1.8.6Physical Chemistry
Gibbs Free-Energy Change, ∆G, and Entropy Change, ∆S — Oxford AQA International A Level Chemistry Revision Notes & Topical Questions
5 min read 6 parts Lesson 45 of 63
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 6The Gibbs Equation
The Gibbs Equation
Equation
- ΔG = ΔH − TΔS combines the enthalpy change, absolute temperature and entropy change of a reaction into one value.
Feasible
- A reaction is thermodynamically feasible when ΔG is negative or zero, which means it can proceed spontaneously.
Units
- ΔG and ΔH are in kJ mol⁻¹, T is in kelvin, and ΔS is in J K⁻¹ mol⁻¹, so ΔS must be divided by 1000.
Kinetics
- Feasibility says nothing about rate, so a feasible reaction with high activation energy can still be extremely slow.
Part 2 of 6Sign Combinations
ΔH negative, ΔS positive
- ΔG is negative at every temperature, so the reaction is always feasible.
ΔH positive, ΔS negative
- ΔG is positive at every temperature, so the reaction is never feasible.
ΔH negative, ΔS negative
- Feasible only at low temperature, where the −TΔS term is small.
ΔH positive, ΔS positive
- Feasible only at high temperature, where the −TΔS term becomes large and negative.
| ΔH | ΔS | Feasible when |
|---|---|---|
| − | + | All temperatures |
| + | − | Never |
| − | − | Low temperature |
| + | + | High temperature |
Part 3 of 6Finding the Temperature
1
At the point where a reaction just becomes feasible, ΔG = 0, so ΔH = TΔS.
2
Rearrange to T = ΔH ÷ ΔS, which gives the minimum temperature for an endothermic reaction with positive ΔS.
3
Convert ΔS from J K⁻¹ mol⁻¹ to kJ K⁻¹ mol⁻¹ by dividing by 1000 before substituting.
4
Give the answer in kelvin and, if asked, subtract 273 to convert to °C.
5
Above this temperature ΔG is negative, so the reaction becomes feasible.
Part 4 of 6Why Entropy Can Drive Reactions
Why Entropy Can Drive Reactions
Endothermic reactions
- Positive ΔH is overcome when TΔS is larger, as in the thermal decomposition of carbonates at high temperature.
Total entropy
- ΔG is linked to the total entropy change of system plus surroundings, which must increase for a spontaneous change.
Temperature dependence
- Raising the temperature increases the importance of entropy relative to enthalpy in deciding feasibility.
Part 5 of 6Worked Example
Question
- CaCO₃(s) → CaO(s) + CO₂(g) has ΔH = +178 kJ mol⁻¹ and ΔS = +160 J K⁻¹ mol⁻¹.
- Find ΔG at 298 K and the minimum temperature for decomposition.
- Convert ΔS160 J K⁻¹ mol⁻¹ = 0.160 kJ K⁻¹ mol⁻¹.
- At 298 K: ΔG = 178 − (298 × 0.160) = 178 − 47.7 = +130.3 kJ mol⁻¹, so not feasible.
- Set ΔG = 0: T = ΔH/ΔS = 178 ÷ 0.160 = 1112.5 K.
Answer
- ΔG = +130 kJ mol⁻¹ at 298 K;
- feasible above about 1110 K (about 840 °C)
Part 6 of 6Exam Checklist & Tips
1.8.6 – Key Points for the Exam
- Gibbs free-energy change ΔG = ΔH − TΔS; negative or zero for a feasible reaction.What the symbols mean: ΔG = Gibbs free energy change (kJ mol−1; feasible when ≤ 0) · ΔH = enthalpy change (kJ mol−1) · T = temperature (K) · ΔS = entropy change (J K−1 mol−1, divide by 1000 to use kJ)
- Feasible Thermodynamically able to occur, regardless of how fast it is.
- Absolute temperature Temperature in kelvin, equal to °C + 273.
- Equilibrium temperature Temperature at which ΔG = 0, found from T = ΔH/ΔS.
Common Mistakes
- Using ΔS in J K⁻¹ mol⁻¹ with ΔH in kJ mol⁻¹ without conversion.
- Substituting a temperature in °C instead of K.
- Claiming a reaction with negative ΔG must be fast.
- Forgetting the negative sign in front of TΔS.
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