lecture 15 natural sulfur, acid rain
Rainout
We mentioned a few of things that may rainout:
1. CH3OOH (CH4 oxidation, low NOx)
2. H2O2 (CO oxidation, low NOx)
3. HNO3 (CH4 and RH oxidation, high NOx)
If hydrocarbons convert to acids during oxidation and dissolve in water acid rain.
lecture 15 natural sulfur, acid rain
History of Acid Rain
In the 19th century, Robert Angus Smith discovered high levels of acidity in rain falling over industrial regions.
In 1950’s and 1960’s, biologists noticed a decline of fish populations in lakes of southern Norway and North America.
Later found that acid rain also affects vegetation, materials, structures.
lecture 15 natural sulfur, acid rain
Two Main Sources of Acid Rain
1. SO2 from industry oxidized to H2SO4
2. NO, NO2 from automobiles oxidized to HNO3
lecture 15 natural sulfur, acid rain
Aqueous Phase Chemical Equilibrium
What happens to a species that dissolves in water?
partly dissociates into ions (bonds toward water ions stronger than to its own atoms)
What happens to water in the process?
partially ionizes
These are reversible reactions that reach equilibrium rapidly:
At equilibrium:
Keq: equilibrium constant
M = mol L-1
-k
kOHHOH
f
r2
K 298at M 1082.1
]OH[
]OH][H[
]OH][H[]OH[
16
weq
2weq
r
f
r2f
K
Kk
k
kk
lecture 15 natural sulfur, acid rain
Concentration of H2O(aq)
[H2O] is very large: 55.5 M virtually constant
Incorporate [H2O] into Keq:
Keq w = [H][OH] = 1.0 x 10-14 M2 at 298 K
Concentration of ions in pure water:
Each water molecule that dissociates produces 1 H and 1 OH:
Concentration of ions is much smaller than [H2O] water has small conductivity.
M 101]OH[]H[ 7
weq K
lecture 15 natural sulfur, acid rain
pH of Pure Water
Definition of pH:
Pure water at 298 K:
acidic: pH < 7
alkaline: pH > 7
neutral: pH = 7
)M](H[logpH 10
0.7101logpH 710
lecture 15 natural sulfur, acid rain
pH of Clean Rainwater
Clean rainwater is not pure water. It equilibrates with CO2:
hydrolysis
ionization – bicarbonate ion
further ionization – carbonate ion
233
332
3222
COHHCO
HCOHCOH
)aq(COHOH(g)CO
lecture 15 natural sulfur, acid rain
Equilibrium Between Gas and Aqueous Phase
Equilibrium for first reaction:
[H2CO3(aq)]: aqueous phase concentration in equilibrium with gas phase
pCO2: partial pressure of gas phase species (atm)
1 atm = 760 mm Hg = 760 Torr = 1013.25 mbar = 1.01325 x 105 Pa (N m-2)
KH: Henry’s law constant (for dilute solutions)
Soluble gases have large KH.
2CO
32
2COHeq
3222
aq)](COH[
aq)(COHOH(g)CO
pKK
lecture 15 natural sulfur, acid rain
What Does [H2CO3(aq)] Depend On?
Does [H2CO3(aq)] depend on amount of liquid water available? No.
Does [H2CO3(aq)] depend on size of droplet? No.
Does [H2CO3(aq)] depend on temperature? Yes.
van’t Hoff equation:
similar to Clausius-Clapeyron equation:
H: reaction enthalpy at constant T and P or heat of dissolution
L: heat of vaporization
KH increases as T decreases gas more soluble at lower T (less energetic molecules on surface, less evaporation, more stays in solution).
2Hln
RT
H
dT
Kd
2
ln
RT
L
dT
ed s
lecture 15 natural sulfur, acid rain
CO2/H2O System
Reactions in CO2/H2O system:
current atm 1ppmv 350
industrial-pre atm 1ppmv 280
M 107.4
M 103.4
atm M 103
]OH][H[
COHHCO (3)
HCOHCOH (2)
OHHOH )aq(COHOH(g)CO (1)
2CO
2CO
11
3eq
7
2eq
12
2COH
weq
23
3eq
3
3
2eq
32
weq
232
2COH
22
p
p
K
K
K
K
K
K
KK
lecture 15 natural sulfur, acid rain
CO2/H2O System cont.
Total dissolved CO2:
2
2CO2COH2eq3eq23
2CO2COH2eq
23
3
23
3eq
2CO2COH2eq
3
2CO2COH
3
32
32eq
2CO2COH32
2CO
32
2COH
]H[]CO[
]H][CO][H[
]HCO[
]CO][H[ )3(
]H[]HCO[
]HCO][H[
)]aq(COH[
]HCO][H[ )2(
)]aq(COH[
)]aq(COH[ )1(
pKKK
pKKK
pKK
pKK
pK
pK
22eq3eq2eq
2CO2COH
23332
Tot2
]H[]H[1
]CO[]HCO[)]aq(COH[)]aq(CO[
KKKpK
lecture 15 natural sulfur, acid rain
Effective Henry’s Law Constant
effective Henry’s Law constant for CO2:
Is greater than or less than ?
Always greater than Total amount of CO2 dissolved always exceeds that predicted by Henry’s Law for CO2 alone (although not by much).
22eq3eq2eq
2COH*
2COH
2CO*
2COH
22eq3eq2eq
2CO2COH
23332
Tot2
]H[]H[1
]H[]H[1
]CO[]HCO[)]aq(COH[)]aq(CO[
KKKKK
pK
KKKpK
*
2COHK 2COHK
lecture 15 natural sulfur, acid rain
Effective Henry’s Law Constant cont.
What does depend on?
T, pH of solution ([H])
As pH increases ([H] decreases), does increase or decrease?
increase
As pH increases, does increase or decrease?
increase
*
2COHK
*
2COHK
)]aq(CO[ Tot2
lecture 15 natural sulfur, acid rain
Calculate the pH of CO2/H2O System – Approximate Method
1. Since KH* is small (compared to 107/108), assume pCO2 ≈ constant.
2. Since Keq w so small, assume it does not contribute to [H].
3. Since Keq 3 so small, assume [CO32] ≈ 0, then every molecule of
H2CO3 that dissociates produces 1H and 1 HCO3:
[H] ≈ [HCO3]
pH of pure rainwater6.5]H[logpH
)atm 110350)(atm M 103M)( 103.4(]H[
]H[]HCO][H[ )2(
10
6127
2CO2COH2eq2
2CO2COH
2
2CO2COH
32eq
pKK
pKpKK
lecture 15 natural sulfur, acid rain
Calculate the pH of CO2/H2O System – More Exact Method
1. Since KH* is small, still assume pCO2 ≈ constant.
2. electroneutrality: concentrations of ions will adjust so that solution is electrically neutral:
(Each CO32 ion contributes charge of 2. Total negative charge is
concentration of ions x 2.)
]CO[2]HCO[]OH[]H[ 233
6.5]H[logpH :]H[for solve
]H[
2
]H[]H[]H[
10
2
2CO2COH2eq3eq2CO2COH2eqweq
pKKKpKKK
lecture 15 natural sulfur, acid rain
CO2 + H2O = H2CO3
H2CO3 = H+ + HCO3- = 2H+ + CO3
=
K1 = 4.3x10-7
K2 = 5.3x10-11
C = [H+] = [HCO3-]
: fraction of concentration in the form of ions
C: concentration
(1-)C = [H2CO3]
1-: fraction of concentration in the form of acid
K1 = [H+][HCO3-] / [H2CO3] = (C)2 / (1 - )C = 2C/(1 - )
Calculate the pH of CO2/H2O System – Third Method
lecture 15 natural sulfur, acid rain
CO2 solubility 0.759 liters@1atm/liter H2OCO2 Partial pressure Pp is 345 ppm
n = VPp/RTn = 0.759 x 345x10-6 / (0.082 x 298) = 1.07x105 mol
In 1 liter C = 1.07 x 105 mol/liter4.3 x 107= 1.07 x 105 2 / (1 - )2 + 0.04- 0.04=0= 0.18
[H+] = 0.18 x 1.07 x 105=1.93x10-5 MpH = log [H+] = 5.7
Calculate the pH of CO2/H2O System – Third Method cont.
lecture 15 natural sulfur, acid rain
SO2/H2O System
Reactions in SO2/H2O system:
bisulfite ion
sulfite ion
ppb 2002.0
M 106.6
M 103.1
K 298at atm M 23.1
]OH][H[
SOHHSO (3)
HSOHSOH (2)
OHHOH )aq(SOHOH(g)SO (1)
2SO
8
3eq
2
2eq
1
2SOH
weq
23
3eq
3
3
2eq
32
weq
232
2SOH
22
p
K
K
K
K
K
K
KK
lecture 15 natural sulfur, acid rain
SO2/H2O System cont.
Total dissolved sulfur:
2
2SO2SOH2eq3eq23
2SO2SOH2eq
23
3
23
3eq
2SO2SOH2eq
3
2SO2SOH
3
32
32eq
2SO2SOH32
2SO
32
2SOH
]H[]SO[
]H][SO][H[
]HSO[
]SO][H[ )3(
]H[]HSO[
]HSO][H[
)]aq(SOH[
]HSO][H[ )2(
)]aq(SOH[
)]aq(SOH[ )1(
pKKK
pKKK
pKK
pKK
pK
pK
]SO[]HSO[)]aq(SOH[)](aqS(IV)[ 23332
Tot
lecture 15 natural sulfur, acid rain
S(IV)
Sulfur occurs in 5 oxidation states in the atmosphere.
Chemical reactivity decreases with sulfur oxidation state.
Water solubility increases with sulfur oxidation state.
Essentially all dissolved species that come from SO2 are in oxidation state 4.
lecture 15 natural sulfur, acid rain
Total Dissolved Sulfur
Total dissolved sulfur:
Effective Henry’s Law constant for SO2:
22eq3eq2eq
2SO2SOH
23332
Tot
]H[]H[1
]SO[]HSO[)]aq(SOH[)]aq()IV(S[
KKKpK
22eq3eq2eq
2SOH*
2SOH
2SO*
2SOHTot
]H[]H[1
)]aq()IV(S[
KKKKK
pK
lecture 15 natural sulfur, acid rain
Effective Henry’s Law Constant of SO2 as a Function of pH
increases by ~7 orders of magnitude with pH Acid-base equilibrium pulls more material into solution.
(Which material? [H2SO3(aq)] does not depend on pH.)
*
2SOHK
lecture 15 natural sulfur, acid rain
If Assume Constant pSO2
Open system: unlimited LWC, unlimited SO2:
[S(IV)Tot(aq)] increases dramatically with pH
lecture 15 natural sulfur, acid rain
If Don’t Assume Constant pSO2
Closed system: supply of SO2 limited Cannot assume pSO2 ≈ constant to calculate concentrations, but can calculate mole fractions as a function of pH:
1
3eq
2eq
1
22eq3eq2eq
2eq
22eq3eq2eq
2SO2SOH
2SO2SOH2eq
Tot3
3HSO
1
22eq3eq2eq
22eq3eq2eq
2SO2SOH
2SO2SOH
Tot32
)aq(3SO2H
]H[1
]H[
]H[]H[1
]H[
]H[]H[1 ]H[
)]aq()IV(S[
]HSO[
]H[]H[1
]H[]H[1
)]aq()IV(S[
)]aq(SOH[
K
K
KKK
K
KKKpK
pKK
KKK
KKKpK
pK
lecture 15 natural sulfur, acid rain
Mole Fractions cont.
1
3eq2eq3eq
2
1
22eq3eq2eq
2eq3eq
2
22eq3eq2eq
2SO2SOH2
2SO2SOH2eq3eq
Tot
23
23SO
1]H[]H[
]H[]H[1
]H[
]H[]H[1 ]H[
)]aq()IV(S[
]SO[
KKK
KKK
KK
KKKpK
pKKK
lecture 15 natural sulfur, acid rain
S(IV) Mole Fractions as a Function of pH
pH < 2: S(IV) mainly in the form of H2SO3(aq)
3 < pH < 6: S(IV) mainly in the form of HSO3
pH > 7: S(IV) mainly in the form of SO32
lecture 15 natural sulfur, acid rain
How Does This Affect Aqueous Phase Reactions?
Since concentrations depend on pH, reaction rates in solution will depend on pH.
Why is this important?
We still have not calculated the pH of the sulfur system with varying pSO2 (closed system).
So far we have: H2SO3(aq), HSO3,SO3
2
We don’t yet have the acid H2SO4.
lecture 15 natural sulfur, acid rain
Sulfuric Acid
What oxidation state is H2SO4 in? 6
We need to convert S(IV) to S(VI) via aqueous phase reactions.
S(IV) reacts with many species in solution:
O3, H2O2, CH3OOH, O2, OH, NO2, HCHO, Mn, Fe, …
Of these, O3 and H2O2 are the most important for converting S(IV) to S(VI).
lecture 15 natural sulfur, acid rain
Aqueous Phase Reaction Rate
S(IV) + A(aq) S(VI) + … rate constant k in M-1 s-1
R = k [S(IV)] [A(aq)] in M s-1 (mol L-1 s-1)
lecture 15 natural sulfur, acid rain
The S(IV)/O3(aq) System
Reactions in the S(IV)/O3(aq) system:
1192
1151
1140
3323231320
23332
23
s M 10)6.05.1(
s M 10)7.07.3(
s M 10)1.14.2(
)]aq(O][S(IV)[)]aq(O[ ]SO[]HSO[)]aq(SOH[]S(IV)[
SO ,HSO ),aq(SOH:S(IV)
)aq(OS(VI))aq(OS(IV)
k
k
k
kkkkdt
d
lecture 15 natural sulfur, acid rain
Rate Constant of the S(IV)/O3(aq) System as a Function of pH
An increase in pH results in an increase in equilibrium [HSO3] and
[SO32] results in an increase in d[S(IV)]/dt.
lecture 15 natural sulfur, acid rain
Self-Limiting Reaction
The strong increase in d[S(IV)]/dt with pH makes the reaction self-limiting. Why?
Production of H2SO4 (acid) lowers the pH and slows further reaction.
The reaction of S(IV) with O3(aq) is a source of cloud water acidification when pH >~ 4 and an important sink of gas phase SO2 when pH >~8 (sea spray).
lecture 15 natural sulfur, acid rain
The S(IV)/H2O2(aq) System
[H2O2(aq)] is ~6 orders of magnitude higher than [O3(aq)]
Reactions in the S(IV)/ H2O2(aq) system:
1.0
s M 106.5
)aq(SOHHOOHSO (2)
OHOOHSO)aq(OHHSO (1)
2
1
1261
422
2
22
1
1223
k
k
k
k
k
k
lecture 15 natural sulfur, acid rain
The S(IV)/H2O2(aq) System cont.
Steady state approximation on SO2OOH:
]H[
]H)][aq(OH][HSO[
]H[ ]H[
)]aq(OH][HSO[]H][OOHSO[
)]aq(SOH[
]H[
)]aq(OH][HSO[]OOHSO[
)]aq(OH][HSO[]H[]OOHSO[
]H][OOH[SO]OOHSO[)]aq(OH][HSO[0]OOHSO[
2
1
2231
21
2231222
42
21
22312
2231212
222122312
k
kk
kk
kkk
dt
d
kk
k
kkk
kkkdt
d
lecture 15 natural sulfur, acid rain
Rate Constant of the S(IV)/H2O2(aq) System as a Function of pH
As pH increases ([H] decreases), first (2) first becomes faster (denominator dominates), then (2) becomes slower (numerator dominates).
The reaction of S(IV) with H2O2(aq) is a source of cloud water acidification when pH <~ 4 and an important sink of gas phase SO2 when pH <~7 (clouds).
lecture 15 natural sulfur, acid rain
pH of the Sulfur System with Varying pSO2 (Closed System)
• SO2(g) becomes depleted less production of S(IV) less acidic?
• O3(aq) and H2O2(aq) become depleted less S(IV) S(VI) less acidic?
• NH3 becomes depleted less neutralization more acidic?
• less S(IV) pH lower (Figure 6.7) faster S(IV) S(VI) via H2O2(aq) more acidic
lecture 15 natural sulfur, acid rain
Assumptions:
1. All sulfate is in the form of H2SO4
2. All the acid is dissolved and dissociated to ions (2H and SO4)
3. Liquid water content is 1.0 gr/m3 (very heavy cloud)
Note: Total water content (vapor+liquid) is 30 gr per m3 at 25oC and 100% humidity……
For [SO4] = 1.0 g/m3 (typical for remote pacific area)
[H] = 2 (moles H per mole H2SO4) x106 (g SO4 per m3 air)
/ 1.0 (g H2O per m3 air) / 98 (g H2SO4 per mole H2SO4)
x 103 (g H2O per L H2O) = 2.04x105 M
pH = log[H] = 4.7
For a thin cloud (0.1 gr/m3);
[H] = 2.04x10-4 M
pH = 3.7
pH of Sulfur System as a Function of Cloud Liquid Water Content
lecture 15 natural sulfur, acid rain
pH of Sulfur System as a Function of Cloud Liquid Water Content cont.
SO4= (g/m3) Cloud Water Content
0.1 0.5 1.0
5 2.8 3.7 4.0
10 2.5 3.4 3.7
30 2.2 2.9 3.2
lecture 15 natural sulfur, acid rain
The HNO3/H2O System
Reactions in HNO3/H2O system:
nitrate ion
very soluble
dissociates quicklyK 298at M 4.15
K 298at atm M 101.2
]OH][H[
NOH)aq(HNO (2)
OHHOH )aq(HNOOH(g)HNO (1)
2eq
15
3HNOH
weq
3
2eq
3
weq
23
3HNOH
23
K
K
K
K
KK
lecture 15 natural sulfur, acid rain
The HNO3/H2O System cont.
Total dissolved nitric acid:
Effective Henry’s law constant for HNO3:
]H[]NO[
]NO][H[
)]aq(HNO[
]NO][H[ )2(
)]aq(HNO[
)]aq(HNO[ )1(
3HNO3HNOH2eq
3
3HNO3HNOH
3
3
32eq
3HNO3HNOH3
3HNO
3
3HNOH
pKK
pKK
pK
pK
]H[1]NO[)]aq(HNO[)](aqHNO[ 2eq
3HNO3HNOH33Tot
3
KpK
]H[1
)](aqHNO[
2eq
3HNOH*
3HNOH
3HNO*
3HNOHTot
3
KKK
pK
lecture 15 natural sulfur, acid rain
If Don’t Assume Constant pHNO3
The Henry’s law constant of HNO3 is very high. Cannot assume pHNO3 ≈ constant to calculate concentrations, but can calculate mole fractions as a function of pH:
]H[1
]H[
]H[1
]H[
]H[1 ]H[
)]aq(HNO[
]NO[
]H[1
]H[1
)]aq(HNO[
)]aq(HNO[
2eq
2eq
1
2eq
1
2eq
2eq
2eq
3HNO3HNOH
3HNO3HNOH2eq
Tot3
3
3NO
1
2eq
2eq
3HNO3HNOH
3HNO3HNOH
Tot3
3)aq(3HNO
K
K
K
K
K
KpK
pKK
K
KpK
pK
lecture 15 natural sulfur, acid rain
HNO3 Mole Fractions as a Function of pH
Since Keq2 is so high, Keq2 /[H] >> 1
Dissolved nitric acid in clouds exists exclusively as nitrate.
Aqueous fraction of nitric acid as a function of pH and cloud LWC:
1~ ,0~)aq(3NO)aq(3HNO
lecture 15 natural sulfur, acid rain
H2SO4(g) and NOX(g)
We looked at the aqueous phase equilibrium of SO2/H2O and HNO3/H2O.
But SO2 (g) H2SO4(g) and NOX (g) HNO3 (g).
What about solubility of H2SO4(g) and NOX (g)?
They are soluble, but not important contributors to acid rain.
Rate of gas phase oxidation of SO2(g) to H2SO4(g) by OH:
0.3%-3% / hr
Rate of gas phase oxidation of NO2(g) to HNO3(g) by OH: 10 x faster.