Gibbs free energyEdexcel A-Level Chemistry: Revision notes
Section 1
Free energy: combining enthalpy and entropy
The Gibbs free energy change, ΔG, combines both factors that decide feasibility:
ΔG = ΔH − TΔS system
units kJ mol⁻¹ (convert ΔS from J to kJ by dividing by 1000, and use T in kelvin). It comes from ΔS total = ΔS system − ΔH/T multiplied by −T.
A reaction is feasible (thermodynamically) when ΔG ≤ 0.
ΔS is in J K⁻¹ mol⁻¹ but ΔH is in kJ mol⁻¹. Divide ΔS by 1000 (or multiply ΔH by 1000) before combining.
Section 2
Predicting feasibility and the temperature of change
Signs of ΔH and ΔS decide how temperature matters:
- ΔH negative, ΔS positive: feasible at all temperatures
- ΔH positive, ΔS negative: never feasible
- ΔH negative, ΔS negative: feasible at low T
- ΔH positive, ΔS positive: feasible at high T
The changeover occurs where ΔG = 0, so T = ΔH / ΔS. Worked example: MgCO₃ decomposition, ΔH = +117 kJ mol⁻¹, ΔS = +176 J K⁻¹ mol⁻¹: T = 117 000 ÷ 176 = 665 K.
Section 3
ΔG and the equilibrium constant
ΔG = −RT ln K (R = 8.31 J K⁻¹ mol⁻¹, T in K, ΔG in J mol⁻¹).
- ΔG negative: ln K positive, K large, products favoured
- ΔG positive: ln K negative, K small, reactants favoured
- ΔG = 0: K = 1
Example: ΔG = −32.7 kJ mol⁻¹ at 298 K: ln K = 32 700 ÷ (8.31 × 298) = 13.2, so K = 5.4 × 10⁵.
Section 4
Why a feasible reaction may not occur
A negative ΔG shows only that a reaction is thermodynamically feasible. It tells you nothing about the rate.
A reaction can be kinetically inhibited if its activation energy is very high, so at room temperature almost no particles have enough energy to react. Examples: diamond to graphite (ΔG = −2.9 kJ mol⁻¹), and hydrogen with oxygen at room temperature (until a spark is applied).
The rate may be increased by heating or by a catalyst. Calculated feasibility also assumes standard conditions and the assumption that ΔH and ΔS do not change with temperature.
Keep the words 'feasible' (thermodynamics, ΔG) and 'fast' (kinetics, activation energy) separate in answers.
That's the notes covered.
Carry on to the next subtopic.
Exam questions on Gibbs free energy
- The Haber process, N₂(g) + 3H₂(g) → 2NH₃(g), has ΔH = −92 kJ mol⁻¹ and ΔS system = −199 J K⁻¹ mol⁻¹. A student uses these values, assumed to be constant with temperature, to investigate how feasible the reaction is at different temperatures.The Haber process has a negative ΔG at 298 K but a mixture of nitrogen and hydrogen does not form ammonia at an observable rate at this temperature. Explain why.2 marks
- Magnesium carbonate decomposes on heating: MgCO₃(s) → MgO(s) + CO₂(g). The reaction has ΔH = +117 kJ mol⁻¹ and ΔS system = +176 J K⁻¹ mol⁻¹. Both values are assumed to be constant with temperature.Explain why the decomposition becomes feasible at high temperatures.2 marks
- A student relates the Gibbs free energy change of a reaction to its equilibrium constant, K, using ΔG = −RT ln K. She uses R = 8.31 J K⁻¹ mol⁻¹ and T = 298 K.A reaction has ΔG = −32.7 kJ mol⁻¹ at 298 K. Calculate the value of K for this reaction.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).