Topic 5.6 Reaction Energy Profile: interpret a potential-energy diagram to identify the activation energy of the forward and reverse reactions, the transition state and the enthalpy of reaction.

What this topic is asking

The College Board (Topic 5.6) wants you to read a potential-energy diagram (reaction energy profile) and identify the activation energy of the forward and reverse reactions, the transition state, and the enthalpy of reaction. These diagrams unify kinetics (the barrier height, which controls rate) and thermodynamics (the energy difference between reactants and products) on one picture.

Reading the diagram

The transition state is the highest-energy point on the path, an unstable arrangement in which old bonds are partly broken and new bonds partly formed. It is not an isolable species and not the same as an intermediate, which sits in a valley between two peaks (Topic 5.10).

Activation energies, forward and reverse

Because both reactions go over the same barrier, the peak is shared. For an exothermic reaction the products are lower than the reactants, so the reverse barrier is taller than the forward one; for an endothermic reaction the forward barrier is taller. Reading off both activation energies from a labelled diagram is a routine FRQ skill.

Enthalpy of reaction

The enthalpy of reaction is the vertical difference between products and reactants:

If the products lie below the reactants the reaction is exothermic ($\Delta H < 0$); if above, endothermic ($\Delta H > 0$). This connects to Unit 6: the height of the barrier (kinetics) is independent of the sign of $\Delta H$ (thermodynamics), so a strongly exothermic reaction can still be slow if its activation energy is high.

The effect of a catalyst

A catalyst provides an alternative pathway with a lower transition-state energy, so it lowers both the forward and reverse activation energies by the same amount and speeds both directions. Crucially it leaves the reactant and product energies untouched, so the enthalpy of reaction and the position of equilibrium are unchanged. The catalyst is drawn as a lower (or split) peak on the same diagram.

Try this

Q1. A reaction has forward $E_a = 90.\ \text{kJ mol}^{-1}$ and $\Delta H = -30.\ \text{kJ mol}^{-1}$. Calculate the reverse activation energy. [2 points]

• Cue. $E_{a,\text{reverse}} = E_{a,\text{forward}} - \Delta H = 90. - (-30.) = 120.\ \text{kJ mol}^{-1}$.

Q2. Explain why a catalyst changes the activation energy but not the enthalpy of reaction. [2 points]

• Cue. It lowers the transition-state energy (a new pathway), but the reactant and product energies are fixed, so $\Delta H = E_\text{products} - E_\text{reactants}$ is unchanged.