Page 200 - Basic physical chemistry for the atmospheric sciences
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1 8 6 Basic physical chemistry
n
7.9. For nitroge , Amax = 0.08 JLm ; for oxygen, Amax = 0.099 JLm.
Both these wavelengths lie in the ultraviolet region .
I
s
7 . I O . 2 . 9 x 0 7 photon .
7 . 1 1 . 34 % .
[N0 2 (g)]
7 . 1 2 . 2.5
[NO(g)]
Solution: From Eq. (7.23)
[N02(g)] 1 . 0 x i o - 2 0 [0 (g)]
[NO(g)] 4.0 x 1 0 - 3 3
3
where [OJ(g)] must be in units of molecules m - . The con
centration of 0 is given in ppmv ; to change this to mole
3
, we must first apply the gas equation in the form
cules m 3 -
(
of Eq. (l 8 g) to air at a pressure of I atm 1 0 1 3 x 0 2 Pa) and
.
1
2
a temperature of 20°C or 293K , which gives n = 2.5 x I 0 5
0
, where n 0 is the concentration of all the
molecules m - 3
molecules in air at 1 atm and 20°C. Since 03(g) occupies
6
0 . 0 40 ppmv of air, [OJ(g)] = (2.5 x 1 025)(0.040 x l 0 - ) =
8
I
1 . 0 x 0 1 molecules m - 3 . Hence,
[NOi(g)] 1 . 0 x 1 0 - 2 0
O
1 8
-
[NO(g)] 4.0 x 1 0 - 3 ( I . x 1 0 ) - 2 . 5
dn 1
7 . 1 3 . dt = 2jan2 - kbn 1 n2nM + jcn3 - kdn n3
1
dn3
-;Jf = kbn 1 n2nM -jcn 3 k dn 1n3
-
7 . 1 4 . Result could have been predicted because n + 2 n + 3n3 is
2
1
s
the sum of the oxygen atom , which cannot change. Hint:
To obtain n 1 + 2n2 + 3n3 = c onstant, add the left and right
sides of the three differential equations given above for the
solution of Exercise 7 1 3 , and then integrate.
.
7 . 1 5 . When X = H
H + 0 3 � OH + 0 2
OH + O � H + 0 2
Net: 0 + 0 3 � 02 + 0