Grade 11 · Chemistry · Lesson 7

Chemical Equilibrium

Understand dynamic equilibrium, write equilibrium expressions, apply Le Chatelier’s principle to predict shifts, and calculate equilibrium constants.

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Reversible Reactions

Many chemical reactions are reversible — they can proceed in both the forward direction (reactants → products) and the reverse direction (products → reactants). Reversible reactions are represented using a double arrow: ⇌

A + B ⇌ C + D

In a closed system, as the forward reaction produces C and D, those products can recombine in the reverse reaction to regenerate A and B. Over time, the system reaches a state where both reactions occur simultaneously.

Dynamic Equilibrium

Dynamic equilibrium is reached when the rate of the forward reaction equals the rate of the reverse reaction. At this point:

Common misconception: At equilibrium, concentrations of reactants and products are constant — but they are NOT necessarily equal. The position of equilibrium depends on the value of Kc. A large Kc means more product is present; a small Kc means more reactant remains.

The Equilibrium Constant Kc

For the general reaction: aA + bB ⇌ cC + dD, the equilibrium constant expression is:

Kc = [C]c[D]d / [A]a[B]b

Important rules:

Kc valueInterpretation
Kc >> 1 (e.g. 10⁶)Products strongly favoured; reaction goes nearly to completion
Kc ≈ 1Neither products nor reactants strongly favoured; significant amounts of both at equilibrium
Kc << 1 (e.g. 10−⁶)Reactants strongly favoured; very little product formed at equilibrium

The Reaction Quotient Qc

The reaction quotient Qc has the same mathematical form as Kc, but uses current (non-equilibrium) concentrations. Comparing Qc to Kc tells us which direction the reaction will shift:

ComparisonDirection of shiftReason
Qc < KcForward (right) →Too many reactants; system needs more products to reach Kc
Qc = KcNo shift — at equilibriumSystem is already at equilibrium
Qc > KcReverse (left) ←Too many products; system needs more reactants to reach Kc

Le Chatelier’s Principle

Le Chatelier’s Principle: If a stress (change in conditions) is applied to a system at equilibrium, the system will shift in the direction that opposes the stress and re-establishes equilibrium.

Effect of Concentration

Stress appliedDirection of shiftEffect on Kc
Increase [reactant]Forward (right) →No change
Decrease [reactant]Reverse (left) ←No change
Increase [product]Reverse (left) ←No change
Decrease [product] (remove product)Forward (right) →No change

Effect of Pressure (Gas-phase reactions)

Effect of Temperature

Unlike concentration and pressure changes, temperature changes alter the value of Kc.

Reaction typeIncrease temperatureDecrease temperature
Exothermic (ΔH < 0)
heat is a product		Shifts LEFT (reverse) ←
Kc decreases		Shifts RIGHT (forward) →
Kc increases
Endothermic (ΔH > 0)
heat is a reactant		Shifts RIGHT (forward) →
Kc increases		Shifts LEFT (reverse) ←
Kc decreases

Effect of a Catalyst

Industrial Applications

Haber Process (synthesis of ammonia):

N₂(g) + 3H₂(g) ⇌ 2NH₃(g)    ΔH = −92 kJ·mol−1
ConditionValue usedJustification
Pressure150–300 atm (high)Fewer moles of gas on product side (2 vs 4); high P favours NH₃ yield
Temperature400–500°C (moderate)Low T favours yield (exothermic) but rate too slow; compromise gives acceptable rate and yield
CatalystIron (Fe) with promotersReaches equilibrium faster; no effect on yield or Kc

Contact Process (synthesis of sulfur trioxide, step in H₂SO₄ manufacture):

2SO₂(g) + O₂(g) ⇌ 2SO₃(g)    ΔH = −196 kJ·mol−1

Similar reasoning: high pressure favours SO₃ (3 moles gas → 2 moles); moderate temperature (450°C); V₂O₅ catalyst.

⭐ IEB Extension — ICE Tables

An ICE table (Initial, Change, Equilibrium) is used to systematically calculate equilibrium concentrations from initial conditions. Example for N₂ + 3H₂ ⇌ 2NH₃:

[N₂][H₂][NH₃]
Initial (I)0.5001.5000
Change (C)−x−3x+2x
Equilibrium (E)0.500−x1.500−3x2x

Substitute equilibrium expressions into the Kc expression and solve for x. Then calculate all equilibrium concentrations.

⭐ IEB Extension — Solubility Product Ksp

The solubility product Ksp is the equilibrium constant for the dissolution of a sparingly soluble ionic compound. For CaCO₃(s) ⇌ Ca²⁺(aq) + CO₃²−(aq):

Ksp = [Ca²⁺][CO₃²−]

The common ion effect: adding a common ion (e.g. Ca²⁺ from CaCl₂) shifts the dissolution equilibrium left, reducing solubility. If the ionic product Q > Ksp, precipitation occurs.

Equilibrium Simulator — Dynamic Approach to Equilibrium

System Controls
300 K
Equilibrium Readouts
[Reactant]
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[Product]
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Kc
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Status
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0/6
Quiz complete! Review the explanations to strengthen your understanding.
IEB Extension Questions
Answer all questions in your notebook. Show all calculations clearly. For Le Chatelier questions, always state the direction of shift AND explain why using the principle.
Question 1 — Writing the Kc Expression
Write the equilibrium constant expression Kc for each of the following reactions. State which species (if any) are excluded and why.

(a) N₂(g) + 3H₂(g) ⇌ 2NH₃(g)
(b) 2SO₂(g) + O₂(g) ⇌ 2SO₃(g)
(c) CaCO₃(s) ⇌ CaO(s) + CO₂(g)
(d) Fe³⁺(aq) + SCN⁻(aq) ⇌ FeSCN²⁺(aq)

(6 marks)
Question 2 — Le Chatelier Predictions
Consider the reaction: CO(g) + 3H₂(g) ⇌ CH₄(g) + H₂O(g)    ΔH = −206 kJ

For each change below, state (i) the direction of shift and (ii) whether Kc increases, decreases, or stays the same:

(a) Increasing the concentration of CO
(b) Removing H₂O from the system
(c) Increasing the total pressure
(d) Increasing the temperature
(e) Adding an iron catalyst

(10 marks — 2 per part)
Question 3 — Kc Calculation
At a certain temperature, the following equilibrium concentrations are measured for the reaction H₂(g) + I₂(g) ⇌ 2HI(g):

[H₂] = 0.40 mol·L−1    [I₂] = 0.20 mol·L−1    [HI] = 1.60 mol·L−1

(a) Write the Kc expression for this reaction.
(b) Calculate the value of Kc.
(c) Do the products or reactants dominate at equilibrium? Explain using your Kc value.
(4 marks)
Question 4 — Haber Process Justification
The Haber process: N₂(g) + 3H₂(g) ⇌ 2NH₃(g)    ΔH = −92 kJ

(a) Explain why high pressure is used in the Haber process. In your answer, refer to the number of moles of gas on each side.
(b) The process uses a temperature of about 450°C. Explain the trade-off involved in choosing this temperature — why not use a very low temperature (for better yield) or a very high temperature (for faster rate)?
(c) Iron catalyst is used. Does the catalyst change the equilibrium position or the value of Kc? Explain.
(d) Unreacted N₂ and H₂ are recycled back into the reactor. Explain, using Le Chatelier’s Principle, why this improves the overall yield.
(8 marks)
Question 5 — Qc vs Kc Prediction
At 700 K, Kc = 54.3 for the reaction H₂(g) + I₂(g) ⇌ 2HI(g).

A mixture is prepared with the following initial concentrations: [H₂] = 0.10 mol·L−1, [I₂] = 0.20 mol·L−1, [HI] = 0.40 mol·L−1.

(a) Calculate Qc for this mixture.
(b) Compare Qc to Kc. In which direction will the reaction proceed to reach equilibrium?
(c) As the reaction shifts, what happens to [HI]? Does it increase or decrease? Explain.
(4 marks)
Question 6 — Reading Concentration Data to Find Kc
The table shows [N₂O₄] and [NO₂] measured at intervals as the reaction N₂O₄(g) ⇌ 2NO₂(g) proceeds in a sealed container at constant temperature:
Time (s)020406080100
[N₂O₄] (mol·L−1)0.1000.0720.0610.0580.0580.058
[NO₂] (mol·L−1)0.0000.0560.0780.0840.0840.084
(a) At what time does the system first reach equilibrium? Explain how you can identify this directly from the data.
(b) Write the Kc expression for this reaction, then calculate the value of Kc using the equilibrium concentrations.
(c) Verify, using the data at t = 40 s, that the amount of N₂O₄ consumed since t = 0 is consistent with the amount of NO₂ formed (remember the mole ratio in the balanced equation).
(d) A learner claims that once equilibrium is reached (constant concentrations), the forward and reverse reactions have both stopped. Is this correct? Explain using the concept of dynamic equilibrium.
(8 marks)