Thermodynamic and Kinetic Control
Later contrast: explain a favorable but slow change using separate free-energy and activation-barrier arrows.
Before this connection: Entropy, free energy, and chemical change: standard free energy; Chemical equilibrium.
AP Chemistry Topic 9.4 · CED printed p. 170 / PDF p. 177. This teaching, practice and feedback were added for the AP edition; original author material remains separately traceable.
Before you start
- Activation barriers
- Reaction profiles
- Equilibrium
Free energy compares the starting and ending states; activation energy concerns the route between them. A product can lie lower in free energy while the path still requires a large uphill barrier. At the observation temperature, too few particles may have the needed energy and arrangement to cross that barrier, so a thermodynamically favored process can be imperceptibly slow. We call that kinetic control. Diamond’s persistence or a fuel mixture awaiting activation can illustrate this distinction conceptually, without implying an experiment. Do not equate “no visible change” with equilibrium: an unchanged reading can describe a kinetically trapped state with a nonzero driving force.
On an energy profile, distinguish the reactant-to-product free-energy difference from the reactant-to-transition-state barrier. A catalyst changes the pathway and can lower the barrier in both directions. At fixed temperature it leaves ΔG° and K unchanged because the same starting and ending states are being compared. Raising temperature may speed reaction, and can also change equilibrium, but those are separate effects. A good explanation identifies the time scale, the pathway barrier, and why the observations do not establish equal forward and reverse rates.
Let's work through it
- A model conversion has reactants at 0 kJ mol−1, products at −20 kJ mol−1, and an uncatalyzed transition configuration at +100 kJ mol−1.
- The forward free-energy difference is −20 kJ mol−1, but the forward barrier is 100 kJ mol−1. These are different distances on the profile.
- An alternative path with its highest configuration at +50 kJ mol−1 can accelerate the process without moving either endpoint.
- The product still lies 20 kJ mol−1 below the reactant. K is the same at the same T, even though equilibrium may now be reached during a measurable time interval.
Try it from memory
Pause & retrieve
Try it before you peek.
Write your answer and the reason from memory. Then compare with the worked answer. Your self-assessment is a study decision, not an automatically graded score.
Question 1
What does a catalyst change on an energy profile?
See every recall question with its answer
These references preserve each prompt and its answer for direct evidence links. Try the interactive questions above before opening the worked reasoning.
Retrieval 1
What does a catalyst change on an energy profile?
Worked answer and reason
The pathway barrier, not the reactant/product energy difference.
Endpoint properties are independent of the chosen route.
Retrieval 2
Is constant concentration enough to prove equilibrium?
Worked answer and reason
No. A large barrier can produce a nearly constant kinetically trapped mixture.
You must distinguish rate limitation from zero driving force.
Catch the tempting mistake
- Negative ΔG° removes any activation barrier.
- The favorable endpoint can still be separated from the reactant by a high barrier.
Put it to work
Give the chemical reason as well as the calculation. Worked solutions are on a separate page.
- Unit 9 problem 5 · ap application · multiple
Scientific calculator workspace
Use this workspace for arithmetic, logarithms and elementary functions. Keep intermediate precision and report final values with appropriate units and significant figures. Graphing and general equation solving need a suitable approved tool.
Scientific calculator
This calculator requires JavaScript. Use the physics formula sheets and show your working manually.
+ − * / ^ ! · pi · e · sqrt · sin, cos, tan · asin, acos, atan · atan2(y,x) · ln, log · abs · exp · hypot(x,y). Use * between factors. Keep units in your written working.
For the teacher
- Ask for separate arrows labeled barrier and state-to-state difference. Require either an actual drawing or a clearly specified profile.
Source objectives and scope
Source edition: ap-chemistry-ced-fall2024-local-b5dfe867. See this topic in the course outline →
- 9.4.A
- Explain, in terms of kinetics, why a thermodynamically favored reaction might not occur at a measurable rate.
- 9.4.A.1
- Many processes that are thermodynamically favored do not occur to any measurable extent, or they occur at extremely slow rates.
- 9.4.A.2
- Processes that are thermodynamically favored, but do not proceed at a measurable rate, are under “kinetic control.” High activation energy is a common reason for a process to be under kinetic control. The fact that a process does not proceed at a noticeable rate does not mean that the chemical system is at equilibrium. If a process is known to be thermodynamically favored, and yet does not occur at a measurable rate, it is reasonable to conclude that the process is under kinetic control.
Science references for this teaching
- AP Chemistry Course and Exam Description, effective Fall 2024 — Required topic-numbered learning objectives and essential knowledge, all continuations and AP exclusions; supplied PDF SHA256 b5dfe8677ef3d88c613865d2e2a3e8d6125d652e2b24c71ef1e8ce4e011094f0.
- OpenStax Chemistry 2e §16.4, Free Energy — Enthalpy–entropy competition and ΔG°/equilibrium connection.
Glossary and deeper reading · 4 refreshers
These editorial refreshers supplement the reading. Definitions and examples are added guidance; original source wording and recorded answers remain in their original context.
-
Gibbs free energy
Gibbs free energy combines enthalpy and entropy contributions to a system at specified temperature and pressure. For a reaction at constant temperature and pressure, a negative actual reaction free-energy change favors its forward direction, while standard and actual values must be distinguished. For example, a favorable reaction can still be slow if its pathway has a large activation barrier.
Read more: Gibbs Free Energy and Thermodynamic Favorability →
-
Activation energy
Activation energy is the energy barrier associated with a reaction pathway in an introductory energy-profile model. It concerns the pathway and is distinct from the reactant-to-product energy change. For example, a thermodynamically favored reaction can remain slow when its activation barrier is large.
-
Catalyst
A catalyst participates in a reaction pathway and is regenerated overall. It can increase the reaction rate by providing a different pathway without changing the reaction's equilibrium constant at fixed temperature. For example, a catalyst can help a mixture reach equilibrium sooner without changing the final equilibrium composition.
-
Chemical equilibrium
Chemical equilibrium is a dynamic state in which forward and reverse reaction rates are equal. Macroscopic composition remains constant in a closed system under fixed conditions, but reactant and product concentrations need not be equal. For example, a reversible reaction can reach equilibrium with much more reactant than product.