SCIENCE · INVESTIGATE AND EXPLAIN

Separate a favorable direction from an accessible rate

Explain kinetic persistence without changing equilibrium thermodynamics.

Learning outcome: Explain kinetic persistence without changing equilibrium thermodynamics.

Starting recap

Activation energy concerns a pathway; reaction free energy concerns states and composition.

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Try

A stipulated reaction has ΔG=−20 kJ/mol but its barrier can be 20–100 kJ/mol.

Open the experiment

Predict

Can two routes have the same free-energy difference and very different rates?

Your prediction and reason

Experiment

At 300 K compare 50 and 80 kJ/mol barriers, then heat the 80 kJ/mol case to 400 K. Keep the assigned thermodynamic statement separate.

Separate a barrier from reaction free energy

Worked teaching model · synthetic values · Illustrative Arrhenius comparison with equal pre-exponential factors and a reference barrier 50 kJ/mol at 300 K. ΔG for the examined state is stipulated −20 kJ/mol. No absolute rate, mechanism or equilibrium composition is inferred.

Activation energy: 80 kJ/mol Temperature: 300 K

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Conditional calculations, not measured observations
Quantity	Value	Unit
ln(k/kref)	-12.02724	dimensionless
k/kref	5.97913e-6	dimensionless
Stipulated ΔG	-20	kJ/mol reaction

A favorable reaction can remain slow when its barrier is high.

The reference ratio uses equal Arrhenius prefactors, not measured rates. A catalyst changes a pathway/barrier, not ΔG°, K or the equilibrium composition at the same temperature.

Predict before changing a control. Save two contrasting captures and explain what is held fixed. Displayed digits are calculation precision, not experimental uncertainty.

A favorable reaction can remain slow when its barrier is high.

The reference ratio uses equal Arrhenius prefactors, not measured rates. A catalyst changes a pathway/barrier, not ΔG°, K or the equilibrium composition at the same temperature.

Observation to attach to a capture

Captures preserve recipes and replay their calculated results. They are synthetic model evidence, not physical measurements.

Understand

A downhill endpoint can be separated by an uphill route

A thermodynamically favored reaction may be imperceptibly slow. The free-energy comparison says which direction lowers the appropriate thermodynamic potential under specified conditions. It does not identify the mechanism, the activation barrier, a collision frequency or a time scale.

In the lab, the reaction Gibbs energy is stipulated to be −20 kJ/mol for the compared state. The kinetic model independently assigns an Arrhenius activation energy. Keeping this distinction explicit lets us test the incorrect inference that a negative ΔG guarantees an immediately observable transformation.

A rate ratio under a declared model

The quantitative Arrhenius comparison below is an optional model extension. The Unit 9 AP target is distinguishing thermodynamic favorability from kinetic accessibility; reproducing this exponential calculation is not a completion requirement.

For an elementary or effective process modeled by k=A exp(−Ea/RT), equal prefactors allow a conditional comparison. Relative to a reference with Ea=50 kJ/mol at 300 K, the displayed log ratio is −Ea/(RT)+50000/(R×300). It is dimensionless. At 300 K, raising the barrier from 50 to 80 kJ/mol gives a ratio exp(−30000/(R×300)), about 0.00000598.

This is not an absolute rate. Without the prefactor and an appropriate rate law, a ratio cannot tell you how many seconds a sample takes to react. A real multistep mechanism may also change its dominant pathway with temperature. The calculation is a controlled teaching comparison, not a universal kinetic predictor.

Diagnose a stationary-looking sample

Suppose repeated observations show no visible composition change. At least two explanations remain: the sample may be at equilibrium, or it may be kinetically trapped on the observation time scale. Knowing Q and K can distinguish a thermodynamic disequilibrium from an equilibrium state, but does not reveal a barrier by itself. A carefully selected perturbation and measured rate evidence are needed for a kinetic claim.

A catalyst offers another route with a different barrier. It speeds forward and reverse approach consistently with the same equilibrium constant at the same temperature. It does not supply missing free energy to make an unfavorable net reaction favorable, and it does not change the final equilibrium composition merely by being a catalyst.

Transfer

Two catalyst candidates have equal measured rates at one temperature. Can you conclude equal activation energies? No: different prefactors may compensate at that temperature. Propose a temperature series and state the assumptions needed for an Arrhenius analysis. In this lesson, keep the equal-prefactor assumption visible rather than silently applying it to a real comparison.

Source scope and continuation

Original teaching explanation, calculations and practice. Framework alignment: AP Chemistry topic 9.4 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 Use K and Q to decide the actual direction.

Apply

Public practice checks a specific claim. It does not establish independent mastery.

Show your calculation or representation, explain the changed case, and separate the stated model from measured evidence. Explain kinetic persistence without changing equilibrium thermodynamics.

Your explanation and limitations

Continue

Completion requires a saved prediction, at least one capture and a reflection. Presence checks do not grade your reasoning.

Skipping retains experiments and awards no mastery.

Previous: Find when enthalpy and entropy favor change Next: Use K and Q to decide the actual direction

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