Question

For the reaction :

NO + O 3 → NO 2 + O 2

the second - order rate constant = 1.8 × 10 - 14 cm 3 molecule - 1 s - 1 at 25ºC. In an unpolluted atmosphere, [NO] = 0.10 ppbv and [ O 3 ] = 15 ppbv.

(a) Calculate these two concentrations in units of molecule s / cm 3.

(b) Calculate the rate of the NO oxidation. Use concentration units of molecule / cm 3.

(c) Show how the rate law may be expressed in pseudo first - order terms and calculate the corresponding pseudo first - order rate constant.

(d) Describe two processes occurring in the above reaction.

Answer 1

a) [NO] = 0.1 ng / cm3 * (1 g / 1x10 ^ 9 ng) * (6.02x10 ^ 23 molecules / 30.01 g) = 2.01x10 ^ 12 molecules / cm3

[O3] = 15 ng / cm3 * (1 g / 1x10 ^ 9 ng) * (6.02x10 ^ 23 molecules / 48 g) = 1.88x10 ^ 14 molecules / cm3

b) V = k * [NO] * [O3] = 1.8x10 ^ -14 cm3 / molecules * s * (2.01x10 ^ 12 molecules / cm3) * (1.88x10 ^ 14 molecules / cm3) = 6.80x10 ^ 12 molecules / cm3 s.

c) In terms of pseudo first order, there is an excess component (O3), this reaction can be expressed:

V = k * [NO] ^ 2

d) Oxidation of the NO and reduction of O3 occurs.