Question

The following information is relevant to conditions observed at an altitude of 35 km and temperature of 235 K.

k(O + O2) = 1.3 × 10−15 cm3 molecule−1 s −1

k(O + O3) = 6.3 × 10−15 cm3 molecule−1 s −1

[O] = 1.0 × 109 molecule cm−3 [O2] = 5.0 × 1016 molecule cm−3

[O3] = 2.0 × 1012 molecule cm−3

Calculate the rate of formation and removal of ozone. Comment on your answers.

Answer 1

from the given data, K(0+02) 21.3x10-15 cand molecules ine. rate constant of ozone formation = 113 x 10-15 cm molecute From the und And K Co.+O₃) = 6.3x10-15 cm² molecules i.e. rate constant of removal of ozone = 6,3x10'une molecute From the unit of the rate constants it is clean that both the equations are second anden. The equations can be written as, OtO₂ , & coton) ... (2) Oto 202, & (0+03) - - - - . (ů)

So, the rate equation of li is, e Rate = k (0+02) [o] [ 0₂] - 1.3x10-15 um molecuté & 1x109 mokank canna x5x1016 molecule un 3 = 7.198 x 10-10 molecul cum 3 g y .2 x 10-10 molecule con 3 g And the rate equation of it is a rate = k (0+03) [o] [03] = 6.3x10-15 and molecule 31 x 1x109 molecule una x 2x1012 molecule am 3 = 1x10-9 molecule un 35-1 Hence, the rate of formation of ozone = 7.20 10-10 molecuk cm-3 g And the rate of removal of ozone 2 1x10 molecule cm 35 (Aus).