Part 4: Gibbs energy, equilibrium and kinetic limits

Combine enthalpy and entropy in ΔG, calculate feasibility temperatures and connect thermodynamic favourability to the size of an equilibrium constant.

One energy quantity combines both entropy contributions

At constant temperature and pressure, ΔG = ΔH − TΔSsystem. Since ΔStotal = ΔSsystem − ΔH/T, this is also ΔG = −TΔStotal. A negative ΔG corresponds to a positive total entropy change for the process at the stated conditions. ΔG = 0 corresponds to equilibrium.

Use one consistent energy unit. If ΔH is in kJ mol⁻¹, convert ΔS from J mol⁻¹ K⁻¹ to kJ mol⁻¹ K⁻¹ before multiplying by kelvin. For the previous ammonia illustration, ΔG = −92.0 − 298(−0.199) = −32.7 kJ mol⁻¹. This agrees with −298 × 0.1097 kJ mol⁻¹ K⁻¹.

The standard Gibbs change ΔG° describes a standard-state reaction. In school questions the degree sign is sometimes omitted, but the distinction matters when connecting it to K: ΔG° = −RT ln K. The actual reaction Gibbs change depends on composition and becomes zero at equilibrium even when ΔG° is negative.

Work out which side of a temperature threshold is favourable

Set ΔG = 0 to obtain T = ΔH/ΔSsystem when both are nonzero. This is a boundary, not automatically “the minimum temperature”. Determine the inequality by the signs or test a temperature on each side.

For ΔH = +48.0 kJ mol⁻¹ and ΔSsystem = +120 J mol⁻¹ K⁻¹, the threshold is 48.0/0.120 = 400 K. At 500 K, ΔG = 48.0 −500(0.120) = −12.0 kJ mol⁻¹, so the forward reaction is favourable above 400 K. At 400 K it is at the thermodynamic boundary.

For ΔH = −36.0 kJ mol⁻¹ and ΔSsystem = −90.0 J mol⁻¹ K⁻¹, the same ratio gives 400 K, but ΔG = −36.0 + 0.0900T. It is negative below 400 K. A memorised “above ΔH/ΔS” rule would give the wrong direction.

These predictions assume ΔH and ΔS vary little over the temperature range and no phase changes alter the data. An extrapolated threshold far outside that range may be a mathematical estimate rather than a reliable physical prediction.

Read a Gibbs-energy graph as a straight-line equation

For approximately constant ΔH and ΔS, a plot of ΔG against T follows y = mx + c. The intercept is ΔH and the gradient is −ΔSsystem. If ΔG is plotted in kJ mol⁻¹ and T in K, the slope has units kJ mol⁻¹ K⁻¹. Multiply by 1000 when reporting entropy in J mol⁻¹ K⁻¹.

A downward slope means positive ΔSsystem; the line crosses zero at a lower-bound feasibility temperature if its intercept is positive. An upward slope means negative ΔSsystem. Mark the region below the ΔG = 0 axis as favourable for the stated forward process.

At a phase transition in equilibrium, such as melting at the melting temperature, ΔG for the phase change is zero and ΔH = TΔS. A line should not be extrapolated unchanged through a phase transition because the relevant enthalpies and entropies change.

Use ΔG° = −RT ln K carefully

In ΔG° = −RT ln K, R = 8.314 J mol⁻¹ K⁻¹, T is in kelvin and ΔG° must be in J mol⁻¹. ln is the natural logarithm. The thermodynamic K inside a logarithm is dimensionless, using standard-state-normalised activities. In an exam calculation, follow the supplied constant convention; do not silently change pressure units in a dimensional Kp value.

If ΔG° < 0 then ln K > 0 and K > 1. More strongly negative ΔG° at fixed temperature gives a larger K and a greater tendency toward products. ΔG° = 0 gives K = 1; this means the equilibrium expression equals one, not that every product and reactant concentration is equal.

For an original example with ΔG° = −8.50 kJ mol⁻¹ at 310 K, ln K = −(−8500)/(8.314 × 310) = 3.2980. K = e³·²⁹⁸⁰ = 27.1. The positive ln K and K greater than one are consistent with a negative standard Gibbs change.

Conversely, for K = 0.0250 at 298 K, ΔG° = −8.314 × 298 × ln(0.0250) = +9140 J mol⁻¹ ≈ +9.14 kJ mol⁻¹. This standard-state reaction favours reactants, although some product can still form and actual direction depends on starting composition.

A very large equilibrium constant does not make the reverse reaction impossible or establish how quickly equilibrium is reached. It expresses the equilibrium ratio for a specific written reaction at that temperature. Reversing the reaction changes the sign of ΔG° and makes K reciprocal.

Feasible does not mean observably fast

A thermodynamically favourable reaction can be extremely slow if the activation energy is high. At the working temperature, only a small fraction of encounters then reach the transition state. For example, a fuel and oxygen can coexist before ignition even though oxidation is favourable.

A catalyst gives an alternative pathway with a lower activation barrier and can make a favourable reaction observable on a useful timescale. It does not change ΔH, ΔSsystem, ΔG° or K at fixed temperature. It accelerates the approach to the same equilibrium.

In evaluating a proposed process, answer two separate questions: is the direction thermodynamically favourable under these conditions, and is its rate useful? A positive standard Gibbs change is not fixed by a catalyst; conditions, composition or coupling to another process would have to change the thermodynamic situation.

Quick checks

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

Sources