Gibbs free-energy change and feasibilityAQA A-Level Chemistry: Revision notes
Section 1
The Gibbs free-energy equation
The balance between enthalpy and entropy determines whether a reaction is feasible. It is expressed by the Gibbs free-energy change:
ΔG = ΔH − TΔS
where ΔH is the enthalpy change, T is the temperature in kelvin and ΔS is the entropy change of the reaction. ΔH is normally in kJ mol⁻¹ but ΔS in J K⁻¹ mol⁻¹, so divide ΔS by 1000 before substituting, giving ΔG in kJ mol⁻¹. (You do not need to derive the equation.)
Using ΔS in J K⁻¹ mol⁻¹ with ΔH in kJ mol⁻¹, or putting the temperature in °C. Convert both first.
Section 2
Feasibility
For a reaction to be feasible, ΔG must be zero or negative: ΔG ≤ 0. A positive ΔG means the reaction is not feasible at that temperature.
Feasible does not mean fast. A reaction with a negative ΔG may be very slow if its activation energy is high, so it appears not to occur. ΔG says whether a reaction can happen, not how quickly.
Saying a reaction with negative ΔG 'will happen'. It is feasible, but kinetics may stop it occurring at a measurable rate.
Section 3
How ΔG varies with temperature
In ΔG = ΔH − TΔS, the sign of ΔH and of ΔS decides how temperature affects ΔG.
- ΔH negative, ΔS positive: ΔG is negative at all temperatures. Always feasible.
- ΔH positive, ΔS negative: ΔG is positive at all temperatures. Never feasible.
- ΔH negative, ΔS negative: feasible at low temperatures only, where the ΔH term dominates.
- ΔH positive, ΔS positive: feasible at high temperatures only, where the TΔS term dominates.
A graph of ΔG against T is a straight line with gradient −ΔS and intercept ΔH.
Section 4
Calculating ΔG
Worked example. At 298 K, N₂O₄(g) → 2NO₂(g) has ΔH = +57.0 kJ mol⁻¹ and ΔS = +176 J K⁻¹ mol⁻¹.
ΔS = 0.176 kJ K⁻¹ mol⁻¹ ΔG = 57.0 − (298 × 0.176) = 57.0 − 52.4 = +4.6 kJ mol⁻¹
ΔG is positive, so the reaction is not feasible at 298 K.
Write ΔS in kJ K⁻¹ mol⁻¹ on its own line before substituting, and give ΔG a sign and the unit kJ mol⁻¹.
Section 5
The temperature at which a reaction becomes feasible
At the point where a reaction just becomes feasible, ΔG = 0, so ΔH = TΔS and
T = ΔH / ΔS
Worked example. CaCO₃(s) → CaO(s) + CO₂(g): ΔH = +178 kJ mol⁻¹, ΔS = +161 J K⁻¹ mol⁻¹.
T = 178 / 0.161 = 1106 K (833 °C)
Both ΔH and ΔS are positive, so the decomposition is feasible above 1106 K. When ΔH and ΔS are both negative the same calculation gives the temperature below which the reaction is feasible.
Check your answer is a sensible temperature in kelvin. If you get a value like 0.4 K, you forgot to convert ΔS.
That's the notes covered.
Carry on to the next subtopic.
Exam questions on Gibbs free-energy change and feasibility
- Whether a reaction is feasible at a given temperature depends on the balance between its enthalpy change and its entropy change, expressed by the equation ΔG = ΔH − TΔS, where T is the temperature in kelvin. A student has values of ΔH in kJ mol⁻¹ and ΔS in J K⁻¹ mol⁻¹ for several reactions.A reaction has a negative ΔH and a positive ΔS. State the sign of ΔG at all temperatures and explain your answer.2 marks
- Dinitrogen tetroxide decomposes reversibly: N₂O₄(g) → 2NO₂(g). For this reaction ΔH = +57.0 kJ mol⁻¹ and ΔS = +176 J K⁻¹ mol⁻¹.Use the equation ΔG = ΔH − TΔS to explain why increasing the temperature makes the decomposition of N₂O₄ more feasible.2 marks
- In the Haber process, N₂(g) + 3H₂(g) → 2NH₃(g), the enthalpy change is ΔH = −92 kJ mol⁻¹ and the entropy change is ΔS = −199 J K⁻¹ mol⁻¹ for the reaction as written.Calculate ΔG for the Haber process at 298 K. State whether the reaction is feasible at this temperature.3 marks
Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).