Page 158 - Basic physical chemistry for the atmospheric sciences

1 4 4 Basic physical chemistry

N
Solution. From Eq. (3 . 1 4 ) the residence time of 0 2 in air under the
stated conditions would be l /j = / ( . 0 x 1 0 - 3 ) = 2 00 s. From Eq . (7. 1 9)
1
5
Rate of production of NO(g) = j [NOi(g)]
5
= . 0 x 1 0 - 3 [N0i(g)]
where the rate of production of NO(g) is in molecules m - 3 s - 1 if
[N0 (g)] is in molecules m - 3 • The total number of molecules in 1 m 3
2
of air (n0) at pressure p (in pascals) and temperature T (in K) is given
by Eq. (l .8g)
p = n 0 k T
1
where k is the Boltzmann constant l . 3 8 1 x 1 0 - 2 3 J deg - molecule - 1 ) .
(
Therefore, a t p = 1 atm = 1 , 0 1 3 mb = 1 . 0 1 3 x 1 0 5 Pa, and T = 293K ,
( l . 3 8 1 . 0 1 3 x 1 0 5
1 0 -
n = 1 x 2 3 ) 293
o
= . 5 x 1 0 25 molecule m - 3
2
Since NOi(g) comprises 0.50 ppbv of air
[N 0 2 (g)] = (0.50 x 1 0 - 9 )(2 . 5 x 1 0 25)
= 1 . 3 x 1 0 1 6 molecule m - 3
Therefore,
Rate of production of NO(g) = (5.0 x 1 0 3 )( 1 . 3 x 1 0 1 6 )
= 6 . 5 x 1 0 1 3 molecule m - 3 s·•
Therefore, in l hour the number of molecules of NO(g) produced in 1
3
m by Reaction (7 .6) is ( I x 60 x 60)(6.5 x 10 1 3 ) = 2 . 3 x 1 0 17·

7.5 Photostationary states
If a chemical system in equilibrium is placed in a beam of em radiation
that is absorbed by one of the reactants, the rate of the forward
reaction will be changed , thereby disturbing the equilibrium of the
system. The system will therefore adjust t self until the forward and
i
reverse rates are again in equilibrium. This is called the photostation­
ary state of the system.
To illustrate the concept of a photostationary state, let us consider
N
the simplest system for the equilibrium of O (g), N02(g) , and 03(g) in
the lower part of the Earth' s atmosphere (i.e . , the troposphere). This
equilibrium is governed by the following three reactions