Energy diagrams — AP Chemistry Study Guide

For: AP Chemistry candidates sitting AP Chemistry.

Covers: Reaction coordinate energy diagrams, identification of endothermic vs exothermic reactions, calculation of activation energy and enthalpy change, identification of transition states and intermediates, and comparison of catalyzed vs uncatalyzed reaction pathways.

You should already know: Definition of enthalpy change of reaction, definition of activation energy, basic collision theory for chemical 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 Energy diagrams?

Energy diagrams (also called reaction coordinate diagrams or potential energy profiles) are graphical representations of how the potential energy of a chemical system changes as reactants are converted to products over the course of a reaction. Per the AP Chemistry CED, this topic is part of Unit 6 Thermodynamics, accounting for roughly 5-7% of the total exam score, and it appears in both multiple-choice (MCQ) and free-response (FRQ) sections, often paired with kinetics or equilibrium concepts.

The standard notation convention uses the y-axis for total potential energy of the system (units are typically kJ/mol), and the x-axis as the reaction coordinate, which represents the progress of the reaction from reactants (left) to products (right). The x-axis is not a time axis; it measures how far the reaction has proceeded along the bond-breaking and bond-forming pathway. Energy diagrams allow chemists to quickly visualize key reaction properties: whether a reaction releases or absorbs energy, the energy barrier that must be overcome, and how a catalyst alters the reaction pathway.

2. Endothermic and Exothermic Energy Diagrams

The relative potential energy of reactants and products on an energy diagram directly tells us whether a reaction is endothermic or exothermic, and lets us calculate the overall enthalpy change $\Delta H$ for the reaction. Enthalpy change is defined as the difference between the potential energy of products and the potential energy of reactants: $$\Delta H=E_{\text{products}}-E_{\text{reactants}}$$

By definition, if $\Delta H<0$, the reaction releases energy to the surroundings, so it is classified as exothermic. On the energy diagram, this means products sit lower on the y-axis than reactants. If $\Delta H>0$, the reaction absorbs energy from the surroundings, so it is classified as endothermic, and products sit higher than reactants on the diagram. This basic classification is the foundation for all more complex energy diagram interpretations, and it is tested in nearly every exam iteration.

Worked Example

An energy diagram for the reaction $2S{O}_3(g)\to 2S{O}_2(g)+O_2(g)$ has reactants at a potential energy of 40 kJ/mol and products at a potential energy of 196 kJ/mol. (a) Is the reaction endothermic or exothermic? (b) Calculate $\Delta H$ for the reaction.

1. Identify the given values: $E_{\text{reactants}}=40$ kJ/mol, $E_{\text{products}}=196$ kJ/mol.
2. Apply the definition of enthalpy change from the energy diagram: $\Delta H=E_{\text{products}}-E_{\text{reactants}}$.
3. Substitute the given values to calculate $\Delta H$: $\Delta H=196-40=+156$ kJ/mol.
4. Interpret the sign: a positive $\Delta H$ means the reaction absorbs energy, so it is endothermic.

Exam tip: Always calculate $\Delta H$ as products minus reactants, never the reverse. A single sign error will flip your endo/exo classification, which is almost always an automatic point deduction on FRQs.

3. Activation Energy and Transition States

Activation energy ($E_a$) is the minimum amount of energy that reactant molecules must have to overcome the energy barrier required to break existing bonds and form new products. On an energy diagram, the highest point along the reaction pathway is the transition state (also called the activated complex), an unstable high-energy species that exists only momentarily as bonds break and form.

The activation energy for the forward reaction is the difference between the energy of the transition state and the energy of the reactants: $$E_{a(\text{forward})}=E_{\text{transition state}}-E_{\text{reactants}}$$ For the reverse reaction (products converting back to reactants), the activation energy is the difference between the transition state energy and the product energy: $$E_{a(\text{reverse})}=E_{\text{transition state}}-E_{\text{products}}$$ Combining these two definitions gives the relationship $\Delta H=E_{a(\text{forward})}-E_{a(\text{reverse})}$, which lets you calculate any one value if you know the other two.

Worked Example

For the reaction $2N{H}_3(g)\rightleftharpoons N_2(g)+3H_2(g)$, $\Delta H=+92$ kJ/mol. If the activation energy of the forward reaction is 335 kJ/mol, what is the activation energy of the reverse reaction?

1. Start with the relationship between $\Delta H$, forward $E_a$, and reverse $E_a$: $\Delta H=E_{a(\text{forward})}-E_{a(\text{reverse})}$.
2. Rearrange the equation to isolate the unknown reverse activation energy: $E_{a(\text{reverse})}=E_{a(\text{forward})}-\Delta H$.
3. Substitute the given values: $E_{a(\text{reverse})}=335\text{ kJ/mol}-92\text{ kJ/mol}=243$ kJ/mol.
4. Verify the result: for this endothermic reaction, the reverse reaction should have a lower activation energy than the forward reaction, which matches our result.

Exam tip: Activation energy is always a positive value, since it is the difference between a peak (transition state) and a valley (reactants or products). If you get a negative $E_a$, you flipped the order of subtraction — go back and check.

4. Multi-Step Reactions, Intermediates, and Catalysis

Most chemical reactions occur in multiple steps, each with its own activation energy barrier and transition state. On an energy diagram for a multi-step reaction, there is one peak (transition state) per reaction step. The slowest (rate-determining) step is always the step with the highest activation energy peak, because that step has the largest energy barrier to overcome.

Reaction intermediates are stable species that are formed in one early step of the reaction and consumed in a later step. On an energy diagram, intermediates are located at local energy minima (valleys) between two transition state peaks, since they are more stable than the transition states before and after them.

Catalysts speed up reactions by providing an entirely new reaction mechanism with a lower overall activation energy. On an energy diagram, a catalyzed pathway has lower activation energy peaks, but the overall enthalpy change $\Delta H$ remains the same, because catalysts do not change the energy of the starting reactants or final products.

Worked Example

A two-step reaction has the following potential energy values relative to reactants: Step 1 transition state = 85 kJ/mol, intermediate after step 1 = 30 kJ/mol, Step 2 transition state = 105 kJ/mol, overall products = 10 kJ/mol. (a) Identify which step is rate-determining, (b) state where the intermediate is located on the diagram, (c) describe the effect of adding a catalyst to this reaction.

1. (a) Calculate the activation energy for each step: Step 1 $E_a=85-0=85$ kJ/mol (reactants are the 0 reference). Step 2 $E_a=105-30=75$ kJ/mol. Step 1 has a higher activation energy, so it is rate-determining.
2. (b) The intermediate is the stable species formed in step 1 and consumed in step 2, so it is located at the local energy minimum (valley) between the two transition state peaks, at 30 kJ/mol above reactant energy.
3. (c) A catalyst will provide an alternate two-step mechanism with lower activation energy for both steps, so both transition state peaks will be lower. The energy of the reactants, intermediate, and products will remain unchanged, so the overall $\Delta H$ of the reaction stays the same.

Exam tip: Do not confuse transition states with intermediates: transition states are at peaks (maxima, unstable), intermediates are at valleys (minima, relatively stable). AP exam questions explicitly test this distinction on a regular basis.

5. Common Pitfalls (and how to avoid them)

• Wrong move: Classifying peaks between steps as intermediates, or valleys between peaks as transition states. Why: Students mix up the definitions of the two species, since both appear between reactants and products on multi-step diagrams. Correct move: On any energy diagram, mark all maxima (peaks) as transition states and all minima (valleys) between peaks as intermediates, and use this rule every time.
• Wrong move: Calculating $\Delta H$ as $E_{\text{reactants}}-E_{\text{products}}$ instead of the reverse. Why: Students are used to "change = initial - final" for most quantities, so they default to the wrong order of subtraction. Correct move: Write the mnemonic "delta H is products minus reactants" at the top of your work before starting any energy diagram calculation.
• Wrong move: Claiming that a catalyst changes the overall $\Delta H$ of a reaction because it lowers activation energy. Why: Students assume that any change to the energy diagram changes the overall energy change. Correct move: Remember catalysts only change the reaction pathway, not the starting and ending energy of reactants and products, so $\Delta H$ is always unchanged by catalysis.
• Wrong move: Interpreting the x-axis (reaction coordinate) as a time axis to answer questions about reaction rate. Why: Students associate reaction progress with time, so they incorrectly use x-axis length to predict rate. Correct move: Always remember the x-axis represents bond breaking/forming progress, not time. All rate information comes from activation energy on the y-axis.
• Wrong move: Calculating the overall activation energy of a multi-step reaction as the sum of activation energies of all steps. Why: Students add energy values out of habit, not remembering that reaction rate is only controlled by the slowest step. Correct move: Overall activation energy is always equal to the activation energy of the rate-determining (highest peak) step only.

6. Practice Questions (AP Chemistry Style)

Question 1 (Multiple Choice)

The potential energy diagram for a reaction has reactants at 35 kJ/mol, a transition state at 110 kJ/mol, and products at 10 kJ/mol. Which of the following gives the correct values for the forward activation energy $E_a$ and enthalpy change $\Delta H$?

A) $E_a=75$ kJ/mol, $\Delta H=-25$ kJ/mol B) $E_a=100$ kJ/mol, $\Delta H=+25$ kJ/mol C) $E_a=75$ kJ/mol, $\Delta H=+25$ kJ/mol D) $E_a=100$ kJ/mol, $\Delta H=-25$ kJ/mol

Worked Solution: First, calculate forward activation energy as the difference between transition state energy and reactant energy: $E_a=110-35=75$ kJ/mol. This eliminates options B and D. Next, calculate enthalpy change as products minus reactants: $\Delta H=10-35=-25$ kJ/mol. This eliminates option C. The correct answer is A.

Question 2 (Free Response)

A three-step reaction has the following potential energy values, relative to an arbitrary zero:

• Reactants: 20 kJ/mol
• Transition state 1 (step 1): 80 kJ/mol
• Intermediate 1: 45 kJ/mol
• Transition state 2 (step 2): 100 kJ/mol
• Intermediate 2: 35 kJ/mol
• Transition state 3 (step 3): 85 kJ/mol
• Products: -10 kJ/mol

(a) Identify the reaction intermediate with the lowest potential energy, and calculate the activation energy for step 2. (b) Identify the rate-determining step of the reaction. Justify your answer. (c) A catalyst is added that lowers the activation energy of all steps by 30 kJ/mol. Will the rate-determining step change? Justify your answer. What effect does the catalyst have on the overall $\Delta H$ of the reaction?

Worked Solution: (a) Intermediates are at local minima between transition states. The intermediate energies are 45 kJ/mol (intermediate 1) and 35 kJ/mol (intermediate 2), so the lowest energy intermediate is intermediate 2. Activation energy for step 2 = energy of transition state 2 minus energy of intermediate 1: $E_{a,2}=100-45=55$ kJ/mol. (b) The rate-determining step is the step with the highest activation energy. Calculate $E_a$ for all steps: Step 1: $80-20=60$ kJ/mol; Step 2: 55 kJ/mol; Step 3: $85-35=50$ kJ/mol. Step 1 has the highest activation energy, so it is rate-determining. (c) After catalysis, activation energies become: Step 1: $60-30=30$ kJ/mol; Step 2: $55-30=25$ kJ/mol; Step 3: $50-30=20$ kJ/mol. Step 1 still has the highest activation energy, so the rate-determining step does not change. Catalysts do not alter the energy of reactants or products, so $\Delta H$ remains unchanged at $\Delta H=-10-20=-30$ kJ/mol.

Question 3 (Application / Real-World Style)

The breakdown of sucrose into glucose and fructose has an overall enthalpy change of -28 kJ/mol. The uncatalyzed reaction has an activation energy of 108 kJ/mol. In the human body, the enzyme sucrase catalyzes this reaction, lowering the activation energy to 8 kJ/mol. By how much does the enzyme reduce the activation energy barrier, and what does this mean for the ability of your digestive system to process sucrose? Describe the key differences between the catalyzed and uncatalyzed energy diagrams.

Worked Solution: Calculate the reduction in activation energy: $108\text{ kJ/mol}-8\text{ kJ/mol}=100$ kJ/mol. Both the catalyzed and uncatalyzed diagrams start with sucrose at higher energy than the glucose + fructose products (consistent with negative $\Delta H$). The uncatalyzed diagram has one high peak at 108 kJ/mol above reactants, while the catalyzed diagram has a much lower peak at 8 kJ/mol above reactants, and both end at the same product energy. This 100 kJ/mol reduction in activation energy means the catalyzed reaction proceeds billions of times faster than the uncatalyzed reaction, allowing your digestive system to break down sucrose into usable sugars quickly enough to provide energy for cellular processes.

7. Quick Reference Cheatsheet

8. What's Next

Mastery of energy diagrams is a critical prerequisite for the remaining topics in AP Chemistry Unit 6, including Hess's law and bond enthalpy calculations, and it also forms the foundation for kinetics topics where you connect activation energy from energy diagrams to rate laws and Arrhenius equation calculations. Without being able to correctly identify activation energy, enthalpy change, and transition states from energy diagrams, you will struggle to connect thermodynamics concepts to reaction rate and equilibrium, which are frequently tested together in multi-concept FRQ questions. Energy diagrams also provide the core visual framework for understanding reaction mechanisms in advanced chemistry. Follow-on topics you should study next: Reaction mechanisms Hess's law Bond enthalpy Arrhenius equation