5.3.7Additional lessons
The Equilibrium Constant and Gibbs Energy (HL) — IB Chemistry HL Revision Notes & Topical Questions
5 min read 5 parts Lesson 115 of 172
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.
Part 1 of 5The Relationship
The relationship connects thermodynamics with equilibrium.
Equation
- Equation The standard Gibbs energy change and the equilibrium constant are related by ΔG° = −RT ln K.
Constants
- Constants R is 8.31 J K⁻¹ mol⁻¹ and T is the temperature in kelvin;
- ΔG° is usually given in kJ mol⁻¹ so a factor of 1000 is needed.
Which K
- Which K K is the equilibrium constant at the same temperature, Kc for solutions or Kp for gases.
Meaning
- Meaning ΔG° describes the spontaneity of the reaction under standard conditions, while K describes where equilibrium lies.
Part 2 of 5Sign and Size
Sign and Size
ΔG° negative
- ΔG° negative Then ln K is positive, so K is greater than 1 and the equilibrium favours products.
ΔG° zero
- ΔG° zero Then ln K is zero, so K equals 1 and reactants and products are present in comparable amounts.
ΔG° positive
- ΔG° positive Then ln K is negative, so K is less than 1 and the equilibrium favours reactants.
Magnitude
- Magnitude A large negative ΔG° gives a huge K, so the reaction goes essentially to completion.
| ΔG° | K | Position |
|---|---|---|
| Negative | > 1 | Products favoured |
| Zero | = 1 | Balanced |
| Positive | < 1 | Reactants favoured |
Part 3 of 5Using the Equation
1
Convert ΔG° from kJ mol⁻¹ into J mol⁻¹ by multiplying by 1000.
2
Convert the temperature to kelvin, for example 25 °C = 298 K.
3
Rearrange to ln K = −ΔG° / RT and evaluate the right-hand side.
4
Calculate K by taking the exponential: K = e^(−ΔG°/RT).
5
If K is known, ΔG° is found from −RT ln K, and the answer carries the sign that shows the favoured direction.
Part 4 of 5Worked Example
Question
- The standard Gibbs energy change for a reaction at 298 K is −10.0 kJ mol⁻¹.
- Calculate K (R = 8.31 J K⁻¹ mol⁻¹).
- ΔG° = −10.0 kJ mol⁻¹ = −10 000 J mol⁻¹.
- ln K = −ΔG° / RT = 10 000 / (8.31 × 298) = 10 000 / 2476.4 = 4.04.
- K = e4.04 ≈ 57.
Answer K ≈ 57, so the equilibrium favours products
Part 5 of 5Exam Checklist & Tips
5.3.7 – Key Points for the Exam
- Gibbs energy change
- A measure of the spontaneity of a reaction;
- a negative value means a spontaneous change.
- A measure of the spontaneity of a reaction;
- ΔG° = −RT ln K The equation linking standard Gibbs energy change and the equilibrium constant.
- Spontaneous Able to occur without continuous energy input, which does not say anything about rate.
- Standard conditions 100 kPa pressure and 1 mol dm⁻³ concentration, usually at 298 K.
Common Mistakes
- Forgetting to convert ΔG° from kJ to J when using R in J K⁻¹ mol⁻¹.
- Using the temperature in °C instead of kelvin.
- Taking log₁₀ instead of the natural logarithm ln.
- Assuming that a spontaneous reaction must be fast.
Last updated:
3 & IGCSE —
Kc
- : Kc! (ΔH < 0) Kc. (ΔH > 0) Kc.
- : ( ).
- : Kc .