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1.8.4Physical Chemistry

Calculations Involving the Equilibrium Constant — AQA AS Chemistry Revision Notes & Topical Questions

5 min read 6 parts Lesson 48 of 102

📚 AQA AS Chemistry Hub / Topical Revision
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Curriculum Equation Key قوانين الرقم الهيدروجيني والاتزان الأيوني - pH & Acid-Base Equilibrium FormulasFast Exam Solution
pH = -log[H⁺] | pOH = -log[OH⁻] | pH + pOH = 14 | [H⁺][OH⁻] = 10⁻¹⁴
Word Equation & Conditions: الرقم الهيدروجيني = -لوغاريتم تركيز أيون الهيدروجين | pH + pOH = 14
Location & Specifications: الاتزان الأيوني، كيمياء 3 ثانوي، التحصيلي، و AP Chemistry.
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Part 1 of 6Writing the Kc Expression

For aA + bB ⇌ cC + dD at a fixed temperature, Kc = [C]ᶜ[D]ᵈ / [A]ᵃ[B]ᵇ, where square brackets mean equilibrium concentration in mol dm⁻³.

Products on top
  • Products on top Equilibrium concentrations of products form the numerator, each raised to the power of its balancing coefficient.
Reactants below
  • Reactants below Equilibrium concentrations of reactants form the denominator, again raised to the power of their coefficients in the equation.
Temperature only
  • Temperature only Kc changes only if the temperature changes, and is unaffected by concentration, pressure or the presence of a catalyst.
Homogeneous systems
  • Homogeneous systems At AS level every species is in the same phase, so every substance appears in the expression.

Part 2 of 6Working Out the Units

Units are found by substituting mol dm⁻³ into the expression and cancelling.

Working Out the Units
Equal moles each side
  • Equal moles each side When the total powers on top and bottom match, the units cancel and Kc has no units, as in H₂ + I₂ ⇌ 2HI.
More moles of reactants
  • More moles of reactants For N₂ + 3H₂ ⇌ 2NH₃ the units are mol⁻² dm⁶, because mol dm⁻³ squared on top divides by the fourth power below.
More moles of products
  • More moles of products For N₂O₄ ⇌ 2NO₂ the powers leave one concentration unit overall, so Kc is quoted in mol dm⁻³.
Always state units
  • Always state units Give units whenever they do not cancel, and write "no units" explicitly when they do.

Part 3 of 6Method for Kc Calculations

A tabulated approach keeps the working organised and avoids lost marks.

1
Write the balanced equation and the Kc expression, then set up rows for initial moles, change in moles and equilibrium moles.
2
Use the reaction ratio to link changes: if x mol of one reactant is used, the others change by x multiplied by their coefficients.
3
Convert equilibrium moles to concentrations by dividing by the volume in dm³, unless the volume cancels because moles of both sides are equal.
4
Substitute equilibrium concentrations into the expression, calculate Kc, and give the units with a sensible number of significant figures.
5
To find an unknown quantity, rearrange the Kc expression after substituting all the known concentrations and the given value of Kc.

Part 4 of 6Typical Data Values

Large Kc
  • Large Kc A value much greater than 1 means the position lies far to the right, so products dominate at equilibrium.
Small Kc
  • Small Kc A value much less than 1 means the position lies far to the left, so reactants dominate at equilibrium.
Kc near 1
  • Kc near 1 A value close to 1 means significant amounts of both reactants and products are present at equilibrium.
EquilibriumKc expressionUnits
H₂ + I₂ ⇌ 2HI[HI]² / [H₂][I₂]none
N₂O₄ ⇌ 2NO₂[NO₂]² / [N₂O₄]mol dm⁻³
N₂ + 3H₂ ⇌ 2NH₃[NH₃]² / [N₂][H₂]³mol⁻² dm⁶

Part 5 of 6Worked Example

Question

  • 1.00 mol H₂ and 1.00 mol I₂ are heated in a 2.00 dm³ vessel until equilibrium is reached: H₂(g) + I₂(g) ⇌ 2HI(g).
  • The equilibrium mixture contains 0.200 mol H₂.
  • Calculate Kc.

  1. Moles of H₂ used = 1.00 − 0.200 = 0.800 mol, so I₂ used is also 0.800 mol and 0.200 mol I₂ remains.
  2. HI formed = 2 × 0.800 = 1.60 mol, because the ratio H₂ : HI is 1 : 2.
  3. Concentrations
    • [H₂] = 0.200 ÷ 2.00 = 0.100;
      • [I₂] = 0.100;
      • [HI] = 1.60 ÷ 2.00 = 0.800 mol dm⁻³.
  4. Kc = [HI]² ÷ ([H₂][I₂]) = 0.800² ÷ (0.100 × 0.100) = 0.640 ÷ 0.0100.
    What the symbols mean: Kc = equilibrium constant in terms of concentration · [ ] = concentration at equilibrium (mol dm−3) · each power is the number in front of that substance in the balanced equation

Answer Kc = 64.0 (no units, since the concentration units cancel)

Part 6 of 6Exam Checklist & Tips

1.8.4 – Key Points for the Exam

  • Kc The equilibrium constant, calculated from equilibrium concentrations at a stated temperature.
  • Dynamic equilibrium A state in which forward and reverse rates are equal and concentrations stay constant in a closed system.
  • Equilibrium concentration The concentration of a species, in mol dm⁻³, once the system has reached equilibrium.
  • Homogeneous equilibrium An equilibrium in which all reactants and products are in the same physical state.

Common Mistakes

  • Using initial concentrations in the Kc expression instead of equilibrium concentrations.
  • Forgetting to divide moles by the volume, which gives the wrong answer whenever the number of moles differs on each side.
  • Omitting the powers from the balancing coefficients, or writing reactants over products.
  • Quoting units that do not match the expression, or writing units when they cancel.

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3 & IGCSE —

Kc

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