Multistep Reaction Energy Profile: Study Guide (Chemistry 5.10)

From one hump to several

In Topic 5.6 you drew the energy profile of a single elementary step. The curve starts at the reactants, climbs to one transition state at the top, then falls to the products. The height of the climb is the activation energy, Ea, and the drop from start to finish is the energy change, ΔH.

Most reactions happen through a mechanism with several elementary steps (Topic 5.7). Each step has its own transition state and its own activation energy. If you know the energetics of every step, you can build the profile of the whole reaction by joining the single-step profiles end to end:

• Step 1 starts at the reactants and ends at the first intermediate.
• Step 2 starts at that intermediate and ends at the next intermediate (or at the products).
• And so on, until the last step ends at the products.

So the profile of a two-step reaction has two humps, a three-step reaction has three humps, and so on.

Anatomy of a multistep profile

Feature on the profile	What it represents	How many?
A peak (maximum)	A transition state: bonds partly broken and partly formed	One per elementary step
A valley between two peaks (local minimum)	An intermediate: a real species, made in one step and used in the next	Steps − 1 (for steps in a chain)
Start of the curve	Reactants of the overall reaction	1
End of the curve	Products of the overall reaction	1

Two points about the valleys:

• An intermediate usually sits above the reactants, because it is less stable. It is still a minimum, so it lasts long enough to collide with something else. A transition state lasts only for the instant of the collision.
• The deeper an intermediate’s valley, the more stable that intermediate is compared with what comes before and after it.

Reading the numbers

Activation energy of each step. Measure from the valley where the step starts up to that step’s peak:

• Ea for step 1 = (energy of TS 1) − (energy of reactants)
• Ea for step 2 = (energy of TS 2) − (energy of the intermediate)

Do not measure every barrier from the reactants. Step 2 starts at the intermediate, so the intermediate is its baseline.

Energy change of each step. ΔH for a step = (energy at its end) − (energy at its start). A step that goes downhill is exothermic (ΔH negative). A step that goes uphill is endothermic (ΔH positive).

Overall energy change. ΔH overall = (energy of products) − (energy of reactants). This always equals the sum of the steps’ ΔH values, because the intermediates cancel out, just as they cancel when you add the steps’ equations. The overall ΔH does not depend on the route.

Reverse barriers. The activation energy for the reverse of a step is measured from the valley after the step, back up to the same peak. For any step, Ea(reverse) = Ea(forward) − ΔH.

Which step is rate-limiting?

A taller barrier means a smaller fraction of collisions have enough energy to get over it (Topic 5.5). So the step with the largest activation energy is the slowest, and it is the rate-limiting step. In Figure 1, step 2 has the larger barrier, so step 2 is rate-limiting.

Background, beyond this topic: the “largest barrier” rule is a simplification. When an intermediate sits very high, only just below the peak before it, chemists also compare the heights of the transition states themselves. In every profile on these pages, both ways of judging pick the same step.

You can also read the profile in the other direction, from a mechanism to a shape:

• First step slow, later steps fast: the first hump is the tallest climb. The rate law comes straight from step 1 (Topic 5.8).
• First step fast and reversible, second step slow: the first hump is low. The intermediate’s valley sits close below TS 1, so the intermediate can easily slide back to the reactants, while the climb to TS 2 is much higher. This is the shape of a pre-equilibrium (Topic 5.9): the intermediate goes back and forth over the first barrier many times before one crosses the second barrier.

Worked example 1: reading a two-step profile

Question. Use Figure 1 (reactants 0, TS 1 at 62, intermediate at 25, TS 2 at 98, products at −40, all in kJ mol⁻¹). Find (a) the number of steps and intermediates, (b) the activation energy and ΔH of each step, (c) the overall ΔH, (d) the rate-limiting step and (e) the activation energy for the reverse of step 2.

1. (a) Two peaks, so two elementary steps. One valley between them, so one intermediate.
2. (b) Step 1: Ea = 62 − 0 = 62 kJ mol⁻¹; ΔH = 25 − 0 = +25 kJ mol⁻¹ (endothermic). Step 2: Ea = 98 − 25 = 73 kJ mol⁻¹; ΔH = −40 − 25 = −65 kJ mol⁻¹ (exothermic).
3. (c) ΔH overall = −40 − 0 = −40 kJ mol⁻¹. Check: +25 + (−65) = −40. The reaction is exothermic overall even though step 1 is endothermic.
4. (d) 73 > 62, so step 2 has the larger barrier and is rate-limiting.
5. (e) The reverse of step 2 starts at the products (−40) and climbs to TS 2 (98): 98 − (−40) = 138 kJ mol⁻¹. Check with the rule: 73 − (−65) = 138.

Common slip. Reading step 2’s barrier as 98 kJ mol⁻¹ (from the reactants) instead of 73 kJ mol⁻¹ (from the intermediate). You still pick step 2 here, but 98 kJ mol⁻¹ is not step 2’s activation energy, so any question that asks for the value loses the mark (see Practice Question 2).

Link to Topic 5.9. The intermediate’s way back over TS 1 is 62 − 25 = 37 kJ mol⁻¹, much lower than its way forward over TS 2 (73 kJ mol⁻¹). So the intermediate returns to reactants far more often than it goes on, which is the situation where a pre-equilibrium approximation is reasonable.

Worked example 2: drawing a profile from step data

Question. A hypothetical reaction has three elementary steps with these energetics (kJ mol⁻¹):

Step	Ea (forward)	ΔH	Speed
1	30	+20	fast, reversible
2	85	−15	slow
3	18	−55	fast

Draw and label the energy profile, taking the reactants as 0 kJ mol⁻¹, and state the overall ΔH.

Step 1: turn the data into heights. Start at 0 and work along, adding Ea to reach each peak and ΔH to reach each valley.

Point	Working	Energy (kJ mol⁻¹)
Reactants	start	0
TS 1	0 + 30	30
Intermediate I₁	0 + 20	20
TS 2	20 + 85	105
Intermediate I₂	20 − 15	5
TS 3	5 + 18	23
Products	5 − 55	−50

Step 2: draw. Three humps; two valleys (I₁ and I₂) between them; the middle hump is by far the tallest.

Step 3: check. Overall ΔH = −50 − 0 = −50 kJ mol⁻¹, and the steps add up: +20 − 15 − 55 = −50. ✓ The largest barrier (85) belongs to step 2, matching “slow” in the table. ✓

What a good sketch must show. On an exam you will usually sketch rather than plot. Get the relative heights right: the right number of humps, the tallest climb on the slow step, each intermediate valley above or below the reactants as the data say, and the products at the correct side of the reactants. Label the axes (energy; reaction coordinate), the transition states, the intermediates, and at least the activation energy and ΔH you are asked for.

Interpretation. Step 1 has a reverse barrier of only 10 kJ mol⁻¹ (30 − 20), so I₁ readily falls back to reactants. That fits step 1 being “fast, reversible”, a pre-equilibrium ahead of the slow step 2. Step 3 comes after the slow step, so it has no effect on the rate law.

Common misconceptions

• “Count the valleys to get the number of steps.” Count the peaks. Each elementary step has exactly one transition state. Valleys between peaks are intermediates.
• “Intermediates and transition states are the same thing.” An intermediate sits in a valley and can, in principle, be detected. A transition state sits at a peak and cannot be isolated.
• “Measure every activation energy from the reactants.” Each step’s barrier is measured from the valley where that step begins.
• “The first step is always the slow one.” The slow step is whichever has the largest barrier. It can be any step.
• “The biggest drop is the slowest step.” A large downhill drop is a large negative ΔH. It says nothing about how fast that step is.
• “More steps means a different overall ΔH.” Overall ΔH depends only on the reactants and products.
• “An exothermic reaction cannot have an endothermic step.” It can: in Worked example 1, step 1 is uphill but the reaction is downhill overall.
• “The x-axis is time.” It is the reaction coordinate: progress through the bond changes.

Where this leads

Topic 5.11 shows how a catalyst changes the profile: it provides a new route, usually with more steps but lower barriers, and new intermediates. The overall ΔH stays the same. Unit 6 then looks at ΔH itself in more detail. Continue to catalysis when you are ready, or revise the pre-equilibrium approximation. Try the practice questions now, then use the revision notes and the checklist to consolidate.