Transcript Atmospheric Chemistry 8

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
Acid Rain Over North America and Norway
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:
H 2O
kf



kr
H   OH -
At equilibrium:
k f [H 2 O]  k r [H  ][OH  ]
kf
[H  ][OH  ]
w 
 K eq
kr
[ H 2 O]
Keq: equilibrium constant
 w  1.82  1016 M at 298 K
K eq
M = mol L-1
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:
[H  ]  [OH  ]  K eq w  1107 M
Concentration of ions is much smaller than [H2O]  water has small
conductivity.
lecture 15 natural sulfur, acid rain
pH of Pure Water
Definition of pH:
pH   log10 [H  ](M)


Pure water at 298 K:
pH   log10 1107   7.0
acidic: pH < 7
alkaline: pH > 7
neutral: pH = 7
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pH of Clean Rainwater
Clean rainwater is not pure water. It equilibrates with CO2:


hydrolysis
CO 2 (g)  H 2 O 
 H 2 CO 3 (aq )


H 2 CO 3  H   HCO 3


HCO 3  H   CO 32 
ionization – bicarbonate ion
further ionization – carbonate ion
lecture 15 natural sulfur, acid rain
Equilibrium Between Gas and Aqueous Phase
Equilibrium for first reaction:


CO 2 (g)  H 2 O 
 H 2 CO 3 (aq)
K eq  K H CO 
2
[H 2 CO 3 (aq)]
pCO 2
[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.
lecture 15 natural sulfur, acid rain
Henry’s Law Constants for Atmospheric Gases
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:
d ln K H DH

dT
RT 2
d ln es
L

similar to Clausius-Clapeyron equation:
dT
RT 2
DH: 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).
lecture 15 natural sulfur, acid rain
Heat of Dissolution for Atmospheric Gases
lecture 15 natural sulfur, acid rain
CO2/H2O System
Reactions in CO2/H2O system:
(1) CO 2 (g)  H 2 O
(2) H 2 CO
K eq
2
 
3  
KH
CO 2
  

  
H 2 CO 3 (aq)
H   HCO 3
K eq 3

  
(3) HCO 3   H   CO 32 
K eq w  [H  ][OH  ]
K H CO  3  10  2 M atm 1
2
K eq 2  4.3  10  7 M
K eq 3  4.7  10 11 M
pCO 2  280 ppmv  1 atm
pre - industrial
pCO 2  350 ppmv  1 atm
current
H 2O
K eq
 w

 
H   OH 
lecture 15 natural sulfur, acid rain
CO2/H2O System cont.
(1) K H CO 
2
[H 2 CO 3 (aq)]
pCO 2
[H 2 CO 3 (aq)]  K H CO pCO 2
2
(2) K eq 2
[H  ][HCO 3 ] [H  ][HCO 3 ]


[H 2 CO 3 (aq)] K H CO pCO 2
2

3
[HCO ] 
(3) K eq 3
K eq 2 K H CO pCO 2

2
[H ]
[H  ][CO 32  ] [H  ][CO 32  ][H  ]



K eq 2 K H CO pCO 2
[HCO 3 ]
2
2
3
[CO ] 
K eq 3 K eq 2 K H CO pCO 2
 2
2
[H ]

2
Total dissolved CO2: [CO Tot
2 (aq)]  [ H 2 CO 3 (aq)]  [ HCO 3 ]  [CO 3 ]
 K H CO
2
 K eq 2 K eq 3 K eq 2
pCO 2 1   
[ H  ]2
 [H ]



lecture 15 natural sulfur, acid rain
Effective Henry’s Law Constant
effective Henry’s Law constant for CO2:

2
[CO Tot
2 (aq )]  [ H 2 CO 3 (aq )]  [ HCO 3 ]  [CO 3 ]
K
*
H CO 2
 K H CO
2
 K H* CO
2
 K eq 2 K eq 3 K eq 2
pCO 2 1   
[ H  ]2
 [H ]
pCO 2
 K eq 2 K eq 3 K eq 2
 K H CO 1   
2
[ H  ]2
 [H ]






Is K H* CO greater than or less than K H CO ?
2
2
Always greater than  Total amount of CO2 dissolved always exceeds
that predicted by Henry’s Law for CO2 alone (although not by much).
lecture 15 natural sulfur, acid rain
Effective Henry’s Law Constant cont.
What does K H* CO depend on?
2
T, pH of solution ([H])
As pH increases ([H] decreases), does K H* CO increase or decrease?
2
increase
Tot
As pH increases, does [CO 2 (aq)] increase or decrease?
increase
lecture 15 natural sulfur, acid rain
Effective Henry’s Law Constant of CO2 as a Function of pH
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]
(2) K eq 2
[H  ][HCO 3 ]
[ H  ]2


K H CO pCO 2
K H CO pCO 2
2
2
[H  ]2  K eq 2 K H CO pCO 2  (4.3  10 7 M)(3  10 2 M atm 1 )(350  10 6  1 atm)
2
pH   log10 [H  ]  5.6
pH of pure rainwater
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:
[H  ]  [OH  ]  [HCO3 ]  2[CO32 ]
(Each CO32 ion contributes charge of 2. Total negative charge is
concentration of ions x 2.)

[H ] 
K eq w
[H  ]

K eq 2 K H CO pCO 2
2
[H  ]

2 K eq 3 K eq 2 K H CO pCO 2
solve for [H  ] : pH   log10 [H  ]  5.6
[ H  ]2
2
lecture 15 natural sulfur, acid rain
Calculate the pH of CO2/H2O System – Third Method
CO2 + H2O = H2CO3
H2CO3 = H+ + HCO3- = 2H+ + CO3=
K1 = 4.3x10-7
K2 = 5.3x10-11
aC = [H+] = [HCO3-]
a: fraction of concentration in the form of ions
C: concentration
(1-a)C = [H2CO3]
1-a: fraction of concentration in the form of acid
K1 = [H+][HCO3-] / [H2CO3] = (aC)2 / (1 - a)C = a2C/(1 - a)
lecture 15 natural sulfur, acid rain
Calculate the pH of CO2/H2O System – Third Method cont.
CO2 solubility 0.759 liters@1atm/liter H2O
CO2 Partial pressure Pp is 345 ppm
n = VPp/RT
n = 0.759 x 345x10-6 / (0.082 x 298) = 1.07x105 mol
In 1 liter C = 1.07 x 105 mol/liter
4.3 x 107= 1.07 x 105 a2 / (1 - a)
a2 + 0.04a - 0.04=0
a = 0.18
[H+] = 0.18 x 1.07 x 105=1.93x10-5 M
pH = log [H+] = 5.7
lecture 15 natural sulfur, acid rain
SO2/H2O System
Reactions in SO2/H2O system:
(1) SO 2 (g)  H 2 O
(2) H 2SO
(3) HSO
K eq 2

 
3  
K eq 3
  
3  
KH
SO 2

  
  
H 2SO 3 (aq)
H   HSO 3
H   SO 32 
K eq w  [H  ][OH  ]
K HSO  1.23 M atm 1 at 298 K
2
K eq 2  1.3  10  2 M
K eq 3  6.6  10 8 M
pSO 2  0.2  200 ppb
H 2O
K eq

 w
 
H   OH 
bisulfite ion
sulfite ion
lecture 15 natural sulfur, acid rain
SO2/H2O System cont.
(1) K HSO 
2
[H 2SO 3 (aq)]
pSO 2
[H 2SO 3 (aq)]  K HSO pSO 2
2
[H  ][HSO 3 ] [H  ][HSO 3 ]


[H 2SO 3 (aq)] K HSO pSO 2
(2) K eq 2
2

3
[HSO ] 
(3) K eq 3
K eq 2 K HSO pSO 2

2
[H ]
[H  ][SO 32  ] [H  ][SO 32  ][H  ]



K eq 2 K HSO pSO 2
[HSO 3 ]
2
2
3
[SO ] 
K eq 3 K eq 2 K HSO pSO 2
 2
2
[H ]
Total dissolved sulfur:
[S(IV) Tot (aq)]  [H 2SO3 (aq)]  [HSO3 ]  [SO32 ]
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
Sulfur Oxidation States
lecture 15 natural sulfur, acid rain
Sulfur Oxidation States cont.
lecture 15 natural sulfur, acid rain
Total Dissolved Sulfur
Total dissolved sulfur:
[S(IV ) Tot (aq)]  [H 2SO 3 (aq)]  [HSO 3 ]  [SO 32  ]
 K HSO
2
 K eq 2 K eq 3 K eq 2
pSO 2 1   
 2
[
H
]
[
H
]

Effective Henry’s Law constant for SO2:
[S(IV ) Tot (aq)]  K H* SO pSO 2
2
K
*
H SO 2
 K eq 2 K eq 3 K eq 2
 K HSO 1   
2
[ H  ]2
 [H ]






lecture 15 natural sulfur, acid rain
Effective Henry’s Law Constant of SO2 as a Function of pH
K H* SO increases by ~7 orders of magnitude with pH  Acid-base
2
equilibrium pulls more material into solution.
(Which material? [H2SO3(aq)] does not depend on pH.)
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:
K HSO pSO 2
[H 2SO 3 (aq)]
2
 H 2SO3 ( aq ) 

[S(IV ) Tot (aq)]
 K eq 2 K eq 3 K eq 2 

K HSO pSO 2 1   
 2
2
[H ] 
 [H ]
 K eq 2 K eq 3 K eq 2
 1   
 2
[
H
]
[
H
]

 HSO 
3
[HSO 3 ]


Tot
[S(IV ) (aq)]



1
K eq 2 K HSO pSO 2
2

[H ] K HSO
[H  ]  K eq 2 K eq 3 K eq 2
1   

 2
K
[
H
]
[
H
]
eq

 2
 [H ]
K eq 3 


1  
K
[H ] 
 eq 2

1
2
 K eq 2 K eq 3 K eq 2
pSO 2 1   
 2
[
H
]
[
H
]




1



lecture 15 natural sulfur, acid rain
Mole Fractions cont.
 SO 2 
3
[SO 32  ]


Tot
[S(IV ) (aq)]
 [ H  ]2

 K eq 3 K eq 2
K eq 3 K eq 2 K HSO pSO 2
2
 2
[H ] K HSO
K eq 3 K eq 2 

1   

 2
[H ] 
 [H ]
K eq 2
 [H ]
[H ] 



1
K K

K
eq
eq
eq


3
2
3
 2
2
 K eq 2 K eq 3 K eq 2
pSO 2 1   
 2
[
H
]
[
H
]


1
1



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 ,SO32.
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:
S(IV)  O 3 (aq) 
 S(VI)  O 2 (aq)
S(IV) : H 2SO 3 (aq), HSO 3 , SO 32 
d [S(IV)]
 k 0 [H 2SO 3 (aq)]  k1[HSO 3 ]  k 2 [SO 32  ][O 3 (aq)]  k[S(IV)][O 3 (aq)]
dt
k 0  (2.4  1.1)  10 4 M 1 s 1

k1  (3.7  0.7)  105 M 1 s 1
k 2  (1.5  0.6)  109 M 1 s 1
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 selflimiting. 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:

3
(1) HSO  H 2 O 2 (aq)
k1



k 1
SO 2 OOH   H 2 O
k2
(2) SO 2 OOH   H  
H 2SO 4 (aq)
k1  5.6  106 M  2 s 1
k 1
 0.1
k2
lecture 15 natural sulfur, acid rain
The S(IV)/H2O2(aq) System cont.
Steady state approximation on SO2OOH:
d [SO 2 OOH  ]
 0  k1[HSO 3 ][H 2 O 2 (aq)]  k 1[SO 2 OOH  ]  k 2 [SO 2 OOH  ][H  ]
dt
[SO 2 OOH  ]k 1  k 2 [H  ]  k1[HSO 3 ][H 2 O 2 (aq)]

k
[
HSO
3 ][ H 2 O 2 (aq )]
[SO 2 OOH  ]  1
k 1  k 2 [H  ]
 k1[HSO 3 ][H 2 O 2 (aq)]  
d [H 2SO 4 (aq)]


 [H ]
 k 2 [SO 2 OOH ][H ]  k 2 

dt
k 1  k 2 [H ]


k1[HSO 3 ][H 2 O 2 (aq)][H  ]

k 1
 [H  ]
k2
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
The Sulfur Cycle
lecture 15 natural sulfur, acid rain
pH of Sulfur System as a Function of Cloud Liquid Water Content
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 mg/m3 (typical for remote pacific area)
[H] = 2 (moles H per mole H2SO4) x106 (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) = 2.04x105 M
pH = log[H] = 4.7
For a thin cloud (0.1 gr/m3);
[H] = 2.04x10-4 M
pH = 3.7
lecture 15 natural sulfur, acid rain
pH of Sulfur System as a Function of Cloud Liquid Water Content
cont.
SO4= (mg/m3)
Cloud Water
Content
1.0
0.1
0.5
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:
(1) HNO 3 (g)  H 2 O
(2) HNO 3 (aq)
KH
HNO 3
   

   

K eq
2
 
 
HNO 3 (aq)
H   NO 3
H 2O
K eq
 w

 
H   OH 
nitrate ion
K eq w  [H  ][OH  ]
K H HNO  2.1 105 M atm 1 at 298 K
very soluble
K eq 2  15.4 M at 298 K
dissociates quickly
3
lecture 15 natural sulfur, acid rain
The HNO3/H2O System cont.
(1) K H HNO 
3
[HNO 3 (aq)]
pHNO3
[HNO 3 (aq)]  K H HNO pHNO3
3
(2) K eq 2
[H  ][ NO 3 ]
[H  ][ NO 3 ]


[HNO 3 (aq)] K H HNO pHNO3
3
K eq 2 K H HNO pHNO3

3
[ NO ] 
3
[H  ]
Total dissolved nitric acid:
[HNO 3
Tot

3
(aq)]  [HNO 3 (aq)]  [ NO ]  K H HNO
Effective Henry’s law constant for HNO3:
Tot
[HNO 3 (aq)]  K H* HNO pHNO3
3
K
*
H HNO 3
 K eq 2 
 K H HNO 1   
3
 [H ] 
3
 K eq 2 
pHNO3 1   
 [H ] 
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:
K H HNO pHNO3
[HNO 3 (aq)]
3
 HNO3 ( aq ) 

Tot
 K eq 2 
[HNO 3 (aq)]
K H HNO pHNO3 1   
3
 [H ] 
 K eq 
 1  2 
 [H ] 
 NO
3
1
[ NO 3 ]


Tot
[HNO 3 (aq)]
K eq 2 K H HNO pHNO3
3
[H ] K H HNO
[H  ]  K eq 2 
1   

 K eq 2  [H ] 
1

1
 [H  ] 
K eq 2
 
 1 

K
K

[
H
]
eq
eq 2


2
3
 K eq 2 
p HNO3 1   
 [H ] 
lecture 15 natural sulfur, acid rain
HNO3 Mole Fractions as a Function of pH
Since Keq2 is so high, Keq2 /[H] >> 1
  HNO3 ( aq ) ~ 0,  NO ( aq ) ~ 1
3
 Dissolved nitric acid in clouds exists exclusively as nitrate.
Aqueous fraction of nitric acid as a function of pH and cloud LWC:
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.
lecture 15 natural sulfur, acid rain
The NH3/H2O System