The equilibrium constant KcAQA A-Level Chemistry: Revision notes
Section 1
Writing the expression for Kc
For a homogeneous reversible reaction aA + bB ⇌ cC + dD at equilibrium:
Kc = [C]ᶜ[D]ᵈ / ([A]ᵃ[B]ᵇ)
Square brackets mean the concentration in mol dm⁻³ at equilibrium. Products go on top, reactants below, each raised to the power of its coefficient in the balanced equation.
Example: N₂(g) + 3H₂(g) ⇌ 2NH₃(g) gives Kc = [NH₃]² / ([N₂][H₂]³).
Use equilibrium concentrations, not amounts in mol, unless the volume cancels. Never include the initial concentrations.
Section 2
Units of Kc
Work out the units by substituting the units into the expression and cancelling.
- N₂ + 3H₂ ⇌ 2NH₃: (mol dm⁻³)² / (mol dm⁻³)⁴ = dm⁶ mol⁻²
- N₂O₄ ⇌ 2NO₂: (mol dm⁻³)² / (mol dm⁻³) = mol dm⁻³
- CH₃COOH + C₂H₅OH ⇌ CH₃COOC₂H₅ + H₂O: no units, because they cancel
Always give units unless they cancel.
Section 3
Calculating Kc
Method: write the expression, convert amounts to concentrations (amount ÷ volume in dm³), substitute and evaluate.
Worked example. At equilibrium, [N₂] = 0.40, [H₂] = 0.20 and [NH₃] = 0.16 mol dm⁻³. Kc = 0.16² / (0.40 × 0.20³) = 0.0256 / 0.0032 = 8.0 dm⁶ mol⁻².
If only initial amounts and one equilibrium amount are given, work out the others from the stoichiometry, as in an equilibrium table (initial, change, equilibrium).
If the number of moles of gas is equal on both sides, the volume cancels and you can use amounts directly.
Section 4
Using Kc to find an unknown amount
Worked example. 1.00 mol of ethanoic acid and 1.00 mol of ethanol reach equilibrium with Kc = 4.0. Let x mol of ester form. Then Kc = x² / (1.00 − x)² = 4.0.
Taking square roots: x / (1.00 − x) = 2.0, so x = 0.67 mol.
The volume cancels because the number of moles is the same on each side.
Section 5
What changes Kc?
Only temperature changes the value of Kc.
- Changing a concentration or the pressure shifts the position of equilibrium, but Kc is unchanged.
- A catalyst does not change Kc.
- For an endothermic reaction, increasing the temperature increases Kc (equilibrium moves right).
- For an exothermic reaction, increasing the temperature decreases Kc (equilibrium moves left).
Do not say that Kc changes when pressure or concentration changes. The equilibrium position moves, but Kc stays the same.
That's the notes covered.
Carry on to the next subtopic.
Exam questions on The equilibrium constant Kc
- Nitrogen and hydrogen are sealed in a vessel at 500 K and form ammonia: N₂(g) + 3H₂(g) ⇌ 2NH₃(g). When equilibrium is reached, the concentrations are [N₂] = 0.40 mol dm⁻³, [H₂] = 0.20 mol dm⁻³ and [NH₃] = 0.16 mol dm⁻³.Calculate the value of Kc for this reaction at 500 K, including its units.2 marks
- Ethanoic acid and ethanol react in the presence of a small amount of sulfuric acid catalyst: CH₃COOH(l) + C₂H₅OH(l) ⇌ CH₃COOC₂H₅(l) + H₂O(l). At room temperature the equilibrium constant Kc for this reaction is 4.0.A mixture of 1.00 mol of ethanoic acid and 1.00 mol of ethanol is left to reach equilibrium at room temperature. Calculate the amount, in mol, of ethyl ethanoate present at equilibrium.2 marks
- Dinitrogen tetraoxide, a colourless gas, is in equilibrium with nitrogen dioxide, a brown gas: N₂O₄(g) ⇌ 2NO₂(g), ΔH = +57 kJ mol⁻¹. A sealed 2.00 dm³ vessel initially contains 0.0600 mol of N₂O₄ only. It is held at a fixed temperature until equilibrium is reached, when 0.0800 mol of NO₂ is present.State and explain the effect of increasing the temperature on the value of Kc, and state the effect on Kc of increasing the pressure by compressing the vessel.3 marks
Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).