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8873 / 2027
Chemical Equilibria

Lesson 2 of 3

Equilibrium quantities

Construct Kc and calculate amounts without quadratic equations.

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

  • I can identify changes that affect the equilibrium constant.Syllabus

    Syllabus 8873, 8(c). Identify changes that affect the equilibrium constant.

  • I can deduce concentration equilibrium expressions.Syllabus

    Syllabus 8873, 8(d). Deduce concentration equilibrium expressions.

  • I can calculate concentration equilibrium constants.Syllabus

    Syllabus 8873, 8(e). Calculate concentration equilibrium constants.

  • I can calculate equilibrium quantities.Syllabus

    Syllabus 8873, 8(f). Calculate equilibrium quantities.

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.

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.