The effect of pore diffusion on the behavior of a slow, irreversible reaction that takes place in parallel with a fast, reversible reaction has been analyzed for the case where the reactions are first order in a common reactant. When the resistance to pore diffusion is significant, the apparent activation energy of the slow reaction generally is not equal to one-half of the true activation energy. Rather, the difference between the true and apparent activation energies of the slow reaction depends on the equilibrium constant and on the enthalpy change of the fast, reversible reaction. If the equilibrium constant is small compared to one, or if the enthalpy change of the fast, reversible reaction is close to zero, a significant resistance to pore diffusion will not cause a falsification of the activation energy of the slow reaction. The true activation energy will be observed in these situations. However, if the equilibrium constant of the reversible reaction is large compared to one and if the enthalpy change of this reaction is large, the observed activation energy for the slow, irreversible reaction in the presence of a significant pore diffusion resistance can range from negative for an endothermic reaction to much greater than the true activation energy for an exothermic reaction. The analysis is applied to recent data on the skeletal isomerization and parallel cracking of n-hexane.