CHE230 ENVIRONMENTAL CHEMISTRY 2010
Problem Set 9
SO2 Chemistry
1) Assuming that the air contains 500 ppbv SO2 and no SO3, calculate the pH of the local
rain based on the following equilibria. (Equilibrium constants (25oC) are in M and bar
units). (4.0)
R1)
R2)
R3)
SO2(g) ⇌ SO2(aq)
SO2(aq) + H2O ⇌ H+(aq) + HSO3-(aq)
HSO3-(aq) ⇌ H+(aq) + SO32-(aq)
KH = 1.2
Ka= 1.7x10-2
Ka2= 1.0x10-7
To find pH, we must find the concentration of H+. R2 would be an appropriate equilibrium to
use, but we don’t know [H2SO3(aq)], and so we combine R2 with R1:
R1) SO2 (g)
⇌ H2SO3 (aq)
R2) H2SO3 (aq) ⇌ H+(aq) + HSO3¯ (aq)
KH = 1.2
Ka = 1.7x10.2
-----------------------------------------------------------------------------R4) SO2 (g) ⇌ H+(aq) + HSO3¯ (aq)
K4 = KHKa = 0.0204
Where
=
[
][
]
Since there is only one value for the concentration of [ ] in the system, we can calculate the
pH of rain using this expression. From R4, the molar ratio of [ ] and [
] is 1. Let
[ ] = [
] = . Now:
=
Now
So
= 500
= 500 × 10
=
×
[
][
×1
]
=
= 5 × 10
= 0.0204 × 5 × 10 = 1.02 × 10
= 1.01 × 10
= − log[ ] = − log[ ] = 3.995 = 4.0
2) Under very humid conditions, SO2 oxidation occurs with a pseudo first order rate
constant of 0.3/hour through the following major reaction pathway:
k= 103 L/mol-s
SO2(aq) + H2O2(aq)  H2SO4(aq)
Assuming that SO2(aq) and SO2(g) are at all times equilibrated, calculate the
concentration of H2O2(aq). Hint: there are 2 rates of reactions provided that can be
related (8.3x10-8 M)
We can write two independent equations to describe the rate of removal of SO2
based on the 30% removal per hour and the equilibrium expression above:
[
]
= 0.3ℎ
×
1ℎ
×[
3600
]
And
[
]
= [
][
] = (10 /
∙ )[
][
]
Equating the two expressions:
0.3
3600
[
×[
] = (10 /
] = 0.833 × 10
∙ )[
][
]
3) Consider the following reaction at standard temperature and pressure (i.e. T=298 K and
P=1 bar)
SO2(g) + OH(g) + M  HSO3(g)
k = 9 x 10-13 cm3·molec-1 s-1
a. What is the second order rate constant for the above reaction? (k=9x10-13
cm3·molec-1 s-1)
The units for a first order rate constant are per-time, e.g. s-1 or h-1. A second order rate constant
has units of per-time per-concentration such as M-1s-1 or cm3· molecule-1s-1.
The rate constant provided is a second order rate constant.
b. What is the true third order rate constant for the above reaction? (k=3.70x10-32
cm6·molec-2 s-1)
The units for a third order rate constant are per-time per-concentration-squared such as
cm6· molec-2s-1. In the second-order constant above, M has been omitted.
We need to find the concentration of M in appropriate units and divide the second order rate
constant by this concentration to obtain the true third order rate constant:
From
=
,
=
=
With P=1atm, T = 298K, and R=0.082L·atm/mol·K:
1
[ ]=
(0.08314 ∙
) × 298
3
In molec/cm , this is
0.0404
1
6 × 10
[ ]=
×
×
1
1000
1
= 0.0404
/
= 2.43 × 10
/
Equating the two rate equations:
[
=
Note that
(9 × 10
2.43 × 10
][
]=
∙
[
)
/
is pressure-dependant, but
=
][
= 3.70 × 10
][ ]
∙
is not. In other words,
[ ]
c. What is the half life of SO2 (g) if [OH (g)] = 1x107 molec/cm3 ? (21.4 h)
/
=
[
ln(2)
][ ]
=
= 7.71 × 10
ln 2
(1 × 10 )(2.45 × 10 )(3.67 × 10
= 21.4 ℎ
Note that this half life applies for any pressure since we used
[
]
)
. Using
/
=
would have made our calculation valid only for the conditions where
was
found.
d. What is the half life of SO2 (g) in Mexico City where T=298K and P = 0.64 bar
assuming that [OH (g)] = 1x107 molec/cm3 ? (33.3 h)
Here, we have to calculate the half-life using the new pressure: [M] has
changed.
Using P=0.64atm, T = 298K, and R=0.08314L·bar/mol/K:
0.64
[ ]=
=
= 0.02583
/
(0.08314 ∙
/
/ ) × 298
Converting units:
0.02583
1
6 × 10
[ ]=
×
×
= 1.555 × 10
1
1000
1
Now, using
=[
][ ]
= 1.20 × 10
=(
×
)( .
= 33.3 ℎ
×
)( .
×
)
/
4) At its creation an ultrafine particle of pure (NH4)2SO4 has a diameter of 2 nm. Assuming
the particle is spherical with a density of 1.8 g/cm3, calculate:
a. The number of atoms of sulphur that it contains. (34)
Before estimating the number of atoms, we need to find the particle’s mass. We have diameter =
2 nm, so for a spherical particle: (r=1 nm)
4
4
=
=
(1 × 10
) = 4.19 × 10
3
3
Which, with density of 1.8 g/cm2, gives a mass of
= ( )
×
(
)
= 1.8
×
× 4.19 × 10
= 7.54 × 10
Now convert this mass to no. of atoms:
7.54 × 10
=
132.17
(
(
)
× 6.022 × 10
)
= 34.4 ≈ 34
b. The number of atoms of nitrogen that this particle contains. (68 atoms)
)
5) (
contains 2 atoms of nitrogen per molecule of ammonium sulphate. Thus it
contains 34 × 2 = 68 atoms of nitrogen.
Aerosol Chemistry
5) Deliquescence
a) Define “deliquescence point”.
The deliquescence point is the relative humidity at which a solid, soluble substance spontaneously
takes up water vapour from its surroundings to form a solution.
This occurs when the vapour pressure of water becomes greater than the vapour pressure of the
resulting saturated solution.
b) What is the significance of deliquescence point to aerosol chemistry.
Aerosol particles have very different physical properties in the crystalline and aqueous phases.
The different phases will scatter radiation (e.g. incoming sunlight, or outgoing heat) differently.
Chemical reactions may proceed differently on crystal particles vs aqueous solutions.
Cloud formation is assisted and affected by deliquescence. On one hand, cloud droplet formation
is assisted by deliquescence of aerosol particles. On the other hand, cloud ice formation is
assisted by crystalline particles.
c) How do you think temperature will affect deliquescence point?
Temperature affects the vapour pressure of the droplet, changing the point at which equilibrium is
established with the surrounding water vapour, and thus changing the deliquescence relative
humidity (DRH) of salts.
As temperature increases, the equilibrium vapour pressure of a droplet of a given concentration
increases. A stable droplet can then exist at higher salt concentrations (lower water
concentrations) and lower RH values.
DRH therefore tends to decrease as temperature increases, with the magnitude of the decrease
varying for each individual salt.
6) The solubility of ammonium nitrate at 298K is 26 moles per kg of water.
a) If a crystalline NH4NO3 particle deliquesces, what is the mole fraction of water in the
concentrated solution produced? (Assume the solution is saturated.) (0.51)
To find the mole fraction of water in this solution, we must first find the number of
moles of water per kg of water.
1000
=
×
=
= 55
18.02 /
Now keep in mind that when ammonium nitrate is dissolved in water, the solution
contains water molecules, ammonium ions, and nitrate ions. Therefore:
=
+
=
+
55
= 0.51
55 + 26 + 26
b) Using Raoults Law, estimate the vapour pressure of water above this concentrated
solution, given that at 298K the vapour pressure of pure water is 0.031 bar. (0.016 bar)
=
×
°
= 0.51 × 0.031
= 0.016
c) Given that deliquescence occurs when the vapour pressure of water is greater than the
vapour pressure of the resulting saturated solution, use your results from (a) and (b) to
calculate the relative humidity at which NH4NO3 deliquesces (its DRH). (51%)
We need two pieces of information to calculate the deliquescence point, and we have
them: the vapour pressure of water and the partial pressure of the saturated solution.
=
°
× 100% =
0.016
0.031
× 100% = 51%
Again, the true DRH of NH4NO3 is 61% as a result of activity
corrections in this highly concentrated solution.
7) A parcel of air containing pure (NH4)2SO4(s) aerosol rises in the air. At an altitute of 2 km,
where T = 273K, vapour pressure of water is 5.96x10-3 atm and the partial pressure of water
is 4.76x10-3 atm, the aerosol deliquesces and a cloud forms. What is the solubility of
(NH4)2SO4(s) in water at 273K?
This question asks the reverse of question 4. First, we use the partial and vapour
pressures of water to calculate the relative humidity expressed as a fraction. As we saw in
question 4, this fraction is the equal to the mole fraction of water in the NH4+, H2O, and
SO42- solution:
=
=
°
=
4.76 × 10
5.96 × 10
= 0.80
(So the DRH is 80%)
Now writing a new expression for mole fraction in the solution:
=
Every mole of (NH4)2SO4 that dissolves yields 2 mols of NH4+ and 1 mol of SO42-.
=
(
)
Per 1 kg of aerosol:
0.80 = (
)
0.80 =
= 3.67
/
(
)