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SOLUTIONSMANUAL TO ACCOMPANY ATKINS' PHYSICAL CHEMISTRY 121
0.000 0.005 0.010 0.015 0.020
0
50
100
150
xB
p B
/(
kP
a)
Figure 5.1
E5A.8(b) Henry’s law gives the partial vapour pressure of a solute B as pB = KBxB,
[5A.24–152]. A test of this law is to make a plot of pB against xB which is
expected to be a straight line with slope KB; such a plot is shown in Fig. 5.1
�e data fall on quite a good straight line, the equation of which is
pHCl/(kPa) = 8.41 × 103 × (xB) − 2.784
If Henry’s law is obeyed the pressure should go to zero as xB goes to zero, and
the graph shows that this is not quite achieved. Overall the conclusion is that
these data obey Henry’s law reasonably well. �e Henry’s law constant KB is
computed from the slope as 8.4 × 103 kPa .
E5A.9(b) In Section 5A.3(b) on page 152 it is explained that for practical applications
Henry’s law is o�en expressed as pB = KBbB, where bB is the molality of the
solute, usually expressed in mol kg−1.�e molality is therefore calculated from
the partial pressure as bB = pB/KB.
Molality is the amount of solute per kg of solvent. �e mass m of a volume V
of solvent is given bym = ρV , where ρ is the mass density of the solvent. If the
amount of solute in volume V is nB, the molar concentration cB is related to
the molality by
cB =
nB
V
= nB
m/ρ
= ρ
bB«
nB
m
= ρbB
Using Henry’s law the concentration is therefore given by
cB = ρbB = ρ
bB«
pB
KB
= ρxBp
KB
where the partial pressure pB is expressed in terms of the mole fraction and the
total pressure p, pB = xBp.