Question

The rate constants for the reactions of atomic chlorine and of hydroxyl radical with ozone are given by 3 x 10-11 e-250/T and 2 x 10-12 e-940/T, where T is the Kelvin temperature. Calculate the rate constant for ozone destruction by chlorine under conditions in the Antarctic ozone hole, when the temperature is about -80°C and the concentration of atomic chlorine increases by a factor of one hundred to about 4 x 105 molecules cm-3.

Answer 1

The ozone distruction by hydroxyl radical and chlorine atom is given by following reactions.

O₃ + OH° -------> O₂ + O₂H

O₃ + Cl -------> ClO + O₂

Chlorine acts as a catalyst in the ozone destruction.One atom of chlorine reduces approx thousand of ozone molecules.Hence on adding catalyst,the rate of reaction increases.On adding the more amount of catalyst the rate constant of reaction changes until all the reacting particals occupy the active sites.when all the particals occupied on catalytic surface,rate and the rate constant of the reaction never changes.