Topic 2 of 3
Equilibrium quantities
Construct Kc and calculate amounts without quadratic equations.
A-Level 8873, revised syllabus (2026-2027)
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.
| Reaction | Kc expression | Units 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.
- [H2] = [I2] = 0.200/2.00 = 0.100 mol dm-3; [HI] = 0.600/2.00 = 0.300 mol dm-3.
- Kc = (0.300)2/(0.100 × 0.100) = 9.00.
- 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.
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
Use the equation to link all the changes
An amount table stops incompatible equilibrium quantities entering the expression.
- Write the balanced equation
Choose either amounts or concentrations for the change table, and keep them consistent.
- Define one change
Every other change follows the coefficients, with opposite signs for consumption and formation.
- Insert equilibrium concentrations
Divide amounts by the final volume, then substitute into the correct Kc expression.
- 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.
- Forming 0.600 mol HI consumes 0.300 mol H2 and 0.300 mol I2, from the 1:1:2 coefficient ratio.
- Equilibrium amounts are 0.200 mol H2, 0.200 mol I2 and 0.600 mol HI.
- Divide by 2.00 dm3: concentrations are 0.100, 0.100 and 0.300 mol dm-3.
- Kc = 0.3002/(0.100 × 0.100) = 9.00.
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.
- Equilibrium concentrations are 0.300 - x, 0.300 - x and 2x mol dm-3.
- 4.00 = (2x)2/(0.300 - x)2. Concentrations are positive, so take the positive square root: 2.00 = 2x/(0.300 - x).
- 0.600 - 2x = 2x, giving x = 0.150 mol dm-3.
- Check: 0.3002/(0.150 × 0.150) = 4.00; all amounts are physically possible.
[H2] = [I2] = 0.150 mol dm-3; [HI] = 0.300 mol dm-3. H1 equilibrium calculations do not require solving quadratic equations.