Lesson 3 of 4

Equilibrium expressions and calculations

Use equilibrium concentrations or partial pressures, with stoichiometry to find missing amounts.

A-Level 9476 (2026-2027)

I can deduce concentration and partial-pressure expressions.Syllabus
Syllabus 9476, 9(d). Deduce concentration and partial-pressure expressions.
I can calculate Kc or Kp from equilibrium data.Syllabus
Syllabus 9476, 9(e). Calculate Kc or Kp from equilibrium data.
I can calculate quantities present at equilibrium.Syllabus
Syllabus 9476, 9(f). Calculate quantities present at equilibrium.

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 K_c = [C]^c[D]^d / ([A]^a[B]^b), using equilibrium concentrations. K_p 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
Reaction	Expression
H₂(g) + I₂(g) ⇌ 2HI(g)	K_c = [HI]² / ([H₂][I₂]); K_p = p(HI)² / (p(H₂)p(I₂)).
N₂(g) + 3H₂(g) ⇌ 2NH₃(g)	K_c = [NH₃]² / ([N₂][H₂]³); K_p = p(NH₃)² / (p(N₂)p(H₂)³).
CaCO₃(s) ⇌ CaO(s) + CO₂(g)	K_p = 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 K_p for N₂O₄ ⇌ 2NO₂.

Try it first. Then show the worked stepsHide the worked steps

Total gas amount = 0.0600 mol.
p(N₂O₄) = (0.0200/0.0600)(150) = 50.0 kPa; p(NO₂) = 100 kPa.
K_p = p(NO₂)² / p(N₂O₄) = 100²/50.0.

Answer

K_p = 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).

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.
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.
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 K_p into K_c is not required in this syllabus.

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 K_c for H₂ + I₂ ⇌ 2HI.

Try it first. Then show the worked stepsHide the worked steps

Forming 0.160 mol HI consumes 0.0800 mol H₂ and 0.0800 mol I₂.
Equilibrium amounts are 0.0200, 0.0200 and 0.160 mol respectively.
Divide each amount by 1.00 dm³ to obtain the equilibrium concentrations.
K_c = 0.160²/(0.0200 × 0.0200).

Answer

K_c = 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.

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.
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.
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
Stage	H₂ / mol	I₂ / mol	HI / mol
Initial	0.100	0.100	0
Change	-x	-x	+2x
Equilibrium	0.100 - x	0.100 - x	2x

Worked exampleFind the equilibrium composition from K

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

Try it first. Then show the worked stepsHide the worked steps

Substitute the equilibrium expressions: 64 = (2x)²/(0.100 - x)².
Concentrations are positive, so take the positive square root: 8 = 2x/(0.100 - x).
8(0.100 - x) = 2x; hence x = 0.0800 mol.
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.