Dynamic equilibrium, Le Chatelier's principle and Kc
Many reactions never go fully to completion. In a closed system they reach a balance where the forward and reverse reactions continue at the same rate. Le Chatelier's principle predicts how that balance shifts, and Kc measures where it lies.
Part 1 of 3: Learn it
In short
- At dynamic equilibrium, forward and reverse rates are equal, so concentrations stay constant.
- If a condition changes, the position of equilibrium shifts to oppose the change.
- For aA + bB ⇌ cC + dD, Kc = [C]ᶜ[D]ᵈ ÷ ([A]ᵃ[B]ᵇ); only temperature changes Kc.
Where this is in your specification
Spec points: AQA 7405 3.1.6.1 and 3.1.6.2, OCR A H432 3.2.3 and 5.1.2, Edexcel 9CH0 Topics 10 and 11
| Board | Topic: Equilibria, Kc and Kp |
|---|---|
| AQA 7405 | 3.1.6 and 3.1.10 |
| Edexcel 9CH0 | Topics 10 and 11 |
| OCR H432 | 3.2.3 and 5.1.2 |
| Higher C813 76 | 3(d) Equilibria |
Dynamic equilibrium
It needs a closed system. Both reactions are still happening, which is what dynamic means, but because their rates are equal the amounts of each substance do not change.
Le Chatelier's principle
| Change | Shift in position | Effect on Kc |
|---|---|---|
| Add a reactant | to the right, using it up | none |
| Increase pressure (gases) | towards the side with fewer gas moles | none |
| Increase temperature | in the endothermic direction | changes |
| Add a catalyst | no shift; equilibrium is reached sooner | none |
The equilibrium constant Kc
- Write products over reactants, each concentration raised to the power of its number in the equation.
- Use equilibrium concentrations in mol dm⁻³, not starting amounts.
- Work out the units by cancelling the mol dm⁻³ terms. If the powers balance, Kc has no units.
- A large Kc means the equilibrium lies to the right (mostly products).
What effect does a catalyst have on the position of equilibrium?
Show the answer
None: it speeds up the forward and reverse reactions equally.
Part 2 of 3: See it worked
Worked examples
Example 1
For H₂(g) + I₂(g) ⇌ 2HI(g), the equilibrium concentrations are [H₂] = 0.10, [I₂] = 0.20 and [HI] = 0.80 mol dm⁻³. Find Kc.
- Kc = [HI]² ÷ ([H₂][I₂])
- = 0.80² ÷ (0.10 × 0.20) = 0.64 ÷ 0.020
- Units: (mol dm⁻³)² ÷ (mol dm⁻³)² cancel
Answer: Kc = 32, with no units.
Example 2
1.00 mol each of A and B are mixed in 1.00 dm³ and react: A + B ⇌ C + D. At equilibrium there is 0.40 mol of C. Find Kc.
- C and D: 0.40 mol each; A and B: 1.00 − 0.40 = 0.60 mol each
- Volume 1.00 dm³, so concentrations equal these amounts
- Kc = (0.40 × 0.40) ÷ (0.60 × 0.60) = 0.16 ÷ 0.36
Answer: 0.44 (2 s.f.), no units.
Common mistakes
- Using starting amounts instead of equilibrium amounts in Kc.
- Saying a catalyst increases the yield.
- Saying that changing pressure or concentration changes Kc.
- Forgetting to divide moles by volume when the volume is not 1 dm³.
For N₂ + 3H₂ ⇌ 2NH₃ (ΔH negative), what happens to the yield of ammonia if the temperature rises?
Show the answer
It falls, because the equilibrium shifts in the endothermic (reverse) direction.
Part 3 of 3: Test yourself
Check yourself
Answer each one in your head or on paper first, then open it to check.
What effect does a catalyst have on the position of equilibrium?
None: it speeds up the forward and reverse reactions equally.
For N₂ + 3H₂ ⇌ 2NH₃ (ΔH negative), what happens to the yield of ammonia if the temperature rises?
It falls, because the equilibrium shifts in the endothermic (reverse) direction.
What are the units of Kc for N₂ + 3H₂ ⇌ 2NH₃?
mol⁻² dm⁶.
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