When an equilibrium question gives amounts in moles, divide each equilibrium amount by the vessel volume in dm³ before substituting into Kc. A shared volume only cancels if the total concentration powers in the numerator and denominator match.
The 2027 H2 Chemistry 9476 syllabus covers equilibrium constants and their calculation from data. This guide shows where the vessel volume goes and how to spot the error that comes from using equilibrium moles directly.
- Write the balanced equation.
- Use its coefficients as concentration powers.
- Convert equilibrium amounts to concentrations.
- Check the units and any volume cancellation.
CHAPTER 1 OF 4 · Back to contents
Write the expression from the equation
For N₂O₄(g) ⇌ 2NO₂(g), write Kc = [NO₂]²/[N₂O₄]. The coefficient 2 means the equilibrium concentration of NO₂ is squared. The numerator therefore carries two factors of inverse volume, while the denominator carries one.
Use the equilibrium amounts. If the question gives only starting amounts and a change, calculate the equilibrium amounts before dividing by the vessel volume.
CHAPTER 2 OF 4 · Back to contents
Worked example | A 2.00 dm³ vessel
At a fixed temperature, suppose equilibrium is reached with 0.200 mol of N₂O₄ and 0.400 mol of NO₂ in a 2.00 dm³ vessel. Their concentrations are 0.100 mol dm⁻³ and 0.200 mol dm⁻³ respectively.
Kc = (0.200)² ÷ 0.100 = 0.400 mol dm⁻³. If you substitute the mole amounts directly, you get 0.800 instead. To see the missing factor, write the expression in terms of amounts n and volume V. It becomes n(NO₂)² ÷ (n(N₂O₄) × V).
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CHAPTER 3 OF 4 · Back to contents
Compare an expression where the volume cancels
For H₂(g) + I₂(g) ⇌ 2HI(g), the expression is [HI]²/([H₂][I₂]). The numerator and denominator each contain two concentration factors. If all amounts are from the same vessel at equilibrium, the factors of 1/V cancel exactly. That cancellation follows from the powers in the expression.
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CHAPTER 4 OF 4 · Back to contents
Frequently asked questions
What happens if the vessel expands at constant temperature?
Immediately after expansion, the concentration ratio can differ from Kc. The composition can then change as the system re-establishes equilibrium. Kc for the same reaction stays constant at the same temperature.
How do I obtain the units in the first example?
Divide the squared concentration units in the numerator by the concentration units in the denominator. One factor of mol dm⁻³ remains.