Find when enthalpy and entropy favor change

Make the system and constraints explicit

At constant temperature and pressure, with only pressure-volume work, a negative reaction Gibbs energy indicates a thermodynamically favored forward direction. For standard-state reactants and products the relevant quantity is ΔrG°. At other compositions, the actual ΔrG can differ. Neither quantity is a speed measurement.

At one temperature, ΔG° = ΔH° − TΔS°. In the original example, 100 J/(mol K) is 0.100 kJ/(mol K). At 300 K, ΔG° = 40−30 = +10 kJ/mol. At 500 K it is 40−50 = −10. At 400 K the assigned terms cancel. Enter all three states and compare your prediction with the ledger.

Derive the four cases rather than memorize a slogan

If ΔH° is negative and ΔS° positive, both contributions make ΔG° negative at positive T. If ΔH° is positive and ΔS° negative, both make it positive. When both are positive, the entropy contribution can dominate at high temperature. When both are negative, the enthalpy benefit dominates at low temperature and the entropy penalty grows with temperature.

The crossover T=ΔH°/ΔS° only makes physical sense when that ratio is positive. Treating H and S as constants over a range is an approximation. Heat capacities, phase changes and changing chemical species can invalidate extrapolation. A table calculated from assigned constants is not a phase diagram measured across unlimited temperatures.

Connect to the surroundings

Under the stated thermal-reservoir conditions, ΔSsurroundings = −ΔHsystem/T, giving ΔStotal = ΔSsystem−ΔHsystem/T = −ΔG/T. Thus a system entropy decrease may be more than compensated by heat released to the surroundings. For the case ΔH°=−40 kJ/mol and ΔS°=−100 J/(mol K) at 300 K, ΔG°=−10 kJ/mol and total entropy change is about +33.33 J/(mol K).

Do not use this simple reservoir expression for a finite surroundings body whose temperature changes substantially without integrating or otherwise accounting for that change.

Formation-data transfer

Gibbs energy is also a state function. For assigned ΔfG° values A=−20 and B=−55 kJ/mol, A→B has ΔrG°=−35 kJ/mol. This alternative ledger must agree with a consistent H/S calculation at the same temperature. Standard formation Gibbs energies of elements in their reference states are zero by convention; that does not make their standard molar entropies zero.

Source scope and continuation

Original teaching explanation, calculations and practice. Framework alignment: AP Chemistry topic 9.3 in the College Board course framework, effective Fall 2024, checked 30 October 2026. These pages do not reproduce restricted examination questions or establish AP authorization. The investigation is a bounded computational model, not a conducted laboratory.

Return to the preceding Unit 9 lesson when its reasoning is unfamiliar.

Continue to Separate a favorable direction from an accessible rate.

Identify the reference state before assigning a standard value

A standard Gibbs reaction energy refers to specified reference states at the stated temperature. Here the gas reference pressure is 1 bar, pure condensed phases use their reference-state activities, and the ideal dilute concentration proxy uses c° = 1 mol/L. One bar is not exactly one atmosphere. A standard-state symbol does not fix the temperature to 298 K; a temperature must still be supplied. Actual solution activity is not generally identical to concentration divided by 1 mol/L. Revisit quotient reference conventions before mixing constants.

Classify: At a stated 350 K, a stipulated ideal gas at 0.20 bar has activity 0.20, not its standard reference activity. A present pure liquid has unit activity in the stated pure-phase approximation; a component of a liquid mixture need not. A 1 mol/L real solution cannot be assigned exact unit activity without an activity model. Explain why none of these classifications follows merely from the temperature being 350 K.

Test named opposing contributions without a handling experiment

Water freezing near its ordinary melting conditions has negative system enthalpy and negative system entropy: heat release favors freezing, while the entropy decrease opposes it. At the melting equilibrium the two contributions balance; lowering temperature can favor freezing and raising it can favor melting under the same phase/pressure conditions. This is not a claim that every supercooled sample freezes promptly.

Sodium nitrate dissolution is a contrasting endothermic example with a favorable entropy contribution under the reference conditions considered in the course framework. The positive enthalpy cost can be outweighed by −TΔS. The named example does not establish a concentration-independent free energy or unlimited solubility; actual composition and saturation still matter.

Original sign transfer: A fictional freezing-like packet has ΔH = −6.0 kJ/mol and ΔS = −0.020 kJ/(mol K), held constant only over a stipulated narrow comparison range. Predict ΔG at 290 and 310 K. A separate fictional dissolution-like packet has ΔH° = +12 kJ/mol and ΔS° = +0.050 kJ/(mol K) at 300 K. Which contribution wins? Do not label these assigned values measured water or sodium nitrate data.

Compare the opposing terms

The first packet gives −0.20 and +0.20 kJ/mol, respectively, with a crossover at 300 K. The second gives 12−15 = −3 kJ/mol in its standard-state comparison. It does not determine actual direction without Q. The examples illustrate opposite pairs of competing signs, not universal phase boundaries or a dissolution procedure.

Reference mechanisms: OpenStax free energy and current AP Chemistry 9.3.A.4, checked 31 October 2026.