5.4.8Physical Chemistry and Transition Elements
Gibbs Free Energy, ΔG
5 min read 5 parts Lesson 28 of 40
Section 1 of 5The Gibbs Equation
Gibbs free energy combines enthalpy and entropy into one test of feasibility.
Equation
- ΔG = ΔH − TΔS, with ΔH in J mol⁻¹, ΔS in J K⁻¹ mol⁻¹ and T in kelvin so that ΔG is in J mol⁻¹.
Link to ΔS(total)
- ΔG = −T × ΔS(total), so the condition ΔS(total) > 0 is the same as ΔG < 0.
Feasible
- A reaction is feasible when ΔG is negative or zero, and it is not feasible when ΔG is positive.
Units
- Convert ΔH to joules, or ΔS to kJ, so that both terms use consistent energy units.
Section 2 of 5Effect of Temperature
The signs of ΔH and ΔS decide how feasibility varies with temperature.
1
If ΔH is negative and ΔS is positive, then ΔG is negative at all temperatures and the reaction is always feasible.
2
If ΔH is positive and ΔS is negative, then ΔG is positive at all temperatures and the reaction is never feasible.
3
If both ΔH and ΔS are positive, the reaction becomes feasible above the temperature T = ΔH/ΔS.
4
If both are negative, the reaction is feasible below T = ΔH/ΔS, as in the Haber process.
5
A graph of ΔG against T is a straight line with gradient −ΔS and intercept ΔH.
Section 3 of 5Using ΔG in Practice
Using ΔG in Practice
Threshold temperature
- Set ΔG = 0, which gives T = ΔH/ΔS, the temperature where the reaction switches between feasible and not feasible.
Extraction of metals
- Carbon reduction of ore is feasible only above a certain temperature, as the ΔG–T lines cross.
Limitations
- ΔG tells you only about feasibility and not about rate, since activation energy is not included.
Non-standard conditions
- Values from tables apply to standard conditions, so real systems at other concentrations may behave differently.
| ΔH | ΔS | Feasible? |
|---|---|---|
| − | + | At all temperatures |
| + | − | At no temperature |
| + | + | Above ΔH/ΔS |
| − | − | Below ΔH/ΔS |
Section 4 of 5Worked Example
Question
- For N₂(g) + 3H₂(g) → 2NH₃(g), ΔH = −92.2 kJ mol⁻¹ and ΔS = −198.8 J K⁻¹ mol⁻¹.
- Calculate ΔG at 298 K and the temperature below which the reaction is feasible.
- ΔH = −92 200 J mol⁻¹.
- TΔS = 298 × (−198.8) = −59 242 J mol⁻¹.
- ΔG = ΔH − TΔS = −92 200 − (−59 242) = −32 958 J mol⁻¹ = −33.0 kJ mol⁻¹,
- so feasible at 298 K.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)
- so feasible at 298 K.
- At ΔG = 0, T = ΔH ÷ ΔS = −92 200 ÷ −198.8 = 464 K.
Answer
- ΔG = −33.0 kJ mol⁻¹ at 298 K;
- feasible below about 464 K.
Section 5 of 5Exam Checklist & Tips
5.4.8 – Key Points for the Exam
- Gibbs free energy change ΔG = ΔH − TΔS, whose sign shows whether a reaction is feasible.
- Feasibility Whether a reaction is thermodynamically possible, judged by ΔG ≤ 0.
- Threshold temperature The temperature T = ΔH/ΔS at which ΔG changes sign.
- Spontaneous A reaction that is feasible and requires no continual input of energy to proceed.
Common Mistakes
- Using ΔH in kJ mol⁻¹ and ΔS in J K⁻¹ mol⁻¹ in the same equation without conversion.
- Entering the temperature in °C instead of K.
- Believing that a negative ΔG guarantees a fast reaction.
- Assuming that endothermic reactions can never be feasible, when a positive ΔS can make ΔG negative at high temperature.