Chemistry · General chemistry II · Concept

Potential energy diagrams for reactions

A potential energy diagram plots a reaction’s energy as the atoms move from reactants to products. The drop or rise between the two ends is $Δ H$ , and the climb from either end to the top of the hump is that direction’s activation energy. A catalyst lowers the hump without moving either end.

A concrete case

A balloon that waits for a spark

A balloon of hydrogen in air can float for hours without burning. Yet $2 H_{2} (g) + O_{2} (g) \to 2 H_{2} O (g)$ releases $483.6 kJ$ for every $2 mol$ of water formed. Touch a flame to it and it explodes.

Two different numbers are at work. The $483.6 kJ$ is how far the energy falls from the reactants to the products. Before any of it is released, the molecules must first climb a barrier, by breaking bonds that hold the $H_{2}$ and $O_{2}$ together. At room temperature almost no collisions carry enough energy to get over it. The flame supplies the climb, and the energy released by the first reactions pushes the next ones over.

Picture a ball in a hollow at the top of a hill. The bottom of the valley is far below, but the ball stays put until something lifts it over the rim.

The rule

What the diagram shows

The horizontal axis is reaction progress, how far the atoms have moved from the reactant arrangement toward the product arrangement. It is not time. The vertical axis is potential energy per mole.

The reactants and products sit on flat levels. $Δ H$ is the product level minus the reactant level. The path between them climbs over a hump, and its top is the transition state, the arrangement with bonds half broken and half formed. The forward activation energy, $E_{a}$ , is the climb from the reactant level to the top.

The reverse reaction climbs from the product level to the same top, so its barrier is the forward one minus $Δ H$ .

Picture it as a line drawn left to right. It runs flat at the reactant level, rises to one peak, falls past the start and runs flat again at a lower product level for an exothermic reaction. The Draw it in Energy diagrams button at the end of this lesson draws such a profile, with and without a catalyst.

Δ H = E_{products} - E_{reactants}, E_{a, rev} = E_{a} - Δ H

\begin{array}{c} Δ H = E_{products} - E_{reactants} \\ E_{a, rev} = E_{a} - Δ H \end{array}

Reading a potential energy diagram
Feature	Measured from	To
$Δ H$	Reactant level	Product level
$E_{a}$ (forward)	Reactant level	Top of the hump
$E_{a}$ (reverse)	Product level	Top of the hump
Transition state	Not a level	The top of a hump
Intermediate	Not a level	A valley between two humps

Exothermic, endothermic, and a catalyst

If the products sit below the reactants, $Δ H$ is negative and the reaction is exothermic. If they sit above, $Δ H$ is positive and the reaction is endothermic. For an endothermic reaction the forward barrier is the larger one, because the top of the hump must also lie above the products.

A catalyst gives the reaction a different path with a lower top. It lowers the forward and reverse barriers by the same amount. It does not move the reactant or product level, so $Δ H$ stays the same.

E_{a} \geq Δ H (endothermic)

Why it works

Why a catalyst cannot change $Δ H$ or $K$

The reactant and product levels belong to the substances themselves. $Δ H$ depends only on where the path starts and ends, not on the route between them, so a catalyst’s new route cannot change it.

Both directions of the reaction cross the same top. Lowering that top by $15 kJ / mol$ lowers both barriers by $15 kJ / mol$ . With the same frequency factor, each rate constant then grows by the same factor, $e^{15, 000 / R T}$ . The equilibrium constant is the ratio of the forward and reverse rate constants, so it does not change. Equilibrium is reached sooner, at the same position.

A mechanism with several steps draws one hump per step. Each valley between two humps is an intermediate, a real substance that lives for a while. When the early steps are fast and reversible, the step whose top stands highest above the starting reactants sets the rate, even if its own climb from the valley is small.

\frac{k_{cat}}{k} = e^{(E_{a} - E_{a, cat}) / R T} (same A)

\begin{array}{c} \frac{k_{cat}}{k} = e^{(E_{a} - E_{a, cat}) / R T} \\ (same A) \end{array}

The method

Reading a diagram

1. Mark the reactant level, the product level and the top of every hump.

2. $Δ H$ is the product level minus the reactant level. Write its sign.

3. The forward $E_{a}$ is the highest top minus the reactant level.

4. The reverse $E_{a}$ is the same top minus the product level, or $E_{a} - Δ H$ .

5. For several humps, find the highest top above the reactants. Its step is the slow one when the earlier steps are fast. Each valley between humps is an intermediate.

6. For a catalyst, draw the new, lower top and leave both end levels where they were.

Where it breaks

The horizontal axis is not time. A tall hump drawn wide does not mean a slow reaction, and a reaction can spend almost all its time in the reactant valley.

A transition state is not an intermediate. The top of a hump lasts about as long as one molecular vibration and can never be bottled. A valley between humps is an intermediate that can sometimes be detected.

The diagram shows energy, not whether a reaction goes. Whether products are favored depends on $Δ G$ , which also counts entropy, so an endothermic reaction can still be favored.

A catalyst cannot make an endothermic reaction’s forward barrier smaller than $Δ H$ . The top of any path must still lie above both ends.

Numbers on a classroom diagram are often schematic. An $E_{a}$ measured from an Arrhenius plot is close to the drawn barrier, but the two are defined differently and need not agree exactly.

Check your answer

Check the gap. The reverse barrier minus the forward barrier must equal $- Δ H$ . With $E_{a} = 50 kJ / mol$ and $E_{a, rev}$ $= 90 kJ / mol$ , the gap is $40 kJ / mol$ and $Δ H$ is $- 40 kJ / mol$ .

Check the signs. Products below reactants means a negative $Δ H$ . A forward barrier smaller than a positive $Δ H$ is impossible.

Check the catalyst. The catalyzed and uncatalyzed paths must start and end on the same levels, and both of their barriers must drop by the same amount.

In a Chemistry box, write $Δ H$ and each $E_{a}$ as a given on its own row, then the relation, such as $E_{a, rev} = E_{a} - Δ H$ , then the values put in. In the worked example’s box the Checker reads that row as “Substitution matches row 4 with the values defined above.”

Which step is slow, or whether a level is a transition state or an intermediate, is a reading of the diagram. A Chemistry box checks the numbers you add and subtract, and the words are yours to give.

Surfaces, Hammond, and free energy

The diagram is a slice through a potential energy surface, which gives the energy for every arrangement of the atoms. The reaction path follows the valley floor, and the transition state is a saddle point, the highest point of the lowest pass between the two valleys.

The Hammond postulate says a transition state resembles whichever neighbor it lies closer to in energy. In a strongly exothermic step the top comes early and looks like the reactants. In a strongly endothermic step it comes late and looks like the products.

Chemists also draw free-energy diagrams, with $Δ G$ on the vertical axis. Their barriers give rate constants through transition-state theory, and the gap between the ends gives the equilibrium constant.

Common mistakes

Reading the horizontal axis as time. It is reaction progress, the change in the atoms’ arrangement. A wide hump is not a slow reaction, and the rate depends on the barrier’s height alone.
Measuring the forward $E_{a}$ from the product level, or from the axis. The forward climb starts at the reactant level, so for reactants at $60 kJ / mol$ and a top at $125 kJ / mol$ , $E_{a}$ is $65 kJ / mol$ , not $125$ .
Saying a catalyst changes $Δ H$ . A catalyst lowers the top of the hump only. The reactant and product levels, and the $40 kJ / mol$ between them, stay where they were.
Saying a catalyst shifts the equilibrium toward products. It speeds the forward and reverse reactions by the same factor, so $K$ is unchanged. Equilibrium is only reached sooner.
Drawing an endothermic reaction with $E_{a}$ smaller than $Δ H$ . With $Δ H = + 30 kJ / mol$ the top must be at least $30 kJ / mol$ above the reactants, because it must lie above the products too.
Calling a transition state an intermediate. The transition state is the top of a hump and cannot be isolated. An intermediate sits in a valley between two humps and lasts long enough to be detected.
Choosing the slow step by the largest single climb. In a two-step path with tops at $50$ and $65 kJ / mol$ above the reactants, the second step is slow even though its own climb, $25 kJ / mol$ , is the smaller one.

Key terms

Exothermic reaction: A reaction that releases heat to its surroundings, so its enthalpy change is negative $(Δ H < 0)$ and the surroundings warm.
Endothermic reaction: A reaction that absorbs heat from its surroundings, so its enthalpy change is positive $(Δ H > 0)$ and the surroundings cool.
Transition state: The highest-energy arrangement of atoms in one elementary step (a single collision or rearrangement in a reaction mechanism), with bonds partly broken and partly formed. It lasts too briefly to isolate, unlike an intermediate, which sits in an energy valley between steps.
Activated complex: The unstable, high-energy arrangement of atoms at the top of the energy barrier, the transition state. Unlike an intermediate, it can’t be isolated.
Catalyst: A substance that speeds up a reaction by giving it a lower-energy pathway and is not used up. It does not change the equilibrium constant or where equilibrium lies.
Arrhenius equation: $k = A \cdot e^{- E_{a} / R T}$ : how a rate constant $k$ grows with temperature, where $A$ is the frequency factor, $E_{a}$ the activation energy, $R$ the gas constant and $T$ the kelvin temperature. A plot of $ln k$ against $1 / T$ is a line with slope $- E_{a} / R$ , which gives the activation energy.
Reaction intermediate: A substance made in one step of a mechanism and used up in a later step, so it does not appear in the overall equation.
Rate-determining step: The slowest step in a mechanism, which limits how fast the whole reaction goes. A fast equilibrium before it also shapes the overall rate law.

Work through an example

A reaction has $Δ H = - 40 kJ / mol$ and $E_{a} = 50 kJ / mol$ . A catalyst lowers the forward barrier to $35 kJ / mol$ . Find the reverse barriers with and without the catalyst, say what happens to $Δ H$ and $K$ , and estimate how much faster the catalyzed reaction is at $298 K$ .

Read a catalyzed potential energy diagram →

Find the slow step on a two-step diagram →

Sources and scope

Authored study material. Tool results depend on the stated inputs and model assumptions.

After reading

Study it, then try it

Practice the key terms as flashcards, then open the example in its tool, change a value and keep a useful result on your board.

Study the 8 key terms Check the reverse barriers in a Chemistry box Draw it in Energy diagrams Open worked example on a board Chemistry formulas: kinetics

Your existing work stays on this device. Examples open as editable copies.