Reaction Energy Profile: Study Guide (Chemistry 5.6)

Bonds break, bonds form

In Topic 5.5 you saw that a successful collision needs enough energy and the right orientation. Now look inside one successful collision. During an elementary step, some bonds in the reactants break and new bonds form. For example, in the step

OH⁻ + CH₃Br → CH₃OH + Br⁻

the C–Br bond breaks while a new C–O bond forms. These changes do not happen at one instant. As the hydroxide ion approaches, the C–O distance shrinks and the C–Br distance grows. The atoms pass through a whole series of arrangements on the way from reactants to products.

The reaction coordinate is the axis along which you plot this progress. It is not time and it is not a single bond length. It is a way of squeezing a complicated set of atomic motions (bonds stretching, angles bending, atoms moving closer) into one line from “reactants” on the left to “products” on the right.

The energy profile

An energy profile (reaction energy diagram) plots the energy of the reacting particles against the reaction coordinate. For one elementary step it has three landmarks:

1. Reactants on the left, at a flat level.
2. The transition state: the single highest point of the curve.
3. Products on the right, at another flat level.

Three energy gaps can be read from any profile:

Gap	From → to	Name
Reactants → transition state	climb on the left	Forward activation energy, Eₐ(forward)
Products → transition state	climb on the right	Reverse activation energy, Eₐ(reverse)
Reactants → products	net change	Overall energy change, ΔH (negative if products are lower)

Because all three are measured to the same three levels, they are linked:

Eₐ(reverse) = Eₐ(forward) − ΔH

If the products lie lower than the reactants (ΔH negative, exothermic), the reverse climb is the bigger one. If they lie higher (ΔH positive, endothermic), the forward climb is bigger.

The transition state

The transition state is the arrangement of atoms at the very top of the peak. Old bonds are partly broken and new bonds are partly formed, so it has the highest energy of any arrangement along the path. For OH⁻ + CH₃Br, the transition state has the O atom and the Br atom on opposite sides of the carbon, each joined to it by a partial bond: [HO···CH₃···Br]⁻, where ··· means a bond that is only partly there.

Two properties matter:

• The transition state is not a substance. It exists for roughly the time of one molecular vibration and cannot be isolated or stored.
• Reaching it is the whole challenge. The activation energy is exactly the energy that colliding particles need to get from reactants to the transition state. Once there, they can roll down to products (or back to reactants).

This links back to Topic 5.5: the area beyond Eₐ on a Maxwell–Boltzmann curve is the fraction of collisions with enough energy to reach the transition state.

Why there is a peak at all

Breaking a bond costs energy; forming a bond releases it. If the old bond broke completely before the new bond began to form, the particles would have to climb the full bond energy, and Eₐ would be enormous. In a real elementary step the two overlap: as the new bond starts to form, it pays back some of the energy spent stretching the old one. The peak is where the cost of weakening the old bond is greatest compared with the payback from the new one. That is why activation energies are usually much smaller than the energy of the bond that breaks.

Drawing a profile: what every sketch needs

When a question asks you to “draw” or “sketch” an energy profile for one elementary step, check for each of these:

• Both axes labelled: energy (kJ mol⁻¹) and reaction coordinate.
• Reactant and product levels labelled with the species, at the correct relative heights (products lower if ΔH is negative).
• One smooth peak, labelled transition state.
• An arrow for Eₐ(forward) from the reactant level to the top of the peak, not from the axis.
• An arrow for ΔH from the reactant level to the product level, if asked.

Worked example 1: reading a profile

Question. Use Figure 1 to find (a) the forward activation energy, (b) the overall energy change, and (c) the reverse activation energy. (d) Is the forward step exothermic or endothermic?

1. Read the three levels: reactants 40, transition state 125, products −15 kJ mol⁻¹.
2. (a) Eₐ(forward) = 125 − 40 = 85 kJ mol⁻¹.
3. (b) ΔH = −15 − 40 = −55 kJ mol⁻¹.
4. (c) Eₐ(reverse) = 125 − (−15) = 140 kJ mol⁻¹.
5. (d) The products are lower than the reactants, so the forward step is exothermic: more energy is released forming the new bond than is used breaking the old one.

Check. Eₐ(forward) − ΔH = 85 − (−55) = 140 kJ mol⁻¹, which matches (c). Notice the trap in the numbers: the transition state is at 125, but the activation energy is the difference of 85. The level of the axis zero is arbitrary; only differences have meaning.

Worked example 2: sketching a profile from data

Question. An elementary step has Eₐ(forward) = 92 kJ mol⁻¹ and ΔH = +35 kJ mol⁻¹. Find Eₐ(reverse) and sketch the energy profile.

1. Eₐ(reverse) = Eₐ(forward) − ΔH = 92 − 35 = 57 kJ mol⁻¹.
2. Set the reactants at 0 for convenience. Then the transition state is at 92 and the products at +35.
3. Sketch: a flat reactant level on the left; one smooth peak at 92; a flat product level on the right, higher than the reactants (endothermic).
4. Label: both axes (energy and reaction coordinate), reactants, products, transition state; an arrow from reactants up to the peak for Eₐ(forward); an arrow from reactants to products for ΔH.

Check. One peak only (it is a single elementary step); the right-hand climb (57) is smaller than the left-hand climb (92), as it must be when the products are higher. A sketch with two humps would describe a multistep reaction, which is a later topic.

Temperature and the Arrhenius equation

Topic 5.5 explained the effect of temperature with Maxwell–Boltzmann curves. The Arrhenius equation puts the same idea into one expression for the rate constant:

k = A e^(−Eₐ/RT)

• e^(−Eₐ/RT) is (approximately) the fraction of collisions with enough energy to reach the transition state. It grows quickly as T rises and shrinks quickly as Eₐ rises.
• A collects the other factors from the collision model: how often the particles collide and what share of collisions have a suitable orientation.
• T is in kelvin, and R is the gas constant (8.314 J mol⁻¹ K⁻¹), so Eₐ must be in J mol⁻¹ when used.

Your exam will not ask you to calculate with this equation. You do need to use it to explain three things:

1. Rate rises with temperature. A larger T makes −Eₐ/RT less negative, so k is larger.
2. A higher peak means a slower step. At the same temperature (and similar A), a larger Eₐ gives a smaller k.
3. A higher peak means a step more sensitive to temperature. When Eₐ is large, the fraction that can reach the transition state is tiny, and a small temperature rise multiplies it by a large factor.

Worked example 3: comparing two steps

Question. Steps P and Q have similar values of A. Step P has Eₐ = 40 kJ mol⁻¹; step Q has Eₐ = 100 kJ mol⁻¹. (a) Which step is faster at 298 K? (b) Which step’s rate increases by the greater factor when the temperature rises to 308 K? Explain both with the energy profile.

1. (a) Step P. Its peak is lower, so a much larger fraction of collisions have enough energy to reach its transition state. With similar A, its k is larger.
2. (b) Step Q. Its transition state sits far out in the tail of the Maxwell–Boltzmann curve. Warming the gas thickens that tail by a larger factor than it thickens the region near P’s lower Eₐ.

Background (not assessed). With equal A, the Arrhenius equation gives k(308 K)/k(298 K) ≈ 1.7 for step P and ≈ 3.7 for step Q, so the same 10 K rise more than triples the rate of the high-barrier step but does not double the other.

Common misconceptions

• “The activation energy is the height of the transition state on the axis.” It is the difference between the reactant level and the transition-state level.
• “An exothermic reaction must be fast.” ΔH says nothing about speed. Only Eₐ (and A) control the rate constant.
• “The transition state is an intermediate.” A transition state is a peak that cannot be isolated; an intermediate is a dip between peaks in a multistep reaction.
• “Heating the reaction lowers the peak.” The profile does not change with temperature. More collisions have enough energy to reach the same peak.
• “The forward and reverse activation energies are equal.” They differ by ΔH: Eₐ(reverse) = Eₐ(forward) − ΔH.
• “The reaction coordinate is time.” It shows progress through the atomic rearrangement, not the clock.
• “Bonds break first, then new bonds form.” In a single elementary step the breaking and forming overlap; that is why the transition state has partial bonds.

Where this leads

Topic 5.7 introduces reaction mechanisms: sequences of elementary steps, each with its own peak on a combined profile. Continue with Introduction to Reaction Mechanisms. Later, catalysts will be shown as a new pathway with a lower peak, and Unit 6 will connect ΔH to bond energies. Now try the practice questions, then use the revision notes and the checklist.