| The reaction steps of the reduction of SO2 and NO by CO on La2O2S, and the effect of H2O on these reactions were studied using temperature-programmed reaction coupled with mass spectrometry, step-change method, steady-state kinetic measurements, XRD and XPS.; The present study shows that the redox mechanism is not the major reaction route for the reduction of SO2 by CO on La2O2S. Rather, the reduction follows a surface reaction between the reaction intermediate COS and pools of SO2 already adsorbed on the oxysulfide.; The reduction of NO on La2O2S is similar to that on La2O3. It starts with the dissociation of NO to N and O. The removal of O can be a rate-limiting step but this step is facilitated by the oxidation reaction with labile sulfur to SO2 and thus, promoting the reduction of NO to N2. Upon the depletion of the labile sulfur, the removal process stops and so does the reduction. The NO reduction is fast. To sustain the reduction, SO2 is needed to regenerate the labile sulfur on the oxysulfide formed by the catalytic reduction of SO2 by CO.; The presence of H2O not only shifts the selectivity of sulfur to H2S but also deactivates the reduction of SO2 and NO. When the ratio NO/SO2 is low (∼0.4) or when NO is absent, the catalyst is partially deactivated. The deactivation and shift in selectivity can be attributed to the competitive adsorption of H2O, reverse Claus reaction between H2O and sulfur in the oxysulfide and hydrolysis of the intermediate COS to H2S. When the ratio is sufficiently high (∼1), the oxysulfide catalyst is completely deactivated and the deactivation is irreversible. The surface of the oxysulfide catalyst is oxidized to oxysulfate, which is stable at the reaction conditions but inactive for the reduction of SO2 and NO. |
