Nighttime loss of NO x in the lower troposphere proceeds by: (1) (2) (3) (4) Rea

Question

Nighttime loss of NO x in the lower troposphere proceeds by:

(1)

(2)

(3)

(4)

Reaction (3) is viewed as an equilibrium process with constant K 3 = [N 2 O 5 ]/([NO 3 ][NO 2 ]) = 3.6x10 -10 cm 3 molecule -1 . Other reactions have rate constants k1 = 3x10-14 cm3 molecule-1 s-1, k 2 = 2x10 -17 cm 3 molecule -1 s -1, and k 4 = 3x10 -4 s -1 . Consider an air parcel with a temperature of 280 K, pressure of 900 hPa, and constant concentrations of 40 ppbv O 3 and 0.1 ppbv NO x .

1. The above mechanism for NOx loss operates only at night. Explain why.

2. At night, almost all of NO x is present as NO 2 (the NO/NO x concentration ratio is negligibly small). Explain why.

3. Let NO 3* represent the chemical family composed of NO 3 and N 2 O 5 , that is, [NO 3* ] = [NO 3 ] + [N 2 O 5 ]. Calculate the lifetime of NO 3* at night.

4. Assuming that NO 3* is in chemical steady state at night (your answer to question 3 should justify this assumption), and that the night lasts 12 hours, calculate the 24-hour average lifetime of NO xagainst oxidation to HNO 3 by the above mechanism. Compare to the typical 1-day lifetime of NOx against oxidation by OH.

Explanation / Answer

Q.1: Because sun light is absent in night and the chemical reaction for the loss of NOx occurs in absence of sunlight and in presence of O3

Q.2: Because in night, in absence of sunlight most of NO reacts with O3 to form NO2 as

NO + O3 ------> NO2 + O2: k1 = 3x10-14 cm3 molecule-1 s-1

At the same time NO2 also reacts with O3 to form NO3 redical as

NO2 + O3 -----> NO3 + O2 : k 2 = 2x10 -17 cm 3 molecule -1 s -1

How ever the rate constant for the first reaction is 1500 times higher. Hence NOx is almost exist as NO2 during night time.

Q.3:

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