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23.4 Gibbs free energy change, ΔG

Syllabus
9701–2028–2029
Topic
23.4
Level
A2

Use the Gibbs equation with kelvin and consistent energy units

ΔG∘=ΔH∘−TΔS∘\Delta G^\circ=\Delta H^\circ-T\Delta S^\circ

Symbol Meaning Typical unit before substitution
ΔG° standard Gibbs free energy change kJ mol⁻¹
ΔH° standard enthalpy change kJ mol⁻¹
T absolute temperature K
ΔS° standard entropy change of the system J K⁻¹ mol⁻¹

Before substitution, convert ΔS° from J K⁻¹ mol⁻¹ to kJ K⁻¹ mol⁻¹ by dividing by 1000 when ΔH° is in kJ mol⁻¹. Then TΔS° has the same kJ mol⁻¹ unit as ΔH°.

The equation combines the enthalpy contribution with the temperature-weighted entropy contribution. Use the signed values exactly as given: a negative ΔS° makes the term −TΔS° positive.

T must be in kelvin, not °C. Do not combine kJ and J, discard a negative sign, or treat T as carrying a stoichiometric coefficient.

Calculate ΔG° from a complete, unit-safe data route

For CH₃OH(l) + HBr(g) → CH₃Br(g) + H₂O(l) at 298 K, ΔH° = −47 kJ mol⁻¹. The standard molar entropies are 240, 99.0, 246 and 70.0 J K⁻¹ mol⁻¹ respectively in reaction order.

ΔS∘=(246+70.0)−(240+99.0)=−23.0 J K−1 mol−1\Delta S^\circ=(246+70.0)-(240+99.0)=-23.0\ \mathrm{J\ K^{-1}\ mol^{-1}}

−23.0 J K−1 mol−1=−0.0230 kJ K−1 mol−1-23.0\ \mathrm{J\ K^{-1}\ mol^{-1}}=-0.0230\ \mathrm{kJ\ K^{-1}\ mol^{-1}}

ΔG∘=−47−[298×(−0.0230)]=−40.146≈−40.1 kJ mol−1\Delta G^\circ=-47-[298\times(-0.0230)]=-40.146\approx-40.1\ \mathrm{kJ\ mol^{-1}}

Calculate ΔS° with products minus reactants if it is not supplied, convert its energy unit, substitute signed quantities, and round only the final answer. The negative result is then interpreted separately as feasible under the stated standard conditions.

A negative entropy does not make TΔS° negative after the Gibbs subtraction is applied twice: −T(negative ΔS°) is positive. Keep brackets around the signed entropy term.

Use the sign of ΔG to state thermodynamic feasibility

Gibbs free energy change Conclusion for the forward process
ΔG < 0 thermodynamically feasible under the stated conditions
ΔG = 0 equilibrium
ΔG > 0 not thermodynamically feasible under the stated conditions

Feasibility is a thermodynamic conclusion, not a rate prediction. A feasible reaction may be imperceptibly slow if its activation energy is high; combustion can have negative ΔG yet still need ignition.

State the direction and the conditions attached to the value. A standard ΔG° conclusion applies to standard states at the specified temperature; changing temperature or composition can change the actual Gibbs free energy change.

Do not translate negative ΔG into 'instantaneous', 'safe', 'goes to completion' or 'high yield'. Those claims require kinetic or equilibrium evidence not supplied by the sign alone.

Predict how temperature changes feasibility from the signs of ΔH° and ΔS°

ΔH° ΔS° Temperature effect on ΔG° Feasibility in the constant-ΔH°/ΔS° model
− + both ΔH° and −TΔS° are negative feasible at all temperatures
+ − both ΔH° and −TΔS° are positive not feasible at any temperature
+ + raising T makes −TΔS° more negative feasible above the crossover temperature
− − raising T makes −TΔS° more positive feasible below the crossover temperature

ΔG∘=0⇒Tcrossover=ΔH∘ΔS∘with consistent units\Delta G^\circ=0\quad\Rightarrow\quad T_{crossover}=\frac{\Delta H^\circ}{\Delta S^\circ}\quad\text{with consistent units}

If ΔH° = +50 kJ mol⁻¹ and ΔS° = +150 J K⁻¹ mol⁻¹, convert ΔS° to +0.150 kJ K⁻¹ mol⁻¹. The crossover is 50/0.150 ≈ 333 K, so the reaction is feasible above 333 K and not feasible below it in this model.

Always explain the change through −TΔS°. Raising T favours a positive ΔS° because the negative entropy term grows in magnitude, but opposes a negative ΔS° because −T(negative ΔS°) becomes increasingly positive.

Do not say that higher temperature always increases feasibility. Inspect both signs, convert units before finding a crossover, and recognise that the four-case table assumes ΔH° and ΔS° do not change appreciably with temperature.