Catalysis — AP Chemistry Study Guide

For: AP Chemistry candidates sitting AP Chemistry.

Covers: definition of catalysts, homogeneous vs heterogeneous catalysis, catalyzed reaction energy profiles, activation energy modification, enzyme catalysis, and the effect of catalysts on reaction rate and thermodynamic equilibrium.

You should already know: Reaction rate laws and the definition of activation energy; the Arrhenius relationship between activation energy and reaction rate; how to interpret reaction coordinate energy diagrams.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Chemistry style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is Catalysis?

Catalysis is the process by which a substance (a catalyst) increases the rate of a chemical reaction without being consumed in the overall reaction. By definition, catalysts do not appear in the overall balanced reaction equation, though they are incorporated into intermediate elementary steps and are fully regenerated by the final step of the reaction, so no net consumption occurs. According to the AP Chemistry CED, catalysis makes up approximately 15-20% of the Kinetics unit (Unit 5), which accounts for 7-9% of the total AP Chemistry exam score. Catalysis questions appear regularly in both multiple-choice (MCQ) and free-response (FRQ) sections of the exam, often combining with energy profile interpretation, rate law determination, or connections to thermodynamics and equilibrium. Catalysts work by providing an alternative reaction mechanism (a different sequence of elementary steps) for the reaction that has a lower overall activation energy ($E_a$) than the uncatalyzed pathway. The AP exam almost exclusively focuses on rate-increasing catalysts; inhibitors (or negative catalysts, which decrease reaction rate) are rarely tested outside of conceptual questions.

2. Homogeneous vs Heterogeneous Catalysis

Catalysis is classified by the relative phase (physical state of matter) of the catalyst and the reaction reactants. Homogeneous catalysis occurs when the catalyst is in the same phase as all reactants. Common examples include acid-catalyzed hydrolysis of esters in aqueous solution, where both the catalyst ($\text{ce}H+$) and reactants are dissolved in water, so all are in the aqueous phase. Heterogeneous catalysis occurs when the catalyst is in a different phase than the reactants. The most common form is solid catalysts for gas-phase or liquid-phase reactions, such as iron in the Haber process for ammonia synthesis, or platinum in automotive catalytic converters.

Heterogeneous catalysis typically works via adsorption: reactant molecules bind to active sites on the surface of the solid catalyst, which weakens existing bonds in the reactants, lowering the activation energy for bond breaking and new bond formation. After reaction, product molecules desorb from the surface, leaving the active sites free for new reactant molecules. Homogeneous catalysis usually involves formation of a temporary intermediate complex between the catalyst and reactant in solution, with the catalyst regenerated at the end of the reaction.

Worked Example

Identify the type of catalysis in each reaction, and justify your answer: (a) The acid-catalyzed decomposition of aqueous sucrose into glucose and fructose, where the catalyst is aqueous sulfuric acid. (b) The conversion of gaseous nitrogen and hydrogen to ammonia, catalyzed by solid iron.

Solution:

1. Recall the core rule: catalysis type depends on whether the catalyst shares the same phase as all reactants.
2. For reaction (a): All reactants (sucrose) are aqueous, and the catalyst $\text{ce}H+$ from sulfuric acid is also aqueous. All species are in the same phase.
3. Therefore, reaction (a) uses homogeneous catalysis.
4. For reaction (b): Reactants $\text{ce}N2$ and $\text{ce}H2$ are gaseous, and the iron catalyst is solid, which is a different phase.
5. Therefore, reaction (b) uses heterogeneous catalysis.

Exam tip: Always check the phase of the catalyst, not just its chemical identity. Even a solid catalyst mixed with liquid reactants counts as heterogeneous, because phase is defined by physical state, not chemical composition.

3. Catalyzed Reaction Energy Profiles

Since catalysts provide an entirely new reaction mechanism between reactants and products, the energy profile (reaction coordinate diagram) for a catalyzed reaction looks very different from the uncatalyzed profile. The uncatalyzed reaction has one activation energy peak per elementary step, with the highest peak representing the overall activation energy $E_a$ for the rate-determining step. The catalyzed reaction, with its new mechanism, has a separate activation energy peak for each elementary step in the new pathway. Critically, the highest peak (overall $E_a$) for the catalyzed pathway is always lower than the highest peak for the uncatalyzed pathway.

From the Arrhenius equation $k=A{e}^{-E_a/RT}$, a lower $E_a$ gives a larger rate constant $k$, which means a faster reaction, matching the definition of a catalyst. A key point heavily tested on the AP exam is that catalysts do not change the energy of the starting reactants or final products, only the pathway between them. Therefore, the overall enthalpy change $\Delta H$ of the reaction is identical for catalyzed and uncatalyzed reactions, and catalysts do not change the equilibrium constant $K$ or the final yield of product at equilibrium.

Worked Example

An endothermic reaction $\text{ce}R->P$ has an uncatalyzed activation energy of 65 kJ/mol and an overall enthalpy change of +22 kJ/mol. Compare the key features of the uncatalyzed and catalyzed energy profiles, if the catalyzed pathway has a maximum activation energy of 30 kJ/mol.

Solution:

1. Start by setting the energy of reactants R to 0 kJ/mol for reference. For the endothermic reaction, the energy of products P is $0+22=22$ kJ/mol, which is higher than reactants for both pathways.
2. The uncatalyzed profile has a single energy peak (for a one-step reaction) with maximum energy at 65 kJ/mol, giving $E_a=65$ kJ/mol, and $\Delta H=+22$ kJ/mol.
3. The catalyzed profile has multiple peaks (one per elementary step in the new mechanism), with the highest peak at 30 kJ/mol, so the overall catalyzed $E_a=30$ kJ/mol < 65 kJ/mol.
4. The overall $\Delta H$ for the catalyzed reaction is still +22 kJ/mol, because catalysts do not change the energy of reactants or products.

Exam tip: On FRQs asking to compare energy profiles, always explicitly state that $\Delta H$ is the same for both pathways; AP graders require this statement to award full credit.

4. Enzyme Catalysis

Enzymes are biological catalysts, almost always globular proteins, that speed up cellular reactions by up to 10 orders of magnitude. They are highly specific to their reactants (called substrates), due to the 3D structure of their active site: the region of the enzyme where the substrate binds to form an enzyme-substrate complex, which then reacts to form product and regenerate free enzyme.

A key kinetic property of enzyme catalysis is saturation kinetics: when enzyme concentration is held constant, the reaction rate depends on substrate concentration as follows: at low substrate concentrations, rate increases linearly with substrate concentration (first order in substrate), because most enzyme active sites are unoccupied, so adding more substrate increases the number of active sites used. At high substrate concentrations, all enzyme active sites are fully occupied (saturated), so the rate becomes independent of substrate concentration (zero order in substrate), because the enzyme is already working at its maximum possible rate. Enzyme activity is also highly dependent on temperature and pH: high temperatures or extreme pH break the 3D structure of the enzyme (a process called denaturation), which destroys the active site shape and eliminates catalytic activity.

Worked Example

An experiment measures the initial rate of an enzyme-catalyzed reaction at different substrate concentrations, with enzyme concentration held constant at 0.1 μmol/L. The data is below:

Determine the order of the reaction with respect to substrate at low [S] and high [S], and explain the observed behavior.

Solution:

1. At low substrate concentrations (0.05 to 0.10 μmol/L), doubling [S] from 0.05 to 0.10 doubles the rate from 0.02 to 0.04 μmol/L·s. Rate is proportional to [S], so the reaction is first order in substrate at low [S].
2. At high substrate concentrations (1.0 to 2.0 μmol/L), doubling [S] from 1.0 to 2.0 leaves the rate unchanged at ~0.20 μmol/L·s. Rate does not depend on [S], so the reaction is zero order in substrate at high [S].
3. Explanation: At low [S], most enzyme active sites are unoccupied, so increasing [S] increases the number of active sites bound to substrate, leading to a proportional increase in rate.
4. At high [S], all active sites are saturated, so the rate is limited by how fast the enzyme can process substrate into product, not by how much substrate is available, so rate does not change when [S] increases.

Exam tip: If an AP question asks for the order of an enzyme-catalyzed reaction when substrate concentration is much higher than enzyme concentration, the answer is almost always zero order.

5. Common Pitfalls (and how to avoid them)

• Wrong move: Claiming that catalysts lower the activation energy of the existing uncatalyzed reaction mechanism. Why: Students confuse lowering activation energy overall with changing the original pathway. Correct move: Always state that catalysts provide an alternative reaction mechanism with a lower overall activation energy, they do not change the activation energy of the original uncatalyzed pathway.
• Wrong move: Stating that catalysts increase the yield of product at equilibrium or change the value of the equilibrium constant $K$. Why: Students mix up the effect on reaction rate with the effect on thermodynamics. Correct move: Remember that catalysts do not change the energy of reactants or products, so $\Delta G$ and $K$ are unchanged; they only speed up how fast equilibrium is reached, so final yield is the same.
• Wrong move: Leaving catalysts in the overall balanced reaction when adding elementary steps. Why: Students forget catalysts are regenerated, so they leave them on the reactant side. Correct move: When summing elementary steps, cancel any species that appears on both sides; catalysts will always cancel fully and do not appear in the overall reaction.
• Wrong move: Claiming that enzymes retain full catalytic activity at all temperatures because catalysts are not consumed. Why: Students generalize the "not consumed" rule to all conditions, forgetting enzymes are proteins. Correct move: Always recall that enzyme activity peaks at an optimal temperature and pH, and drops to near zero at high temperatures or extreme pH due to denaturation of the active site.
• Wrong move: Classifying a solid catalyst reacting with liquid reactants as homogeneous. Why: Students confuse chemical identity with physical phase. Correct move: Always compare the physical state (phase) of the catalyst and reactants; any difference in phase means heterogeneous catalysis.
• Wrong move: Drawing a catalyzed energy profile with the same number of peaks as the uncatalyzed profile, just shifted down. Why: Students forget the alternative mechanism has multiple elementary steps. Correct move: A catalyzed mechanism will always have more activation energy peaks than the uncatalyzed mechanism (one per step), all lower than the original maximum activation energy.

6. Practice Questions (AP Chemistry Style)

Question 1 (Multiple Choice)

Which of the following changes will always occur when a catalyst is added to a reaction system, with all other conditions held constant? A) The rate constant $k$ increases, and the equilibrium constant $K$ increases B) The overall activation energy of the reaction decreases, and the enthalpy change $\Delta H$ remains the same C) The number of elementary steps in the mechanism decreases, and the overall rate increases D) The frequency of collisions between reactant molecules increases, leading to a faster rate

Worked Solution: First, recall the core properties of catalysts: they provide an alternative mechanism with lower activation energy, but do not change the energy of reactants or products, so thermodynamic properties like $\Delta H$ and $K$ are unchanged. Eliminate incorrect options: Option A is wrong because $K$ does not change. Option C is wrong because the catalyzed alternative mechanism almost always has more elementary steps than the uncatalyzed pathway, not fewer. Option D is wrong because collision frequency depends only on temperature and concentration, which are held constant; catalysts only change the fraction of collisions with enough energy to react. Only Option B is correct. Correct Answer: B

Question 2 (Free Response)

The oxidation of sulfur dioxide to sulfur trioxide in the contact process for sulfuric acid production uses nitrogen dioxide as a homogeneous catalyst. The mechanism is: Step 1 (slow): $\text{ce}NO2(g)+SO2(g)->NO(g)+SO3(g)$ Step 2 (fast): $\text{ce}2NO(g)+O2(g)->2NO2(g)$

(a) Identify the catalyst and the reaction intermediate in this mechanism. Justify each answer. (b) Write the overall balanced reaction for this process. (c) How does the addition of the catalyst change the value of the equilibrium constant for this reaction? Explain.

Worked Solution: (a) The catalyst is $\text{ce}NO2(g)$. It is consumed in step 1 and regenerated in step 2, with no net consumption, which matches the definition of a catalyst. The intermediate is $\text{ce}NO(g)$. It is produced in step 1 and consumed in step 2, so it does not appear in the overall reaction, which matches the definition of an intermediate. (b) Multiply step 1 by 2 to balance electrons and add to step 2: $2\text{ce}NO2(g)+2\text{ce}SO2(g)+2\text{ce}NO(g)+\text{ce}O2(g)->2\text{ce}NO(g)+2\text{ce}SO3(g)+2\text{ce}NO2(g)$. Cancel species common to both sides to get the overall reaction: $$2\text{ce}SO2(g)+\text{ce}O2(g)->2\text{ce}SO3(g)$$ (c) The equilibrium constant does not change. Catalysts only provide an alternative reaction mechanism with lower activation energy; they do not change the energy of the reactants or products, so the overall Gibbs free energy change $\Delta G$ and equilibrium constant $K$ are identical to the uncatalyzed reaction.

Question 3 (Application / Real-World Style)

Food manufacturers use enzyme catalysis to break down starch into glucose for high-fructose corn syrup. Two enzyme preparations are tested: Preparation A has 1 mg of active enzyme, and Preparation B has 1 mg of enzyme that has been denatured by heating to 100°C before use. The reaction is run at 35°C (the optimal temperature for this enzyme) with excess starch substrate in both reactions. Predict which preparation will have a faster reaction rate, and explain your answer in terms of enzyme structure and catalysis.

Worked Solution: Preparation A will have a much faster reaction rate. Enzyme catalysis depends on the specific 3D structure of the enzyme's active site, which allows the starch substrate to bind and react. Denaturation by heating breaks the noncovalent interactions that hold the enzyme's 3D structure together, so the active site loses its shape and can no longer bind the starch substrate. Preparation A has the correctly folded active enzyme, so all active sites are available to bind starch and catalyze the reaction, while Preparation B has no functional active sites, so it cannot catalyze the reaction. In context, this means only Preparation A will produce glucose at a useful rate for industrial manufacturing.

7. Quick Reference Cheatsheet

8. What's Next

After mastering catalysis in Unit 5 Kinetics, you will next apply this understanding to reaction mechanism analysis and rate-determining step identification, which builds directly on the ability to distinguish catalysts from reaction intermediates in multi-step mechanisms. Without mastering the core rules of catalysis, you cannot correctly derive rate laws for multi-step reactions, which is a high-weight FRQ skill on the AP exam. Beyond Kinetics, catalysis is a foundational concept for understanding biological enzyme reactions, and it connects to equilibrium and thermodynamics by reinforcing that reaction pathway does not change overall reaction thermodynamics. The skills you learn here for interpreting catalyzed energy profiles also transfer to organic reaction analysis, where most synthetic reactions rely on catalysts.

Reaction Mechanisms Arrhenius Equation and Activation Energy Enthalpy and Reaction Energy Profiles Chemical Equilibrium