Page 21 - Basic physical chemistry for the atmospheric sciences

Chemical equilibrium 7

is given by n0 in Eq. ( 1 . 8 g) with p = 1 atm = 1 0 1 3 x 1 0 2 Pa, T= 273K
and k = 1. 38 1 x 1 0 - 23 J de - 1 molecule - 1• Therefore,
g
1 0 1 3 x 1 0 2
Loschmidt number = 23 2.69 x 1 0 2 5 molecule m - 3
( 1 . 3 8 1 x 10_ )273
Since, at the same temperature and pressure, the volumes occupied
i
by gases are proportional to the numbers of molecules n the gases , we
can write
3
Volume occupied by C02 molecules in air Number of C02 molecules in I m of air
3
Voume occupied by air Total number of molecules in I m of air
Therefore,
3
_6_ Number of C0 2 molecules in I m of air
354 x 10 -
2 . 6 9 x 1 0 25
3
Hence, the number of C02 molecules in I m of air is (354 x 10-6) x
2
(2.69 x 1 0 5) = 9.52 x 1017•
We can now derive an expression for the equilibrium constant for a
chemical reaction involving only gases in terms of the partial pressures
(
of the gase . From Eqs . ( 1 .6) and l .8i)
s
�R·nan
[pG/Rc*J1gfpH/Rc*7]h . . . P g "tth • • •
K
c fpA/R�1]afpB/R�1]b... p'.!.�··· c
or,
R
K =K ( *niln ( l.9a)
c p c
where,

( l .9b)

and ,
an= (a+ b + )-( g + h + . . . ) ( l . 9c)

.
.
.
P
K is generally used as the equilibrium constant in problems involv ­
ing gaseous reactions. As in the case of Kc, terms for pure liquids and
solids do not appear in the expression for K ' and the coefficients for
P
these terms are taken to be zero in the expression for an. Note that
e
the units, as well as the numerical valu s , of K and K may differ.
c
P
(
For example , for Reaction 1 . 3 ) the units of Kc are those of