MCAT content library / Chemistry & physics / Free energy and equilibrium: distinguish standard from actual conditions
Determine a mixture’s favored direction from its composition, and recognize exactly what becomes zero at chemical equilibrium.
About 14 minutes, including the worked examples and checks.
Useful foundations
On this page
- Concept: a standard value and a mixture’s driving force are different quantities
- Pattern: build Q for the reaction as written
- Application: one temperature, four compositions
- Read the energy diagram: the minimum is not zero energy
- Rule: equilibrium is dynamic, and rate requires other information
- Check your reasoning
- Continue learning
- Sources
Concept: a standard value and a mixture’s driving force are different quantities
The standard reaction Gibbs energy, ΔrG°, compares specified standard states at a stated temperature. The actual reaction Gibbs energy, ΔrG, describes the local thermodynamic driving force for forward reaction at the mixture’s current composition. Many introductory problems shorten these symbols to ΔG° and ΔG. The subscript r here helps distinguish a reaction’s driving force from the total Gibbs energy G of the entire mixture.2
The connection is ΔrG = ΔrG° + RT ln Q. At equilibrium, ΔrG = 0 and Q = K, so ΔrG° = −RT ln K. Combining the equations gives ΔrG = RT ln(Q/K). For positive absolute temperature, Q < K gives a negative forward driving force; Q > K favors net reverse reaction; Q = K gives no net thermodynamic preference.2
Pattern: build Q for the reaction as written
Q uses the current activities of products divided by those of reactants, with stoichiometric coefficients as exponents. Thermodynamic activities are dimensionless. Introductory ideal approximations use concentration divided by its reference concentration or partial pressure divided by its reference pressure; the appropriate standard-state convention must stay consistent.3
For the 1:1 isomerization A ⇌ B in an ideal mixture with pure-component reference states, Q = aB/aA = xB/xA. In a different balanced reaction, the expression can have different exponents and a different relationship to composition. A large equilibrium constant is not a universal statement that every product concentration exceeds every reactant concentration.
Application: one temperature, four compositions
Consider an original idealized reaction between isomers A and B at fixed temperature and pressure. The total amount is 1.00 mol, K = 4.00, and the supplied rounded value of RT is 2.50 kJ/mol. The standard reaction Gibbs energy is −2.50 ln 4 = −3.47 kJ/mol. Each row below describes a separately prepared mixture at that same temperature.
| Moles B / moles A | Q | ΔrG (kJ/mol) | Favored net direction |
|---|---|---|---|
| 0.20 / 0.80 | 0.25 | 2.50 ln(0.25/4) = −6.93 | Forward |
| 0.50 / 0.50 | 1.00 | 2.50 ln(1/4) = −3.47 | Forward |
| 0.80 / 0.20 | 4.00 | 2.50 ln(4/4) = 0 | Neither net direction |
| 0.90 / 0.10 | 9.00 | 2.50 ln(9/4) = +2.03 | Reverse |
The 0.90/0.10 mixture has negative ΔrG° but positive actual ΔrG. It contains more B relative to A than its equilibrium ratio permits, so net conversion toward A lowers the mixture’s Gibbs energy. The standard value did not change; the composition contribution did. When Q = 1, the logarithmic correction is zero and the actual reaction value equals the standard one, even if individual activities are not all one.
At equilibrium, xB/xA = 4 and xA + xB = 1. Thus xB = 0.80 and xA = 0.20. A negative standard reaction Gibbs energy has produced a product-enriched equilibrium, not complete disappearance of reactant.
Read the energy diagram: the minimum is not zero energy
The figure plots total G for the one-mole mixture as A is converted to B. For this original ideal-mixture model, choose the standard chemical potential of A as zero and that of B as −RT ln 4. Combining the ideal mixing contribution with those reference values produces the curve. Its mathematical form is supplied to define the drawing; you do not need to memorize it to interpret the minimum.4
With x = moles B divided by 1.00 mol, the displayed one-mole model is G = (1.00 mol){x[−RT ln 4] + RT[(1 − x)ln(1 − x) + x ln x]}, using the zero limit for 0 ln 0. The minimum is approximately −4.02 kJ at x = 0.80. Pure A has G = 0 by the chosen reference, while pure B has G ≈ −3.47 kJ. Neither reference level says that the substances contain no energy.
The slope of G with respect to forward reaction extent at fixed temperature and pressure is ΔrG. Left of the minimum, forward conversion moves downhill; right of it, reverse conversion moves downhill. At the minimum, the slope is zero. The total G itself remains approximately −4.02 kJ in this reference system. A finite change between two distant compositions is a difference of total G values, not the initial ΔrG multiplied indiscriminately by the full amount converted.
Rule: equilibrium is dynamic, and rate requires other information
At chemical equilibrium, forward and reverse reactions continue at equal rates in the usual dynamic picture. The species’ concentrations remain constant over time, but they need not be equal to each other. No net concentration change therefore does not mean molecular exchange has stopped.5
A catalyst can help a system approach the same equilibrium faster without changing K at that temperature. It changes the route toward equilibrium rather than the relative Gibbs energies of the same initial and final states.7
The plotted composition curve is not a molecular activation-energy profile. Its minimum describes an equilibrium mixture; it does not show a transition-state barrier or how long the system takes to arrive there. A negative actual reaction Gibbs energy can coexist with a very slow rate.6
Prioritize whether the question asks for standard energy, current direction, equilibrium composition, or rate. Paraphrase Q as the current ratio and K as the equilibrium ratio for this equation. Prognose which direction moves Q toward K. Pick the sign consistent with RT ln(Q/K). Post-mortem claims that equilibrium means zero total energy, equal concentrations, or no exchange. This is the free-energy and equilibrium reasoning in AAMC category 5E.1
ACTIVE PRACTICE · THERMODYNAMIC REASONING
Distinguish the Reaction Driving Force From the Mixture’s Energy
The lesson’s idealized isomerization is A ⇌ B, with 1.00 total mol, K = 4.00, and rounded RT = 2.50 kJ/mol at fixed temperature. At the reference state, ΔrG° = −RT ln 4 ≈ −3.47 kJ/mol. All activities are modeled as ideal mole fractions.
Current ratio Q = xB/xA =; K = 4.00.
Actual ΔrG = kJ/mol. Total mixture G (selected reference) = kJ per 1.00-mol mixture.
The reaction has a negative forward driving force at this composition.
This is the original ideal, fixed-temperature model. The plotted total mixture G is not the reaction ΔG, an activation-energy barrier, or a reaction-time plot. Zero of total G is a chosen reference.
Check your reasoning
Choose an answer before opening its feedback. Each option explains the reasoning behind that choice.
Check 1
For a reaction as written at a positive absolute temperature, the dimensionless equilibrium constant is K = 0.050. What is the sign of ΔrG°?
A. Negative
Review this choice. For K below one, ln K is negative. The minus sign in −RT ln K makes the standard reaction value positive.
B. Zero
Review this choice. Zero standard reaction Gibbs energy corresponds to K = 1, not K = 0.050.
C. Positive
Correct. ln(0.050) is negative, and −RT ln K is therefore positive at positive absolute temperature.
D. Cannot be determined without the initial mixture composition
Review this choice. The initial composition is needed for the actual ΔrG through Q. The standard value’s sign follows from K and positive T.
The principle to carry forward
Use K for the standard reaction quantity; use Q as well for the actual mixture.
Check 2
At one temperature, an ideal reaction A ⇌ B has K = 5.0. A prepared mixture has Q = 10.0. Which conclusion follows before any composition change occurs?
A. Net forward reaction is favored because K exceeds one.
Review this choice. K greater than one makes the standard reaction value negative, but the actual mixture is more product-rich than equilibrium and favors net reverse reaction.
B. The mixture is at equilibrium because both ratios exceed one.
Review this choice. Equilibrium requires Q = K, not merely that both numbers lie on the same side of one.
C. The standard reaction Gibbs energy must have become positive.
Review this choice. At the same temperature the standard reaction value remains −RT ln 5. Composition changes the actual driving force, not that standard value.
D. Net reverse reaction is favored because Q exceeds K.
Correct. RT ln(Q/K) = RT ln 2 is positive for the forward reaction. Net reverse reaction lowers Q toward 5.
The principle to carry forward
A favorable standard forward reaction can have a reverse-favored prepared mixture.
Check 3
A reversible reaction reaches equilibrium at fixed temperature and pressure. Which quantity is zero at that equilibrium without requiring a special choice of energy reference?
A. The mixture’s total Gibbs energy G
Review this choice. The total G reaches an allowed minimum, whose numerical value depends on the reference. It need not be zero.
B. The actual reaction Gibbs energy ΔrG
Correct. At Q = K, ΔrG = RT ln(Q/K) = 0. This is the local reaction driving force, corresponding to zero slope at the Gibbs-energy minimum.
C. The concentrations of all reactants
Review this choice. An equilibrium commonly retains reactants. Zero net change does not imply that they have disappeared.
D. The forward and reverse reaction rates individually
Review this choice. The forward and reverse rates are equal in dynamic equilibrium; they are not generally both zero.
The principle to carry forward
Equilibrium sets the reaction driving force and net rate to zero, not the total energy or all molecular exchange.
Continue learning
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Sources
1. AAMC: Chemical and Physical Foundations, Content Category 5E
2. OpenStax Chemistry 2e, 16.4 Free Energy
3. OpenStax Chemistry 2e, 13.2 Equilibrium Constants
4. MIT OpenCourseWare, 3.012: ideal-solution thermodynamics problem-set solution
5. OpenStax Chemistry 2e, 13.1 Chemical Equilibria
6. OpenStax Chemistry 2e, 16.1 Spontaneity
7. OpenStax Chemistry 2e, 13.3 Shifting Equilibria: Le Châtelier’s Principle