4.4.4Reactivity 1 — What drives chemical reactions?
Spontaneous Reactions (HL) — IB Chemistry HL Revision Notes & Topical Questions
5 min read 5 parts Lesson 8 of 12 Unit quiz
Curriculum Equation Key Gibbs Free Energy FormulaFast Exam Solution
ΔG = ΔH - TΔS | (ΔG < 0: تلقائي Spontaneous, ΔG > 0: غير تلقائي, ΔG = 0: اتزان)
Part 1 of 5Four Sign Combinations
Negative ΔH, positive ΔS
- ΔG is negative at all temperatures, so the reaction is always spontaneous.
Positive ΔH, negative ΔS
- ΔG is positive at all temperatures, so the reaction is never spontaneous.
Negative ΔH, negative ΔS
- Spontaneous only at low temperature, where the enthalpy term outweighs the entropy term.
Positive ΔH, positive ΔS
- Spontaneous only at high temperature, where the TΔS term outweighs the positive ΔH.
| ΔH | ΔS | Spontaneous when |
|---|---|---|
| − | + | All T |
| + | − | Never |
| − | − | Low T |
| + | + | High T |
Part 2 of 5Finding the Switching Temperature
1
At the boundary between spontaneous and non-spontaneous behaviour ΔG⦵ is zero.
2
Set ΔH⦵ − TΔS⦵ = 0 and rearrange to give T = ΔH⦵ ÷ ΔS⦵.
3
Convert ΔS⦵ into kJ K⁻¹ mol⁻¹ so that its units match ΔH⦵.
4
Calculate T in kelvin and convert to °C only if the question asks.
5
State that the reaction is spontaneous above this temperature if both ΔH and ΔS are positive, or below it if both are negative.
Part 3 of 5Chemical Examples
Chemical Examples
Combustion
- Strongly exothermic with positive or small ΔS, so spontaneous at normal temperatures.
Thermal decomposition
- Endothermic with large positive ΔS, so it needs heating, for example CaCO₃(s) → CaO(s) + CO₂(g).
Freezing water
- Negative ΔH and negative ΔS, so it is spontaneous only below 273 K.
Dissolving
- Some endothermic dissolving processes occur because the increase in entropy compensates for the energy needed.
Part 4 of 5Worked Example
Question
- CaCO₃(s) → CaO(s) + CO₂(g) has ΔH⦵ = +178 kJ mol⁻¹ and ΔS⦵ = +161 J K⁻¹ mol⁻¹.
- Calculate the minimum temperature at which decomposition is spontaneous.
- Convert ΔS⦵ = 161 J K⁻¹ mol⁻¹ = 0.161 kJ K⁻¹ mol⁻¹.
- At the switching point ΔG⦵ = 0, so T = ΔH⦵ ÷ ΔS⦵.
- T = 178 ÷ 0.161 = 1106 K (about 833 °C).
Answer
- T ≈ 1.11 × 10³ K;
- spontaneous above this temperature
Part 5 of 5Exam Checklist & Tips
4.4.4 – Key Points for the Exam
- Spontaneous process A process with negative ΔG that can occur without continuous energy input.
- Switching temperature The temperature at which ΔG changes sign, equal to ΔH ÷ ΔS.
- Exothermic A reaction with a negative ΔH.
- Entropy-driven A change that occurs mainly because of an increase in entropy.
Common Mistakes
- Assuming every exothermic reaction is spontaneous at every temperature.
- Forgetting to convert ΔS⦵ to kJ before calculating the switching temperature.
- Reporting the switching temperature in °C when the answer was found in kelvin without converting.
- Stating that a positive ΔG means the reaction cannot happen under any conditions.
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