AP Chemistry Activation Energy

Reading reaction energy profiles for activation energy, ΔH, catalysts, and the rate-determining step, in four worked examples.

Questions about activation energy almost always come with a graph, and the graph already contains the answer: you only need to know which distances to measure. The Unit 5 review introduces the energy profile in one section; this page is the practice for it, with numbers you can check, a two-step profile, and the temperature explanation that free-response questions like to ask for.

How to Read Activation Energy and ΔH From an Energy Profile

An energy profile plots potential energy on the vertical axis against reaction progress on the horizontal axis. Three measurements cover nearly every question:

Forward activation energy: E_a(forward) = energy of the transition state − energy of the reactants.
Reverse activation energy: E_a(reverse) = energy of the transition state − energy of the products.
Enthalpy change: ΔH = energy of the products − energy of the reactants. It is negative when the products are lower (exothermic) and positive when they are higher (endothermic), as in the Unit 6 review.

The peak is the transition state, which cannot be isolated; a valley between two peaks is an intermediate, a real species that forms and is then consumed. In a multistep reaction the step with the highest barrier is the rate-determining step, the slowest step. A useful check on your three measurements: E_a(reverse) must equal E_a(forward) − ΔH.

Worked Example 1: Finding Both Activation Energies and ΔH From a One-Step Profile

(a) A reaction profile has the reactants at 50 kJ/mol, the transition state at 130 kJ/mol, and the products at 20 kJ/mol. Find E_a(forward), E_a(reverse), and ΔH.
(b) Repeat for reactants at 20 kJ/mol, transition state at 90 kJ/mol, and products at 60 kJ/mol.

(a) E_a(forward) = 130 − 50 = 80 kJ/mol. E_a(reverse) = 130 − 20 = 110 kJ/mol. ΔH = 20 − 50 = −30 kJ/mol, exothermic. Check: 80 − (−30) = 110. ✓
(b) E_a(forward) = 90 − 20 = 70 kJ/mol. E_a(reverse) = 90 − 60 = 30 kJ/mol. ΔH = 60 − 20 = +40 kJ/mol, endothermic. Check: 70 − 40 = 30. ✓

Notice that the reverse barrier in (a) is larger than the forward one. An exothermic reaction has the steeper climb going backward.

Worked Example 2: Showing What a Catalyst Changes and What It Does Not

A catalyst gives the reaction in Example 1(a) a new pathway with a transition state at 100 kJ/mol. Find the new activation energies and ΔH, and say what changed.

E_a(forward) = 100 − 50 = 50 kJ/mol (was 80).
E_a(reverse) = 100 − 20 = 80 kJ/mol (was 110).
ΔH = 20 − 50 = −30 kJ/mol (unchanged).

The catalyst lowered both barriers by the same 30 kJ/mol and left ΔH alone, because the reactants and products sit at the same energies as before. Both the forward and reverse reactions speed up equally, so a catalyst helps a reaction reach equilibrium sooner but does not shift where that equilibrium lies, the point behind Le Chatelier's principle problems that mention a catalyst.

Worked Example 3: Reading a Two-Step Profile for the Intermediate and Rate-Determining Step

For the profile below, identify the intermediate, state each step's activation energy, find the overall ΔH, and name the rate-determining step.

Intermediate: the valley at 70 kJ/mol between the two peaks.
Step 1: E_a = 110 − 40 = 70 kJ/mol (reactants to transition state 1).
Step 2: E_a = 90 − 70 = 20 kJ/mol (intermediate to transition state 2).
Overall ΔH: 20 − 40 = −20 kJ/mol. The intermediate's energy does not enter the overall change, only the start and end.
Rate-determining step: step 1, with the larger barrier and the highest peak.

The rate law then reflects step 1, which is the link to the mechanism questions in the Unit 5 review.

Worked Example 4: Explaining Why a Higher Temperature Speeds Up a Reaction

A reaction is run at 300 K and again at 350 K. Explain, without any calculation, why it is faster at 350 K, and say whether the activation energy changes.

Particles in a sample have a spread of kinetic energies, and only collisions with at least the activation energy can lead to products. At 350 K the distribution is shifted and broadened, so a larger fraction of collisions reaches that threshold. More collisions per second are successful, so the rate rises.
The activation energy does not change. Temperature changes how many collisions get over the barrier, not how high the barrier is.

This is the same distribution shown in the diagram on the kinetic molecular theory page, where the high-energy tail of the hotter curve holds far more particles. A full-credit answer names both parts: the fraction of collisions with sufficient energy, and the unchanged barrier.

What to Know About the Arrhenius Equation for the AP Exam

The Arrhenius equation, k = Ae^−E_a/RT, ties the rate constant to the activation energy and temperature. The course expects you to understand it qualitatively: a smaller E_a or a higher T gives a larger k. Calculations that use it, such as the two-temperature form or plots of ln k against 1/T, are not assessed, a point also noted in the Unit 5 review. If a practice problem asks you to compute an activation energy from rate constants, it goes beyond the exam.

Common Activation Energy Mistakes

Measuring from the bottom of the graph. Activation energy is measured from the reactants, not from zero.
Using the same barrier for both directions. The reverse barrier is measured from the products and differs unless ΔH = 0.
Saying a catalyst changes ΔH or the equilibrium. It lowers the barrier only, for both directions.
Calling an intermediate a transition state. Peaks are transition states; valleys are intermediates.
Saying temperature lowers the activation energy. It raises the fraction of collisions that exceed it.

If you are measuring the ΔH that these diagrams show, the lab version is covered in calorimetry. For a place to turn your practice into an estimated score, try the AP Chem Score Calculator when you finish the unit.

Frequently Asked Questions

What is activation energy?

Activation energy (Ea) is the minimum energy colliding particles need for a collision to produce products. On an energy profile it is the height from the reactants up to the top of the barrier, the transition state.

How do you find the activation energy of the reverse reaction?

Measure from the products up to the transition state: Ea(reverse) = energy of the transition state minus energy of the products. It always equals Ea(forward) minus ΔH.

Does a catalyst change ΔH?

No. A catalyst lowers the activation energy by providing a different pathway, but the reactants and products are at the same energies, so ΔH is unchanged. It lowers the forward and reverse barriers by the same amount.

What is the difference between an intermediate and a transition state?

A transition state is the highest-energy point of a step, the peak of the curve, and cannot be isolated. An intermediate is a species formed in one step and used up in the next; it sits in a valley between two peaks.

Why does raising the temperature speed up a reaction if the activation energy stays the same?

At a higher temperature a larger fraction of collisions has at least the activation energy, so more collisions are successful. The barrier itself is unchanged.

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