Geoscience Reference
In-Depth Information
Solubility of SO
2
in Pure Water: Equilibrium Data
Concentration SO
2
(g of SO
2
per 100 g/H
2
O)
p
(Partial Pressure of SO
2
)
0.5
6 kPa (42 mmHg)
1.0
11.6 kPa (85 mmHg)
1.5
18.3 kPa (129 mmHg)
2.0
24.3 kPa (176 mmHg)
2.5
30.0 kPa (224 mmHg)
3.0
36.4 kPa (273 mmHg)
Solution:
The data must first be converted to mole fraction units. The mole fraction in the gas phase
(
y
) is obtained by dividing the partial pressure of SO
2
by the total pressure of the system. From the
first entry in the data table:
y
=
p/P
= (6 kPa/101.3 kPa) = 0.06 kPa
The mole fraction in the liquid phase (
x
) is obtained by dividing the moles of SO
2
by the total moles
of liquid:
MolesofSO nsolution
MolesofSO ns
2
x
=
olutionMoles HO
2
+
2
For the first entry (
x
) of the data table:
x
= 0.0078/(0.0078 + 5.55) = 0.0014
The table on the next page has been completed. The data from this table are plotted in Figure 16.4.
Henry's law applies in the given concentration range with Henry's law constant equal to 42.7 mole
fraction SO
2
in air/mole fraction SO
2
in water.
50
30
10
0.004
0.008
0.012
x, Mole Fraction of SO
2
Absorption Data
FIGURE 16.4
Sulfur dioxide absorption data for Example 16.1. (Adapted from USEPA,
APTI Course 415:
Control of Gaseous Emissions
, EPA 450/2-81-005, U.S. Environmental Protection Agency Air Pollution
Training Institute, Washington, DC, 1981, p. 4-8.)
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