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R2.3.2 The equilibrium law and constant

Exploring the Equilibrium Constant K

  1. You're designing a chemical process to produce ammonia in the Haber process.
  2. You know the reaction is reversible, meaning it won't go entirely in one direction.
  3. At equilibrium, the forward and reverse reactions balance each other out. But how much ammonia can you expect at equilibrium?
This is where the equilibrium constant K becomes invaluable.

What is the Equilibrium Constant?

Definition

Equilibrium constant

The equilibrium constant K is a ratio that provides a snapshot of the relative concentrations of products and reactants at equilibrium for a reversible chemical reaction.

  • It is derived from the stoichiometry of the reaction and remains constant as long as the temperature is unchanged.
  • Consider the general reaction:

aA+bBcC+dD

  • At equilibrium, the concentrations of A, B, C, and D remain constant. The equilibrium constant K is expressed as:

K=[C]c[D]d[A]a[B]b

where:

  • [X] represents the equilibrium concentration of species X in mol dm3.
  • a,b,c,d are the stoichiometric coefficients from the balanced chemical equation.

Tip

Before writing the K expression, ensure the chemical equation is balanced. The stoichiometric coefficients directly determine the exponents in the K expression.

Deriving the Expression for K: Step-by-Step

Let’s illustrate this with an example: the synthesis of ammonia.

N2(g)+3H2(g)2NH3(g)

  1. Identify the Reactants and Products:
    • Reactants: N2 and H2
    • Product: NH3
  2. Write the General Form of K:
    According to the equilibrium law, the concentrations of the products are placed in the numerator, while the concentrations of the reactants are placed in the denominator. Each species is raised to the power of its stoichiometric coefficient:

K=[NH3]2[N2][H2]3

  1. Interpret the Expression:
    • If K is large (K>1), the numerator dominates, indicating that the equilibrium mixture contains more products than reactants.
    • If K is small (K<1), the denominator dominates, meaning the equilibrium mixture contains more reactants than products.
Example question

Write the K expression for the reaction:2SO2(g)+O2(g)2SO3(g)

Solution

The equilibrium constant expression is:K=[SO3]2[SO2]2[O2]

Properties of the Equilibrium Constant

  1. Temperature Dependence:
    • K remains constant only at a fixed temperature. Altering the temperature shifts the equilibrium position and changes the value of K.
  2. No Units:
    • K is often treated as a dimensionless quantity because the units of concentration cancel out when the expression is written correctly.
  3. Homogeneous vs. Heterogeneous Equilibria:
    • For homogeneous equilibria (all species in the same phase), all reactants and products appear in the K expression.
    • For heterogeneous equilibria (species in different phases), the concentrations of pure solids and liquids are omitted because they are constant.

Note

In aqueous reactions, the concentration of water is often omitted from the K expression if it is the solvent and present in large excess.

Interpreting K: What Does It Tell Us?

The value of K provides insight into the extent of the reaction at equilibrium:

  1. K>1:
    • Products are favored.
    • The reaction proceeds significantly in the forward direction.
  2. K<1:
    • Reactants are favored.
    • The reaction does not proceed significantly in the forward direction.
  3. K=1:
    • Neither reactants nor products are favored.
    • The concentrations of reactants and products are comparable.

Common Mistake

Students often confuse K with the reaction quotient Q. Remember, K applies only at equilibrium, while Q can be calculated at any point during the reaction.

Example question

Calculating K

The equilibrium concentrations for the reaction N2(g)+3H2(g)2NH3(g) at 475 K are:

  • [N2]=0.50mol dm3,
  • [H2]=1.50mol dm3,
  • [NH3]=1.00mol dm3.

Calculate K for the reaction.

Solution

  1. Write the K expression:
    K=[NH3]2[N2][H2]3
  2. Substitute the equilibrium concentrations:
    K=(1.00)2(0.50)(1.50)3
  3. Perform the calculation:
    K=1.000.50×3.375=1.001.68750.59

Answer: K=0.59

Self review

Can you calculate K for the reverse reaction? Hint: The K value for the reverse reaction is the reciprocal of the forward reaction’s K.

Reflection and Practice

Self review

  1. Why is the equilibrium constant independent of initial concentrations?
  2. How does K change when the reaction equation is reversed or its stoichiometric coefficients are altered?
  3. Consider the reaction H2(g)+I2(g)2HI(g). If K=50 at 298 K, what does this tell you about the relative concentrations of reactants and products at equilibrium?

Theory of Knowledge

How does the equilibrium constant help us balance economic and environmental considerations in industrial chemistry?

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Questions

Recap questions

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Question 1

If the equilibrium constant KK for a reaction is known to be 0.25, what can you infer about the concentrations of reactants and products at equilibrium?

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What does a large equilibrium constant (K>1K > 1) indicate about the reaction?

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Note

Chemical Equilibrium and the Equilibrium Law

A reversible reaction is one where the products can react to reform the reactants. At equilibrium, the rate of the forward reaction equals the rate of the reverse reaction, and the concentrations of reactants and products remain constant. This balance is dynamic, meaning the reactions continue to occur, but at equal rates.

Analogy

Think of a busy highway with cars entering and leaving a city at the same rate. The number of cars in the city remains constant, but the movement continues.

Example

In the reaction N2O4(g)2NO2(g)\text{N}_2\text{O}_4(g) \rightleftharpoons 2\text{NO}_2(g), the pale yellow color of N2O4\text{N}_2\text{O}_4 and the brown color of NO2\text{NO}_2 reach a constant intensity at equilibrium.