Part 4: Entropy, Gibbs energy and feasibility

Calculate entropy and Gibbs energy, interpret temperature thresholds and separate thermodynamic feasibility from reaction rate.

Count arrangements, then use data

Entropy measures how widely energy can be distributed among possible arrangements. “Disorder” is a useful introductory shorthand. For the same substance, gas generally has greater entropy than liquid, which has greater entropy than solid. Heating normally increases entropy; melting and boiling produce increases at the transitions.

Producing more moles of gas often increases reaction entropy. Compare gaseous coefficients, not just the total number of symbols in an equation. Equal gas amounts do not prove ΔS = 0: different molecules have different absolute entropies. Standard elemental entropies at 298 K are not zero; this differs from standard formation enthalpies.

Worked entropy calculation

For CaCO₃(s) → CaO(s) + CO₂(g), use rounded illustrative S° values 93, 40 and 214 J K⁻¹ mol⁻¹ respectively. ΔS° = 40 + 214 − 93 = +161 J K⁻¹ mol⁻¹. Gas formation is consistent with the positive sign.

Multiply every absolute entropy by its equation coefficient. Doubling the entire reaction doubles ΔH°, ΔS° and ΔG° together; it does not change the predicted threshold temperature.

Balance heat and entropy

At constant temperature and pressure, negative ΔG favours the forward change; zero is the equilibrium boundary. AQA uses zero or negative for the feasibility criterion. With standard data, ΔG° describes the standard-state comparison. Actual mixtures can behave differently as their composition changes.

Use kelvin and compatible units. If ΔH is in kJ mol⁻¹, divide ΔS in J K⁻¹ mol⁻¹ by 1000. A negative ΔG does not guarantee a fast reaction: a large activation barrier can make a thermodynamically favoured reaction imperceptibly slow.

Decide the direction of the inequality

Find the boundary with ΔG = 0, then test one temperature to establish the correct side. If ΔS is zero, ΔG equals ΔH in this approximation. A negative algebraic threshold is not a physically accessible negative kelvin temperature.

Extract ΔH and ΔS from a graph

Plot ΔG on the vertical axis and T in kelvin on the horizontal axis. In ΔG = −ΔS × T + ΔH, the gradient is −ΔS and the vertical intercept is ΔH. If the vertical axis is kJ mol⁻¹, the gradient’s unit is kJ K⁻¹ mol⁻¹.

For a line through (300 K, −24.0 kJ mol⁻¹) and (500 K, −12.0 kJ mol⁻¹), gradient = +0.0600 kJ K⁻¹ mol⁻¹. Thus ΔS = −60.0 J K⁻¹ mol⁻¹ and ΔH = −24.0 − 300 × 0.0600 = −42.0 kJ mol⁻¹. The line crosses ΔG = 0 at 700 K.

A change of phase can change the slope and invalidate a single straight-line extrapolation. For a reversible phase transition at its transition temperature, ΔG = 0 and ΔS = ΔH/T. A kettle estimate of vaporisation enthalpy uses measured energy and evaporated mass, but losses to the surroundings make raw electrical input an overestimate.

Quick checks

Original Finesse questions. Reveal the indicative worked solutions after attempting each question; these are not official AQA mark allocations.

Sources