A negative free-energy change does not tell you how quickly a reaction occurs. A positive standard free-energy change does not tell you its direction at every composition. And two reactions with a favourable sum do not become coupled merely because someone writes them beside one another.
This lesson separates those three questions using original fictional reactions. The AAMC outline includes free energy, equilibrium, concentration effects, coupled reactions and enzyme catalysis. The calculations below are teaching examples, not official items or claims about a particular cell. AAMC content outline
Name the conditions behind the symbol
For an ordinary chemical reaction, the composition-dependent relation is ΔG = ΔG° + RT ln Q. Biochemical passages often specify a buffered pH and use a transformed standard value, ΔG°′. We will consistently write the corresponding current value as ΔG′:
ΔG′ = ΔG°′ + RT ln Q′.
The prime preserves the fixed-pH convention; the degree symbol identifies the standard reference. A standard value is not automatically the value at the current concentrations. The IUBMB recommendations make this distinction explicit. IUBMB biochemical thermodynamics, sections 2 and 7
For all calculations here, assume constant temperature and pressure, buffered pH 7, constant ionic conditions and consistent definitions of the reactants. Use the supplied rounded value RT = 2.50 kJ/mol. Activities are dimensionless; for these ideal dilute models, approximate each solute activity by its concentration divided by 1 mol/L. No membrane potential is included.
The free energies refer to one mole of reaction as written. Multiplying the entire reaction multiplies its free-energy change; reversing it reverses the sign.
Worked example: a positive standard value with a favourable direction
Consider the fictional reaction A ⇌ B, with ΔG°′ = +5.0 kJ/mol. Because the stoichiometry is one to one, Q′ = aB/aA, approximated here by [B]/[A]. Equal concentration units cancel.
| Mixture | [A] | [B] | Q′ = [B]/[A] | Current ΔG′ for A → B |
|---|---|---|---|---|
| 1 | 1.0 mM | 0.010 mM | 0.010 | −6.51 kJ/mol |
| 2 | 1.0 mM | 1.0 mM | 1.0 | +5.00 kJ/mol |
| 3 | 1.0 mM | 10 mM | 10 | +10.76 kJ/mol |
For mixture 1, use ln(0.010) ≈ −4.605:
ΔG′ = 5.0 + 2.50(−4.605) ≈ −6.51 kJ/mol.
A → B is thermodynamically favourable at that composition despite its positive standard value. The low product-to-reactant ratio contributes sufficiently negatively to overcome +5.0.
Mixture 2 gives ln(1) = 0. Mixture 3 gives ln(10) ≈ 2.3026, so its composition term reinforces the positive standard value. Under those conditions, the reverse direction is favourable.
These are comparisons at specified compositions. They do not say that a closed mixture will keep the same driving force as it reacts. As A becomes B, Q′ changes. Nor does “favourable” mean that all A must disappear.
Compare Q′ with equilibrium, not merely with one
At equilibrium, ΔG′ = 0 and Q′ = K′. Rearranging gives ΔG°′ = −RT ln K′. IUBMB recommendations, equations 7-7 and 7-9
Here, ln K′ = −5.0/2.50 = −2, so K′ ≈ 0.135. Mixture 1 has Q′ below that value and favours A → B. Mixtures 2 and 3 have Q′ above it and favour B → A.
This gives a useful counterexample to a shortcut: Q′ does not have to exceed one for the forward direction to become unfavourable. At Q′ = 0.50, ΔG′ = 5.0 + 2.50 ln(0.50) ≈ +3.27 kJ/mol. Product is still less concentrated than reactant, but its ratio already exceeds K′.
The equilibrium condition concerns the absence of net driving force. It does not imply zero molecules or no molecular interconversion.
A catalyst changes the route, not the endpoints
An enzyme can lower the activation barrier for a reaction without changing its free-energy difference or equilibrium constant under the same conditions. That is the distinction between the reaction's kinetic barrier and its thermodynamic driving force. OpenStax, Enzymes
Suppose mixture 1 contains no effective catalyst and changes imperceptibly during a short observation. That observation would not contradict its calculated negative ΔG′. The calculation did not specify a barrier, enzyme amount or rate law.
Conversely, adding a catalyst to mixture 3 does not turn its +10.76 kJ/mol forward change into a negative one. A catalyst may accelerate approach to equilibrium; it does not provide an independent energy source.
Be equally careful with metabolic regulation. Changing enzyme activity can change pathway flux and eventually metabolite concentrations. Those later concentration changes may alter Q′ and therefore ΔG′. That is an indirect change of conditions, not a catalyst rewriting the standard free energy of the same reaction.
Coupling requires a connected process
Now introduce a fictional driving conversion, X ⇌ Y, with ΔG°′ = −25.0 kJ/mol. Stipulate an enzyme-mediated route in which each A-to-B conversion is linked to one X-to-Y conversion through a shared intermediate I:
A + X ⇌ I
I ⇌ B + Y
The labels stand for chemically consistent reactant groups; the pathway and one-to-one linkage are supplied model conditions. Cancelling I gives the net reaction:
A + X ⇌ B + Y, with ΔG°′net = +5.0 − 25.0 = −20.0 kJ/mol.
Adding free energies is valid for the combined stoichiometry. Real coupling also requires a physical reaction pathway, such as the linked chemistry through activated intermediates discussed in Molecular Biology of the Cell. Alberts and colleagues, relevant indexed chapter passages
Use mixture 3's aB/aA = 10 and set aY/aX = 1. The net quotient is:
Q′net = (aB × aY)/(aA × aX) = 10 × 1 = 10.
Therefore ΔG′net = −20.0 + 2.50 ln(10) ≈ −14.24 kJ/mol. At the same composition, the uncoupled target change is +10.76 and the driving change is −25.0; their sum agrees.
The target conversion can therefore be part of a favourable combined process. This does not establish its speed or prove that every proposed intermediate step has suitable kinetics.
As a counterexample, place A and B in one compartment and X and Y in an isolated compartment, with no mechanism transmitting chemical or mechanical work between them. Their numerical free-energy sum remains calculable. It does not establish that X → Y drives A → B. The connection is a physical premise, not an arithmetic operation.
Transfer exercise: weaken the driving reaction
Keep the linked pathway, standard values, temperature and aB/aA = 10 unchanged. Now set aY/aX = 10,000. Use ln(10) ≈ 2.3026.
Which conclusion is supported?
A. The combined forward change remains −20.0 kJ/mol because the standard values have not changed.
B. The combined forward change is approximately +8.8 kJ/mol; the stipulated coupling no longer makes that direction favourable at this composition.
C. X → Y still has a negative current free-energy change, so it must be sufficient to drive A → B.
D. More enzyme must make the combined free-energy change negative.
Feedback: B. Q′net = 10 × 10,000 = 100,000. Thus ln Q′net = 5 ln(10) ≈ 11.513, and:
ΔG′net ≈ −20.0 + 2.50(11.513) = +8.78 kJ/mol.
A discards the current composition. C overlooks magnitude: the driving conversion now has ΔG′ ≈ −25.0 + 2.50 × 4 ln(10) = −1.97 kJ/mol, insufficient to offset +10.76. D confuses catalysis with changing thermodynamic conditions.
The reverse combined direction is favourable under the stipulated conditions, but neither its rate nor the behaviour of a complete cellular network is calculated here.
Use three questions on your next passage
Write the reaction and current quotient, identify what controls the rate, then trace the proposed coupling mechanism. Keep those answers separate before connecting them. A measured steady concentration is not, by itself, proof of equilibrium: ongoing production and consumption can balance while a pathway carries flux.
This method helps explain exactly which conclusion a calculation supports, without using a favourable free-energy sum as a substitute for a mechanism or rate measurement.
Sources and further reading
- AAMC: What Is on the MCAT Exam? Content Outline.
- IUBMB: Recommendations for Terminology and Databases for Biochemical Thermodynamics.
- OpenStax Biology 2e: Enzymes.
- Alberts et al., Molecular Biology of the Cell: Catalysis and the Use of Energy by Cells. Relevant indexed passages were accessible; direct chapter retrieval presented a browser challenge.