Page 188 - Basic physical chemistry for the atmospheric sciences

1 7 4 Basic physical chemistry

l . 1 9 . Solution:
H20(1) � H20(g)

Since we can write [H 0(1)] = l in Eq. ( l . 6) and H o m = l in
2
P 2
Eq . ( l . 9 b)

and ,

KP = p
where p is the partial pressure exerted by the water vapor
in air.
q
From E . ( l .9c)
6.n 0 - l = - l
=
Therefore, from Eq. ( l .9a)

l . 20. H20(g) would decrease . Hint: Apply LeChatelier's prin­
ciple .
.
P
l . 2 1 . 4 f1- 1 -p· H . mt: Th e partia pressure exerte d b y a gas m a mix-
.
. I
ture of gases is equal to the fractional contribution that the
gas makes to the total number of moles in the mixture
e
multiplied by the total pressure x erted by the mixtur .
e
1 . 22. 0 . 6 93 at m .
Solution:
(A)
From Eq. (l 9 b , remembering that the partial pressures of
.
)
solids can be equated to zero in this equation,
(B)
where p and p 2 are the partial pressures of NH3(g) and
1
H 2S(g), respectivel y , that are in equilibrium with NH4HS(s).
The total pressure p produced by these gases is
(C)

For every mole of NH 3 (g) released by reaction (A), a mole
of H2S(g) is released. Therefore, the amounts of the two