Lecture 15 natural sulfur, acid rain Rainout We mentioned a few of things that may rainout: 1.CH 3 OOH (CH 4 oxidation, low NO x ) 2.H 2 O 2 (CO oxidation,

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.


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.


Acid Rain Over North America and Norway


Two Main Sources of Acid Rain

1. SO2 from industry oxidized to H2SO4

2. NO, NO2 from automobiles oxidized to HNO3


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


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


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


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


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


Henry’s Law Constants for Atmospheric Gases


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


Heat of Dissolution for Atmospheric Gases


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


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


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


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


Effective Henry’s Law Constant of CO2 as a Function of pH


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


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


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


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.


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


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


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.


Sulfur Oxidation States


Sulfur Oxidation States cont.


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


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


If Assume Constant pSO2

Open system: unlimited LWC, unlimited SO2:

[S(IV)Tot(aq)] increases dramatically with pH


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


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


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


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.


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).


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)


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


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.


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).


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


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


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).


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


The Sulfur Cycle


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


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


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


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


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


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


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.


The NH3/H2O System

Documents

Lecture 15 natural sulfur, acid rain Rainout We mentioned a few of things that may rainout: 1.CH 3 OOH (CH 4 oxidation, low NO x ) 2.H 2 O 2 (CO oxidation,