Master AP Chemistry Topic 5.6 with this free practice quiz. Test your skills on energy profiles, transition state, and enthalpy of reaction to prepare for your AP exam.
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Reading the Diagram
A reaction energy profile plots potential energy on the vertical axis against reaction progress on the horizontal axis. Reaction progress is the journey from reactants to products; it is not time. Reactants sit at the left, products at the right.
The peak of the curve is the transition state, also called the activated complex: the highest-energy arrangement of the atoms as old bonds break and new ones form. It exists for an instant and cannot be isolated. On a multi-step diagram each elementary step contributes its own peak.
A valley between two peaks is an intermediate, a real species formed in one step and consumed in the next. It sits in a local energy minimum, which is why it can exist briefly while a transition state cannot. Count the peaks to count the steps, and count the valleys to count the intermediates: a mechanism with n steps shows n peaks and n − 1 valleys.
Activation Energy, ΔH, and the Reverse Reaction
The activation energy of the forward reaction, Eₐ, is the climb from the reactants up to the peak: Eₐ = E(peak) − E(reactants). It is always measured from the level the step starts at, never as the absolute height of the peak.
The enthalpy change compares only the two ends of the diagram: ΔH = E(products) − E(reactants). If the products sit lower than the reactants, ΔH is negative and the reaction is exothermic. If the products sit higher, ΔH is positive and the reaction is endothermic. The peaks and valleys in between contribute nothing to ΔH.
For the reverse reaction, the old products become the starting level, so the reverse activation energy is Eₐ(reverse) = E(peak) − E(products). The three quantities obey one relation: Eₐ(forward) − Eₐ(reverse) = ΔH. Reversing an exothermic reaction gives an endothermic one with a larger activation energy, because the climb back starts from the lower product level.
Raising the temperature changes none of the labeled energies: reactant level, product level, peak and ΔH all stay put. Temperature changes only how many particles can cross the barrier.
Catalysts and Multi-Step Diagrams
A catalyst provides an alternative pathway with a lower peak. Drawn on the same axes, the catalyzed and uncatalyzed routes share identical reactant and product levels, so ΔH is exactly the same for both. Only the barrier changes, which is why the catalyzed reaction is faster. The catalyzed route may pass through a different number of steps and different intermediates; that does not affect ΔH.
In a multi-step diagram, each step’s activation energy is measured from the level immediately before its peak: the reactant level for step 1, and the valley of the preceding intermediate for every later step. The rate-determining step is the step with the largest such barrier, because that climb is crossed most slowly and throttles everything behind it.
Worked Examples
Example 1: one step, three quantities
A diagram shows reactants at 20 kJ/mol, a peak at 80 kJ/mol and products at 50 kJ/mol.
Eₐ(forward) = 80 − 20 = 60 kJ/mol. ΔH = 50 − 20 = +30 kJ/mol, so the reaction is endothermic. Eₐ(reverse) = 80 − 50 = 30 kJ/mol. Check with the relation: 60 − 30 = +30 = ΔH. Consistent.
Example 2: a two-step profile
Reactants sit at 0 kJ/mol, the first peak at 90 kJ/mol, the intermediate valley at 40 kJ/mol, the second peak at 150 kJ/mol and the products at −20 kJ/mol.
Step 1 barrier: 90 − 0 = 90 kJ/mol. Step 2 barrier: 150 − 40 = 110 kJ/mol. Step 2 has the larger barrier, so step 2 is rate-determining. ΔH = −20 − 0 = −20 kJ/mol; the reaction is exothermic overall despite the large barriers.
Traps Specific to This Topic
Reading ΔH off the peak is the most common misread. The peak height relative to reactants is Eₐ. ΔH uses products minus reactants and ignores the peak entirely, so a reaction can have a huge Eₐ and a ΔH of either sign.
Measuring a later step’s barrier from the reactant level overstates or understates it. Step 2 starts at the intermediate valley, so its barrier is peak minus valley. In Example 2, measuring from 0 would give 150 kJ/mol instead of the correct 110 kJ/mol.
A catalyst does not make ΔH more negative. It lowers the peak only; reactant and product energies are properties of the substances, not of the path, so both routes on one diagram must share the same two end levels.
Peaks and valleys are not interchangeable labels. The peak is a transition state that cannot be isolated; the valley is an intermediate that can, in principle, be detected. Calling the peak an intermediate reverses both definitions.
The first step is not automatically rate-determining. Compare every barrier, each measured from its own starting level, and pick the largest.