Skip to main content
United StatesChemistrySyllabus dot point

Why can a thermodynamically favorable reaction still fail to proceed at a noticeable rate?

Topic 9.4 Thermodynamic and Kinetic Control: distinguish thermodynamic favourability (sign of delta G) from kinetic feasibility (rate), and explain why a favorable reaction may be slow.

A focused answer to AP Chemistry Topic 9.4, covering the distinction between thermodynamic favourability and kinetic feasibility, why a favorable reaction can be slow due to a high activation energy, and the role of catalysts, with full worked examples.

Generated by Claude Opus 4.89 min answer

Reviewed by: AI editorial process; not yet individually human-reviewed

Have a quick question? Jump to the Q&A page

Jump to a section
  1. What this topic is asking
  2. Two different questions
  3. Why a favorable reaction can be slow
  4. The role of a catalyst
  5. Try this

What this topic is asking

The College Board (Topic 9.4) wants you to distinguish thermodynamic favourability (the sign of ΔG\Delta G) from kinetic feasibility (the rate), and to explain why a favorable reaction may be slow. This resolves the apparent paradox of reactions that should happen but do not, by separating whether a reaction can occur from how fast it occurs.

Two different questions

A negative ΔG\Delta G says the products are lower in free energy than the reactants, so the reaction is downhill and can happen. But whether it actually happens at a useful rate is a separate question, answered by kinetics. A reaction can be favorable and fast, favorable and slow, or unfavorable entirely.

Why a favorable reaction can be slow

This is why many favorable reactions do not visibly occur at room temperature: a fuel and oxygen are thermodynamically poised to react, but the mixture is stable until a spark provides the activation energy. The energy profile of Unit 5 makes this concrete: ΔG\Delta G is the difference between the start and end levels, but the rate depends on the height of the barrier between them.

The role of a catalyst

A catalyst lowers the activation energy, providing a faster pathway, so it can make a kinetically hindered but favorable reaction proceed at a useful rate. Crucially, the catalyst does not change ΔG\Delta G or the favourability, because it does not alter the free energies of the reactants or products. It only changes how fast the system reaches equilibrium. Raising the temperature has a similar kinetic effect (more collisions clear the barrier), though temperature can also change ΔG\Delta G through the TΔST\Delta S term.

Try this

Q1. A reaction has ΔG<0\Delta G < 0 but does not proceed at room temperature. State whether the problem is thermodynamic or kinetic, and explain. [2 points]

  • Cue. Kinetic; the reaction is favorable (ΔG<0\Delta G < 0) but has a high activation energy, so the rate is negligible.

Q2. State whether a catalyst changes the favourability of a reaction, and explain. [2 points]

  • Cue. No; a catalyst lowers the activation energy and speeds the reaction but does not change ΔG\Delta G or the favourability.

Exam-style practice questions

Practice questions written in the style of College Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AP 2022 (style)4 marksSection II (long FRQ, part). The reaction C(diamond)→C(graphite)\text{C}(\text{diamond}) \rightarrow \text{C}(\text{graphite}) has ΔG∘<0\Delta G^\circ < 0, yet diamonds do not noticeably turn to graphite. (a) State whether the reaction is thermodynamically favorable, and justify. (b) Explain why diamond persists despite this. (c) Explain the role of activation energy in this behavior. (d) Describe how a catalyst would change the rate but not the favourability.
Show worked answer →

A 4-point conceptual FRQ on kinetic control.

(a) Favourability (1 point): ΔG∘<0\Delta G^\circ < 0, so the conversion of diamond to graphite is thermodynamically favorable (spontaneous in the thermodynamic sense).
(b) Persistence (1 point): the reaction is favorable but kinetically hindered; the activation energy is enormous, so the rate is immeasurably slow at room temperature and diamond persists.
(c) Activation energy (1 point): a high activation energy means almost no collisions have enough energy to rearrange the carbon lattice, so even a favorable reaction proceeds at a negligible rate.
(d) Catalyst (1 point): a catalyst would lower the activation energy and speed the reaction, but it would not change ΔG∘\Delta G^\circ or the favourability, only how fast equilibrium is approached.

Markers reward the favorable conclusion, the kinetic-hindrance reasoning, the activation-energy explanation, and the catalyst distinction.

AP 2021 (style)1 marksSection I (multiple choice). A reaction has ΔG<0\Delta G < 0 but proceeds extremely slowly. The most likely reason is (A) ΔG\Delta G is wrong (B) the reaction has a high activation energy (kinetic control) (C) the reaction is endothermic (D) the entropy decreases. Justify your choice.
Show worked answer →

A 1-point conceptual MCQ. The answer is (B).

A favorable reaction (ΔG<0\Delta G < 0) can still be slow if it has a high activation energy, so it is kinetically controlled; thermodynamics says it can happen, kinetics says how fast. The trap is (A): a negative ΔG\Delta G is consistent with a slow reaction; the two describe different things.

Related dot points

Sources & how we know this