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Equilibrium Composition Calculations

Use ICE tables, approximations and sanity checks.

  • GCE A-Level H2 Chemistry 9476-2027

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Equilibrium composition questions are “ICE tables under pressure”: set up the changes with coefficients, substitute into K_c or Kₚ, solve for x, then sanity-check the result against the size of K.

Anchor this lesson with Equilibrium Constants (Kc, Kp) and the Chemical Equilibria hub so you can move between concept and calculation questions.

What this page is really testing

  • Can you set up an ICE table with correct coefficient-based changes?
  • Can you translate that table into the correct K_c/Kₚ expression?
  • Can you defend your final value with a quick physical sanity check?

Definitions (Must Know)

A. Equilibrium composition

Equilibrium composition means the amounts/concentrations/partial pressures of species at equilibrium.

B. ICE table

An ICE table tracks:

  • Initial values
  • Change (using x and coefficients)
  • Equilibrium values

Key Ideas (What Earns Marks)

  • Use an ICE (Initial–Change–Equilibrium) table.
  • Only make an approximation if you can justify it (and check it afterwards).
  • Sanity checks: concentrations can’t be negative; results should fit the size of K.
  • This syllabus does not require solving quadratic equations in equilibrium calculations (your setup should lead to linear algebra or a valid approximation).
Quick Recall (Approximation Check)

If you assume a-x ≈ a, check x/a < 0.05 afterwards.

ICE Table Result: Initial vs Equilibrium (Example)Example A ⇌ B with Kc = 0.200: initial [A] = 0.500, [B] = 0.000; equilibrium [A] ≈ 0.417 and [B] ≈ 0.0833.ICE Table Result: Initial vs Equilibrium (Example)SpeciesConcentration (mol dm^-3)KeyInitialInitialEquilibriumEquilibrium
Example A ⇌ B with Kc = 0.200: initial [A] = 0.500, [B] = 0.000; equilibrium [A] ≈ 0.417 and [B] ≈ 0.0833.
Data table
SpeciesInitialEquilibrium
A0.50.4167
B00.0833

Detailed Explanations

A. ICE table method (workflow)

  1. Write the balanced equation.
  2. Decide whether you are using concentrations (K_c) or partial pressures (Kₚ).
  3. Fill in the ICE table using x and coefficients (e.g. -2x if the coefficient is 2).
  4. Write the K_c or Kₚ expression and substitute the equilibrium row.
  5. Solve for x, then calculate the equilibrium composition asked for.

B. Approximation sanity check

If you assume a-x ≈ a, check:

x/a < 0.05

If the check fails, do not force an approximation. In this syllabus, questions are set so that:

  • either the approximation is valid, or
  • the algebra simplifies without needing a quadratic (e.g. 1:1 stoichiometry, or one equilibrium value is given).

C. Worked method (full-mark layout)

Use this layout in your script:

  1. Balanced equation.
  2. ICE table with symbols and units.
  3. K expression with powers from coefficients.
  4. Substitution line (equilibrium row only).
  5. Solve for x.
  6. Convert x into the exact quantity asked (concentration/partial pressure/amount).
  7. Sanity check: sign, magnitude, and compatibility with K.

Worked Examples

Modelled example 1

(ICE setup)

Core

Problem

For A(g) ⇌ B(g), initially [A] = 0.50 mol dm⁻³ and [B] = 0. At equilibrium, [B] = x. Write K_c in terms of x.
Study the worked solution

  1. Build the change row

    Method

    Use the 1:1 coefficients to pair a decrease of x in A with an increase of x in B.

    Reason

    Reaction progress changes species in their stoichiometric ratio.

    Working

    Δ[A] = -x; Δ[B] = +x.

  2. Write the equilibrium row

    Method

    Add each change to its initial concentration.

    Reason

    Only equilibrium values may be inserted into K_c.

    Working

    [A]_eq = 0.50-x; [B]_eq = x.

  3. Substitute symbolically

    Method

    Insert the equilibrium row into products over reactants.

    Reason

    Both coefficients are 1.

    Working

    K_c = [B]/[A] = x/(0.50-x).

Guided practice 2

Solve a 1:1 Kc composition

About 8 min

Problem

For A(g) ⇌ B(g) at 298 K, K_c = 0.200. Initially, [A] = 0.500 mol dm⁻³ and [B] = 0. Calculate the equilibrium concentration of B.

Try this before viewing the solution

Hints

Hint 1: write equilibrium values
For the 1:1 equation, use [A]_eq = 0.500-x and [B]_eq = x.
Hint 2: form one equation
Set x/(0.500-x) equal to 0.200 and solve without rounding early.
View solution step by step

  1. Set up ICE

    Method

    Apply equal and opposite changes to A and B.

    Reason

    The balanced ratio is 1:1.

    Working

    [A]_eq = 0.500-x; [B]_eq = x.

  2. Substitute into Kc

    Method

    Use the equilibrium concentrations in the expression.

    Reason

    K_c = [B]/[A] for this equation.

    Working

    x/(0.500-x) = 0.200.

  3. Solve and report

    Method

    Rearrange the linear equation and identify x as B’s equilibrium concentration.

    Reason

    B starts at zero and increases by x.

    Working

    1.200x = 0.100, so [B]_eq = x = 0.0833 mol dm⁻³.

Common misconception 3

Check a small-change approximation

Find and correct the mistake

Proposed method

For A(g) ⇌ B(g), K_c = 0.0100, initially [A] = 1.00 mol dm⁻³ and [B] = 0. A learner uses 1.00-x ≈ 1.00. Estimate [B]_eq and decide whether the approximation is justified.

Test the approximation

View solution step by step

  1. Estimate x

    Method

    Replace 1.00-x by 1.00 in the denominator.

    Reason

    The proposed approximation treats the reactant decrease as small.

    Working

    0.0100 ≈ x/1.00, so x ≈ 0.0100 mol dm⁻³.

  2. Test the assumption

    Method

    Compare the estimated change with the initial concentration.

    Reason

    The stated small-change check is x/a < 0.05.

    Working

    x/1.00 = 0.0100 < 0.05, so the approximation is valid and [B]_eq ≈ 0.0100 mol dm⁻³.

Examiner practice 4

Calculate Kc from equilibrium data

3 marks

Problem

For H₂(g) + I₂(g) ⇌ 2HI(g), equilibrium concentrations are [H₂] = 0.20, [I₂] = 0.20 and [HI] = 0.60 mol dm⁻³. Calculate K_c. [3 marks]

Try this before viewing the solution

View solution step by step

  1. Write the expression

    1 mark

    Method

    Use the coefficient 2 as the power on HI.

    Reason

    Coefficient powers are part of the Kc definition.

    Working

    K_c = [HI]²/[H₂][I₂].

  2. Substitute equilibrium data

    1 mark

    Method

    Insert the three measured equilibrium concentrations.

    Reason

    No ICE deduction is needed when every required equilibrium value is supplied.

    Working

    K_c = (0.60)²/(0.20)(0.20) = 0.36/0.040.

  3. Evaluate

    1 mark

    Method

    Complete the ratio.

    Reason

    The overall concentration powers cancel for this equation.

    Working

    K_c = 9.0.

Challenge 5

Transfer ICE reasoning to Kp

Minimal support

Problem

For A(g) ⇌ B(g) at 298 K, Kₚ = 0.500. Initially, p_A = 80.0 kPa and p_B = 0. Calculate p_B at equilibrium.

Try this before viewing the solution

Hints

Hint 1: change the representation
Use the same 1:1 ICE logic, but write partial pressures rather than concentrations.
Hint 2: solve for the product pressure
Set x/(80.0-x) = 0.500 and remember that x is p_B.
View solution step by step

  1. Build the pressure ICE row

    Method

    Decrease A by x kPa and increase B by x kPa.

    Reason

    The gaseous stoichiometric ratio is 1:1.

    Working

    p_(A,eq) = 80.0-x; p_(B,eq) = x.

  2. Apply Kp

    Method

    Substitute equilibrium partial pressures into Kₚ = p_B/p_A.

    Reason

    Kp uses partial pressures of the gaseous species.

    Working

    x/(80.0-x) = 0.500.

  3. Solve and identify the target

    Method

    Solve the linear equation and report the product pressure.

    Reason

    B starts at zero, so its equilibrium partial pressure equals x.

    Working

    1.50x = 40.0, so p_B = 26.7 kPa.

Common Exam Traps

  • Using “moles” directly in a K_c expression without converting to concentration.
  • Forgetting coefficients in changes (e.g. -2x for 2).
  • Making an approximation when K is not extreme (it often fails).
  • Solving for x correctly but reporting the wrong species.
  • Forgetting to square terms when a coefficient is 2 in the K expression.

Use the Chemical Equilibria topic check to practise and check your understanding.

Phrasing and Keyword Reminders

  • Write the ICE table clearly: it’s often method marks.
  • If you approximate, state it and then show a check like x/a.
  • If asked to “justify”, write one line: “x/a = ... < 0.05, so the approximation is valid.”
  • If no approximation is used, still include “equilibrium concentrations are physically valid (all positive).”

Mind Stretchers

Connect This To

Mind stretcher 1Extension

Explain why a “sanity check” is needed even if your algebra is correct.

Show Hint

Build the change row from one variable and the balanced coefficients before inserting equilibrium values into K.

Show Answer

Mark scheme:

  • Approximations can give numerically plausible but physically impossible results.
  • Algebra mistakes can yield negative concentrations or values inconsistent with the size of K.
  • A sanity check confirms the result is physically meaningful and matches the expected equilibrium position.

Mind stretcher 2: Rejecting an impossible rootExtension

Question. For an ICE table with initial [A] = 0.40 mol dm⁻³ and change -x, algebra gives x = 0.52 or x = 0.18 mol dm⁻³. Select the physical root and justify it.

Show Hint
No equilibrium concentration may be negative.
Show Answer

x = 0.52 would give [A]_eq = -0.12 mol dm⁻³ and is impossible. The physical root is x = 0.18, giving [A]_eq = 0.22 mol dm⁻³.

Syllabus and review details

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