Reaction energy profile — AP Chemistry Study Guide

For: AP Chemistry candidates sitting AP Chemistry.

Covers: Construction and interpretation of reaction energy profiles (reaction coordinate diagrams), including activation energy, enthalpy change, transition states, intermediates, catalyzed profiles, and rate-determining step identification for multi-step reactions.

You should already know: Enthalpy change and endothermic/exothermic reaction definitions; collision theory and activation energy basics; rate-determining step concept for multi-step reactions.

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 Reaction energy profile?

A reaction energy profile (also called a reaction coordinate diagram or potential energy profile) is a graphical plot that maps changes in potential energy of a chemical system as reactants convert to products along the reaction pathway. The x-axis is always the reaction coordinate, which tracks the progress of bond breaking and formation from reactants (left) to products (right), not elapsed time. The y-axis is always the potential energy of the system, typically reported in kJ/mol. This topic makes up approximately 8-10% of Unit 5 (Kinetics) weight in the AP Chemistry CED, and appears regularly on both multiple-choice (MCQ) and free-response (FRQ) sections of the exam. AP questions often combine energy profile interpretation with concepts from thermodynamics and reaction mechanisms, so mastery of this topic is critical for full credit on cross-unit questions. Common question tasks include labeling profile components, calculating energy values, identifying rate-determining steps, and comparing catalyzed and uncatalyzed pathways.

2. Key Components of Single-Step Reaction Profiles

For a single-step (elementary) reaction, the energy profile has three core features: reactants at the left starting point, a single peak corresponding to the transition state (activated complex), and products at the right ending point. The transition state is the highest-energy, unstable intermediate state along the pathway where old bonds are partially broken and new bonds are partially formed. Three key energy values are defined for all single-step reversible profiles:

1. Forward activation energy ($E_{a,\text{fwd}}$): The energy input required to go from reactants to the transition state: $$E_{a,\text{fwd}}=E_{\text{transition}}-E_{\text{reactants}}$$
2. Reverse activation energy ($E_{a,\text{rev}}$): The energy input required to go from products back to the transition state: $$E_{a,\text{rev}}=E_{\text{transition}}-E_{\text{products}}$$
3. Enthalpy change of reaction ($\Delta H$): The overall change in potential energy for the forward reaction: $$\Delta H=E_{\text{products}}-E_{\text{reactants}}$$ A negative $\Delta H$ means the reaction is exothermic (products have lower energy than reactants), while a positive $\Delta H$ means the reaction is endothermic.

Worked Example

For a single-step reaction, the potential energy of reactants is 25 kJ/mol, the transition state energy is 145 kJ/mol, and the potential energy of products is 70 kJ/mol. (a) Calculate $E_{a,\text{fwd}}$, $E_{a,\text{rev}}$, and $\Delta H$. (b) Classify the reaction as endothermic or exothermic.

1. Calculate forward activation energy: $E_{a,\text{fwd}}=145\text{kJ/mol}-25\text{kJ/mol}=120\text{kJ/mol}$.
2. Calculate reverse activation energy: $E_{a,\text{rev}}=145\text{kJ/mol}-70\text{kJ/mol}=75\text{kJ/mol}$.
3. Calculate enthalpy change: $\Delta H=70\text{kJ/mol}-25\text{kJ/mol}=+45\text{kJ/mol}$.
4. Classification: A positive $\Delta H$ means the reaction absorbs energy from the surroundings, so it is endothermic.

Exam tip: AP FRQ questions always require the correct sign for $\Delta H$. Missing the positive/negative sign will almost always cost you a point, so double-check the sign before moving on.

3. Multi-Step Profiles and Rate-Determining Step Identification

A multi-step reaction has one elementary step per activation energy barrier, so the profile will have one peak (transition state) per elementary step. Valleys between adjacent peaks correspond to reaction intermediates: species that are formed in one early elementary step and consumed in a later step, so they do not appear in the overall balanced reaction. Intermediates are stable enough to be detected experimentally, so they occupy energy valleys that are lower than adjacent transition states, but usually higher than the initial reactants or final products. The rate-determining step (RDS), the slowest step that limits the overall reaction rate, is always the step with the highest activation energy barrier. On an energy profile, this corresponds to the transition state with the highest energy relative to the initial starting reactants. This is because the overall rate depends on the fraction of molecules that can reach the highest-energy transition state, so the highest barrier controls the rate.

Worked Example

A three-step reaction has transition state energies (relative to initial reactants at 0 kJ/mol) of 35 kJ/mol (Step 1), 82 kJ/mol (Step 2), and 48 kJ/mol (Step 3). (a) Identify the rate-determining step. (b) How many reaction intermediates are present? Justify your answer.

1. For part (a): The RDS is the step with the highest transition state energy relative to initial reactants. Comparing the three values, 82 kJ/mol is the highest, so Step 2 is the rate-determining step.
2. For part (b): A three-step reaction has 3 transition state peaks. Intermediates occupy the valleys between peaks, so the number of intermediates equals the number of peaks minus 1.
3. 3 peaks - 1 = 2 intermediates, so there are two reaction intermediates in this mechanism.

Exam tip: Never mislabel the x-axis as "time" on FRQ drawn responses. The x-axis is the reaction coordinate (progress of bond changes), not elapsed time, so labeling it "time" will cost you a point.

4. Catalyzed Reaction Energy Profiles

A catalyst speeds up a reaction by providing an entirely alternative reaction mechanism (different reaction pathway) with a lower overall activation energy than the uncatalyzed reaction. A lower activation energy means a larger fraction of reactant molecules have enough kinetic energy to overcome the energy barrier at a given temperature, which increases the reaction rate. A common misconception is that catalysts change the energy of reactants or products. In reality, catalysts do not affect the potential energy of the starting reactants or final products, so the overall enthalpy change $\Delta H$ of the reaction is identical for catalyzed and uncatalyzed reactions. Catalyzed pathways often have more elementary steps (and thus more peaks) than the original uncatalyzed pathway, but the highest peak (maximum activation energy) of the catalyzed pathway is always lower than the highest peak of the uncatalyzed pathway.

Worked Example

An uncatalyzed single-step reaction has $E_{a,\text{fwd}}=150$ kJ/mol and $\Delta H=-30$ kJ/mol. A catalyst is added that provides an alternative two-step pathway with a maximum activation energy of 85 kJ/mol. What is the $\Delta H$ of the catalyzed reaction, and how does the reaction rate compare to the uncatalyzed rate? Justify your answer.

1. Recall that catalysts only change the reaction pathway, not the initial energy of reactants or final energy of products. Enthalpy change is the difference between product and reactant energy, so $\Delta H$ does not change.
2. Therefore, $\Delta H$ for the catalyzed reaction is still $-30$ kJ/mol, the same as the uncatalyzed reaction.
3. The catalyzed pathway has a lower maximum activation energy (85 kJ/mol < 150 kJ/mol), so more reactant molecules have sufficient kinetic energy to overcome the barrier at the same temperature.
4. This leads to a faster overall reaction rate, so the catalyzed reaction is significantly faster than the uncatalyzed reaction.

Exam tip: If a question asks how a catalyst affects $\Delta H$, the answer is always no change. Never state that a catalyst lowers $\Delta H$ — this is one of the most common errors tested on AP exams.

5. Common Pitfalls (and how to avoid them)

• Wrong move: Calculating $\Delta H$ as $E_{\text{reactants}}-E_{\text{products}}$ instead of $E_{\text{products}}-E_{\text{reactants}}$. Why: Students mix up the "final minus initial" rule for enthalpy with the "peak minus starting" rule for activation energy, leading to inverted signs. Correct move: Always write $\Delta H=E_{\text{products}}-E_{\text{reactants}}$ at the top of your work before starting any calculation to avoid inversion.
• Wrong move: Confusing intermediates with transition states, or counting one intermediate per elementary step. Why: Students assume every step produces an intermediate that persists to the end of the reaction. Correct move: Remember one transition state (peak) per step, one intermediate per valley between peaks, so number of intermediates = number of steps - 1.
• Wrong move: Claiming a catalyst increases the amount of product formed at equilibrium. Why: Students confuse faster reaction rate with higher equilibrium yield. Correct move: Recall that catalysts speed up forward and reverse reactions equally, so they do not change equilibrium yield or the amount of product formed.
• Wrong move: Calculating $E_{a,\text{rev}}$ using a shortcut that ignores the sign of $\Delta H$, leading to incorrect values. Why: Students rely on $E_{a,\text{rev}}=E_{a,\text{fwd}}-\Delta H$ without checking the sign of $\Delta H$ for exothermic reactions. Correct move: Always calculate $E_{a,\text{rev}}$ directly as $E_{\text{transition}}-E_{\text{products}}$ to avoid sign errors from shortcuts.
• Wrong move: Identifying the RDS as the step with the highest energy relative to the previous intermediate, not the initial reactants. Why: Students confuse activation energy of the individual step with the overall activation energy for the entire reaction. Correct move: Always compare transition state energies relative to the initial starting reactants to find the RDS.
• Wrong move: Labeling the x-axis of a reaction energy profile as "time". Why: Students intuitively associate reaction progress with elapsed time. Correct move: Always label the x-axis "reaction coordinate" or "reaction progress" for full credit on drawn FRQ responses.

6. Practice Questions (AP Chemistry Style)

Question 1 (Multiple Choice)

For an uncatalyzed reversible reaction, reactants have a potential energy of 10 kJ/mol, the transition state has a potential energy of 105 kJ/mol, and products have a potential energy of 50 kJ/mol. Which of the following gives the correct values for $E_{a,\text{rev}}$ (reverse activation energy) and $\Delta H$ (forward reaction enthalpy change)?

A) $E_{a,\text{rev}}=55$ kJ/mol, $\Delta H=+40$ kJ/mol B) $E_{a,\text{rev}}=55$ kJ/mol, $\Delta H=-40$ kJ/mol C) $E_{a,\text{rev}}=95$ kJ/mol, $\Delta H=+40$ kJ/mol D) $E_{a,\text{rev}}=95$ kJ/mol, $\Delta H=-40$ kJ/mol

Worked Solution: First, calculate $\Delta H$ using the definition $\Delta H=E_{\text{products}}-E_{\text{reactants}}=50\text{kJ/mol}-10\text{kJ/mol}=+40\text{kJ/mol}$. This eliminates options B and D, which have incorrect negative $\Delta H$. Next, $E_{a,\text{rev}}$ is the energy difference between the transition state and the products: $E_{a,\text{rev}}=105\text{kJ/mol}-50\text{kJ/mol}=55\text{kJ/mol}$. This matches option A, eliminating option C. The correct answer is A.

Question 2 (Free Response)

The overall reaction ${\text{NO}}_2(g)+\text{CO}(g)\to \text{NO}(g)+{\text{CO}}_2(g)$ follows a two-step mechanism: Step 1: ${\text{NO}}_2+{\text{NO}}_2\to {\text{NO}}_3+\text{NO}$ Step 2: ${\text{NO}}_3+\text{CO}\to {\text{NO}}_2+{\text{CO}}_2$

All energy values below are in kJ/mol, with initial reactants set to 0 kJ/mol: Step 1 transition state = 78, intermediate ${\text{NO}}_3$ = 42, Step 2 transition state = 112, final products = -215.

(a) Identify the rate-determining step for this reaction. Justify your answer. (b) Calculate the overall enthalpy change $\Delta H$ for the reaction, and classify it as endothermic or exothermic. (c) A catalyst lowers the activation energy of Step 1 to 32 kJ/mol and the activation energy of Step 2 to 84 kJ/mol. Will the catalyst change the identity of the rate-determining step? Justify your answer.

Worked Solution: (a) The rate-determining step is the step with the highest transition state energy relative to the initial reactants, which gives the highest activation energy. Step 1 has a transition state energy of 78 kJ/mol, while Step 2 has a transition state energy of 112 kJ/mol. Since 112 > 78, Step 2 is the rate-determining step. (b) $\Delta H=E_{\text{products}}-E_{\text{reactants}}=-215\text{kJ/mol}-0\text{kJ/mol}=-215\text{kJ/mol}$. The negative $\Delta H$ means the reaction is exothermic. (c) After catalysis, Step 1 has an activation energy of 32 kJ/mol and Step 2 has an activation energy of 84 kJ/mol. Step 2 still has a higher activation energy than Step 1 (84 > 32), so the identity of the rate-determining step does not change.

Question 3 (Application / Real-World Style)

The uncatalyzed decomposition of toxic hydrogen peroxide in human cells ($2{\text{H}}_2{\text{O}}_2(aq)\to 2{\text{H}}_2\text{O}(l)+{\text{O}}_2(g)$) has a forward activation energy of 72 kJ/mol and an overall enthalpy change of $-196$ kJ/mol. The enzyme catalase provides a catalyzed pathway with a forward activation energy of 8 kJ/mol. (a) Calculate the activation energy of the reverse reaction for the uncatalyzed decomposition. (b) Explain why catalase is critical for preventing cell damage from accumulated hydrogen peroxide.

Worked Solution: (a) For a single-step reaction, the relationship between the values is $\Delta H=E_{a,\text{fwd}}-E_{a,\text{rev}}$. Rearranging to solve for $E_{a,\text{rev}}$ gives: $$E_{a,\text{rev}}=E_{a,\text{fwd}}-\Delta H=72\text{kJ/mol}-(-196\text{kJ/mol})=268\text{kJ/mol}$$ (b) By lowering the activation energy of decomposition from 72 kJ/mol to 8 kJ/mol, catalase dramatically increases the rate of hydrogen peroxide breakdown at body temperature. This prevents toxic hydrogen peroxide from accumulating in cells, which would otherwise damage cell membranes and DNA.

7. Quick Reference Cheatsheet

8. What's Next

Reaction energy profiles are the foundational graphical tool connecting reaction kinetics to thermodynamics and reaction mechanism design, core themes across the AP Chemistry course. Next, you will apply the activation energy concepts you learned here to the Arrhenius equation, which allows you to calculate rate constants at different temperatures and quantify how temperature changes alter reaction rate. Without mastering how to identify and calculate activation energy from a reaction energy profile, you cannot correctly interpret Arrhenius plots or solve for activation energy from experimental rate data. This topic also directly feeds into the study of reaction mechanisms, where you will use energy profile features to confirm or reject proposed mechanisms based on experimental rate laws.

Arrhenius Equation Reaction Mechanisms and Rate-Determining Steps Enthalpy Change of Reaction