DisturbanceEquilibrium responseReason
Add H2 at fixed volume and temperatureShift towards NH3.Consume some of the added reactant.
Remove NH3Shift towards NH3.Replace some of the removed product.
Reduce volume at fixed temperatureShift towards NH3.There are fewer gas moles on the product side: two rather than four.
Raise temperatureShift towards N2 and H2.The endothermic reverse direction absorbs some added energy.

Compression affects a gaseous equilibrium through the changed partial pressures. If both sides have the same total gaseous coefficient, as in H2(g) + I2(g) ⇌ 2HI(g), compression does not shift the ideal-gas equilibrium position. Pure solids and liquids are not counted as gaseous particles.

Adding an inert gas at constant volume and temperature raises total pressure but leaves each reacting gas partial pressure unchanged, so it causes no shift in the ideal-gas model. At constant total pressure, adding inert gas expands the mixture and lowers reacting partial pressures; the side with more gas moles is then favoured. State the constraint rather than using total pressure alone.

Worked exampleInterpret colour after a temperature change

N2O4(g) ⇌ 2NO2(g) has an endothermic forward direction. N2O4 is colourless and NO2 is brown. What does warming a sealed sample favour?

Try it first. Then show the worked stepsHide the worked steps
  1. Heating favours the endothermic direction, which absorbs energy.
  2. More NO2 is present at the new equilibrium.
  3. At a controlled optical path length and with the volume constraint specified, increased NO2 concentration gives a darker brown colour.
Answer

Warming favours dissociation to NO2. For a rigid sealed vessel, the equilibrium colour becomes darker.

On a concentration-time graph, adding a species produces an immediate jump in its concentration, followed by a gradual reaction-driven adjustment. A volume decrease immediately increases all gaseous concentrations before any shift occurs. Distinguish that instantaneous physical change from the subsequent change in composition.

03

Temperature changes the constant; other disturbances change the route back to it

For a fixed written reaction, K is fixed at a specified temperature.

Make a guess. You are not marked.

A catalyst is added to a reaction already at equilibrium. What happens to the equilibrium constant K?

  1. K decreases, because the catalyst is used up
  2. K increases, because products form faster
  3. K stays the same, as both directions speed up
Show the answer

K stays the same, as both directions speed up

A catalyst speeds up the forward and reverse reactions equally, so it changes how fast equilibrium is reached, not where it lies.

Equilibrium position and equilibrium constant are different ideas
ChangeCan the equilibrium composition change?Does K change?
Concentration or reacting-gas pressureYes, when the change disturbs the equilibrium ratio.No, at constant temperature.
TemperatureUsually yes.Yes. Heating increases K for an endothermic forward reaction and decreases K for an exothermic forward reaction.
CatalystNo change to the final equilibrium composition at the same temperature.No. Both directions approach equilibrium faster.
Amount of a pure solid already presentNot merely from adding more of the same solid phase, provided relevant phases remain present.No, at constant temperature.

A catalyst changes kinetics through an alternative mechanism. It does not alter the thermodynamic free-energy difference between the reactants and products and therefore does not change K. An industrial catalyst may allow a different operating temperature; it is that temperature change, not the catalyst itself, that can alter the equilibrium yield.

Check your understanding

Compressing an equilibrium mixture increases the amount of ammonia. Has Kp increased?

  1. No. Kp falls on compression, which is why more ammonia forms to restore it.
  2. Yes. More ammonia at equilibrium means a larger equilibrium constant.
  3. No. At the same temperature, the mixture shifts until it again matches the same Kp.
Show the answer

No. At the same temperature, the mixture shifts until it again matches the same Kp.

Compression changes every partial pressure at first, so the ratio no longer equals Kp. The equilibrium shifts towards fewer gas moles until the ratio equals Kp again. Only temperature changes Kp.

04

Build K from the written equation and equilibrium quantities

Products go above reactants; stoichiometric coefficients become powers.

For aA + bB ⇌ cC + dD, the concentration expression is Kc = [C]c[D]d / ([A]a[B]b), using equilibrium concentrations. Kp uses equilibrium partial pressures of gaseous species in the same pattern. Pure solids and pure liquids are omitted because their activities are constant; a dissolved species is not omitted merely because it is in a liquid solution.

Expressions for common equilibrium forms
ReactionExpression
H2(g) + I2(g) ⇌ 2HI(g)Kc = [HI]2 / ([H2][I2]); Kp = p(HI)2 / (p(H2)p(I2)).
N2(g) + 3H2(g) ⇌ 2NH3(g)Kc = [NH3]2 / ([N2][H2]3); Kp = p(NH3)2 / (p(N2)p(H2)3).
CaCO3(s) ⇌ CaO(s) + CO2(g)Kp = p(CO2), with both pure solid phases present.
CH3COOH(aq) + C2H5OH(aq) ⇌ CH3COOC2H5(aq) + H2O(l)In a dilute aqueous treatment, water as the solvent is incorporated into the constant; retain the dissolved reacting species. Read the stated medium.

In conventional examination expressions, derive any units from the concentration or pressure powers and use one consistent pressure unit. For N2O4(g) ⇌ 2NO2(g), the concentration expression has units mol dm-3 and a pressure expression in kPa has units kPa. The numerical pressure-based value changes if the pressure unit changes, so report the unit used.

Worked exampleFind Kp from a gas mixture

At equilibrium, a mixture contains 0.0200 mol N2O4 and 0.0400 mol NO2 at total pressure 150 kPa. Calculate Kp for N2O4 ⇌ 2NO2.

Try it first. Then show the worked stepsHide the worked steps
  1. Total gas amount = 0.0600 mol.
  2. p(N2O4) = (0.0200/0.0600)(150) = 50.0 kPa; p(NO2) = 100 kPa.
  3. Kp = p(NO2)2 / p(N2O4) = 1002/50.0.
Answer

Kp = 200 kPa for the stated pressure convention and temperature.

Try it step by step

Work it out in steps: Total amount of gas, then Partial pressure of N2O4, then Kp = p(NO2)2 / p(N2O4).

  1. Problem 1. An equilibrium mixture for N2O4 <=> 2NO2 holds 0.100 mol N2O4 and 0.200 mol NO2 at a total pressure of 300 kPa. Find Kp in kPa to 3 significant figures.
  2. Problem 2. An equilibrium mixture holds 0.300 mol N2O4 and 0.100 mol NO2 at a total pressure of 200 kPa. Find Kp in kPa to 3 significant figures.
  3. Problem 3. An equilibrium mixture holds 0.0500 mol N2O4 and 0.150 mol NO2 at a total pressure of 120 kPa. Give each answer to 3 significant figures.
Show the answer

Problem 1.

Total amount of gas: Total = 0.100 + 0.200 = 0.300 mol.

Partial pressure of N2O4: p(N2O4) = (0.100/0.300) x 300 = 100 kPa, and p(NO2) = 200 kPa.

Kp = p(NO2)2 / p(N2O4): Kp = 2002 / 100 = 400 kPa.

Problem 2.

Total amount of gas: Total = 0.400 mol.

Partial pressure of N2O4: p(N2O4) = (0.300/0.400) x 200 = 150 kPa, and p(NO2) = 50.0 kPa.

Kp = p(NO2)2 / p(N2O4): Kp = 50.02 / 150 = 16.7 kPa.

Problem 3.

Total amount of gas: Total = 0.0500 + 0.150 = 0.200 mol.

Partial pressure of N2O4: p(N2O4) = (0.0500/0.200) x 120 = 30.0 kPa, and p(NO2) = 90.0 kPa.

Kp = p(NO2)2 / p(N2O4): Kp = 90.02 / 30.0 = 270 kPa.

Reversing a reaction takes the reciprocal of K; multiplying all coefficients by a factor raises K to that power. A large K favours products relative to reactants in its expression, but gives no direct information about the speed of reaching equilibrium. The relationship converting Kp into Kc is not required in this syllabus.

05

Use stoichiometry to complete the equilibrium row

Amounts change in the coefficient ratio before they are converted to concentration or partial pressure.

Worked exampleCalculate Kc from an observed equilibrium amount

A 1.00 dm3 vessel initially contains 0.100 mol H2 and 0.100 mol I2, with no HI. At equilibrium, it contains 0.160 mol HI. Find Kc for H2 + I2 ⇌ 2HI.

Try it first. Then show the worked stepsHide the worked steps
  1. Forming 0.160 mol HI consumes 0.0800 mol H2 and 0.0800 mol I2.
  2. Equilibrium amounts are 0.0200, 0.0200 and 0.160 mol respectively.
  3. Divide each amount by 1.00 dm3 to obtain the equilibrium concentrations.
  4. Kc = 0.1602/(0.0200 × 0.0200).
Answer

Kc = 64.0, with no net concentration unit for this expression.

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 0.400 mol H2 and 0.400 mol I2. At equilibrium it holds 0.600 mol HI. H2 + I2 <=> 2HI. Find Kc to 3 significant figures.
  2. Problem 2. A 1.00 dm3 vessel starts with 0.500 mol PCl5. At equilibrium it holds 0.200 mol Cl2. PCl5 <=> PCl3 + Cl2. Find Kc in mol dm-3 to 3 significant figures.
  3. Problem 3. A 2.00 dm3 vessel starts with 1.00 mol N2O4. At equilibrium it holds 0.400 mol NO2. N2O4 <=> 2NO2. Give each answer to 3 significant figures, with Kc in mol dm-3.
Show the answer

Problem 1.

Amount of reactant used up: Forming 0.600 mol HI uses 0.300 mol H2 and 0.300 mol I2 (ratio 1 : 1 : 2).

Equilibrium amount of reactant: Equilibrium H2 = I2 = 0.400 - 0.300 = 0.100 mol. Divide by 2.00 dm3: [H2] = [I2] = 0.0500 and [HI] = 0.300 mol dm-3.

Kc: Kc = 0.3002 / (0.0500 x 0.0500) = 36.0. The units cancel for this equation.

Problem 2.

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

Equilibrium amount of reactant: Equilibrium PCl5 = 0.500 - 0.200 = 0.300 mol, so [PCl5] = 0.300, [PCl3] = [Cl2] = 0.200 mol dm-3.

Kc: Kc = (0.200 x 0.200) / 0.300 = 0.133 mol dm-3.

Problem 3.

Amount of reactant used up: Forming 0.400 mol NO2 uses 0.400 / 2 = 0.200 mol N2O4.

Equilibrium amount of reactant: Equilibrium N2O4 = 1.00 - 0.200 = 0.800 mol, so [N2O4] = 0.400 and [NO2] = 0.200 mol dm-3.

Kc: Kc = 0.2002 / 0.400 = 0.100 mol dm-3.

An amount table for the same reaction, initially 0.100 mol of each reactant
StageH2 / molI2 / molHI / mol
Initial0.1000.1000
Change-x-x+2x
Equilibrium0.100 - x0.100 - x2x

Worked exampleFind the equilibrium composition from K

For the same initial amounts in 1.00 dm3, suppose Kc = 64.0 is supplied instead. Calculate the equilibrium amounts without solving a quadratic.

Try it first. Then show the worked stepsHide the worked steps
  1. Substitute the equilibrium expressions: 64 = (2x)2/(0.100 - x)2.
  2. Concentrations are positive, so take the positive square root: 8 = 2x/(0.100 - x).
  3. 8(0.100 - x) = 2x; hence x = 0.0800 mol.
  4. Substitute back and check that no amount is negative.
Answer

H2 = I2 = 0.0200 mol and HI = 0.160 mol. Substitution into K verifies the result.

A useful table is not a substitute for checking the chemistry. Include any initial product, account for unequal coefficients, and use the actual equilibrium volume. When gaseous mole number changes, total pressure cannot be split using initial mole fractions. The required syllabus calculations do not need quadratic-equation solutions.

06

A useful ammonia process needs both yield and speed

The operating conditions are an economic compromise supported by equilibrium and kinetics.

N2(g) + 3H2(g) ⇌ 2NH3(g) is exothermic, with ΔH about -92 kJ for the equation as written. Lower temperature favours the equilibrium amount of ammonia; higher pressure favours the side with fewer gaseous particles. Neither statement alone specifies the best operating plant.

Typical teaching conditions: approximately 450 degrees C, 200 atm and an iron catalyst
ChoiceBenefitCompromise or limit
Moderately high temperatureGives a useful rate for breaking and rearranging strong bonds.A higher temperature lowers the equilibrium ammonia yield; a very low temperature is too slow.
High pressureFavours two product gas moles over four reactant gas moles and increases collision frequency.Compression and pressure-resistant equipment have costs; unlimited pressure is not economical.
Iron catalystProvides a lower-barrier surface mechanism, allowing a useful rate.Does not change K or the equilibrium yield at a fixed temperature.
Cool the outlet mixtureCondense and remove ammonia.Unreacted nitrogen and hydrogen are separated from the liquid product and recycled.
A recycle loop improves overall conversion
  1. Feed nitrogen and hydrogen

    Use an appropriate 1:3 stoichiometric supply and pass the gases over the catalyst under controlled conditions.

  2. Allow partial conversion

    Each pass gives a mixture limited by both equilibrium and residence time.

  3. Condense ammonia and recycle gases

    Removing product and returning unreacted gases increases total resource use efficiency without claiming complete conversion in one pass.

The quoted conditions are representative, not a universal specification for every plant. In a data question, compare the actual equilibrium-yield curve, reaction-rate information, energy use and equipment constraints supplied. Keep the explanations separate: temperature affects both K and rate; a catalyst affects rate; recycling improves overall conversion.

Check your understanding

Why is the Haber process run at about 450 degrees C, although a lower temperature gives a higher equilibrium yield?

  1. The iron catalyst works when it is molten, which needs over 450 degrees C.
  2. The forward reaction is endothermic, so heating raises the yield.
  3. At lower temperatures equilibrium is reached too slowly to be economic.
Show the answer

At lower temperatures equilibrium is reached too slowly to be economic.

The forward reaction is exothermic, so cooling raises the yield but slows the rate. About 450 degrees C with an iron catalyst is a compromise between yield and rate.

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.

Dynamic equilibrium means equal forward and reverse rates. A concentration or pressure disturbance can change composition while K stays fixed. Only temperature changes K for the same written reaction; a catalyst speeds approach without changing the endpoint.

Write K from the balanced equation, use equilibrium values and respect phases. Build an amount-change table before converting to concentrations or partial pressures. For Haber, explain yield, rate, compression cost and recycling as distinct considerations.

Scope

Learning outcomes

9. Chemical Equilibria. Use the outcome map to find the explanation for a particular syllabus requirement.

See the learning outcome map

  1. 9(a) Explain reversibility and dynamic equilibrium through rates.

    • Forward and reverse reactions
    • Equal nonzero rates
    • Constant but not necessarily equal amounts
    • Closed system

    Equal rates give constant composition, not equal amounts

  2. 9(b) Apply Le Chatelier Principle to disturbances.

    • Concentration
    • Pressure with stated constraints
    • Temperature and reaction enthalpy

    A disturbed equilibrium responds against the imposed change

  3. 9(c) Distinguish changes in K from changes in composition.

    • Concentration, pressure, temperature
    • Catalyst
    • Temperature dependence of K

    Temperature changes the constant; other disturbances change the route back to it

  4. 9(d) Deduce concentration and partial-pressure expressions.

    • Kc
    • Kp
    • Coefficients as powers and treatment of phases
    • Kp-Kc relationship not required

    Build K from the written equation and equilibrium quantities

  5. 9(e) Calculate Kc or Kp from equilibrium data.

    • Equilibrium concentrations
    • Equilibrium partial pressures
    • Stoichiometry and units

    Build K from the written equation and equilibrium quantitiesUse stoichiometry to complete the equilibrium row

  6. 9(f) Calculate quantities present at equilibrium.

    • Initial/change/equilibrium amounts
    • Given equilibrium constant and data
    • No quadratic-equation solution required

    Use stoichiometry to complete the equilibrium row

  7. 9(g) Explain industrial Haber-process conditions.

    • Equilibrium yield and reaction-rate compromise
    • Pressure, temperature, catalyst
    • Product removal and recycling

    A useful ammonia process needs both yield and speed