How Does a Catalyst Speed Up Both Forward and Reverse Reactions?

A catalyst makes a reaction reach equilibrium faster; it does not make the equilibrium favor products or reactants. The apparent puzzle is that a catalyst can speed the forward and reverse reactions at once. The key is that reaction rate depends on the height of the barrier between states, while equilibrium depends on the free-energy difference between those states.

A catalyst changes the pathway, not the destination

For a reversible reaction such as

$$\mathrm{A} \rightleftharpoons \mathrm{B},$$

A and B have a free-energy difference, and converting one into the other requires crossing an activation barrier. A catalyst offers a different sequence of steps—often involving temporary catalyst-bound intermediates—that reaches the same products and regenerates the catalyst overall.

The catalyst does not need to lower the energy of A or B. Instead, it lowers the barrier along the alternative route. Since the pathway can be traversed from A to B or from B to A, it provides a lower-barrier route for both directions.

A useful energy diagram has two horizontal levels for A and B and a peak between them. A catalyst lowers the peak, but leaves the vertical separation between A and B unchanged. That separation determines the reaction’s thermodynamic preference; the peak heights determine how quickly the reaction proceeds.

Why both rates increase

In transition-state theory, a rate constant is approximately proportional to an exponential of the negative activation free energy:

$$k \propto \exp\!\left(-\frac{\Delta G^{\ddagger}}{RT}\right),$$

where $\Delta G^{\ddagger}$ is the activation free energy, $R$ is the gas constant, and $T$ is the absolute temperature. Lowering the barrier therefore increases the rate constant, often substantially.

For a single reversible elementary step that passes through the same transition state in both directions, the forward and reverse activation barriers are related to the free-energy change of the reaction. Using a consistent standard-state convention,

$$\Delta G^{\ddagger}_{\mathrm{fwd}}-\Delta G^{\ddagger}_{\mathrm{rev}}=\Delta G^{\circ}_{\mathrm{rxn}}.$$

Because a catalyst does not change $\Delta G^{\circ}_{\mathrm{rxn}}$, it cannot lower one barrier while changing their difference. If the transition state is lowered relative to both endpoints, both barriers decrease. The exact observed rate changes can depend on the mechanism and conditions, but catalysis does not create a thermodynamic push in one direction.

Worked example with barrier heights

Suppose B is more stable than A by $10\ \mathrm{kJ\,mol^{-1}}$, so $\Delta G^{\circ}_{\mathrm{rxn}}=-10\ \mathrm{kJ\,mol^{-1}}$. If the uncatalyzed forward barrier is $80\ \mathrm{kJ\,mol^{-1}}$, the reverse barrier is $90\ \mathrm{kJ\,mol^{-1}}$. Their difference is $-10\ \mathrm{kJ\,mol^{-1}}$, as required.

Now suppose a catalyst offers a pathway that lowers the transition-state free energy by $30\ \mathrm{kJ\,mol^{-1}}$ relative to the same reactant and product states. The barriers become $50\ \mathrm{kJ\,mol^{-1}}$ forward and $60\ \mathrm{kJ\,mol^{-1}}$ in reverse. Both reactions are faster, but their barrier difference is still $-10\ \mathrm{kJ\,mol^{-1}}$.

At $298\ \mathrm{K}$, if the prefactors are unchanged, lowering a barrier by $30\ \mathrm{kJ\,mol^{-1}}$ changes its rate constant by approximately

$$\exp\!\left(\frac{30{,}000}{8.314\times 298}\right) \approx 1.8\times 10^5.$$

This factor applies to both directions in this simplified example. In a real catalytic mechanism, the rate changes need not match exactly because several steps, intermediate concentrations, and catalyst availability can affect the measured rates.

Why equilibrium does not shift

For a simple reversible elementary reaction, the equilibrium constant is related to the forward and reverse rate constants by $K=k_{\mathrm{fwd}}/k_{\mathrm{rev}}$ under the appropriate standard-state convention. If catalysis increases both rate constants by the same factor, their ratio—and therefore $K$—stays unchanged.

More generally, equilibrium is determined by thermodynamics: the relative free energies of the states. A catalyst changes the route and the time needed to reach equilibrium, not those relative free energies. If a mixture starts with excess A, the forward reaction may initially be faster overall; if it starts with excess B, the reverse reaction may be faster. The catalyst speeds the relaxation from either starting mixture toward the same equilibrium composition.

Common mistakes

• Saying a catalyst favors the forward reaction. It may make products form faster at first when reactants are abundant, but it also accelerates the reverse reaction. It does not alter the equilibrium constant.
• Thinking the catalyst lowers the energy of the products. A catalyst changes the pathway’s barriers; it does not change the net reaction’s thermodynamic free-energy difference.
• Assuming the catalyst must remain unchanged at every moment. It can bind reactants or form intermediates and be temporarily changed. What matters is that it is regenerated in the overall catalytic cycle.
• Assuming every measured rate increases by exactly the same amount. Equal barrier reductions give equal exponential factors in the simplified model with unchanged prefactors. Real mechanisms can be more complex, while still leaving the equilibrium position unchanged.

A catalyst can therefore speed both directions without contradiction: it makes it easier to cross the barrier, but does not change which side is thermodynamically favored.