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4.4.5Additional lessons

Gibbs Free Energy and Equilibrium Constant (HL) — IB Chemistry HL Revision Notes & Topical Questions

5 min read 5 parts Lesson 88 of 172

📚 IB Chemistry HL Hub / Topical Revision
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Curriculum Equation Key قوانين الرقم الهيدروجيني والاتزان الأيوني - pH & Acid-Base Equilibrium FormulasFast Exam Solution
pH = -log[H⁺] | pOH = -log[OH⁻] | pH + pOH = 14 | [H⁺][OH⁻] = 10⁻¹⁴
Word Equation & Conditions: الرقم الهيدروجيني = -لوغاريتم تركيز أيون الهيدروجين | pH + pOH = 14
Location & Specifications: الاتزان الأيوني، كيمياء 3 ثانوي، التحصيلي، و AP Chemistry.
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Part 1 of 5The Link Between ΔG⦵ and K

Equation
  • Equation ΔG⦵ = −RT ln K, where R is 8.31 J K⁻¹ mol⁻¹ and T is the temperature in kelvin.
Large K
  • Large K If ΔG⦵ is strongly negative, K is much greater than 1 and products dominate at equilibrium.
Small K
  • Small K If ΔG⦵ is strongly positive, K is much less than 1 and reactants dominate at equilibrium.
ΔG⦵ = 0
  • ΔG⦵ = 0 At exactly zero, ln K is zero, so K equals 1 and neither side is strongly favoured.

Part 2 of 5Using the Equation

The most frequent slip is forgetting that R is in joules while ΔG⦵ is usually given in kilojoules.

1
Write the equation with the correct units, converting ΔG⦵ from kJ mol⁻¹ into J mol⁻¹ by multiplying by 1000.
2
Rearrange to ln K = −ΔG⦵ ÷ RT and substitute the temperature in kelvin.
3
Calculate ln K and then K using the exponential function, K = e^(ln K).
4
For the reverse process, change the sign of ΔG⦵, and K becomes 1 ÷ K.
5
Comment on whether the equilibrium mixture contains mainly products or mainly reactants.

Part 3 of 5Wider Connections

Wider Connections
Temperature
  • Temperature Because ΔG⦵ depends on T, the value of K also changes with temperature.
Exothermic reactions
  • Exothermic reactions Raising the temperature makes ΔG⦵ less negative and lowers K.
Reaction quotient
  • Reaction quotient Away from equilibrium, ΔG = ΔG⦵ + RT ln Q shows the direction of net change.
Not about speed
  • Not about speed K tells us the position of equilibrium, not how quickly it is reached.

Part 4 of 5Worked Example

Question

  • For N₂(g) + 3H₂(g) ⇌ 2NH₃(g), ΔG⦵ = −32.8 kJ mol⁻¹ at 298 K.
  • Calculate K (R = 8.31 J K⁻¹ mol⁻¹).

  1. Convert ΔG⦵ to joules−32 800 J mol⁻¹.
  2. ln K = −ΔG⦵ ÷ RT = 32 800 ÷ (8.31 × 298) = 13.25.
  3. K = e13.25 = 5.7 × 10⁵.

Answer K ≈ 5.7 × 10⁵ at 298 K (products strongly favoured)

Part 5 of 5Exam Checklist & Tips

4.4.5 – Key Points for the Exam

  • Equilibrium constant A ratio of product to reactant concentrations at equilibrium, each raised to its coefficient.
  • ln K The natural logarithm of the equilibrium constant.
  • Gas constant R 8.31 J K⁻¹ mol⁻¹, used in thermodynamic and kinetic equations.
  • Reaction quotient The same expression as K but evaluated at any moment, not only at equilibrium.

Common Mistakes

  • Using ΔG⦵ in kJ and R in J without converting the units.
  • Dropping the minus sign in the equation, so K appears to rise as ΔG⦵ becomes more positive.
  • Using log₁₀ instead of the natural logarithm.
  • Saying a large K proves the reaction is fast.

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3 & IGCSE —

Kc

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