Reaction Energy Profile

AP Chemistry· AP Chemistry CED — Kinetics· 14 min read

1. Introduction to Reaction Energy Profiles★★☆☆☆⏱ 3 min

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 represents the reaction coordinate, which tracks the progress of bond breaking and formation from reactants (left) to products (right), not elapsed time. The y-axis always represents the potential energy of the system, typically reported in kJ/mol.

This topic appears regularly on both multiple-choice (MCQ) and free-response (FRQ) sections of the AP Chemistry exam, and questions often combine energy profile interpretation with thermodynamics and reaction mechanism concepts. Common exam 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★★☆☆☆⏱ 4 min

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 state along the pathway where old bonds are partially broken and new bonds are partially formed.

Forward activation energy ( $E_{a, fwd}$ ): Energy required to go from reactants to the transition state
Reverse activation energy ( $E_{a, rev}$ ): Energy required to go from products back to the transition state
Enthalpy change of reaction ( $Δ H$ ): Overall change in potential energy for the forward reaction

E_{a,\text{fwd}} = E_{\text{transition}} - E_{\text{reactants}}

E_{a,\text{rev}} = E_{\text{transition}} - E_{\text{products}}

\Delta H = E_{\text{products}} - E_{\text{reactants}}

A negative $Δ H$ means the reaction is exothermic (products have lower energy than reactants), while a positive $Δ H$ 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 $E_{a, fwd}$ , $E_{a, rev}$ , and $Δ H$ . (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 $Δ H$ means the reaction absorbs energy from the surroundings, so it is endothermic.

Exam tip:

AP FRQ questions always require the correct sign for $Δ H$ . 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★★★☆☆⏱ 3 min

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.

📐 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★★★☆☆⏱ 3 min

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 $Δ H$ 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 $E_{a, fwd} = 150$ kJ/mol and $Δ H = - 30$ 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 $Δ H$ 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 $Δ H$ does not change.
2
Therefore, $Δ H$ 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 $Δ H$ , the answer is always no change. Never state that a catalyst lowers $Δ H$ — this is one of the most common errors tested on AP exams.

5. AP-Style Practice Problems★★★★☆⏱ 5 min

📐 Worked Example

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 $E_{a, rev}$ (reverse activation energy) and $Δ H$ (forward reaction enthalpy change)?

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

1
First calculate $Δ H$ using the definition:
$\Delta H = E_{\text{products}} - E_{\text{reactants}} = 50\ \text{kJ/mol} - 10\ \text{kJ/mol} = +40\ \text{kJ/mol}$
2
This eliminates options B and D. Next calculate $E_{a, rev}$ , the energy difference between transition state and products:
$E_{a,\text{rev}} = 105\ \text{kJ/mol} - 50\ \text{kJ/mol} = 55\ \text{kJ/mol}$
3
This matches option A, so the correct answer is A.

📐 Worked Example

Free Response: The overall reaction $NO_{2} (g) + CO (g) \to NO (g) + CO_{2} (g)$ follows a two-step mechanism: Step 1: $NO_{2} + NO_{2} \to NO_{3} + NO$ Step 2: $NO_{3} + CO \to NO_{2} + CO_{2}$

All energy values (kJ/mol, initial reactants = 0): Step 1 transition state = 78, intermediate $NO_{3}$ = 42, Step 2 transition state = 112, final products = -215.

(a) Identify the rate-determining step. Justify. (b) Calculate overall $Δ H$ and classify the reaction. (c) A catalyst lowers Step 1 $E_{a}$ to 32 kJ/mol and Step 2 $E_{a}$ to 84 kJ/mol. Does the RDS identity change? Justify.

1
(a) RDS is the step with the highest transition state energy relative to initial reactants. 112 kJ/mol > 78 kJ/mol, so Step 2 is the rate-determining step.
2
(b) Calculate $Δ H$ :
$\Delta H = -215\ \text{kJ/mol} - 0\ \text{kJ/mol} = -215\ \text{kJ/mol}$
3
Negative $Δ H$ means the reaction is exothermic.
4
(c) After catalysis, 84 kJ/mol > 32 kJ/mol, so Step 2 still has higher activation energy. The identity of the RDS does not change.

6. Common Pitfalls

Wrong move:

Calculating $Δ H$ as $E_{reactants} - E_{products}$ instead of $E_{products} - E_{reactants}$

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 $Δ H = E_{products} - E_{reactants}$ 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 $E_{a, rev}$ using a shortcut that ignores the sign of $Δ H$ , leading to incorrect values

Why:

Students rely on $E_{a, rev} = E_{a, fwd} - Δ H$ without checking the sign of $Δ H$ for exothermic reactions

Correct move:

Always calculate $E_{a, rev}$ directly as $E_{transition} - E_{products}$ 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

7. Quick Reference Cheatsheet

Category	Formula / Rule	Notes
Forward Activation Energy	$E_{a, fwd} = E_{TS} - E_{reactants}$	Always positive, measured from reactants to transition state
Reverse Activation Energy	$E_{a, rev} = E_{TS} - E_{products}$	For multi-step reactions, use the highest energy overall transition state
Enthalpy Change	$Δ H = E_{products} - E_{reactants}$	Negative = exothermic, positive = endothermic; unchanged by catalyst
Energy Relationship	$Δ H = E_{a, fwd} - E_{a, rev}$	Only applies to single-step reversible reactions
Number of Intermediates	$Intermediates = Number of steps - 1$	Intermediates occupy valleys between transition state peaks
Rate-Determining Step	Step with highest transition state energy (relative to initial reactants)	Slowest step, limits overall reaction rate
Catalyst Effect	Provides alternative lower $E_{a}$ pathway	Does not change $Δ H$ , reactant energy, or product energy
Transition State vs Intermediate	Transition state = peak; Intermediate = valley	Transition states are unstable; intermediates are detectable

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2023 · MCQ
Interpret catalyzed energy profile
2022 · FRQ
Identify RDS from reaction profile

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. Mastery of this topic is critical for answering cross-unit questions that combine kinetics and thermodynamics concepts, which appear regularly on both MCQ and FRQ sections of the exam. 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 deeper study of reaction mechanisms, where you will use energy profile features to confirm or reject proposed mechanisms based on experimental rate laws.