Energy Diagrams

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

1. Core Structure of Energy Diagrams★★☆☆☆⏱ 3 min

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. This topic accounts for roughly 5-7% of the total AP Chemistry exam score, and 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.

2. Endothermic and Exothermic Energy Diagrams★★☆☆☆⏱ 4 min

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 $Δ H$ for the reaction.

📘 Definition

Enthalpy Change (ΔH)

\Delta H

The total change in potential energy of the system over the course of a reaction, equal to the difference between the potential energy of products and reactants.

Example:

Positive ΔH = endothermic, negative ΔH = exothermic

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

By definition, if $Δ H < 0$ , 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 $Δ H > 0$ , the reaction absorbs energy from the surroundings, so it is classified as endothermic, and products sit higher than reactants on the diagram.

📐 Worked Example

An energy diagram for the reaction $2 S O_{3} (g) \to 2 S O_{2} (g) + O_{2} (g)$ 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 $Δ H$ for the reaction.

1
Identify the given values:
2
$E_{\text{reactants}} = 40 \text{ kJ/mol}, \quad E_{\text{products}} = 196 \text{ kJ/mol}$
3
Apply the definition of enthalpy change from the energy diagram:
4
$\Delta H = E_{\text{products}} - E_{\text{reactants}}$
5
Substitute the given values to calculate $Δ H$ :
6
$\Delta H = 196 - 40 = +156 \text{ kJ/mol}$
7
Interpret the sign: a positive $Δ H$ means the reaction absorbs energy, so it is endothermic.

Exam tip:

Always calculate $Δ H$ 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★★★☆☆⏱ 4 min

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.

📘 Definition

Activation Energy

\(E_a\)

The minimum energy required to overcome the energy barrier for a reaction step, measured as the difference between transition state energy and the energy of the starting species for that step.

The activation energy for the forward reaction is calculated as:

E_{a(\text{forward})} = E_{\text{transition state}} - E_{\text{reactants}}

For the reverse reaction (products converting back to reactants), the activation energy is:

E_{a(\text{reverse})} = E_{\text{transition state}} - E_{\text{products}}

Combining these definitions gives the relationship that lets you calculate any unknown value if you know the other two:

\Delta H = E_{a(\text{forward})} - E_{a(\text{reverse})}

📐 Worked Example

For the reaction $2 N H_{3} (g) ⇌ N_{2} (g) + 3 H_{2} (g)$ , $Δ H = + 92$ 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 $Δ H$ , forward $E_{a}$ , and reverse $E_{a}$ :
2
$\Delta H = E_{a(\text{forward})} - E_{a(\text{reverse})}$
3
Rearrange the equation to isolate the unknown reverse activation energy:
4
$E_{a(\text{reverse})} = E_{a(\text{forward})} - \Delta H$
5
Substitute the given values:
6
$E_{a(\text{reverse})} = 335 \text{ kJ/mol} - 92 \text{ kJ/mol} = 243 \text{ kJ/mol}$
7
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★★★★☆⏱ 3 min

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.

📘 Definition

Reaction Intermediate

A stable species that is formed in one early step of a multi-step reaction and consumed in a later step. On an energy diagram, intermediates are located at local energy minima (valleys) between two transition state peaks.

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 $Δ H$ 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 (with reactants as the 0 energy reference):
2
$E_{a,\text{Step 1}} = 85 - 0 = 85 \text{ kJ/mol}, \quad E_{a,\text{Step 2}} = 105 - 30 = 75 \text{ kJ/mol}$
3
Step 1 has a higher activation energy, so it is rate-determining.
4
(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.
5
(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 $Δ H$ of the reaction stays the same.

✓ Quick check

Test your understanding with these AP-style questions:

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 $Δ H$ ?
- A) $E_{a} = 75$ kJ/mol, $Δ H = - 25$ kJ/mol
- B) $E_{a} = 100$ kJ/mol, $Δ H = + 25$ kJ/mol
- C) $E_{a} = 75$ kJ/mol, $Δ H = + 25$ kJ/mol
- D) $E_{a} = 100$ kJ/mol, $Δ H = - 25$ kJ/mol

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

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 $Δ H$ as $E_{reactants} - E_{products}$ 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 $Δ H$ 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 $Δ H$ 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. Quick Reference Cheatsheet

Category	Formula/Rule	Notes
Overall Enthalpy Change	$Δ H = E_{products} - E_{reactants}$	$Δ H < 0$ = exothermic (products lower energy), $Δ H > 0$ = endothermic (products higher energy)
Forward Activation Energy	$E_{a (forward)} = E_{transition state} - E_{reactants}$	Always positive; overall $E_{a}$ equals $E_{a}$ of the rate-determining step
Reverse Activation Energy	$E_{a (reverse)} = E_{a (forward)} - Δ H$	Calculated from forward $E_{a}$ and overall enthalpy change
Transition State Identification	Local maximum (peak) on energy diagram	Unstable, cannot be isolated; one per reaction step
Intermediate Identification	Local minimum (valley) between two peaks	Stable, formed in one step and consumed in a later step
Catalyst Effect	$E_{a (catalyzed)} < E_{a (uncatalyzed)}, Δ H_{catalyzed} = Δ H_{uncatalyzed}$	Catalysts provide an alternate reaction pathway, do not change overall reaction energy

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
Identify intermediate on multi-step diagram
2022 · FRQ
Calculate ΔH and Ea from energy values

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.