Atmosphere Chemistry

Topics: Water, Trigraph Pages: 57 (5365 words) Published: December 19, 2012
Atmosphere –Water Interactions
C. P. Huang

Stumm W. and J. J. Morgan. Chapter 5. Aquatic Chemistry
1

Environmental Systems

2

5.2 Atmospheric generation of acidity

3

Genesis of acid rain

4

Chemical processes and Chemical
Composition of Water Droplets

5

uM
ueq/L

NO360
60

SO4244
88

Cl 25
25

S(X)
173

Mg2+
4.5
9

Ca2+
16
32

NH4+
85
85

K+
2
2

Na +
5
5

H+
50
50

S(M)
183

CH3COOH
5
5

6

7

8

Total Annual Atmospheric Deposition

9

5.3 Gas-Water Partitioning:
Henry’s Law

10

Henry’s Law
A (g)  A(l)
[A(aq)] = H [A(g)]

(mol/L)
[A(aq)] [A]w= PA/(RT) (mol/L)
[A(aq)] = KHRT [A(g)]
H = KH RT

(A(aq)) = [A(aq)] q
mol/m3

mol/L

L/m3

KH= M/atm

R= 0.082057 (L-atm/K)
[A(aq)] = H [A(g)]

q: content of water per m3 of the entire system
q = 5x10-5 – 5x10-4 L-m-3 for fog
q = 1x10-4 – 1x10-3 L-m-3 for clouds

11

Henry’s Law

[A(aq)] = H [A(g)]
(A(aq)) = [A(g)] q

[A(aq] = KHRT [A(g)]
(A(aq)) = (A)w
[A(aq)] = [A]w

mol/m3

mol/L

L/m3

(A)T = (A)g + (A)w = pA/RT +[A]w q

[A(aq)] = H [A(g)]
H = KH RT (L(g)/L(aq)
R = 0.082057 (L-atm-mol-1-K-1)

(A)T = (A)g + KHRT (A)gq = (A)g (1+KHRTq)
(A)T = (A)g (1+ Hq)
fg = (A)g/(A)T = 1/(1+qH)
fw = (A)w/(A)T =qH/(1+qH)

R = 8.2057x10-5 (m3-atm-mol-1-K-1)
q: content of water per m3 of the entire system
H = KH RT (m3/L)
-5 – 5x10-4 L-m-3 for fog
q = 5x10
q = 1x10-4 – 1x10-3 L-m-3 for clouds

12

Solubility of oxygen
O2(g) = O2(aq); H = 0.038
[O2(aq)] = KH p(O2) = (H/RT)[O2(g)]

RT = 0.024465 (m3-atm-mol-1)
RT = 6.132946 (L-atm-mol-1)

KH(O2) = H/(RT) =0.038/6.1632946 = 1.26x10-3 M-atm-1
In freshwater KH(O2) = 3.1x10-3 M-atm-1
In 10 o% estuarine water KH(O2) = 3.73x10-3 M-atm-1
In 35 o% sea water KH(O2) = 3.32x10-3 M-atm-1
13

Important Gas-Water Equilibria

14

15

5.4 Gas-water equilibria in closed
and open systems

16

5.4 gas-water equilibrium in closed and
open systems
Open system: water is in contact with an unlimited amount
of a gas; partial pressure is constant; surface water in
equilibrium with air; rainwater
Closed system: a limited amount of a gas becomes
distributed between the gas and the water; equilibrium
concentration is governed by the Henrys law.; distribution
of aqueous and gas phase concentration is dependent on
the ratio of water to air; fog,

17

Example 5.2
COS (carbonyl sulfide) = 500 ppt in the atmosphere
What is the partial pressure and its concentration in g m-3 at 0oC? 1 ppm = 10-6 atm or 1 ppt =10-12 atm
R = 8.2057x10-5 m3 atm K-1 mol-1
PCOS = 500 x 10-12 atm = 5x10-10 atm
pV=nRT

[n/V]=P/RT

At 0oC

RT = 8.2057 x10-5 (atm-m3-mol-1-K-1) 273.15 (K)=2.24x10-2 (mol/m3-atm) [COS] (mol-m-3) = 5x10-10/(2.24x10-2) = 2.24x10-8 (mol-m-3)
[COS] = 2.24x10-8 (mol/m3) x 60 (g/mol) = 1.36x10-6 (g/m3)
18

Example 5.3a (closed system)
Dissolution of H2O2 and O3 in the atmosphere
KH(H2O2) =1x105 M-atm-1
(A)T = (A)g + (A)w
(A)T = (A)g + H (A)g

KH(O3) = 9.4x10-3 (M-atm-1)

H = KHRT (m3/L)
R = 8.2057x10-5 (atm-m3/(mol-K)

19

O3

H2O2

[ H2O]
q(L/m^3)
KH.RT.q
-1 244.6857683
-2 24.46857683
-3 2.446857683
-4 0.244685768
-5
2.45E-02
-6 0.002446858
-7 0.000244686
KH(H2O2)
1.00E+05
KH(O3)
9.40E-03
R
8.21E-05
T
298.19
KH(RT)(H2O2)
2.45E+03
KH(RT)(O3)
2.30E-04

f(water)
0.99592976
0.960735929
0.709880682
0.196584371
0.023884165
0.002440885
0.000244626
(M/atm)
(M/atm)
(m^3-atm)/(mol-K)
K
(m^3/L)
(m^3/L)

f(gas)
0.00407
0.03926
0.29012
0.80342
0.97612
0.99756
0.99976

[O3]
KH.RT.q
f(water)
f(gas)
2.3E-05
2.3E-05 0.99998
2.3E-06
2.3E-06
1
2.3E-07
2.3E-07
1
2.3E-08
2.3E-08
1
2.3E-09
2.3E-09
1
0.0002313 0.00023125 0.99977
2.3E-11
2.3E-11
1

20

Fraction of H2O2 and O3 in atmospheric
waster

21

Example 5.3b
Distribution between water and atmosphere
One liter of water containing...
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