The uncatalysed isomerization Y → Z is elementary with an activation energy of

CHEMISTRY · Single correct · AP · Chemistry · Kinetics · Kinetics

Problem

The uncatalysed isomerization Y → Z is elementary with an activation energy of 130 kJ mol−1130\,\text{kJ mol}^{-1}. In the presence of a solid catalyst, the reaction proceeds through adsorption of Y (Ea1=65 kJ mol−1E_{a1}=65\,\text{kJ mol}^{-1}), surface rearrangement (Ea2=80 kJ mol−1E_{a2}=80\,\text{kJ mol}^{-1}), and desorption of Z with negligible additional barrier. Which statement best explains how the catalyst changes the rate-determining step? (a) Adsorption remains rate determining since it occurs first and therefore controls how fast surface intermediates can form. (b) Because the two surface barriers add up to 145 kJ mol−1145\,\text{kJ mol}^{-1}, the catalysed route is overall slower and there is effectively no change in the rate-determining step. (c) The surface rearrangement step is rate determining, because its 80 kJ mol−180\,\text{kJ mol}^{-1} activation energy is the highest barrier in the catalysed pathway. (d) Desorption becomes rate determining; product release from the surface is typically the slowest event in heterogeneous catalysis.

Related theory

Browse the problem library