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Kc Equilibrium Calculations

Learn and apply Kc Equilibrium Calculations in the published Chemistry course sequence.

  • GCE A-Level H1 Chemistry 8873-2027
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H1 Kc Equilibrium Calculations: Orientation

Kc composition calculations are stoichiometry problems constrained by equilibrium. A clear ICE table keeps the balanced coefficients, change variable and equilibrium concentrations connected.

H1 8873 scope
  • Calculate Kc or an equilibrium concentration using concentration data.
  • 8873 questions are constructed so solving a quadratic equation is not required; Kp calculations are excluded.

Definitions (Must Know)

  • An ICE table records initial concentration, stoichiometric change and equilibrium concentration.
  • The change variable, x, represents one reaction-progress amount in concentration terms.
  • A physical solution gives no negative concentration and does not consume more material than was initially available.

Detailed Explanations

A. Coefficient discipline

For A ⇌ 2B, if A decreases by x, B increases by 2x. The change row follows the equation, not convenience.

B. Amount versus concentration

Convert n/V before substituting into K_c unless volume cancellation has been explicitly established.

C. Finding Kc

When equilibrium concentrations are known, insert them directly into the correct expression and retain units implied by the overall concentration power.

D. Finding composition

Use a supplied Kc and enough initial/equilibrium evidence to isolate the unknown without quadratic solving. Reject any value that makes a concentration negative.

Worked Examples

Modelled example 1

Using a measured equilibrium concentration

Core

Problem

For A ⇌ 2B, [A] falls from 1.00 to 0.80 mol dm⁻³ and B starts at zero. Find [B]_eq.
Study the worked solution

  1. Find A's change

    Method

    Subtract A’s equilibrium concentration from its initial concentration.

    Reason

    The measured decrease fixes the reaction progress.

    Working

    Δ[A] = 0.80-1.00 = -0.20 mol dm⁻³.

  2. Apply the coefficient ratio

    Method

    Multiply the magnitude of A’s decrease by 2 for B’s increase.

    Reason

    The equation forms two B for every one A consumed.

    Working

    Δ[B] = 2(0.20) = 0.40 mol dm⁻³.

  3. State the equilibrium value

    Method

    Add the increase to B’s initial concentration of zero.

    Reason

    Equilibrium concentration equals initial plus change.

    Working

    [B]_eq = 0.40 mol dm⁻³.

Guided practice 2

Calculating Kc

About 6 min

Problem

For A ⇌ B, equilibrium concentrations are [A] = 0.25 and [B] = 0.75 mol dm⁻³. Find K_c.

Try this before viewing the solution

Hints

Hint 1: write the expression
For the equation as written, put B over A.
Hint 2: use equilibrium values
Both supplied concentrations are already equilibrium data, so no ICE deduction is required.
View solution step by step

  1. Write Kc

    Method

    Place the product concentration over the reactant concentration.

    Reason

    Both balanced coefficients are 1.

    Working

    K_c = [B]/[A].

  2. Substitute and evaluate

    Method

    Insert the supplied equilibrium values.

    Reason

    Initial concentrations must not replace equilibrium concentrations in Kc.

    Working

    K_c = 0.75/0.25 = 3.0.

Common misconception 3

Correct an equal-change ICE row

Find and correct the mistake

Learner row

For A ⇌ 3B, a learner writes changes -x for A and + x for B. Identify the error and write the correct changes.

Choose B's change

If A changes by -x, B changes by

View solution step by step

  1. Read the stoichiometry

    Method

    Compare the coefficients 1 for A and 3 for B.

    Reason

    Every one A consumed forms three B.

    Working

    Δ[A]:Δ[B] = -1: + 3.

  2. Correct the row

    Method

    Keep -x for A and replace + x by + 3x for B.

    Reason

    The signs show consumption and formation; the magnitudes follow coefficients.

    Working

    Δ[A] = -x; Δ[B] = +3x.

Examiner practice 4

Convert equilibrium amounts before Kc

4 marks

Problem

For A ⇌ B, a 2.00 dm³ equilibrium mixture contains 0.400 mol A and 1.20 mol B. Calculate K_c. [4 marks]

Try this before viewing the solution

View solution step by step

  1. Convert A

    1 mark

    Method

    Divide A’s equilibrium amount by the mixture volume.

    Reason

    Kc requires concentration rather than amount.

    Working

    [A] = 0.400/2.00 = 0.200 mol dm⁻³.

  2. Convert B

    1 mark

    Method

    Apply the same volume conversion to B.

    Reason

    Both concentrations refer to the same equilibrium mixture.

    Working

    [B] = 1.20/2.00 = 0.600 mol dm⁻³.

  3. Write the expression

    1 mark

    Method

    Use products over reactants for the 1:1 equation.

    Reason

    Both coefficients are 1.

    Working

    K_c = [B]/[A].

  4. Evaluate

    1 mark

    Method

    Substitute the concentrations.

    Reason

    The concentration units cancel in this ratio.

    Working

    K_c = 0.600/0.200 = 3.00.

Challenge 5

Solve and audit a Kc composition

Minimal support

Problem

For A ⇌ B, initially [A] = 0.400 mol dm⁻³ and [B] = 0. At the stated temperature, K_c = 3.00. Calculate both equilibrium concentrations without a quadratic equation and audit the result physically.

Try this before viewing the solution

Hints

Hint 1: write the equilibrium row
Use [A]_eq = 0.400-x and [B]_eq = x.
Hint 2: keep it linear
Set x/(0.400-x) = 3.00; cross-multiplication gives a linear equation.
View solution step by step

  1. Set up and substitute

    Method

    Use the 1:1 ICE changes in the Kc expression.

    Reason

    B starts at zero and both species change by x.

    Working

    K_c = x/(0.400-x) = 3.00.

  2. Solve the linear equation

    Method

    Cross-multiply and collect the x terms.

    Reason

    The chosen stoichiometry needs no quadratic method.

    Working

    x = 1.20-3x, so 4x = 1.20 and x = 0.300 mol dm⁻³.

  3. Report and audit

    Method

    Calculate both equilibrium concentrations and test their sign and ratio.

    Reason

    A physical solution must keep concentrations non-negative and reproduce Kc.

    Working

    [A]_eq = 0.100, [B]_eq = 0.300 mol dm⁻³; 0.300/0.100 = 3.00.

Mind Stretchers

Attempt each unfamiliar application before opening the hint, then compare your reasoning with the solution.

Mind stretcher 1: Coefficient-aware deductionExtension

Question. For 2A ⇌ B, A falls by 0.30 mol dm⁻³. By how much does B rise?

Show Hint

Compare the coefficients: two A are consumed per one B formed.

Show Answer

[B] rises by 0.30/2 = 0.15 mol dm⁻³.

Mind stretcher 2: Physical auditExtension

Question. A calculation says 0.60 mol dm⁻³ A reacts when only 0.40 mol dm⁻³ was initially present. Evaluate it.

Show Hint

Compute the resulting equilibrium concentration.

Show Answer

It would give a negative equilibrium concentration, so it is physically impossible and the set-up or selected solution must be rejected.

Mind stretcher 3: A squared term and its unitsExtension

Question. For A ⇌ 2B, the initial concentrations are [A] = 0.60 and [B] = 0.10 mol dm⁻³. At equilibrium, [A] = 0.40 mol dm⁻³. Calculate K_c, including its units.

Show Hint

B already has a starting concentration, and it rises by twice the fall in A. Then square [B] in the expression.

Show Answer

A falls by 0.20 mol dm⁻³, so B rises by 2(0.20) = 0.40 mol dm⁻³ and [B]_eq = 0.10 + 0.40 = 0.50 mol dm⁻³. Then K_c = [B]²/[A] = (0.50)²/0.40 = 0.625, which is 0.63 mol dm⁻³ to the two significant figures of the data. The units are (mol dm⁻³)²/(mol dm⁻³) = mol dm⁻³.