Consider the changes in the energy of the system over the course of the reaction - we typically visualise it as a curve, starting with the energy substrates (at 0% conversion), progressing over a peak (e.g. the energy barrier to initiate the reaction, with its maximum corresponding to the energy of the reactive intermediate; the difference between the two levels is equal to the activation energy) and towards the energy of the products (at 100% conversion). Enclosed is a link to a diagram illustrating this.

Now, when you doped the metal oxide, the energy of the substrates changed. Consequently, unless the energy of the reaction intermediate state changes in the same way as the energy of the substrates, the difference between the two (i.e. the observed activation energy) will also change. The change in the energy of the substrates (i.e. your metal oxide) is caused by the introduction of either a deficit or surplus of electrons (depending whether n-doping or p-doping was carried out) into the metal oxide, changing its Fermi level, as Justin mentioned above.

Furthermore, while we can reasonably *expect* that the mechanism of the reaction doesn't change, this needs not be the case and a completely different transition state may be formed due to the presence of the dopant.

http://4.bp.blogspot.com/_Obwvpc-KSC4/S5RD0I3NynI/AAAAAAAAARs/G1WGjY6rOvI/s400/activation+energy+transition+state.JPG

Yes Dr. Ausama

Metal oxide normally formed in two structures: n-type containing metal excess and p-type containing metal deficient.

 In n-type metal oxide the doping of lower valence metallic ions increases the concentration of interstitial metallic ions and decreases the number of excess electrons. In this case the diffusion controlled oxidation rate increased then the activation energy decreased, vice versa; the doping of higher valence metallic ions decreases the concentration of interstitial metallic ions and increases the number of excess electrons. In this case the diffusion controlled oxidation rate decreased then the activation energy increased. 

In p-type metal oxide, the doping with lower valence cations, decreases the concentration of cations vacancies and increases the number of electron holes then the diffusion controlled oxidation rate decreased and the activation energy increased. Vice versa, the doping with higher valence cations increases vacancies concentration and decreases electron holes concentration then the diffusion controlled oxidation rate increased and the activation energy decreased.

Best regards

Yes Prof. Braun, simply it is the minimum energy required to start the metal/oxygen reaction. Now I am pleased to know more from you.

Best regards

it depend on type of semiconductor