Chemical Equilibrium

30min Part 2 / Ch4 / Lesson 2

Prerequisites: 2-4-1

Objectives

Understand chemical equilibrium and the equilibrium constant
Use Le Chatelier's principle to predict equilibrium shifts
Calculate equilibrium constants

What Is Chemical Equilibrium?

Chemical equilibrium: the state where the forward and reverse reactions proceed at the same rate, so concentrations appear constant. Both reactions continue — this is dynamic equilibrium.

Reaction rates approaching equilibrium

Equilibrium Constant K

$a\mathrm{A} + b\mathrm{B} \rightleftharpoons c\mathrm{C} + d\mathrm{D}$

Concentration equilibrium constant:

$K_c = \frac{[\mathrm{C}]^c [\mathrm{D}]^d}{[\mathrm{A}]^a [\mathrm{B}]^b}$

Large $K$ → products favored (reaction goes nearly to completion)
Small $K$ → reactants favored (reaction barely proceeds)
$K$ depends on temperature only (not concentration or pressure)

Le Chatelier’s Principle

Le Chatelier’s principle: when a system at equilibrium is disturbed, it shifts to counteract the change.

Change	Equilibrium shifts toward
Increase reactant concentration	→ Products (forward)
Increase product concentration	← Reactants (reverse)
Increase temperature	Endothermic direction
Increase pressure	Side with fewer gas moles
Add a catalyst	No shift (speeds both directions equally)

Worked Example

Example: Calculating K

For $\mathrm{H_2(g) + I_2(g) \rightleftharpoons 2HI(g)}$ , at equilibrium: $[\mathrm{H_2}] = 0.10$ , $[\mathrm{I_2}] = 0.10$ , $[\mathrm{HI}] = 0.80\ \mathrm{mol/L}$ . Find $K_c$ .

$K_c = \frac{[\mathrm{HI}]^2}{[\mathrm{H_2}][\mathrm{I_2}]}$

$K_c = \frac{(0.80)^2}{(0.10)(0.10)} = \frac{0.64}{0.010} = 64$

Check Your Understanding

Q1 Equilibrium means:

Q2 N₂ + 3H₂ ⇌ 2NH₃ (exothermic). What happens if temperature is raised?

Q3 Adding a catalyst to a system at equilibrium:

Exercises

Q1. For $\mathrm{N_2O_4(g) \rightleftharpoons 2NO_2(g)}$ (endothermic), predict the direction of equilibrium shift for:

(a) Adding $\mathrm{NO_2}$ 　(b) Increasing pressure　(c) Increasing temperature

Solution

(a) Reverse (← toward $\mathrm{N_2O_4}$ ): Adding $\mathrm{NO_2}$ increases its concentration, so equilibrium shifts to reduce it.

(b) Reverse (← toward $\mathrm{N_2O_4}$ ): The side with fewer gas moles is favored ( $\mathrm{N_2O_4}$ : 1 mol vs $\mathrm{NO_2}$ : 2 mol).