7.4 Calculating K

ICE From Experiment

Set initial concentrations and one measured equilibrium value. Stoichiometry auto-fills the Change and Equilibrium rows, and K appears substitution-by-substitution below

ICE Table4 ReactionsK_cStoichiometry
Topic 7.4

Calculating the Equilibrium Constant

Calculate Kc or Kp based on experimental observations of concentrations or pressures at equilibrium.

Equilibrium constants are determined from experimental measurements of the concentrations or partial pressures of the reactants and products at equilibrium.

The direct case is trivial: if the problem hands you every equilibrium concentration, substitute into the K expression and evaluate.

The realistic case gives you initial amounts plus one equilibrium measurement, and asks you to reconstruct the rest. That is an ICE table run in reverse:

  1. Write the balanced equation and the K expression.
  2. Fill in the Initial row.
  3. Use the one known equilibrium value to determine x, the extent of reaction.
  4. Fill in the Change row using x scaled by the coefficients.
  5. Complete the Equilibrium row and substitute into K.

The critical step is 4. Changes must be in the stoichiometric ratio. For N₂ + 3 H₂ ⇌ 2 NH₃, if x mol/L of N₂ is consumed, then 3x of H₂ is consumed and 2x of NH₃ is produced.

Common data formats: the problem may report percent decomposition, the total pressure at equilibrium, or the equilibrium amount of a single species. Convert whatever you are given into the x that the change row needs.

Key points

  • K comes from equilibrium values only — never substitute initial concentrations.
  • Changes must follow the coefficient ratio.
  • One measured equilibrium concentration is enough to determine the whole table.
  • K is unitless on the AP Exam, and depends only on temperature.

Common mistakes

  • Substituting initial concentrations into K. That gives Q, not K.
  • Ignoring the coefficient ratio in the Change row.
  • Forgetting to exclude solids and pure liquids.
  • Using moles instead of concentrations when the volume is not 1.00 L.

Worked example

A 2.00 L vessel is charged with 0.800 mol of N₂ and 0.800 mol of H₂. At equilibrium 0.120 mol of NH₃ is present. Calculate Kc for N₂(g) + 3 H₂(g) ⇌ 2 NH₃(g).

Convert to concentrations (V = 2.00 L):
[N₂]₀ = 0.400 M, [H₂]₀ = 0.400 M, [NH₃]eq = 0.120/2.00 = 0.0600 M

Find x. The change row for NH₃ is +2x, and NH₃ started at 0, so
2x = 0.0600 → x = 0.0300

ICE table (M):

N₂3 H₂2 NH₃
I0.4000.4000
C−0.0300−0.0900+0.0600
E0.3700.3100.0600

Substitute:
Kc = [NH₃]² / ([N₂][H₂]³) = (0.0600)² / [(0.370)(0.310)³]
Kc = 3.60 × 10⁻³ / (0.370 × 0.02979) = 3.60 × 10⁻³ / 0.011022
Kc = 0.327

K is near 1, consistent with substantial amounts of both reactants and products remaining.

Full notes for topic 7.4 →