This section examines the relationship between thermodynamic favorability and kinetics, specifically the shortcomings of Gibbs Free Energy in predicting reaction behavior. Despite being spontaneous, many reactions do not occur at an observable rate because they are under kinetic control. In this guide, we will explore what kinetic control means and why it occurs.
Brief Review of Kinetics
Before getting into the thermodynamics of kinetic control, we need to briefly review topics from Unit 5. If you feel confident with kinetics, feel free to move past this section.
Recall that kinetics is the study of the rate of a reaction—essentially, how fast a reaction occurs. We measure the rate of a reaction by measuring the change in the concentration of reactants over time: R = Δ[A]/Δt or R = -d[A]/dt (for those familiar with calculus). Higher R values indicate a quicker loss of reactants and formation of products.
Rates can also be described using rate laws, where initial reactant concentration is directly tied to the rate of a reaction as follows:
R = k[A]^n[B]^m
where [A]^n, [B]^m, etc., are various reactants raised to their reaction order (how much of an impact their concentration has on the rate).
Take a look at the following rate law as an example:
rate = k[A]
These laws help us describe how quickly a reaction will occur, with a higher rate implying a faster reaction overall.
Most important to our study of kinetic control is understanding activation energy. Due to the kinetic molecular theory, chemical reactions occur when molecules hit each other at the right angle and speed/energy. This activation energy is the energy required for a chemical reaction to actually occur. The higher the activation energy, the harder it is for the reaction to occur at an observable rate.
The following diagram visually shows the concept of activation energy:
A Shortcoming of Gibbs Free Energy
A common misconception when looking at thermodynamic favorability is that a thermodynamically favorable reaction occurs quickly. Many reactions that are spontaneous occur incredibly slowly. A good example of this concept that can be applied to the real world is the conversion from diamonds into graphite (represented as Cdiamond(s) → Cgraphite(s)).
For this reaction, ΔG° = -3 kJ. This tells us that the reaction for graphite formation occurs spontaneously and does not require the input of any external energy to occur. However, take a look at the nearest diamond (because people have those around the house…right?). Is it suddenly morphing into graphite? No! (I hope not. Fiveable is not responsible for diamond graphitification…).
The conversion of diamond into graphite is incredibly slow, as in thousands of years slow. We say that this reaction is in kinetic control because it is driven by the slowness of the reaction. These types of reactions are often slow because they have a high activation energy.
A spontaneous process may take either the thermodynamically controlled or the kinetic controlled pathway. A kinetically controlled path like the one above is driven by a high activation energy. A thermodynamically controlled reaction is driven by the difference in free energy between the products and reactants, the type of reaction we saw in the last section.
Reasons For Kinetic Control
As we mentioned before, the primary reason for a reaction to be under kinetic control is because of a high activation energy. Because of this, even if the reaction is thermodynamically favorable, it may not continue at a measurable rate. There is a way around this, however, and that is through the use of a catalyst! Catalysts change the mechanism behind a reaction in order to decrease the activation energy and make reactions quicker. By reducing the activation energy, the reaction can proceed at a measurable rate.
For example, the decomposition of hydrogen peroxide usually occurs at an unmeasurable rate. However, when iodide ions are used as a catalyst, it creates the famous "Elephant’s Toothpaste" reaction, which you may be familiar with.
This example shows us how catalysts can transform a reaction from one at an immeasurable rate to one at a measurable rate. The concept of catalysis can be applied to kinetic control by noticing that catalysts can help get a reaction out of kinetic control by lowering the activation energy. For example, if there was a catalyst for the reaction: Cdiamond(s) → Cgraphite(s), the reaction would be able to proceed at a measurable rate. However, without one, it cannot because the rate of the reaction is incredibly slow.
Vocabulary
The following words are mentioned explicitly in the College Board Course and Exam Description for this topic.
| Term | Definition |
|---|---|
| activation energy | The minimum energy required for reactants to overcome the energy barrier and proceed to products in a chemical reaction. |
| equilibrium | The state in which the forward and reverse reaction rates are equal, resulting in constant concentrations or partial pressures of reactants and products. |
| kinetic control | A situation where a thermodynamically favored process does not proceed at a noticeable rate due to a high activation energy barrier. |
| kinetics | The study of the rate at which a chemical reaction occurs and the factors that influence this rate. |
| measurable rate | A reaction rate that is fast enough to be observed and quantified within a reasonable time frame. |
| thermodynamically favored | A reaction or process that has a negative Gibbs free energy (ΔG < 0) and is spontaneous under given conditions. |
Frequently Asked Questions
What's the difference between thermodynamic control and kinetic control?
Thermodynamic control vs kinetic control—short version: thermodynamic control picks the product with the lowest Gibbs free energy (most stable, ΔG most negative) if the system can reach equilibrium; kinetic control picks the product that forms fastest (lowest activation energy, Ea) even if it’s less stable. On a reaction-coordinate diagram thermodynamic product is the deeper well (lower ΔG), while the kinetically controlled product has the smaller energy barrier (smaller Ea) and forms faster. A thermodynamically favored reaction might not happen at a measurable rate if Ea is large (kinetic control); adding a catalyst lowers Ea and can let the thermodynamic pathway proceed. Also expect metastable states under kinetic control and use Arrhenius/rate constant k to connect Ea to rate. For AP exam items, explain using ΔG, Ea, transition state, and possibly Hammond postulate. For a topic review check the Fiveable study guide (https://library.fiveable.me/ap-chemistry/unit-9/thermodynamic-kinetic-control/study-guide/hRZ0V3goVueXCw1JeUdA) and practice problems (https://library.fiveable.me/practice/ap-chemistry).
Why doesn't a thermodynamically favored reaction always happen right away?
Thermodynamically favored just means ΔG is negative—the products are more stable overall—but that doesn’t tell you how fast the system gets there. A high activation energy (big energy barrier on a reaction-coordinate diagram) can make the rate extremely slow: very few molecules have enough energy to reach the transition state, so the reaction is under kinetic control (it’s “stuck” in a metastable state). Arrhenius and the rate constant k show temperature and catalysts matter: raising T or adding a catalyst lowers the effective barrier (or increases the fraction of molecules that can cross it), so the reaction speeds up. For AP-style explanations, name activation energy, transition state, kinetic control vs thermodynamic control, and mention catalysis (CED 9.4.A). For a quick review, see the Topic 9.4 study guide (https://library.fiveable.me/ap-chemistry/unit-9/thermodynamic-kinetic-control/study-guide/hRZ0V3goVueXCw1JeUdA) and try practice problems (https://library.fiveable.me/practice/ap-chemistry).
Can someone explain activation energy in simple terms and how it relates to kinetic control?
Activation energy (Ea) is the energy “hill” reactants must climb to become products—it’s the extra energy needed to reach the transition state on a reaction-coordinate diagram. Think: even if a reaction is downhill overall (ΔG is negative, thermodynamically favored), it still might sit behind a tall hill. If that hill (Ea) is large, very few molecules have enough energy to get over it, so the reaction is extremely slow—that’s kinetic control (CED 9.4.A.1–9.4.A.2). Under kinetic control you don’t have equilibrium happening on lab timescales; the system is stuck in a metastable state. The Arrhenius equation shows k depends exponentially on −Ea/RT, so lowering Ea (with a catalyst) or raising T increases the rate and can let the thermodynamically favored path occur. For AP review on this topic, see Fiveable’s study guide (https://library.fiveable.me/ap-chemistry/unit-9/thermodynamic-kinetic-control/study-guide/hRZ0V3goVueXCw1JeUdA) and more practice problems (https://library.fiveable.me/practice/ap-chemistry).
I'm confused about how a reaction can be favorable but still not occur - doesn't that contradict itself?
No contradiction—“favorable” (ΔG < 0) means the products are lower in free energy, so the reaction is thermodynamically allowed. It doesn’t say how fast the reaction happens. Kinetics controls the rate: a high activation energy (Ea) creates an energy barrier and makes the rate constant k very small (k = A·e^(−Ea/RT)), so the reaction can be essentially unobservable even though ΔG is negative. On a reaction-coordinate diagram you’d see products lower than reactants but a tall peak (transition state) between them—that’s a metastable system under kinetic control. Catalysts lower Ea (and the transition-state energy) without changing ΔG, letting the reaction proceed faster. AP exam questions on Topic 9.4 often ask you to explain exactly this (CED 9.4.A: relate thermodynamic favorability to slow or non-observable rates via activation energy). For a short study refresher, see the Topic 9.4 study guide (https://library.fiveable.me/ap-chemistry/unit-9/thermodynamic-kinetic-control/study-guide/hRZ0V3goVueXCw1JeUdA) and try practice problems (https://library.fiveable.me/practice/ap-chemistry).
What does it mean when we say a process is "under kinetic control"?
Saying a process is "under kinetic control" means the reaction is thermodynamically favored (ΔG < 0) but doesn’t proceed at a measurable rate because of a large energy barrier—a high activation energy. In other words, products may be more stable, but the pathway to get there is slow (small k from the Arrhenius equation, k = A e^(−Ea/RT)). A system under kinetic control can be trapped in a metastable state (not at equilibrium) until enough energy or a catalyst lowers Ea (or temperature/time changes) so the reaction can proceed. On the AP exam you might be asked to explain this idea using activation energy, reaction coordinate/transition state, and how catalysis or temperature affect rates (CED 9.4.A). For a focused review, see the Topic 9.4 study guide (https://library.fiveable.me/ap-chemistry/unit-9/thermodynamic-kinetic-control/study-guide/hRZ0V3goVueXCw1JeUdA) and try related practice problems (https://library.fiveable.me/practice/ap-chemistry).
How do I know if a reaction is thermodynamically favored but kinetically slow?
You know a reaction is thermodynamically favored but kinetically slow when ΔG° (or ΔG) is negative—so products are more stable—but the reaction proceeds negligibly because the activation energy (Ea) is large. Practically: thermodynamic check = calculate/compare ΔG (or use ΔH and ΔS with ΔG = ΔH − TΔS); kinetic check = look at rates, reaction coordinate diagram, or measure rate constant k (Arrhenius: k = A e^(−Ea/RT)). If ΔG < 0 but k is tiny (or the reaction time is extremely long), the process is under kinetic control (CED 9.4.A: high Ea and metastable states). A catalyst lowers Ea and can make the thermodynamically favored path observable without changing ΔG. For AP-style reasoning, draw a reaction-coordinate diagram showing deep product well (ΔG < 0) but a very high barrier (large Ea) and cite Arrhenius behavior. For a focused review, see the Topic 9.4 study guide (https://library.fiveable.me/ap-chemistry/unit-9/thermodynamic-kinetic-control/study-guide/hRZ0V3goVueXCw1JeUdA); for more unit review and practice, see (https://library.fiveable.me/ap-chemistry/unit-9) and practice problems (https://library.fiveable.me/practice/ap-chemistry).
Why do some reactions need a catalyst even if they're already thermodynamically favorable?
Thermodynamically favorable (ΔG < 0) just means the products are more stable than the reactants—it doesn’t tell you how fast the reaction goes. If the activation energy (Ea) is large, the reaction sits behind a big energy barrier and proceeds extremely slowly (kinetic control / metastable state). A catalyst gives an alternate reaction pathway with a lower Ea and a different (lower-energy) transition state so more molecules can get over the barrier (Arrhenius: k increases when Ea decreases). Catalysts don’t change ΔH, ΔS, or ΔG for the overall reaction—they only change the rate. On the AP CED this is LO 9.4.A (activation energy, transition state, reaction-coordinate diagram); you might see questions asking you to explain why a favored process is slow or how a catalyst affects the energy diagram. For a quick review, check the Topic 9.4 study guide (https://library.fiveable.me/ap-chemistry/unit-9/thermodynamic-kinetic-control/study-guide/hRZ0V3goVueXCw1JeUdA) and practice problems (https://library.fiveable.me/practice/ap-chemistry).
What's the relationship between activation energy and reaction rate?
Activation energy (Ea) is the energy barrier reactants must overcome to reach the transition state; the higher Ea is, the fewer molecules have enough energy to react at a given temperature, so the reaction rate is slower. In AP terms: Ea appears on a reaction-coordinate diagram between reactants and the activated complex (transition state). The Arrhenius relation, k = A e^(−Ea/RT), shows that as Ea increases, the rate constant k decreases exponentially (at fixed T), so the reaction proceeds more slowly. That’s why many thermodynamically favored (ΔG < 0) reactions don’t happen measurably—they’re under kinetic control because Ea is large (a metastable state). Catalysts lower Ea (without changing ΔG), raising k and speeding the reaction. For more examples and diagrams tied to Topic 9.4, see the Fiveable study guide (https://library.fiveable.me/ap-chemistry/unit-9/thermodynamic-kinetic-control/study-guide/hRZ0V3goVueXCw1JeUdA) and try practice problems (https://library.fiveable.me/practice/ap-chemistry).
I don't understand how equilibrium is different from kinetic control - can someone help?
Equilibrium (thermodynamic control) is about where a system wants to end up: the lowest ΔG (most stable products) when forward and reverse rates balance. Kinetic control is about how fast you get there. A reaction can be thermodynamically favored (ΔG < 0) but still not happen at a measurable rate if the activation energy is huge—it’s stuck behind an energy barrier and effectively “metastable.” On a reaction-coordinate diagram, thermodynamic control picks the deepest well (lowest product energy); kinetic control picks the product that’s easiest to form (lowest activation barrier) even if it’s higher in energy. Arrhenius and rate constants explain the temperature dependence: raising T or adding a catalyst lowers the effective kinetic barrier and can switch control from kinetic to thermodynamic. This distinction is exactly what AP CED 9.4.A emphasizes (activation energy, transition state, catalysis). For a short study guide and practice problems, check the Topic 9.4 study guide (https://library.fiveable.me/ap-chemistry/unit-9/thermodynamic-kinetic-control/study-guide/hRZ0V3goVueXCw1JeUdA) and the Unit 9 page (https://library.fiveable.me/ap-chemistry/unit-9). Practice more with problems at (https://library.fiveable.me/practice/ap-chemistry).
How does temperature affect whether a reaction is under thermodynamic or kinetic control?
Temperature changes the balance between kinetics (rate) and thermodynamics (stability). At low temperature a reaction is more likely under kinetic control: the product that forms fastest (lowest activation energy, Ea) dominates because molecules don’t have enough energy to cross larger barriers. At higher temperature you give molecules enough energy to surmount bigger Ea’s, so the system can reach equilibrium and the thermodynamically favored product (lowest ΔG) predominates. Use reaction-coordinate diagrams to see this: small Ea → fast (kinetic product); deeper product well (more negative ΔG) → thermodynamic product. Arrhenius shows k increases with T (k = A e^(−Ea/RT)), so raising T speeds pathways with higher Ea more, shifting outcomes. Catalysts lower Ea and can convert a kinetically trapped (metastable) system to the thermodynamic product without changing ΔG. For AP Chem, this ties to EO 9.4.A and keywords like activation energy, reaction coordinate, ΔG, and Arrhenius (see the Topic 9.4 study guide on Fiveable: https://library.fiveable.me/ap-chemistry/unit-9/thermodynamic-kinetic-control/study-guide/hRZ0V3goVueXCw1JeUdA). For practice, try problems at https://library.fiveable.me/practice/ap-chemistry.
What are some real world examples of reactions that are thermodynamically favored but kinetically slow?
Common real-world examples: - Diamond → graphite: thermodynamically favored (graphite is lower G) but the solid-state rearrangement has a huge activation energy, so diamonds persist (metastable). - Iron rusting (Fe → Fe2O3): overall favorable, but the initial oxidation is slow without moisture/oxygen and often proceeds under kinetic control; catalysts and salt speed it up. - Ester hydrolysis in neutral water: formation of carboxylic acid + alcohol is thermodynamically allowed, but uncatalyzed hydrolysis is very slow—acid or base catalysis lowers the activation energy. - Formation of many organic reactions (e.g., rearrangements or bond-breaking in stable molecules): favorable ΔG but blocked by large energy barriers unless heated or catalyzed. Explain these on the exam by invoking activation energy, reaction coordinate diagrams, and kinetic vs thermodynamic control (CED 9.4.A, keywords: activation energy, transition state, metastable). For review, see the Topic 9.4 study guide (https://library.fiveable.me/ap-chemistry/unit-9/thermodynamic-kinetic-control/study-guide/hRZ0V3goVueXCw1JeUdA) and practice problems (https://library.fiveable.me/practice/ap-chemistry).
Why do diamonds exist if graphite is more thermodynamically stable?
Because graphite has lower Gibbs free energy, it's the thermodynamically favored form of carbon—but diamond is a kinetic (metastable) form. The C–C bonds in diamond are in a rigid 3-D network, so converting diamond to graphite requires breaking many strong bonds and passing over a very large activation energy (energy barrier). At room temperature that barrier makes the rate essentially zero (process under kinetic control), so diamonds persist for geologic timescales even though ΔG for diamond → graphite is negative. Raise the temperature or add a catalyst (or extreme pressure changes) and the rate increases (Arrhenius behavior), allowing the conversion to happen faster. For AP Chem, this is a classic 9.4.A example: thermodynamically favored ≠ fast—high activation energy and a metastable state explain why both forms exist (see Topic 9.4 study guide for more: https://library.fiveable.me/ap-chemistry/unit-9/thermodynamic-kinetic-control/study-guide/hRZ0V3goVueXCw1JeUdA). For extra practice, try problems at https://library.fiveable.me/practice/ap-chemistry.
How do enzymes help overcome kinetic control in biological systems?
If a thermodynamically favored reaction doesn’t happen because of a huge activation energy (kinetic control), enzymes act as biological catalysts to lower that barrier so the reaction proceeds at a measurable rate. They do this by providing an alternative reaction pathway and stabilizing the transition state (lowering EA), increasing the rate constant k (Arrhenius), and thus raising reaction rate without changing ΔG° (thermodynamics stay the same). Mechanisms include orienting substrates correctly, creating a favorable microenvironment, using induced fit to strain bonds, and providing catalytic groups or cofactors. By Hammond-postulate logic, stabilizing the transition state lowers its energy and speeds the reaction—converting a metastable but slow process into a fast, biologically useful one. For a focused review, see the Topic 9.4 study guide (https://library.fiveable.me/ap-chemistry/unit-9/thermodynamic-kinetic-control/study-guide/hRZ0V3goVueXCw1JeUdA), the Unit 9 overview (https://library.fiveable.me/ap-chemistry/unit-9), and practice problems (https://library.fiveable.me/practice/ap-chemistry).
What happens when you lower the activation energy of a thermodynamically favored reaction?
Lowering the activation energy (Ea) of a reaction speeds it up—it raises the rate constant k (Arrhenius: k = A·e^(−Ea/RT)), often by a large exponential amount—so a thermodynamically favored reaction (ΔG < 0) that was previously too slow because of a high Ea will proceed faster and reach equilibrium sooner. Lowering Ea (for example with a catalyst) does not change ΔH, ΔS, or ΔG, and it doesn’t change the equilibrium position—it only helps the system get to that equilibrium more quickly. If a process is thermodynamically favored but slow, it’s under kinetic control (metastable); lowering Ea can move it out of kinetic control. This distinction (kinetic vs thermodynamic control, role of Ea, Arrhenius) is exactly what Topic 9.4 tests in the CED (LO 9.4.A). For a short study refresher, see the Topic 9.4 study guide (https://library.fiveable.me/ap-chemistry/unit-9/thermodynamic-kinetic-control/study-guide/hRZ0V3goVueXCw1JeUdA) and practice problems (https://library.fiveable.me/practice/ap-chemistry).
I missed the lecture on this - what's the main difference between a reaction being possible vs actually happening?
Short answer: “Possible” means thermodynamically favored (ΔG < 0)—the products are lower in free energy so the reaction is allowed by thermodynamics. “Actually happening” depends on kinetics: the reaction needs a pathway with a low enough rate constant k (low activation energy, Ea) so you see it at a measurable rate. A large Ea (high energy barrier, slow formation of the transition state) makes a thermodynamically favored process kinetically trapped or metastable. Catalysts lower Ea and increase k (Arrhenius equation), letting the reaction proceed even if it was previously too slow. On the AP exam this is Topic 9.4 (learning objective 9.4.A): you should be able to explain why a favored ΔG doesn’t guarantee observable reaction and use terms like activation energy, transition state, kinetic control, and catalysis. For a quick review, see the Topic 9.4 study guide (https://library.fiveable.me/ap-chemistry/unit-9/thermodynamic-kinetic-control/study-guide/hRZ0V3goVueXCw1JeUdA) and practice problems (https://library.fiveable.me/practice/ap-chemistry).