Reaction Energy Diagram & Catalysis

The graph plots potential energy in kJ/mol against reaction progress. A flat reactant plateau rises to a single peak labelled transition state and falls to a product plateau. Three vertical arrows mark Ea(fwd), Ea(rev) and $\Delta H$.

With the catalyst on, the simulator lowers the peak partway toward the higher plateau; the exact amount is a modelling choice, not a property you need to predict. What matters is what the catalyst does and does not change.

The enthalpy change is the energy of the products minus the energy of the reactants: $$\Delta H = E_{\text{products}} - E_{\text{reactants}}$$ Negative means exothermic (products lower); positive means endothermic (products higher).

The activation energy is measured from the starting plateau up to the transition state, in whichever direction you are going. Since both barriers share the same peak, $$E_{a,\text{rev}} = E_{a,\text{fwd}} - \Delta H$$

A catalyst provides a different reaction pathway with a lower-energy transition state. Because the peak drops while both plateaus stay where they are, $E_{a,\text{fwd}}$ and $E_{a,\text{rev}}$ fall by the same amount and $\Delta H$ is unchanged. A lower barrier means a larger fraction of collisions have enough energy to react, so the rate constant increases for both the forward and reverse reactions. The equilibrium constant is not changed.

Question. An exothermic reaction has $E_{a,\text{fwd}} = 60$ kJ/mol and $\Delta H = -20$ kJ/mol. (a) Find $E_{a,\text{rev}}$. (b) A catalyst lowers the forward barrier to 30 kJ/mol. Find the new $E_{a,\text{rev}}$ and $\Delta H$.

(a) $E_{a,\text{rev}} = E_{a,\text{fwd}} - \Delta H = 60 - (-20) = 80$ kJ/mol. The reverse reaction has to climb from the lower product plateau, so its barrier is larger.

(b) The catalyst lowered the peak by $60 - 30 = 30$ kJ/mol, so the reverse barrier drops by the same 30: $80 - 30 = 50$ kJ/mol. $\Delta H$ stays $-20$ kJ/mol, since the reactant and product energies have not moved. Check: $50 = 30 - (-20)$.

Check it in the simulator. Reset, then set E_a to 60 and leave |$\Delta H$| at 20. The readout shows E_a forward 60 kJ, reverse 80 kJ, $\Delta H$ $-20$ kJ. Tick Add catalyst: forward 30 kJ, reverse 50 kJ, $\Delta H$ still $-20$ kJ.

Endothermic version. Switch the menu to Endothermic with the catalyst off: forward 60 kJ, reverse 40 kJ, $\Delta H = +20$ kJ. Now the reverse barrier is the smaller one. With the catalyst on, both barriers drop by 20 kJ, to 40 kJ and 20 kJ.

Energy profiles appear in Unit 5 (Topic 5.6 for single steps, Topic 5.11 for catalysis) and Unit 6. Expect to label E_a and $\Delta H$ on a given diagram, sketch the effect of a catalyst, and, for multistep mechanisms, identify intermediates and the rate-determining step as the step with the highest barrier. The simulator draws one elementary step; a multistep profile is several humps joined by valleys.