Chemical Reactions & Equilibrium
Chemical Equilibrium: Le Chatelier's Principle and Equilibrium Constant (Kc)
5th Year · 6th Year (Leaving Cert)
- ✓By the end of this lesson students will be able to define chemical equilibrium and dynamic equilibrium.
- ✓By the end of this lesson students will be able to state and apply Le Chatelier's Principle to predict shifts in equilibrium.
- ✓By the end of this lesson students will be able to derive and use the expression for the equilibrium constant (Kc) for homogeneous and heterogeneous reactions (HL).
- ✓By the end of this lesson students will be able to explain the industrial applications of Le Chatelier's Principle in the Haber process and the Contact process.
Key concepts
Chemical equilibrium is a state in a reversible reaction where the rate of the forward reaction is equal to the rate of the reverse reaction. At equilibrium, the concentrations of reactants and products remain constant, but the reactions are still occurring. This is known as dynamic equilibrium.
Le Chatelier's Principle states that if a system at equilibrium is subjected to a change, the system will adjust itself to counteract the effect of the change and establish a new equilibrium. Changes can include alterations in concentration, pressure (for gaseous reactions), or temperature.
For a general reversible reaction aA + bB ⇌ cC + dD, where a, b, c, and d are the stoichiometric coefficients and A, B, C, and D are the chemical species, the equilibrium constant, Kc, is a value that expresses the ratio of product concentrations to reactant concentrations, each raised to the power of their stoichiometric coefficients, at a given temperature. Pure solids and liquids are not included in the Kc expression.
The Haber process is the industrial synthesis of ammonia from nitrogen and hydrogen gases. It is a key example of applying Le Chatelier's Principle to optimise yield. The reaction is N₂(g) + 3H₂(g) ⇌ 2NH₃(g), which is exothermic (ΔH = -92 kJ mol⁻¹). High pressure favours the forward reaction (fewer moles of gas on product side), and a moderate temperature (around 400-450 °C) is used to balance yield (favoured by lower temperature) with reaction rate (favoured by higher temperature). An iron catalyst is used to increase the rate of reaction.
The Contact process is the industrial method for producing sulfuric acid. The key equilibrium step is the oxidation of sulfur dioxide to sulfur trioxide: 2SO₂(g) + O₂(g) ⇌ 2SO₃(g), which is also exothermic (ΔH = -197 kJ mol⁻¹). High pressure favours the forward reaction (fewer moles of gas on product side), and a moderate temperature (around 450 °C) is used. A vanadium(V) oxide (V₂O₅) catalyst is used to increase the rate of reaction.
Key facts to remember
- 1Chemical equilibrium is a dynamic state where forward and reverse reaction rates are equal.
- 2Le Chatelier's Principle predicts how an equilibrium system responds to changes in conditions.
- 3A catalyst increases the rate of both forward and reverse reactions equally, thus it does not affect the position of equilibrium or the value of Kc.
- 4The value of Kc for a given reaction is constant only at a specific temperature; changing the temperature changes Kc.
- 5Pressure changes only affect the equilibrium position of reactions involving gases where there is a change in the total number of moles of gas.
- 6Pure solids and pure liquids have constant concentrations and are therefore omitted from the Kc expression.
Worked examples
Example 1
Consider the Haber process reaction: N₂(g) + 3H₂(g) ⇌ 2NH₃(g) ΔH = -92 kJ mol⁻¹. Predict the effect of the following changes on the position of equilibrium:
Answer
a) Equilibrium shifts to the left (less NH₃ formed). b) Equilibrium shifts to the right (more NH₃ formed). c) Equilibrium shifts to the right (more NH₃ formed).
Example 2
For the reaction H₂(g) + I₂(g) ⇌ 2HI(g), at a certain temperature, the equilibrium concentrations are [H₂] = 0.10 mol L⁻¹, [I₂] = 0.20 mol L⁻¹, and [HI] = 0.40 mol L⁻¹. Calculate the equilibrium constant, Kc.
Answer
Kc = 8.0
In this specific reaction, the units for Kc cancel out, so it is dimensionless.
Example 3
2.0 mol of SO₂ and 1.0 mol of O₂ are mixed in a 1.0 L container and allowed to reach equilibrium at a certain temperature. At equilibrium, 0.8 mol of SO₃ is formed. Calculate Kc for the reaction 2SO₂(g) + O₂(g) ⇌ 2SO₃(g).
Answer
Kc = 0.74 L mol⁻¹ (to 2 decimal places)
Always remember to include units for Kc unless they cancel out completely.
Common mistakes
- ✗Confusing the effect of a catalyst on reaction rate with its effect on equilibrium position.
- ✗Incorrectly applying Le Chatelier's Principle to pressure changes for reactions where the number of moles of gas does not change (e.g., H₂(g) + I₂(g) ⇌ 2HI(g)).
- ✗Including concentrations of pure solids or liquids in the Kc expression.
- ✗Forgetting to raise concentrations to the power of their stoichiometric coefficients when writing or calculating Kc.
- ✗Incorrectly determining the units of Kc, or omitting them when they are required.
Exam tips
- ★Always state Le Chatelier's Principle clearly before applying it to predict shifts in equilibrium.
- ★For Kc calculations, always write the balanced chemical equation and the correct Kc expression as your first step.
- ★Pay close attention to the units of concentration and ensure the correct units for Kc are derived and stated in your final answer.
- ★Practice using ICE (Initial, Change, Equilibrium) tables for Kc calculations where initial concentrations are given, as these are common in Higher Level exams.
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