Mastering Equilibrium Constants: A Step-by-Step Guide

Expert reviewed • 22 November 2024 • 5 minute read

Introduction

Chemical equilibrium is a fundamental concept in chemistry where forward and reverse reactions occur at equal rates. The equilibrium constant (KeqK_{eq}) is a crucial value that helps us understand and predict the behavior of these reactions.

Understanding the Equilibrium Constant

The equilibrium constant (KeqK_{eq}) provides a quantitative measure of a reaction at equilibrium. For a general reaction:

aA+bB⇌cC+dDaA + bB \rightleftharpoons cC + dD

The equilibrium constant expression is:

Keq=[C]c[D]d[A]a[B]bK_{eq} = \frac{[C]^c[D]^d}{[A]^a[B]^b}

Key characteristics of KeqK_{eq}:

  • Always positive (Keq>0K_{eq} > 0)
  • Larger values indicate more products at equilibrium
  • Smaller values indicate more reactants at equilibrium
  • Only includes aqueous and gaseous species

The Effect of Reaction Manipulation

When we modify a balanced equation, the equilibrium constant changes predictably:

  • Reverse Reactions: When a reaction is reversed, the new equilibrium constant is the reciprocal of the original: Keq(reverse)=1Keq(forward)K_{eq(\text{reverse})} = \frac{1}{K_{eq(\text{forward})}}

  • Multiplying Equations: When stoichiometric coefficients are multiplied by n, the new equilibrium constant is raised to that power: Keq(new)=(Keq)nK_{eq(\text{new})} = (K_{eq})^n

The ICE Table Method

ICE tables provide a systematic approach to equilibrium calculations:

  • I: Initial concentrations
  • C: Change in concentrations
  • E: Equilibrium concentrations

Example Problem: Hydrogen Iodide Equilibrium

Consider the following reaction at 450°C: H2(g)+I2(g)⇌2HI(g)H_2(g) + I_2(g) \rightleftharpoons 2HI(g)

Given:

  • Initial H2H_2 = 5.00 moles in 5.00 L (1.00 M)
  • Initial I2I_2 = 10.0 moles in 5.00 L (2.00 M)
  • Equilibrium [HI] = 1.87 M

Solution Steps:

  • Set up ICE table:

    • Convert all amounts to concentrations
    • Let x be the change in concentration of H2H_2 and I2I_2
    • Express [HI] in terms of 2x

  • Solve for x: 2x=1.87 M2x = 1.87 \text{ M} x=0.935 Mx = 0.935 \text{ M}

  • Calculate KeqK_{eq}: Keq=[HI]2[H2][I2]K_{eq} = \frac{[HI]^2}{[H_2][I_2]} Keq=(1.87)2(1.00−0.935)(2.00−0.935)=50.5K_{eq} = \frac{(1.87)^2}{(1.00-0.935)(2.00-0.935)} \\= 50.5

The large KeqK_{eq} value (50.5) indicates that at equilibrium, the products (HI) are favored over the reactants (H2H_2 and I2I_2).

Up Next

Chemical Equilibrium: Understanding Reaction Quotients and Temperature Effects

Return to Module 5: Equilibrium and Acid Reactions