Chemical Equilibria

Introduction

As-salamu alaykum, students. I am Dr. Fatima Malik. In my years with Cambridge, I have seen that Chemical Equilibria is a topic that separates the A* students from the rest. It is not merely about memorising definitions; it is about developing a deep, intuitive understanding of how chemical systems balance themselves. Many reactions, from the industrial synthesis of fertilisers that feed our nation to the complex biochemical systems within our own bodies, do not go to completion. Instead, they exist in a state of dynamic equilibrium, a delicate dance between reactants and products.

This chapter is fundamental. It connects thermodynamics (why a reaction happens) with kinetics (how fast it happens) and explains the 'how much' – what is the final composition of a reaction mixture? Understanding Le Chatelier's principle will empower you to think like a chemical engineer, manipulating conditions to maximise the yield of a desired product. From calculating the pH of a buffer solution in a lab to understanding the industrial conditions at Engro's fertiliser plants, the principles you learn here are universally applicable and frequently tested.

Core Theory

At the heart of this topic is the concept of a reversible reaction, symbolised by the '⇌' arrows. In a closed system, as products are formed, they can also react to re-form the original reactants. Eventually, the rate of the forward reaction becomes equal to the rate of the reverse reaction. This state is called dynamic equilibrium. It is 'dynamic' because reactions are still occurring at the molecular level, but 'equilibrium' because there is no net change in the concentrations of reactants and products. Macroscopic properties like colour, concentration, and pressure become constant.

Le Chatelier's Principle

This is your predictive tool. It states that if a change of condition is applied to a system in equilibrium, the system will shift in a direction that opposes the change.

1. Concentration: If you increase the concentration of a reactant, the equilibrium will shift to the right (→) to use up the extra reactant and form more products. If you remove a product, the equilibrium will also shift right to replace it.

* Example: N₂(g) + 3H₂(g) ⇌ 2NH₃(g). Adding more N₂ will shift the equilibrium to the right, producing more NH₃.

1. Pressure (for gases only): An increase in pressure favours the side with fewer moles of gas. This is because the system tries to reduce the pressure by occupying less volume.

* Example: N₂(g) + 3H₂(g) ⇌ 2NH₃(g). There are 4 moles of gas on the left and 2 on the right. Increasing pressure will shift the equilibrium to the right, favouring the formation of ammonia.

1. Temperature: This is the only factor that changes the value of the equilibrium constant (Kc or Kp).

* If the forward reaction is exothermic (ΔH is negative), an increase in temperature will shift the equilibrium to the left (←) in the endothermic direction to absorb the added heat. Kc/Kp will decrease.

* If the forward reaction is endothermic (ΔH is positive), an increase in temperature will shift the equilibrium to the right (→) to absorb the added heat. Kc/Kp will increase.

* Example: The Haber process is exothermic (ΔH = -92 kJ/mol). Therefore, a high temperature shifts the equilibrium to the left, decreasing the yield of NH₃. A compromise temperature is used industrially to balance yield and rate.

The Equilibrium Constant, Kc

For a general reaction: aA + bB ⇌ cC + dD

The expression for Kc is:

Kc = ([C]ᶜ[D]ᵈ) / ([A]ᵃ[B]ᵇ)

Where [X] denotes the equilibrium concentration of substance X in mol dm⁻³. Note that pure solids and liquids (including solvents like water in many cases) are omitted from the expression as their 'concentration' is constant.

The Equilibrium Constant, Kp

For reactions involving gases, it is often more convenient to use partial pressures.

Kp = (p(C)ᶜ * p(D)ᵈ) / (p(A)ᵃ * p(B)ᵇ)

Where p(X) is the partial pressure of gas X.

Partial Pressure = Mole Fraction × Total Pressure

Mole Fraction of A = (Number of moles of A) / (Total number of moles of all gases)

Acid-Base Equilibria

* Ka and pKa: For a weak acid, HA ⇌ H⁺ + A⁻, the acid dissociation constant is Ka = ([H⁺][A⁻]) / ([HA]). A larger Ka means a stronger acid. pKa = -log₁₀(Ka). A smaller pKa means a stronger acid.

* Kw: The ionic product of water, Kw = [H⁺][OH⁻] = 1.0 × 10⁻¹⁴ mol² dm⁻⁶ at 298K.

* Calculating pH of a weak acid: We assume the dissociation is small, so [HA] at equilibrium ≈ [HA] initial. We also assume [H⁺] from water is negligible, so [H⁺] ≈ [A⁻].

This simplifies to Ka ≈ [H⁺]² / [HA]initial, allowing you to solve for [H⁺] and then find pH.

Buffer Solutions

A buffer solution resists changes in pH upon the addition of small amounts of acid or alkali. It consists of a mixture of a weak acid and its conjugate base (e.g., CH₃COOH and CH₃COONa) or a weak base and its conjugate acid.

To calculate the pH of an acidic buffer, we use the Ka expression:

Ka = ([H⁺][A⁻]) / ([HA])

Rearranging for [H⁺]: [H⁺] = Ka × ([HA] / [A⁻])

We assume that [HA] ≈ initial concentration of the weak acid and [A⁻] ≈ initial concentration of the salt.

Solubility Product, Ksp

For a sparingly soluble salt, an equilibrium exists between the undissolved solid and its aqueous ions.

Example: AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq)

The solubility product expression is Ksp = [Ag⁺(aq)][Cl⁻(aq)]. The solid AgCl is omitted. The value of Ksp indicates how soluble a salt is; a smaller Ksp means lower solubility.

The Common Ion Effect

This is an application of Le Chatelier's principle to solubility. The solubility of a sparingly soluble salt is reduced by the presence of a solution containing a common ion. For example, AgCl is less soluble in a solution of NaCl than in pure water because the added Cl⁻ ions shift the equilibrium AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq) to the left, causing more AgCl to precipitate.

Key Definitions

* Dynamic Equilibrium: The state reached in a closed system where the rate of the forward reaction is equal to the rate of the reverse reaction, and the concentrations of reactants and products remain constant.

* Le Chatelier's Principle: When a change is made to the conditions of a system at equilibrium, the system responds by shifting the equilibrium position to counteract the change.

* Equilibrium Constant (Kc): A ratio of product concentrations to reactant concentrations at equilibrium, with each concentration raised to the power of its stoichiometric coefficient. Its value is constant at a constant temperature.

* Equilibrium Constant (Kp): An equilibrium constant for gas-phase reactions expressed in terms of the partial pressures of reactants and products.

* Partial Pressure: The pressure that an individual gas in a mixture of gases would exert if it alone occupied the entire volume of the container at the same temperature.

* Acid Dissociation Constant (Ka): An equilibrium constant that measures the extent of dissociation of a weak acid. pKa = -log₁₀(Ka).

* Buffer Solution: A solution that minimises pH changes on the addition of small amounts of acid or alkali, typically composed of a weak acid and its conjugate base.

* Solubility Product (Ksp): The equilibrium constant for the dissolution of a sparingly soluble ionic compound in water.

* Common Ion Effect: The reduction in the solubility of a sparingly soluble salt when a solution containing an ion in common with the salt is added.

Worked Examples (Pakistani Context)

Example 1: Kc Calculation for Esterification

In a Karachi-based chemical plant producing flavourings, ethyl ethanoate is synthesised. 1.20 mol of ethanol and 1.00 mol of ethanoic acid are mixed in a sealed vessel. At equilibrium, it is found that 0.85 mol of ethyl ethanoate has been formed. The total volume is 500 cm³. Calculate Kc for this reaction.

CH₃COOH(l) + C₂H₅OH(l) ⇌ CH₃COOC₂H₅(l) + H₂O(l)

Step 1: Set up an ICE table (Initial, Change, Equilibrium) using moles.

| | CH₃COOH | C₂H₅OH | CH₃COOC₂H₅ | H₂O |

|-------------|---------|--------|------------|-----|

| Initial (mol) | 1.00 | 1.20 | 0 | 0 |

| Change (mol) | -0.85 | -0.85 | +0.85 | +0.85 |

| Equil. (mol) | 0.15 | 0.35 | 0.85 | 0.85 |

*Rationale: The stoichiometry is 1:1:1:1. If 0.85 mol of ester is formed, then 0.85 mol of each reactant must have been consumed, and 0.85 mol of water must have been formed.*

Step 2: Calculate equilibrium concentrations.

Volume = 500 cm³ = 0.500 dm³

[CH₃COOH] = 0.15 mol / 0.500 dm³ = 0.30 mol dm⁻³

[C₂H₅OH] = 0.35 mol / 0.500 dm³ = 0.70 mol dm⁻³

[CH₃COOC₂H₅] = 0.85 mol / 0.500 dm³ = 1.70 mol dm⁻³

[H₂O] = 0.85 mol / 0.500 dm³ = 1.70 mol dm⁻³

Step 3: Write the Kc expression and substitute values.

Kc = ([CH₃COOC₂H₅][H₂O]) / ([CH₃COOH][C₂H₅OH])

Kc = (1.70 * 1.70) / (0.30 * 0.70)

Kc = 2.89 / 0.21

Kc = 13.76

Step 4: Determine the units.

Units = (mol dm⁻³)(mol dm⁻³) / (mol dm⁻³)(mol dm⁻³) = No units.

Final Answer: Kc = 13.8 (to 3 s.f.)

Example 2: Le Chatelier's Principle at Engro Fertilisers, Daharki

Engro Fertilisers is one of Pakistan's largest producers of urea, a critical component for our agricultural economy. The first step is the Haber-Bosch process to produce ammonia (NH₃).

N₂(g) + 3H₂(g) ⇌ 2NH₃(g) ΔH = -92 kJ mol⁻¹

Explain, using Le Chatelier's principle, the conditions of pressure and temperature used to maximise the yield of ammonia. Why is a catalyst used?

Pressure:

* Le Chatelier's Principle: The forward reaction proceeds with a decrease in the number of moles of gas (from 4 moles on the left to 2 moles on the right). According to Le Chatelier's principle, an increase in pressure will cause the equilibrium to shift to the side with fewer moles to counteract the change.

* Application: The equilibrium shifts to the right, favouring the formation of ammonia. Therefore, a high pressure (typically 150-250 atm) is used at the Daharki plant to achieve a high equilibrium yield of NH₃.

Temperature:

* Le Chatelier's Principle: The forward reaction is exothermic (ΔH is negative). According to Le Chatelier's principle, an increase in temperature will shift the equilibrium in the endothermic (reverse) direction to absorb the added heat. A decrease in temperature would shift it to the right.

* Application & Compromise: To get the highest yield, a very low temperature should be used. However, at low temperatures, the rate of reaction is extremely slow, which is not economically viable for Engro. A compromise temperature (around 400-450 °C) is used. This temperature is high enough to achieve a satisfactory rate of reaction but low enough to produce a reasonable equilibrium yield.

Catalyst:

* A finely divided iron catalyst is used. A catalyst increases the rate of both the forward and reverse reactions equally. It does not change the position of the equilibrium or the value of Kc/Kp. Its purpose is purely kinetic: to allow the system to reach equilibrium much faster at the compromise temperature, making the process economically feasible.

Exam Technique

Paper 2 (AS Structured):

* Definitions: Must be precise. For dynamic equilibrium, state *'rate of forward reaction equals rate of reverse reaction'* and *'concentrations of reactants and products are constant'*.

* Kc/Kp Expressions: Always use square brackets [ ] for concentration (Kc) and p( ) for partial pressure (Kp). Do not forget to include the powers from the stoichiometry. Check state symbols; omit (s) and (l).

* Calculations: Show your working clearly. An ICE table is highly recommended. Always calculate the units for Kc/Kp unless there are none. A common mistake is forgetting to use equilibrium concentrations, not initial ones.

* Le Chatelier's Principle: Use a three-step explanation: 1. State the change. 2. State how the system opposes the change according to the principle. 3. State the direction of the shift (e.g., 'equilibrium shifts to the right').

Paper 4 (A2 Structured):

* Buffers: When explaining how a buffer works, write two separate equations. One showing the weak acid reacting with added OH⁻, and one showing the conjugate base reacting with added H⁺.

* pH Calculations: For weak acids, always state your assumptions: [HA]eqm ≈ [HA]initial and [H⁺] = [A⁻].

* Ksp: Be careful with stoichiometry. For CaF₂(s) ⇌ Ca²⁺(aq) + 2F⁻(aq), the expression is Ksp = [Ca²⁺][F⁻]². If the solubility of CaF₂ is 's', then [Ca²⁺] = s and [F⁻] = 2s, so Ksp = (s)(2s)² = 4s³. This is a very common trap.

* Common Mistakes: Confusing Kc with Kp. Forgetting to convert volumes to dm³. Incorrectly calculating mole fractions. Stating that a catalyst changes the equilibrium position (it does not!). Thinking that pressure or concentration changes alter the value of K (only temperature does).