Multistep reaction energy profile — AP Chemistry Study Guide

For: AP Chemistry candidates sitting AP Chemistry.

Covers: Identification of transition states, reaction intermediates, rate-determining steps, overall reaction enthalpy, and connection of energy profiles to multistep reaction mechanisms for kinetics problems.

You should already know: Definition of elementary reactions in a reaction mechanism. Relationship between activation energy and reaction rate. Definition of enthalpy change for a chemical reaction.

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 Multistep reaction energy profile?

A multistep reaction energy profile (also called a reaction coordinate diagram for multistep mechanisms) is a graph that plots the potential energy of all species along a reaction pathway against the reaction coordinate, a qualitative measure of reaction progress from initial reactants to final products. Unlike single-step reactions, which have only one energy peak, multistep profiles have one energy peak per elementary reaction step, with valleys between peaks corresponding to reaction intermediates. Per the AP Chemistry Course and Exam Description (CED), this topic is part of Unit 5: Kinetics, which accounts for 7–9% of the overall AP exam score. Concepts from this topic appear in both multiple-choice (MCQ) and free-response (FRQ) sections: MCQ typically ask for identification of key features, while FRQ ask to connect the profile to reaction mechanism and rate behavior. Mastery of this topic is required for all mechanism-related kinetics questions on the exam.

2. Core Structural Features of Multistep Energy Profiles

A multistep energy profile is defined by two axes: the y-axis is always potential energy (units typically kJ/mol), and the x-axis is the reaction coordinate (progress from reactants to products, no fixed scale required). For a reaction with $n$ elementary steps, the profile will always have exactly $n$ peaks and $n-1$ valleys. Peaks correspond to transition states (also called activated complexes): high-energy, unstable species that exist only fleetingly as bonds break and form, and cannot be isolated. Valleys between the initial reactant peak and final product valley correspond to reaction intermediates: species that are formed in one elementary step and consumed in a later step. Intermediates are stable enough to be detected experimentally, even if they are short-lived. The overall enthalpy change of the reaction is calculated as the difference between the potential energy of the final products and the initial reactants: $$\Delta H_{\text{rxn}}=E_{\text{products}}-E_{\text{reactants}}$$ If $\Delta H_{\text{rxn}}<0$, the reaction is exothermic (products have lower energy than reactants); if $\Delta H_{\text{rxn}}>0$, it is endothermic. $\Delta H_{\text{rxn}}$ does not depend on the number of steps in the mechanism.

Worked Example

A two-step reaction has the following potential energies (all in kJ/mol): reactants = 12, first transition state = 52, intermediate = 18, second transition state = 70, products = 28. (a) How many transition states and intermediates are present? (b) Calculate $\Delta H_{\text{rxn}}$ and classify the reaction as endothermic or exothermic.

1. Use the core rule: number of transition states = number of elementary steps = 2. Number of intermediates = number of steps - 1 = 1.
2. Write the formula for overall enthalpy: $\Delta H_{\text{rxn}}=E_{\text{products}}-E_{\text{reactants}}$.
3. Substitute the given values: $\Delta H_{\text{rxn}}=28-12=+16\text{kJ/mol}$.
4. A positive $\Delta H_{\text{rxn}}$ means the reaction is endothermic.

Exam tip: Never count the initial reactant or final product as an intermediate. Only energy minima between the start and end points count as intermediates, regardless of their energy value.

3. Identifying the Rate-Determining Step

The rate-determining step (RDS, or rate-limiting step) is the slowest elementary step in a multistep mechanism, and it dictates the overall rate of the entire reaction. This follows the logic that a reaction can only proceed as fast as its slowest step, similar to how a bottleneck dictates traffic flow through a highway. To find the RDS on an energy profile, you first calculate the activation energy $E_a$ for each individual step. Activation energy for a step is the difference between the energy of the step's transition state (peak) and the energy of the species that starts the step (either the initial reactant for the first step, or the intermediate from the previous valley for subsequent steps): $$E_{a,\text{step }i}=E_{{\text{TS}}_i}-E_{{\text{start}}_i}$$ The step with the highest activation energy is always the rate-determining step, because a higher $E_a$ means fewer molecules have enough kinetic energy to overcome the barrier, so the step proceeds slower.

Worked Example

Use the energy values from the previous example: reactants (12 kJ/mol), TS1 (52 kJ/mol), intermediate (18 kJ/mol), TS2 (70 kJ/mol), products (28 kJ/mol). Identify the rate-determining step and calculate its activation energy.

1. Calculate $E_a$ for the first step: $E_{a1}=E_{\text{TS1}}-E_{\text{reactants}}=52-12=40\text{kJ/mol}$.
2. Calculate $E_a$ for the second step: $E_{a2}=E_{\text{TS2}}-E_{\text{intermediate}}=70-18=52\text{kJ/mol}$.
3. Compare the two activation energies: $E_{a2}(52\text{kJ/mol})>E_{a1}(40\text{kJ/mol})$.
4. The second step is the rate-determining step, with an activation energy of 52 kJ/mol.

Exam tip: Do not just pick the peak that is highest relative to the initial reactant y-axis as the RDS. Always calculate $E_a$ relative to the starting point of the individual step, because intermediates can be higher in energy than the original reactants.

4. Connecting Energy Profiles to Reaction Mechanisms

A core AP Chemistry skill is matching an energy profile to a proposed mechanism, and vice versa. Every elementary step in a mechanism corresponds to exactly one energy peak (transition state) on the profile, so the number of peaks directly tells you the number of steps in the mechanism. Intermediates, which are formed in one step and consumed in another, are never part of the overall balanced reaction, and correspond to the valleys between peaks. Catalysts also modify multistep energy profiles: a catalyst works by providing an alternate reaction mechanism with a lower activation energy for the rate-determining step. Catalysts do not change the overall energy of reactants or products, so $\Delta H_{\text{rxn}}$ remains identical for catalyzed and uncatalyzed reactions. Only the height of the activation energy barrier for the RDS changes.

Worked Example

A reaction follows the three-step mechanism below: Step 1 (fast): $\text{ce}A+B->AB$ Step 2 (slow): $\text{ce}AB+C->ABC$ Step 3 (fast): $\text{ce}ABC->AD+E$ What key features would you expect to see on the energy profile for this mechanism? Identify the reaction intermediate(s) and the location of the highest energy peak.

1. Count the number of elementary steps: 3 steps, so the profile will have 3 peaks (transition states).
2. Number of intermediates = 3 - 1 = 2 intermediates: $\text{ce}AB$ and $\text{ce}ABC$, which correspond to the two valleys between the three peaks.
3. Step 2 is given as the slow step, so it is the rate-determining step, meaning it will have the highest activation energy, so its peak will be the highest energy peak on the profile.
4. Confirm: Intermediates $\text{ce}AB$ and $\text{ce}ABC$ cancel out when adding the steps, so they do not appear in the overall reaction, matching their definition.

Exam tip: If you are asked to draw an energy profile from a mechanism, always draw the RDS peak as the highest peak, regardless of where it falls in the reaction sequence. AP exam graders look for this key feature.

5. Common Pitfalls (and how to avoid them)

• Wrong move: Counting the starting reactant or final product as a reaction intermediate. Why: Students confuse any energy minimum on the graph with an intermediate, forgetting intermediates are formed and consumed during the reaction. Correct move: Only count energy minima between the initial reactant and final product as intermediates, ignore the starting and end points.
• Wrong move: Identifying the RDS by comparing peak height relative to the initial reactant energy, rather than activation energy for the individual step. Why: Students see the tallest peak relative to the y-axis origin and automatically pick it, but forget intermediates can be higher in energy than reactants. Correct move: For every step, calculate $E_a=E_{\text{TS of step}}-E_{\text{starting species of step}}$, then select the step with the largest $E_a$ as RDS.
• Wrong move: Claiming a transition state is the same as a reaction intermediate because it is a high-energy species. Why: Students mix up the definitions of unstable transition states and stable (short-lived) intermediates. Correct move: Memorize the rule: peaks = transition states, valleys = intermediates; transition states cannot be isolated, intermediates can be detected.
• Wrong move: Calculating $\Delta H_{\text{rxn}}$ as the difference between the highest transition state energy and initial reactant energy. Why: Students confuse activation energy of the RDS with overall enthalpy change. Correct move: Always calculate $\Delta H_{\text{rxn}}$ as the energy of final products minus energy of initial reactants, regardless of the number of peaks in between.
• Wrong move: Stating that a catalyst changes the overall enthalpy of reaction because it lowers the activation energy. Why: Students associate lower energy barriers with lower product energy, forgetting catalysts work by changing the mechanism, not the starting or ending energy. Correct move: Remember: catalysts only change activation energies of elementary steps, they do not change $\Delta H_{\text{rxn}}$, so products and reactants have the same energy in catalyzed and uncatalyzed profiles.
• Wrong move: Stating that a 3-step reaction has 3 intermediates. Why: Students match the number of intermediates to the number of steps, instead of steps minus one. Correct move: Recall that every step after the first consumes the intermediate from the previous step, so number of intermediates = number of elementary steps - 1.

6. Practice Questions (AP Chemistry Style)

Question 1 (Multiple Choice)

The energy profile for a three-step reaction has the following energy values (all in kJ/mol): Reactants: 0, TS1: 50, Int1: 20, TS2: 80, Int2: 30, TS3: 65, Products: 10. Which of the following statements is correct? A) The overall reaction is endothermic, and the third step is the rate-determining step. B) The overall reaction is endothermic, and the second step is the rate-determining step. C) The overall reaction is exothermic, and the second step is the rate-determining step. D) The overall reaction is exothermic, and the third step is the rate-determining step.

Worked Solution: First, calculate the overall enthalpy change: $\Delta H_{\text{rxn}}=E_{\text{products}}-E_{\text{reactants}}=10-0=+10\text{kJ/mol}$. A positive $\Delta H$ means the reaction is endothermic, eliminating options C and D. Next, calculate the activation energy for each step: $E_{a1}=50-0=50\text{kJ/mol}$, $E_{a2}=80-20=60\text{kJ/mol}$, $E_{a3}=65-30=35\text{kJ/mol}$. The highest activation energy is for the second step, so the second step is the rate-determining step. Correct answer: B.

Question 2 (Free Response)

Ozone decomposition in the upper atmosphere follows a two-step uncatalyzed mechanism: Step 1: $\text{ce}O3(g)->O2(g)+O(g)$ $E_a=15\text{kJ/mol}$ Step 2: $\text{ce}O(g)+O3(g)->2O2(g)$ $E_a=4\text{kJ/mol}$ The overall reaction has $\Delta H_{\text{rxn}}=-28\text{kJ/mol}$, and the initial $\text{ce}O3$ reactant energy is defined as 0 kJ/mol. The oxygen intermediate is at 10 kJ/mol after step 1. (a) Calculate the energy of TS1, TS2, and the final products. (b) Identify the rate-determining step and justify your answer. (c) Chlorine radicals catalyze this decomposition. Explain how the catalyzed energy profile will differ from the uncatalyzed profile.

Worked Solution: (a)

• Energy of TS1: $E_{\text{TS1}}=E_{\text{reactants}}+E_{a1}=0+15=15\text{kJ/mol}$.
• Energy of TS2: $E_{\text{TS2}}=E_{\text{intermediate}}+E_{a2}=10+4=14\text{kJ/mol}$.
• Energy of products: $E_{\text{products}}=E_{\text{reactants}}+\Delta H_{\text{rxn}}=0+(-28)=-28\text{kJ/mol}$.

(b) The rate-determining step is step 1. Justification: The RDS is the step with the highest activation energy. $E_{a1}=15\text{kJ/mol}$ is greater than $E_{a2}=4\text{kJ/mol}$, so step 1 is slower and determines the overall reaction rate.

(c) The catalyst provides an alternate two-step mechanism with a lower activation energy for the rate-determining step (step 1). The energy of initial reactants and final products does not change, so $\Delta H_{\text{rxn}}$ remains $-28\text{kJ/mol}$. Only the activation energy barrier for step 1 is lowered, so the peak for step 1 will be lower on the catalyzed energy profile.

Question 3 (Application / Real-World Style)

A chemical plant produces nitrogen dioxide via the two-step reaction below, with the following energy values (kJ/mol): reactants at 0, TS1 at 80, intermediate at 20, TS2 at 95, products at 60. Chemists test a new catalyst that specifically lowers the activation energy of the first step by 20 kJ/mol. Does adding this catalyst significantly increase the overall rate of $\text{ce}NO2$ production? Justify your answer with activation energy calculations.

Worked Solution: First, calculate the original activation energies for each step: $E_{a1}=80-0=80\text{kJ/mol}$, $E_{a2}=95-20=75\text{kJ/mol}$. The original RDS is step 1, with a maximum activation energy of 80 kJ/mol. After adding the catalyst, the new $E_{a1}=80-20=60\text{kJ/mol}$. The new maximum activation energy across steps is 75 kJ/mol for step 2, which is 5 kJ/mol lower than the original 80 kJ/mol. Yes, the catalyst does increase the overall reaction rate. In context, lowering the highest activation energy in the mechanism increases the fraction of molecules with enough energy to cross the rate-limiting barrier, leading to faster production of $\text{ce}NO2$.

7. Quick Reference Cheatsheet

8. What's Next

This topic is the foundational link between reaction mechanisms and reaction kinetics, the core of AP Chemistry Unit 5. Next, you will connect the rate-determining step you identify from energy profiles to writing rate laws for multistep reactions, a skill heavily tested on AP FRQ. Without being able to correctly identify the RDS and intermediates from an energy profile, you will not be able to write a consistent rate law that matches a proposed mechanism, a common AP prompt. This topic also connects to thermodynamics in Unit 6, where you will use Hess's law to confirm that overall enthalpy is independent of reaction pathway, matching the relationship you learned here.

• Rate laws for reaction mechanisms
• Elementary reactions
• Catalysis and rate modification
• Hess's law for enthalpy calculation