MCAT content library / Chemistry & physics / Entropy and temperature: explain when a process is favored
Account for the system and surroundings, then explain how temperature changes the enthalpy–entropy balance under stated conditions.
About 14 minutes, including the worked examples and checks.
Useful foundations
On this page
- Concept: entropy accounting includes the surroundings
- Application: the same heat transfer has two entropy terms
- Pattern: temperature weights the entropy contribution
- Application: compare two temperature-dependent processes
- Rule: use the signs, then test the relevant range
- Thermodynamic favorability does not specify speed
- Check your reasoning
- Continue learning
- Sources
Concept: entropy accounting includes the surroundings
Entropy is a state function connected to the number of accessible microscopic arrangements consistent with a macroscopic state. For the same substance under comparable conditions, a gas usually has more accessible arrangements than a liquid, and a liquid more than a solid. Melting and vaporization therefore generally increase the substance’s entropy. A visual impression of “messiness” is not a sufficient molecular explanation.2
A process can decrease the entropy of a chosen system while still being thermodynamically favored. The second-law comparison uses ΔStotal = ΔSsystem + ΔSsurroundings. A spontaneous irreversible change increases this total. A reversible limiting process gives zero total entropy change. Energy is conserved across an isolated combined system; entropy can be generated by an irreversible process.3
Application: the same heat transfer has two entropy terms
In an original model, 600 J passes from a large reservoir held at 400 K to another held at 300 K, with no other relevant transfer. Each reservoir changes negligibly in temperature. Its entropy change can be evaluated along an equivalent reversible path at that fixed temperature: the hot reservoir changes by −600/400 = −1.50 J/K and the cold reservoir by +600/300 = +2.00 J/K. The combined change is +0.50 J/K.
The hot reservoir’s negative contribution does not violate the second law, because the cold reservoir gains more entropy than the hot one loses. Reversing this heat transfer without additional changes elsewhere would give a negative total and is not spontaneous. A refrigerator can move heat from cold to hot because its energy input and additional transfers must be included in the full accounting.3
Use ΔS = qrev/T for an isothermal reversible path, or an equivalent reversible path between the same states. It is not a rule to divide any measured heat by any convenient temperature. If temperature changes substantially, the reversible-path entropy contributions must be accumulated across the changing temperatures rather than replacing them with an unjustified single value.2
Pattern: temperature weights the entropy contribution
For the usual constant-temperature, constant-pressure chemical comparison with only pressure–volume work, ΔG = ΔH − TΔS connects the system’s Gibbs-energy change to the total entropy criterion. Under the corresponding reservoir assumptions, ΔG = −TΔStotal. A negative actual change favors that direction. T must be in kelvins, and ΔH and TΔS must use the same energy units.4
Standard-state values instead give ΔG° = ΔH° − TΔS°. They describe a reference reaction at a specified temperature. The sign of ΔG° does not by itself determine the favored direction in every possible mixture; composition also matters. Here the temperature plots describe the standard reaction quantity. Treating ΔH° and ΔS° as constants across a range is an approximation that must be stated, especially when extrapolation or a phase transition is involved.4
Application: compare two temperature-dependent processes
Consider the original hypothetical processes A and B below. Their values are held constant from 250 to 450 K for this exercise, and no phase transition occurs. Convert the entropy values to kJ/(mol·K) before calculating: +120 J/(mol·K) becomes +0.120 kJ/(mol·K), and −60 J/(mol·K) becomes −0.060 kJ/(mol·K).
| Process | ΔH° (kJ/mol) | ΔS° (J/mol·K) |
|---|---|---|
| A | +36 | +120 |
| B | −24 | −60 |
| Temperature (K) | A: ΔG° (kJ/mol) | B: ΔG° (kJ/mol) |
|---|---|---|
| 250 | +6 | -9 |
| 300 | +0 | -6 |
| 350 | -6 | -3 |
| 400 | -12 | +0 |
| 450 | -18 | +3 |
For A at 350 K, ΔG° = 36 − 350(0.120) = −6 kJ/mol. Its positive entropy change makes the −TΔS° contribution increasingly negative as temperature rises. The zero crossing occurs at 36/0.120 = 300 K. For B at 350 K, ΔG° = −24 − 350(−0.060) = −3 kJ/mol. Its negative entropy change makes increasing temperature less favorable for the standard forward reaction; it reaches zero at 400 K.
The two lines have slopes −ΔS°, so A slopes downward and B upward. At 350 K both standard values are negative, but their slopes have opposite signs. “Negative now” and “becomes more favorable when heated” are different conclusions. At a reaction’s ΔG° = 0 crossover, its equilibrium constant is one; this does not mean an arbitrarily prepared mixture is already at equilibrium.
Rule: use the signs, then test the relevant range
Within a range where the values can be treated as constant, negative ΔH° with positive ΔS° gives negative ΔG° for every positive T in that range. Positive ΔH° with negative ΔS° gives positive ΔG°. When both signs are positive, the standard forward reaction becomes favorable above ΔH°/ΔS°. When both are negative, it is favorable below that positive crossover. An extrapolated crossover outside the valid range is not established by the approximation.
A phase transition has its own enthalpy and entropy changes. At coexistence under the relevant pressure, the actual Gibbs-energy difference between the phases is zero. Heating within one phase and converting between phases are distinct processes, so a straight-line model cannot automatically continue unchanged through every transition.3
Thermodynamic favorability does not specify speed
A process with a negative actual Gibbs-energy change can remain slow on the observation timescale because of its kinetic barriers. The data above contain no activation energies or measured rates, so they cannot establish which process finishes first. Changing temperature can affect both thermodynamics and kinetics, but a ΔG°-versus-temperature slope is not an activation-energy measurement.5
Prioritize whether the problem asks about the system, total entropy, actual ΔG, or standard ΔG°. Paraphrase the signs and convert units. Prognose how multiplying ΔS by a larger T changes the balance. Pick the calculation within its stated assumptions. Post-mortem any unsupported claim about rate, equilibrium composition, or extrapolation. The AAMC places these entropy and free-energy relationships in content category 5E.1
ACTIVE PRACTICE · EXPERIMENT AND TRANSFER
Compare Temperature Dependence Without Confusing Standard and Actual ΔG
The source supplies two hypothetical standard reactions with constant ΔH° and ΔS° over 250–450 K, without phase changes. A: +36 kJ/mol, +120 J/(mol·K); B: −24 kJ/mol, −60 J/(mol·K). These are standard reaction quantities, not reaction rates.
ΔG° = ΔH° − TΔS° = .
Standard forward direction is favored at this temperature.
Standard crossover for this process:.
Negative ΔG° does not prove an arbitrary nonstandard mixture is already at equilibrium or say how fast its reaction occurs. Crossover ΔG°=0 corresponds to K=1, not automatic equilibrium in a sample with unspecified Q.
Check your reasoning
Choose an answer before opening its feedback. Each option explains the reasoning behind that choice.
Check 1
A hypothetical process has ΔH° = +21 kJ/mol and ΔS° = +80 J/(mol·K), treated as constant over the relevant range. What is ΔG° at 350 K?
A. −28 kJ/mol
Review this choice. The entropy contribution TΔS° is +28 kJ/mol. You must subtract it from the +21 kJ/mol enthalpy contribution, not omit the latter.
B. −7 kJ/mol
Correct. Convert 80 J/(mol·K) to 0.080 kJ/(mol·K), then calculate 21 − 350(0.080) = −7 kJ/mol.
C. +7 kJ/mol
Review this choice. This reverses the final sign. The 28 kJ/mol entropic contribution exceeds the +21 kJ/mol enthalpy term.
D. +49 kJ/mol
Review this choice. This adds TΔS° to ΔH°. The relation is ΔG° = ΔH° − TΔS°.
The principle to carry forward
Convert entropy units before comparing the two energy contributions.
Check 2
For a different process, ΔH° = −54 kJ/mol and ΔS° = −120 J/(mol·K). Both are approximately constant from 300 to 600 K. At what temperature in this range does ΔG° become zero?
A. 350 K
Review this choice. At 350 K, −54 + 0.120(350) = −12 kJ/mol, so the zero crossing has not been reached.
B. 400 K
Review this choice. At 400 K, the value remains −6 kJ/mol. The entropic penalty is still smaller than the favorable enthalpy magnitude.
C. 450 K
Correct. With ΔS° = −0.120 kJ/(mol·K), T = ΔH°/ΔS° = (−54)/(−0.120) = 450 K, within the supplied validity range.
D. 500 K
Review this choice. At 500 K, the value is +6 kJ/mol, so the crossover has already been passed.
The principle to carry forward
A positive crossover can arise from two negative values; verify the approximation covers it.
Check 3
At a specified temperature and composition, a reaction has negative actual reaction Gibbs energy. No kinetic measurements are supplied. Which additional observation would be needed to compare its speed with another reaction under the same conditions?
A. Time-resolved concentration changes for both reactions
Correct. Concentration versus time provides rate information. A thermodynamic driving force alone does not determine the speed.
B. The sign of each reaction’s standard formation enthalpy
Review this choice. Formation enthalpies describe thermodynamic reference changes, not the measured rate at the specified composition.
C. The number of negative terms in each entropy calculation
Review this choice. Counting signs does not determine either the numerical total entropy change or a kinetic rate.
D. Confirmation that both reactions conserve energy
Review this choice. Energy conservation is necessary physical bookkeeping but does not rank reaction speeds.
The principle to carry forward
A claim about speed requires kinetic evidence, not only thermodynamic signs.
Continue learning
Related lessons
- Free energy and equilibrium: distinguish standard from actual conditions
- Biological energy: use coupling, concentrations and carriers
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Sources
1. AAMC: Chemical and Physical Foundations, Content Category 5E
2. OpenStax Chemistry 2e, 16.2 Entropy
3. OpenStax Chemistry 2e, 16.3 The Second and Third Laws of Thermodynamics