Reversible Reactions
What is a Reversible Reaction?
A reversible reaction is one that can proceed in both the forward and reverse directions. It is shown with a double arrow (⇌):
The forward reaction converts reactants (A, B) to products (C, D). The reverse reaction converts products back to reactants. Not all reactions are reversible — irreversible reactions go essentially to completion (single arrow →).
Examples of Reversible Reactions
Test for reversibility: cobalt chloride colour change — CoCl₂·6H₂O (pink) ⇌ CoCl₂ (blue) + 6H₂O. Heating drives forward (blue); cooling/adding water drives reverse (pink).
Dynamic Equilibrium
Key Features of Dynamic Equilibrium
- Closed system required — no products or reactants can escape
- Constant macroscopic properties — concentrations, colour, pressure, and density do not change
- Dynamic at the molecular level — forward and reverse reactions both ongoing; rates are equal (not zero)
- Can be reached from either direction — same equilibrium state is reached whether you start with reactants or products
- Temperature dependent — the equilibrium position changes with temperature; a new equilibrium is established
Le Châtelier's Principle
Why "Partially Oppose"?
The system never fully neutralises the disturbance — it only partially counteracts it. For example, if you add more reactant, the equilibrium shifts forward to use some of it up — but not all. A new equilibrium is established with higher concentrations of both reactants and products than the original equilibrium, but with a higher proportion of products.
Effects of Changes on Equilibrium Position
1. Effect of Concentration
For: aA + bB ⇌ cC + dD
| Change | Shift | Effect on concentrations |
|---|---|---|
| Increase [A] or [B] | → Forward (right) | [C] and [D] increase; [A] and [B] partially decrease |
| Decrease [A] or [B] | ← Reverse (left) | [C] and [D] decrease; [A] and [B] partially increase |
| Increase [C] or [D] | ← Reverse (left) | [A] and [B] increase; [C] and [D] partially decrease |
| Remove product C or D | → Forward (right) | More products formed — used in synthesis to improve yield |
2. Effect of Pressure (for gaseous equilibria)
Pressure changes only affect equilibria involving gases. The effect depends on the total moles of gas on each side:
| Change | Shift | Condition |
|---|---|---|
| Increase pressure | → Side with fewer moles of gas | Reduces the "pressure" by producing fewer gas molecules |
| Decrease pressure | → Side with more moles of gas | Increases gas molecules to partially restore pressure |
| Equal moles of gas on both sides | No shift | Pressure change has no effect on position |
For N₂(g) + 3H₂(g) ⇌ 2NH₃(g) — left side: 4 mol gas; right side: 2 mol gas. Increasing pressure → shifts right (fewer moles gas) → more NH₃ produced. Decreasing pressure → shifts left → less NH₃.
For H₂(g) + I₂(g) ⇌ 2HI(g) — left: 2 mol; right: 2 mol. Pressure change has NO effect on the equilibrium position (Kc unchanged).
3. Effect of Temperature
Temperature is the only factor that changes the value of Kc (or Kp) — not just the position:
| Reaction type | Increase temperature | Decrease temperature | Effect on Kc |
|---|---|---|---|
| Exothermic (ΔH < 0): A ⇌ B + heat | Shifts left (← reverse) — opposes heat input | Shifts right (→ forward) — generates heat | Increase T → Kc decreases (less product at equilibrium) |
| Endothermic (ΔH > 0): A + heat ⇌ B | Shifts right (→ forward) — uses up added heat | Shifts left (← reverse) | Increase T → Kc increases (more product at equilibrium) |
4. Effect of a Catalyst
A catalyst speeds up both the forward and reverse reactions equally by providing an alternative pathway with lower activation energy. Therefore:
- Equilibrium is reached faster
- The equilibrium position is unchanged (same concentrations at equilibrium)
- Kc is unchanged
- A catalyst does NOT improve the yield — it only saves time
Equilibrium Constants: Kc and Kp
The Equilibrium Law and Kc
For a reversible reaction: aA + bB ⇌ cC + dD, the equilibrium constant Kc is:
Kc is a constant at a given temperature for a given reaction. It does not depend on initial concentrations, pressure, or the presence of a catalyst.
Magnitude of Kc — What It Tells Us
| Value of Kc | Interpretation | Example |
|---|---|---|
| Kc >> 1 (e.g. 10¹⁰) | Equilibrium lies far to the right — mostly products at equilibrium; reaction almost complete | H₂ + ½O₂ ⇌ H₂O: Kc = 10⁴⁰ |
| Kc ≈ 1 | Significant amounts of both reactants and products at equilibrium | H₂ + I₂ ⇌ 2HI: Kc = 46 at 450°C |
| Kc << 1 (e.g. 10⁻¹⁰) | Equilibrium lies far to the left — mostly reactants; reaction barely proceeds | N₂ + O₂ ⇌ 2NO: Kc = 10⁻³⁰ at 25°C |
Kp — Equilibrium Constant in Terms of Partial Pressures
For gaseous equilibria, Kp uses partial pressures instead of concentrations:
Q1: Write the Kc expression for: 2SO₂(g) + O₂(g) ⇌ 2SO₃(g)
Kc = [SO₃]² / ([SO₂]²[O₂])
Q2: At equilibrium in a 2.0 dm³ vessel: [N₂] = 0.50 mol dm⁻³, [H₂] = 0.20 mol dm⁻³, [NH₃] = 0.30 mol dm⁻³. Calculate Kc for N₂ + 3H₂ ⇌ 2NH₃.
Kc = [NH₃]² / ([N₂][H₂]³) = (0.30)² / (0.50 × (0.20)³)
= 0.090 / (0.50 × 0.0080) = 0.090 / 0.0040 = 22.5 mol⁻² dm⁶
Q3: Write Kc for the reverse: 2NH₃ ⇌ N₂ + 3H₂
Kc(reverse) = [N₂][H₂]³ / [NH₃]² = 1/22.5 = 0.044 mol² dm⁻⁶
Industrial Applications of Equilibrium
The Haber Process — A Le Châtelier Case Study
N₂(g) + 3H₂(g) ⇌ 2NH₃(g) ΔH = −92 kJ mol⁻¹
| Factor | Favours yield (Le Châtelier) | Favours rate | Industrial compromise |
|---|---|---|---|
| Temperature | Lower T → more NH₃ (exothermic → low T shifts right) | Higher T → faster | ~450°C — acceptable rate AND ~15% conversion per pass |
| Pressure | Higher P → more NH₃ (4 mol → 2 mol gas) | Higher P → faster | ~200 atm — balance of yield, cost, and safety |
| Catalyst (Fe) | No effect on Kc or equilibrium position | Dramatically increases rate | Iron + K₂O/Al₂O₃ promoters — essential |
| Remove NH₃ | Shifts right → more NH₃ | — | NH₃ condensed and removed continuously; N₂/H₂ recycled |
The Contact Process — Another Equilibrium Example
2SO₂(g) + O₂(g) ⇌ 2SO₃(g) ΔH = −197 kJ mol⁻¹
- Conditions: 450°C, V₂O₅ catalyst, 1–2 atm (slightly above atmospheric)
- Temperature: 450°C is a compromise — lower T would give more SO₃ (exothermic) but reaction too slow; 450°C gives ~99.5% conversion with V₂O₅
- Pressure: Relatively low pressure used (3 mol gas → 2 mol gas, high P would help, but low P already gives good yield and saves costs)
- Excess air: Excess O₂ shifts equilibrium right, increasing SO₃ yield
Esterification Equilibrium
CH₃COOH + C₂H₅OH ⇌ CH₃COOC₂H₅ + H₂O (acid catalyst, Kc ≈ 4 at 25°C)
To improve yield of ester (ethyl ethanoate):
- Use excess of one reactant (ethanol cheaper — use in excess)
- Remove water as it forms (e.g. by distillation or drying agent) → shifts right
- Remove ester by distillation → shifts right
- H₂SO₄ catalyst increases rate but does NOT change Kc or equilibrium position
Solubility Product (Ksp) and Partition Equilibria
Solubility Product Ksp
When a sparingly soluble salt dissolves: AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq)
The solid AgCl is not included in the expression (pure solid, constant "concentration"):
The common ion effect: adding a common ion (e.g. adding NaCl to AgCl solution) increases [Cl⁻] → ionic product [Ag⁺][Cl⁻] > Ksp → more AgCl precipitates → solubility decreases.
Partition Equilibrium
When a substance distributes between two immiscible solvents (e.g. water and organic solvent), a partition (distribution) equilibrium is established:
This is the basis of solvent extraction: using an organic solvent to extract a substance from an aqueous solution. Multiple small-volume extractions are more efficient than one large-volume extraction.
No videos added yet for this unit.
Exercises
- Define dynamic equilibrium. State three characteristics that distinguish it from a state where no reaction is occurring.
Dynamic equilibrium: a state in a closed system where the rate of the forward reaction equals the rate of the reverse reaction, and the concentrations of all species remain constant. Three characteristics: (1) Both forward and reverse reactions are occurring simultaneously at equal rates — neither has stopped. (2) The system is dynamic at the molecular level even though macroscopic properties (concentration, colour, pressure) appear constant. (3) Can only exist in a closed system — if products escape, the equilibrium is destroyed and reaction goes to completion. (vs. no reaction: if no reaction, concentrations are also constant but rates of both reactions are zero).
- For the reaction PCl₅(g) ⇌ PCl₃(g) + Cl₂(g) (ΔH = +88 kJ mol⁻¹), predict the effect on the equilibrium position of: (a) increasing temperature, (b) increasing pressure, (c) adding a catalyst, (d) removing Cl₂.
(a) Increasing T: reaction is endothermic → adding heat → system opposes by using heat → shifts right (forward) → more PCl₃ and Cl₂ formed → Kc increases (endothermic reaction, higher T → larger Kc). (b) Increasing pressure: left side = 1 mol gas; right side = 2 mol gas. Higher pressure → shifts LEFT (fewer gas moles on left) → more PCl₅ forms. Kc unchanged. (c) Adding catalyst: no effect on equilibrium position or Kc — both forward and reverse rates increase equally → equilibrium reached faster only. (d) Removing Cl₂: [Cl₂] decreases → system opposes by producing more Cl₂ → shifts RIGHT → more PCl₃ and Cl₂ formed, more PCl₅ used. Kc unchanged.
- Write the Kc expression for: (a) 2NO(g) + O₂(g) ⇌ 2NO₂(g) (b) CaCO₃(s) ⇌ CaO(s) + CO₂(g) (c) Fe³⁺(aq) + SCN⁻(aq) ⇌ [FeSCN]²⁺(aq)
(a) Kc = [NO₂]² / ([NO]²[O₂]) — units: mol⁻¹ dm³. (b) Kc = [CO₂] — CaCO₃ and CaO are pure solids, not included in Kc expression. Units: mol dm⁻³. (c) Kc = [[FeSCN]²⁺] / ([Fe³⁺][SCN⁻]) — units: dm³ mol⁻¹. The blood-red colour of [FeSCN]²⁺ intensifies when more Fe³⁺ or SCN⁻ is added (equilibrium shifts right), and fades when these are diluted or complexed.
- At 500°C, the equilibrium constant for N₂(g) + 3H₂(g) ⇌ 2NH₃(g) is Kc = 6.0 × 10⁻² mol⁻² dm⁶. At equilibrium, [N₂] = 0.50 mol dm⁻³ and [H₂] = 0.60 mol dm⁻³. Calculate [NH₃] at equilibrium.
Kc = [NH₃]² / ([N₂][H₂]³) = 6.0 × 10⁻². [N₂] = 0.50, [H₂] = 0.60. [NH₃]² = Kc × [N₂] × [H₂]³ = 6.0×10⁻² × 0.50 × (0.60)³ = 6.0×10⁻² × 0.50 × 0.216 = 6.0×10⁻² × 0.108 = 6.48×10⁻³. [NH₃] = √(6.48×10⁻³) = 0.0805 mol dm⁻³ ≈ 0.080 mol dm⁻³.
- Explain why a catalyst does not change the equilibrium yield of a product, but is still used industrially. Use the Haber Process as your example.
A catalyst provides an alternative reaction pathway with lower activation energy. It speeds up both the forward and reverse reactions by the same factor → both rates increase equally → equilibrium is reached faster → equilibrium position is unchanged → same concentrations of NH₃ at equilibrium. Kc is unchanged. However, in industry, the Fe catalyst in the Haber Process allows equilibrium to be reached at 450°C in a reasonable time. Without the catalyst, the reaction would be too slow at 450°C. To achieve a usable rate without the catalyst, temperatures of ~700°C+ would be needed — but this would give a very poor yield (exothermic reaction, higher T → less NH₃). The catalyst thus allows the best of both: acceptable rate AND acceptable yield at 450°C. It saves energy and reduces process time, which is economically essential.
- The Ksp of AgCl at 25°C is 1.8 × 10⁻¹⁰ mol² dm⁻⁶. (a) Calculate the molar solubility of AgCl in pure water. (b) Calculate the molar solubility of AgCl in 0.10 mol dm⁻³ NaCl solution (common ion effect).
(a) AgCl ⇌ Ag⁺ + Cl⁻. Let s = molar solubility. [Ag⁺] = [Cl⁻] = s. Ksp = s² = 1.8×10⁻¹⁰ → s = √(1.8×10⁻¹⁰) = 1.34×10⁻⁵ mol dm⁻³. (b) In 0.10 mol dm⁻³ NaCl: [Cl⁻] = 0.10 + s ≈ 0.10 (s << 0.10). Ksp = [Ag⁺][Cl⁻] = s × 0.10 = 1.8×10⁻¹⁰ → s = 1.8×10⁻⁹ mol dm⁻³. The common ion effect reduces the solubility by a factor of ~7500 (from 1.34×10⁻⁵ to 1.8×10⁻⁹ mol dm⁻³). Adding NaCl shifts AgCl ⇌ Ag⁺ + Cl⁻ to the left → less AgCl dissolves.
Multiple Choice Quiz — 25 Questions
Unit 15: Chemical Equilibrium
25 QuestionsAt dynamic equilibrium, which statement is correct?
Le Châtelier's Principle states that when a system at equilibrium is disturbed, it will:
For 2SO₂(g) + O₂(g) ⇌ 2SO₃(g), increasing pressure will:
Which of the following changes will alter the VALUE of the equilibrium constant Kc?
The Kc expression for N₂(g) + 3H₂(g) ⇌ 2NH₃(g) is:
For the exothermic reaction A ⇌ B (ΔH = −50 kJ mol⁻¹), increasing temperature:
Adding a catalyst to a reaction at equilibrium:
For N₂O₄(g) ⇌ 2NO₂(g) (endothermic, ΔH = +57 kJ/mol), a mixture of N₂O₄ and NO₂ is placed in a syringe and rapidly compressed. The immediate change is:
If Kc for a reaction is very large (e.g. 10¹⁰), this means:
For CaCO₃(s) ⇌ CaO(s) + CO₂(g), the Kc expression is:
In the Haber Process (N₂ + 3H₂ ⇌ 2NH₃, exothermic), which set of conditions gives the best YIELD of NH₃?
The reaction CH₃COOH + C₂H₅OH ⇌ CH₃COOC₂H₅ + H₂O has Kc ≈ 4. To increase the yield of ester, you could:
Adding iron catalyst to the Haber Process at equilibrium:
For the equilibrium: Fe³⁺(aq) + SCN⁻(aq) ⇌ [FeSCN]²⁺(aq) (blood red), adding more Fe(NO₃)₃ solution will cause:
The units of Kc for the reaction N₂(g) + 3H₂(g) ⇌ 2NH₃(g) are:
For H₂(g) + I₂(g) ⇌ 2HI(g), increasing pressure has:
Ksp for AgBr = 5.0 × 10⁻¹³ mol² dm⁻⁶. The molar solubility of AgBr in pure water is:
The equilibrium constant for A + B ⇌ C + D is Kc. If the equation is reversed (C + D ⇌ A + B), the new Kc is:
Why must a reversible reaction be carried out in a closed system to achieve equilibrium?
In the Contact Process (2SO₂ + O₂ ⇌ 2SO₃, exothermic), which change would DECREASE the yield of SO₃?
The Kp for 2NO₂(g) ⇌ N₂O₄(g) at 25°C is 8.8 atm⁻¹. This means:
At equilibrium, [A] = 0.20, [B] = 0.30, [C] = 0.60 mol dm⁻³ for A + B ⇌ C. Calculate Kc.
Adding a common ion reduces the solubility of a sparingly soluble salt. This is because:
The equilibrium H₂(g) + I₂(g) ⇌ 2HI(g) has Kc = 46 at 450°C. Starting with 0.50 mol H₂ and 0.50 mol I₂ in a 1.0 dm³ vessel, the [HI] at equilibrium is approximately:
In an industrial ammonia synthesis plant, NH₃ is continuously removed by condensation as it is formed. The effect of this is:
Unit Test — 50 Marks
Section A — Short Answer
30 marksDefine dynamic equilibrium. Describe an experiment to demonstrate that both the forward and reverse reactions are still occurring at equilibrium, using the N₂O₄/NO₂ system as your example. [5]
For the reaction: 2SO₂(g) + O₂(g) ⇌ 2SO₃(g), ΔH = −197 kJ mol⁻¹
Predict and explain the effect on the equilibrium position AND Kc of: (a) increasing temperature (b) increasing pressure (c) adding V₂O₅ catalyst (d) adding more SO₂. [5]
Write the Kc expressions for the following reactions and state the units: (a) 2HI(g) ⇌ H₂(g) + I₂(g) (b) PCl₅(g) ⇌ PCl₃(g) + Cl₂(g) (c) Fe³⁺(aq) + SCN⁻(aq) ⇌ [FeSCN]²⁺(aq) (d) CH₃COOH(aq) + C₂H₅OH(aq) ⇌ CH₃COOC₂H₅(aq) + H₂O(l). [5]
In the esterification: CH₃COOH + C₂H₅OH ⇌ CH₃COOC₂H₅ + H₂O, Kc = 4.0 at 60°C. 1.0 mol of each reactant is mixed in a 1.0 dm³ flask. Calculate the equilibrium concentrations of all species. [5]
The Ksp of CaF₂ is 3.9 × 10⁻¹¹ mol³ dm⁻⁹ at 25°C. (a) Write the dissolution equation and Ksp expression. (b) Calculate the molar solubility of CaF₂ in pure water. (c) Explain what happens when NaF solution is added to a saturated CaF₂ solution. [5]
Explain the industrial significance of the following equilibrium strategies: (a) recycling unreacted feedstock (b) removing products as they form (c) using high pressure in the Haber Process (d) why equilibrium yield and rate considerations conflict in industrial processes. [5]
Section B — Extended Answer
20 marks(a) Le Châtelier's Principle is essential to understanding industrial chemical processes. Using the Haber Process as your main example, describe how each factor (temperature, pressure, catalyst, and product removal) is optimised, explaining the reasoning for each industrial choice. [6]
(b) The equilibrium constant Kc has different units for different reactions. Explain why, and derive the units for: (i) N₂ + 3H₂ ⇌ 2NH₃ (ii) 2NO(g) ⇌ N₂(g) + O₂(g). [4]
(a) Explain the difference between thermodynamic feasibility (Kc) and kinetic feasibility (rate) of a reaction, using two examples where a reaction has very large Kc but is still very slow without special conditions. [4]
(b) Describe the relationship between temperature, Kc, and ΔH for equilibrium reactions. Explain how van 't Hoff's equation qualitatively describes this: ln(Kc) changes linearly with 1/T (plot gives −ΔH/R as slope). Include what happens to Kc as T → ∞ and T → 0 for exothermic and endothermic reactions. [6]