Factors affecting equilibrium and KEdexcel International A Level Chemistry: Revision notes
Section 1
Pressure, concentration and catalysts
If a system at equilibrium is disturbed, the position shifts to oppose the change (Le Chatelier's principle).
- Pressure: a higher pressure moves the position towards the side with fewer moles of gas. N₂ + 3H₂ ⇌ 2NH₃ has 4 mol on the left and 2 mol on the right, so a high pressure increases the yield. If the numbers of moles are equal, pressure has no effect on composition.
- Catalyst: speeds up the forward and reverse reactions equally, so equilibrium is reached faster but the composition does not change.
- Concentration: adding a reactant moves the position to the right, but K is unchanged as the system re-adjusts.
For a heterogeneous equilibrium, adding or removing a pure solid does not shift the position.
Saying that a catalyst increases the yield. It speeds up reaching equilibrium and does not change the equilibrium composition.
Section 2
What changes the value of K
The value of Kc or Kp depends only on temperature.
- Concentration changes: K unchanged (the position moves to restore the same K).
- Pressure changes: K unchanged.
- A catalyst: K unchanged.
- Temperature changes: K changes.
Example: for CaCO₃(s) ⇌ CaO(s) + CO₂(g), Kp = p(CO₂). At a fixed temperature the equilibrium pressure of CO₂ is fixed, so compressing the vessel just makes some CO₂ react until p(CO₂) returns to its original value.
Section 3
Temperature, K and the position of equilibrium
- Exothermic forward reaction (ΔH negative): raising the temperature decreases K and moves the position towards the reactants.
- Endothermic forward reaction (ΔH positive): raising the temperature increases K and moves the position towards the products.
In both cases the position moves in the endothermic direction, which opposes the temperature rise.
The pressure-based Kp and concentration-based Kc both change in the same direction with temperature.
Say what happens to K, then what happens to the position, then why in terms of the sign of ΔH.
Section 4
The link to ΔStotal: ΔStotal = R ln K
ΔStotal = R ln K, with R = 8.31 J K⁻¹ mol⁻¹. A larger ΔStotal gives a larger K.
For an exothermic reaction, ΔSsurroundings = −ΔH/T is positive but gets smaller as T rises, so ΔStotal falls and K falls. For an endothermic reaction ΔSsurroundings is negative and gets less negative as T rises, so ΔStotal rises and K rises.
Example: ΔH = −196 kJ mol⁻¹, ΔSsystem = −188 J K⁻¹ mol⁻¹. At 700 K, ΔStotal = 280.0 − 188 = +92.0, ln K = 11.07 and K = 6.4 × 10⁴. At 900 K, ΔStotal = 217.8 − 188 = +29.8 and K = 36.
Section 5
Predicting the extent of reaction
The size of K shows how far a reaction goes at that temperature.
- K much greater than 1 (for example 10⁶): the reaction goes almost to completion, with mainly products.
- K about 1: significant amounts of both reactants and products.
- K much less than 1 (for example 10⁻⁶): hardly any reaction, with mainly reactants.
In industry, conditions are a compromise between yield (equilibrium position), rate and cost. A catalyst allows a reasonable rate at a lower temperature, where the equilibrium yield is better for an exothermic reaction.
That's the notes covered.
Carry on to the next subtopic.
Exam questions on Factors affecting equilibrium and K
- Ammonia is manufactured by the Haber process: N₂(g) + 3H₂(g) ⇌ 2NH₃(g), ΔH = −92 kJ mol⁻¹. The reaction is carried out at about 450 °C and 200 atm using an iron catalyst.Explain why using the iron catalyst does not change the equilibrium yield of ammonia.2 marks
- Limestone is heated in a sealed kiln in which this equilibrium is established: CaCO₃(s) ⇌ CaO(s) + CO₂(g), ΔH = +178 kJ mol⁻¹. For this equilibrium Kp = p(CO₂).The volume of the sealed kiln is halved at constant temperature. Explain why the equilibrium pressure of carbon dioxide returns to its original value.2 marks
- The Contact process makes sulfur trioxide: 2SO₂(g) + O₂(g) ⇌ 2SO₃(g), ΔH = −196 kJ mol⁻¹. The entropy change of the system is −188 J K⁻¹ mol⁻¹, and both this and ΔH may be assumed constant with temperature. Use R = 8.31 J K⁻¹ mol⁻¹.Calculate ΔStotal at 700 K and hence the value of K at 700 K.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).