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Chemical Equilibrium

Subject: Chemistry
Topic: 8
Cambridge Code: 0620 / 0971 / 5070

Reversible Reactions​

Reversible reaction - Proceeds in both directions

A+B⇌C+DA + B ⇌ C + D

(Forward and reverse reactions occur simultaneously)

Irreversible Reactions​

  • Proceed in one direction to completion
  • One product escapes (gas), one product insoluble
  • Example: Combustion

Reversible vs Irreversible​

FeatureReversibleIrreversible
DirectionBoth waysOne way
EquilibriumReachedComplete reaction
ReactantsSome remainAll consumed
Symbol⇌→

Dynamic Equilibrium​

Dynamic equilibrium - Forward and reverse rates equal

Characteristics​

  1. Macroscopic: Concentrations appear constant
  2. Microscopic: Reactions continue (molecules still reacting)
  3. At equilibrium: Rate forward = Rate reverse
  4. Closed system required: No material enters/leaves

Reaching Equilibrium​

Time 0 → Forward rate > Reverse ↓ (time passes) ↓ Reverse rate increases, Forward rate decreases ↓ (time passes) ↓ Equilibrium: Rate forward = Rate reverse

Equilibrium Constant (K)​

Equilibrium constant (Kc) - Ratio of product to reactant concentrations at equilibrium

For Reaction: aA + bB ⇌ cC + dD​

Kc=[C]c[D]d[A]a[B]bK_c = \frac{[C]^c[D]^d}{[A]^a[B]^b}

Notes:

  • Exponents = stoichiometric coefficients
  • Only products over only reactants
  • [#] = concentration in mol/dm³
  • Solids and pure liquids NOT included

K Value Interpretation​

  • K >> 1: Equilibrium favors products (right)
  • K >> 1: Equilibrium favors reactants (left)
  • K = 1: Equal amounts at equilibrium

Example​

Reaction: CO(g) + Cl₂(g) ⇌ COCl₂(g)

Kc=[COCl2][CO][Cl2]K_c = \frac{[COCl_2]}{[CO][Cl_2]}

At equilibrium: [COCl₂] = 1.2 mol/dm³, [CO] = 0.4, [Cl₂] = 0.3

Kc=1.20.4×0.3=1.20.12=10K_c = \frac{1.2}{0.4 \times 0.3} = \frac{1.2}{0.12} = 10

Le Chatelier's Principle​

Le Chatelier - System responds to counteract changes

Effects of Changes​

1. Concentration Change​

Increase reactant concentration:

  • Equilibrium shifts RIGHT (forward)
  • More products formed
  • Reduces excess reactant

Increase product concentration:

  • Equilibrium shifts LEFT (reverse)
  • More reactants formed
  • Reduces excess product

Remove reactant:

  • Equilibrium shifts LEFT

Remove product:

  • Equilibrium shifts RIGHT

2. Temperature Change​

For exothermic reaction (ΔH < 0):

  • Heat is product: A + B ⇌ C + D + heat
  • Increase temperature: Equilibrium shifts LEFT (endothermic direction)
    • Rate increases but K decreases
  • Decrease temperature: Equilibrium shifts RIGHT
    • K increases

For endothermic reaction (ΔH > 0):

  • Heat is reactant: A + B + heat ⇌ C + D
  • Increase temperature: Equilibrium shifts RIGHT
  • Decrease temperature: Equilibrium shifts LEFT

Temperature only factor that changes K

3. Pressure Change​

For gas reactions only

Increase pressure:

  • Equilibrium shifts toward side with fewer moles of gas
  • Reduces pressure by favoring smaller volume

Example: N₂ + 3H₂ ⇌ 2NH₃

  • Left side: 4 moles gas
  • Right side: 2 moles gas
  • Increase pressure → shifts RIGHT

Decrease pressure:

  • Equilibrium shifts toward side with more moles of gas

4. Catalyst​

Catalyst effect:

  • Increases rate in BOTH directions equally
  • Does NOT shift equilibrium position
  • Reaches equilibrium faster
  • K unchanged

Effect Summary Table​

ChangeEquilibriumKRate
[Reactant]↑Right-↑
[Product]↑Left-↑
T↑ (exothermic)Left↓↑
T↑ (endothermic)Right↑↑
P↑ (gas, fewer right)Right-↑
Catalyst--↑ both

Industrial Applications​

Haber Process​

N2(g)+3H2(g)⇌2NH3(g),ΔH=−92 kJN_2(g) + 3H_2(g) ⇌ 2NH_3(g), \quad ΔH = -92 \text{ kJ}

Conditions used:

  • High pressure: Shifts right (4→2 moles)
  • Low temperature: Shifts right (exothermic)
  • Catalyst: Iron, speeds up reaction

Compromise:

  • 200 atm pressure (not super high due to cost)
  • 450°C temperature (not low due to rate)

Contact Process​

2SO2(g)+O2(g)⇌2SO3(g),ΔH=−198 kJ2SO_2(g) + O_2(g) ⇌ 2SO_3(g), \quad ΔH = -198 \text{ kJ}

Conditions:

  • High pressure: Shifts right (3→2 moles)
  • Moderate temperature: Balance rate and position
  • Catalyst: Vanadium(V) oxide

Key Points​

  1. Reversible reactions go both ways
  2. Dynamic equilibrium: rates equal at molecular level
  3. K indicates position of equilibrium
  4. Le Chatelier: System counteracts changes
  5. Concentration changes shift equilibrium
  6. Temperature changes shift equilibrium AND change K
  7. Pressure affects gas equilibria
  8. Catalysts increase rate but don't shift position

Practice Questions​

  1. Calculate equilibrium constant
  2. Predict equilibrium position changes
  3. Apply Le Chatelier principles
  4. Analyze industrial conditions
  5. Compare K values
  6. Predict effects of changes

Revision Tips​

  • Know equilibrium expression format
  • Understand dynamic equilibrium concept
  • Learn Le Chatelier principle
  • Know temperature effects on K
  • Pressure effects (moles of gas)
  • Catalyst doesn't shift equilibrium
  • Industrial applications
  • Practice equilibrium calculations