Energy Profile Diagrams

30 min

Downhill, but it waits

Hydrogen peroxide decomposes exothermically into water and oxygen, releasing 98 kJ for every mole. Yet a bottle of dilute hydrogen peroxide from a pharmacy lasts for months. Add a little potassium iodide and the same liquid foams furiously, fast enough to push a column of foam out of a bottle. The reaction was always "downhill" in energy; what changed is the size of the hill that had to be climbed first. An energy profile diagram shows both: the overall energy change and the barrier in the way.

Definition

Energy profile diagram (reaction profile): a graph of the enthalpy (energy) of the reacting particles against the reaction coordinate, the progress of the reaction from reactants to products.

Activation energy, $E_{a}$ : the minimum energy that colliding particles must have for a reaction to occur; on the diagram, the height of the peak above the reactants.

Transition state (activated complex): the arrangement of atoms at the top of the energy barrier, in which old bonds are partly broken and new bonds partly formed. It cannot be isolated.

Catalyst: a substance that increases the rate of a reaction by providing an alternative route with a lower activation energy, and is chemically unchanged at the end.

An exothermic profile, drawn to scale

Figure 17.4: Energy profile for $\mathrm{H_2O_2(l) \rightarrow H_2O(l) + \frac{1}{2}O_2(g)}$ , in kJ mol⁻¹, drawn to scale. ΔH from enthalpies of formation; activation energies from LibreTexts (data only).

The products lie below the reactants: ΔH is negative (−98.0 kJ mol⁻¹).
The activation energy is measured from the reactants up to the peak: 75.3 kJ mol⁻¹ without a catalyst.
With iodide ions the peak is lower (56.5 kJ mol⁻¹), so many more collisions have enough energy and the reaction is much faster. The start and end levels do not move: a catalyst does not change ΔH.
The activation energy of the reverse reaction is measured from the products to the peak: $E_a(\text{reverse}) = E_a(\text{forward}) - \Delta H = 75.3 - (-98.0) = 173.3\ \mathrm{kJ\,mol^{-1}}$ so the reverse reaction is very much harder.

An endothermic profile, drawn to scale

Figure 17.5: Energy profile for $\mathrm{2HI(g) \rightarrow H_2(g) + I_2(g)}$ , in kJ per mole of equation, drawn to scale. ΔH from enthalpies of formation; activation energies from Doc Brown's Chemistry (data only).

The products lie above the reactants: ΔH is positive, here only +9.4 kJ for the equation as written, from $\Delta H = \Delta_f H(\mathrm{I_2,g}) - 2\Delta_f H(\mathrm{HI,g}) = 62.4 - 2(26.5)$ .
The barrier is far larger than the energy change: 183 kJ without a catalyst. On a platinum surface it falls to 58 kJ, and on gold to 105 kJ.
For an endothermic reaction $E_{a}$ is always at least as large as ΔH, because the peak must be at least as high as the products.

How to draw an energy profile diagram

Draw axes: enthalpy (kJ mol⁻¹) up; reaction coordinate across.
Draw the reactant level on the left and the product level on the right, at the right heights for ΔH (lower for exothermic, higher for endothermic).
Join them with a smooth hump whose top is $E_{a}$ above the reactants; label the peak "transition state".
Draw vertical arrows for $E_{a}$ (reactants to peak) and ΔH (reactants to products), with values and signs.
For a catalysed reaction, add a lower hump between the same two levels.

Reactions that happen in several steps have one hump for each step; the dips between humps are intermediates, which you will study with reaction mechanisms in Senior 6.

Worked example 1: reading a profile

Use Figure 17.4 to give (a) ΔH, (b) $E_{a}$ for the uncatalysed forward reaction, (c) $E_{a}$ for the reverse reaction with the catalyst present, and (d) the effect of iodide ions on ΔH.

Solution. (a) Products are 98.0 kJ below reactants: ΔH = −98.0 kJ mol⁻¹. (b) 75.3 kJ mol⁻¹. (c) The catalysed peak is 56.5 kJ above the reactants, so it is $56.5 + 98.0 = 154.5\ \mathrm{kJ\,mol^{-1}}$ above the products. (d) None: the catalyst lowers both barriers by the same amount (18.8 kJ) but leaves the reactant and product levels where they are.

Worked example 2: from data to a diagram

For $\mathrm{N_2(g) + 3H_2(g) \rightarrow 2NH_3(g)}$ , $\Delta_f H(\mathrm{NH_3,g}) = -46.1\ \mathrm{kJ\,mol^{-1}}$ . Without a catalyst $E_a \approx 350\ \mathrm{kJ\,mol^{-1}}$ ; on a tungsten catalyst it is about 162 kJ mol⁻¹ (Doc Brown's Chemistry). Find ΔH, describe the profile, and calculate $E_{a}$ for the reverse reaction on tungsten.

Solution. $\Delta H = 2(-46.1) - 0 = -92.2\ \mathrm{kJ}$ Exothermic: products 92.2 kJ below reactants; a tall hump 350 kJ above the reactants, and a lower catalysed hump 162 kJ above them. Reverse on tungsten: $E_a(\text{reverse}) = 162 - (-92.2) = 254\ \mathrm{kJ}$ The very high uncatalysed barrier (the N≡N triple bond must be weakened) is why the Haber process needs an iron catalyst even though the reaction is exothermic.

Warning

Common mistakes. (1) Measuring $E_{a}$ from the zero of the axis or from the products: it is measured from the reactants to the peak. (2) Drawing a catalysed profile that ends at a different product level: a catalyst cannot change ΔH. (3) Confusing an enthalpy level diagram (start and end only) with an energy profile (which shows the barrier). (4) Thinking a large negative ΔH means a fast reaction: speed depends on $E_{a}$ , not on ΔH.

Safety. The elephant's toothpaste demonstration uses concentrated hydrogen peroxide (often 30%), which burns the skin and eyes, and the foam is hot: it is done only by a teacher, wearing gloves and goggles, on a tray, with students standing back. School practicals use dilute (3–6%) hydrogen peroxide.

Tip

"Ea from the start; ΔH from start to finish." Exothermic: finish lower. Endothermic: finish higher. Catalyst: lower hump, same start and finish. Reverse barrier = Ea − ΔH.

The Organic Chemistry Tutor: Potential Energy Diagrams, Catalyst, Endothermic and Exothermic Reactions (12 min). Watch how ΔH, activation energy and the effect of a catalyst are read from each diagram.

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Practice

Define activation energy and transition state.
Sketch (and label) the energy profile for an exothermic reaction, showing $E_{a}$ , ΔH and the transition state.
Explain why a catalyst speeds up a reaction but does not change ΔH.
A reaction has ΔH = −60 kJ mol⁻¹ and $E_a = 45\ \mathrm{kJ\,mol^{-1}}$ . Calculate $E_{a}$ for the reverse reaction.
A reaction has ΔH = +50 kJ mol⁻¹. Explain why its activation energy cannot be 30 kJ mol⁻¹.
Use Figure 17.5 to give $E_{a}$ for $\mathrm{H_2(g) + I_2(g) \rightarrow 2HI(g)}$ on a platinum surface.
Methane in air does not burn until a spark is applied, although its combustion is very exothermic. Explain with an energy profile.
Using $\Delta_f H$ values ( $\mathrm{CaCO_3}$ −1206.9, CaO −635.1, $\mathrm{CO_2}$ −393.5 kJ mol⁻¹), calculate ΔH for $\mathrm{CaCO_3 \rightarrow CaO + CO_2}$ and state the least possible value of $E_{a}$ .

Show answersHide answers

$E_{a}$ : the minimum energy colliding particles need to react (height of the barrier above the reactants). Transition state: the highest-energy arrangement on the reaction path, with bonds partly broken and partly formed.
Reactant level high on the left, product level lower on the right, a hump between; $E_{a}$ arrow from reactants to peak; ΔH arrow down from reactants to products; peak labelled transition state.
It provides a different route with a lower peak, so more collisions have enough energy; but the reactants and products, and so their enthalpies, are the same, so ΔH is unchanged.
$E_a(\text{reverse}) = 45 - (-60) = 105\ \mathrm{kJ\,mol^{-1}}$ .
The products are 50 kJ above the reactants, so the peak must be at least 50 kJ above the reactants: $E_a \ge 50\ \mathrm{kJ\,mol^{-1}}$ .
$58 - 9.4 = 48.6\ \mathrm{kJ}$ (from the products' level to the catalysed peak).
Methane and oxygen must first climb a large activation-energy barrier; at room temperature too few collisions have that energy. The spark supplies it to a few molecules; the heat released then lets neighbouring molecules over the barrier, so the burning continues.
$\Delta H = (-635.1 - 393.5) - (-1206.9) = +178.3\ \mathrm{kJ\,mol^{-1}}$ ; $E_{a}$ must be at least 178.3 kJ mol⁻¹.

Quick check

On an energy profile diagram, how is the activation energy measured?

Key points

An energy profile plots enthalpy against reaction progress: reactant level, a hump (transition state), product level.
$E_{a}$ = peak − reactants; ΔH = products − reactants (negative if products are lower, positive if higher); $E_a(\text{reverse}) = E_a - \Delta H$ .
A catalyst gives a lower hump between the same two levels: lower $E_{a}$ , same ΔH. Iodide lowers $E_{a}$ for $\mathrm{H_2O_2}$ decomposition from 75.3 to 56.5 kJ mol⁻¹; platinum lowers it for HI from 183 to 58 kJ.
For an endothermic reaction, $E_a \ge \Delta H$ . The rate depends on $E_{a}$ , not on ΔH.

1Downhill, but it waits

2An exothermic profile, drawn to scale

3An endothermic profile, drawn to scale

4How to draw an energy profile diagram

Downhill, but it waits

An exothermic profile, drawn to scale

An endothermic profile, drawn to scale

How to draw an energy profile diagram