Question

The rate constants for the reactions of atomic chlorine and of hydroxyl radical with ozone are given by 3 × 10-11 e-250/T and 2 × 10-12 e-940/T, where T is the Kelvin temperature. Calculate the ratio of the rates of ozone destruction by these catalysts at 20 km, given that at this altitude the average concentration of OH is about 100 times that of Cl and that the temperature is about -50 °C. 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 × 105 molecules cm-3.

Answer 1

Answer:

Calculate the ratio of the rates of ozone destruction by these catalysts at 20 km, given that at this altitude the average concentration of OH is about 100 times that of Cl and that the temperature is about -50 °C

Knowing

Rate constants for the reactions of atomic chlorine and of hydroxyl radical with ozone are given by 3x$10^{-11} e^{-255/T}$  and 2x$10^{-12} e^{-940/T}$  

T = -50 °C = 223 K

The reaction rate will be given by [Cl] [O3] 3x$10^{-11} e^{-255/223} = 9.78^{-12} [Cl] [O3]$  

Than, the reaction rate of OH with O3 is

Rate = [OH] [O3] 2x$10^{-12} e^{-940/223} = 2.95^{-14} [OH] [O3]$

Considering these 2 rates we can realize the ratio of the reaction with Cl to the reaction with OH is 330 * [Cl] / [OH]

Than, the concentration of OH is approximately 100 times of Cl, and the result will be that the reaction with Cl is 3.3 times faster than the  reaction with OH

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 × 105 molecules cm-3

Knowing

Rate constants for the reactions of atomic chlorine and of hydroxyl radical with ozone are given by 3x$10^{-11} e^{-255/T}$  and 2x$10^{-12} e^{-940/T}$  

T = -80 °C = 193 K

The reaction rate will be given by [Cl] [O3] 3x$10^{-11} e^{-255/193} = 8.21^{-12} [Cl] [O3]$  

Than, the reaction rate of OH with O3 is

Rate = [OH] [O3] 2x$10^{-12} e^{-940/193} = 1.53^{-14} [OH] [O3]$

Considering these 2 rates we can realize the ratio of the reaction with Cl to the reaction with OH is 535 * [Cl] / [OH]

Than, considering the concentration of Cl increases by a factor of 100 to about 4 × $10^{5}$ molecules $cm^{-3}$, the result will be that the reaction with OH will be 535 + (100 to about 4 × $10^{5}$ molecules $cm^{-3}$) times faster than the  reaction with Cl

Explanation: