Atmospheric Chemistry and Physics From Air Pollution to Climate Change, 3rd Edition John H Seinfeld

ATMOSPHERIC CHEMISTRY AND PHYSICS: FROM AIR POLLUTION TO CLIMATE CHANGE Third Edition John H. Seinfeld California Institute of Technology Spyros N. Pandis University of Patras and Carnegie Mellon University John Wiley 2015

1.4

CH3SCH3 concentrations range from 250 to 500 ng m-3 (0.25 to 0.50 μg m-3).

ξ CH SCH (ppm) = 3

3

=

RT pM CH 3SCH 3

[CH3SCH 3 ] (μg m −3 )

(8.314) (298) (0.25) = 9.86 × 10-5 ppm

(1.0133 × 10 ) 5

= 0.0986 ppb At 0.5 μg m-3, ξ CH 3SCH 3 = 0.197 ppb

1.5 ⎛ 3.7879 L ⎞ ⎛ 1000 cm 3 ⎞ ⎛ 0.85 g ⎞ ⎟⎟ ⎜ ⎟⎟ ⎜⎜ 500 gallons ⎜⎜ = 1,609,857.5 g Gasoline 3 ⎟ L ⎝ gallon ⎠ ⎝ ⎠ ⎝ cm ⎠

C: 12.01 g/mol H: 1.01 g/mol O: 16.00 g/mol gasoline (C8H18) = [8(12.01) + 18(1.01)] g/mol = 114.26 g/mol CO2: (12.01 + 2·16.00) g/mol = 44.01 g/mol ⎞ ⎛ 44.01 g CO 2 ⎞ ⎛ 1 kg ⎞ ⎟⎟ ⎜⎜ ⎟⎟ ⎟⎟ ⎜⎜ 1 mol gasoline 1 mol CO 1000 g ⎠ ⎠⎝ ⎠⎝ 2 ⎠⎝ ⎞⎛

⎛

(1,609,857.5 g )⎜⎜ 1 mol gasoline ⎟⎟ ⎜⎜ 8 mol CO 2 ⎝

= 4960 kg CO2

114.26 g

1.6 Mass of the atmosphere = 5.148 × 1018 kg = 5.148 × 1021 g Moles of the atmosphere = 5.148/29 × 1021 = 1.77 × 1020 mol air Moles of 1 ppm CO2 = (10-6) (1.77 × 1020) = 1.77 × 1014 mol CO2 Mass of 1 ppm CO2 = (1.77 × 1014 mol CO2) (44 g/mol) = 7.8 × 1015 Mass of carbon = (7.8 × 1015) (12/44) = 2.1 × 1015 g = 2.1 Pg C Current mass of C in atmosphere = (2.1) (400) = 840 Pg C

1.7

The scale heights for the atmospheres of Venus and Mars are: Venus Mars

15.85 km 10.89 km

Although the temperature on Mars is colder than that of Earth and the molecular weight of its atmosphere is slightly higher than that of our atmosphere, the scale height on Mars is larger than Earth’s owing to the much smaller gravity on Mars.

Chapter 2

2.1 The mean lifetime of CH4 in the atmosphere is estimated as the ratio of the amount of CH4 in the atmosphere (Tg) to its source or sink strength (~ 600 Tg yr-1).

To determine the abundance of CH4 we need to convert a mixing ratio of 1745 ppb into total atmospheric mass. The total number of moles of air in the atmosphere = 1.62 × 1020. So Moles of CH4 = (1745 × 10-9) (1.62 × 1020) = 2.83 × 1014 Mass of CH4 = (2.83 × 1014 moles) (16 g mol-1) = 45.3 × 1014 g = 4530 Tg Mean atmospheric lifetime of CH4 = 4530/600 ≅ 7.6 years For CO: Mass of CO = (100 × 10-9) (1.62 × 1020) (28) = 45.4 × 1013 g = 454 Tg Mean atmospheric lifetime of CO = 454/2780 = 0.16 year ≅ 2 months Thus the explanation lies in the much longer atmospheric lifetime of CH4.

2.2 z (km)

ξ O (ppm)

cO3 (molec cm-3)

cair (molec cm-3)

cair (ppm)

5 10 20 25 30 35 40 50

0.0326 0.0581 1.0283 5.7632 8.8874 9.6667 6.0214 4.6858

5.00 × 1011 5.00 × 1011 1.90 × 1012 4.80 × 1012 3.40 × 1012 1.70 × 1012 5.00 × 1011 1.00 × 1011

1.53024 × 1011 8.59253 × 1018 1.84755 × 1018 8.32862 × 1017 3.8256 × 1017 1.75861 × 1017 8.30369 × 1016 2.13411 × 1016

736 413.5 88.91 40.08 18.41 8.463 3.996 1.027

3

Mixing ratio is cO3 /cair , and cair is not a constant with altitude. Absolute O3 concentration peaks

at about 25 km; O3 mixing ratio peaks at about 35 km.

2.3 Total amount of N2O in the atmosphere = (315 × 10-9) (1.62 × 1020 moles) (44 g mol-1) = 2245 Tg N2O = 2245 (28/44) = 1429 Tg N Mean N2O lifetime = 1429 Tg N/12.6 Tg N yr-1 = 113 years.

2.4

nO3 = (3 × 1012 molecules cm-3) (15 × 105 cm) = 4.5 × 1018 molecules cm-2 Then

h = (3.72 × 10-20) (4.5 × 1018) = 0.167 cm = 1.67 mm = 167 DU

2.5

2.6

a.

4.2 × 1015 and 2.5 × 1015

b.

5.3 months

9.0 months

5.0 months

6.6 months

1 ppb CH4 = 2.78 Tg CH4 46 Tg y-1/2.78 Tg ppb-1 = 16.5 ppb y-1

2.7

Two compartment model for CH3Br in the atmosphere.

mSH

kOH

kT

kT

mNH

kOH

S

The mass balances on CH3Br for the two hemispheres are:

dmNH = S + kT mSH − kT mNH − kOH mNH dt dmSH = kT mNH − kT mSH − kOH mSH dt

Assume steady state:

dmNH dmSH

= 0 = 0 dt dt

Obtain

O = = S + kT mSH − kT mNH − kOH mNH

kT mNH − kT mSH − kOH mSH O =

a) To get mNH /mSH consider (2),

kT ( mNH − mSH ) = kOH mSH

(1)

(2)

mNH k = 1+ OH mSH kT

= 1+ b)

0.667 = 1.74 0.09

The observed mNH /mSH ratio is 1.3. This suggests that there are sources of CH3Br in the SH, possibly from the ocean.

2.8

a.

For the 3 species at 273 K:

CHBr3: CH2Br2: C2H5Br:

j = 1.4 × 10-­‐6 s-­‐1 j = 1.2 × 10-­‐8 s-­‐1 j = 2.1 × 10-­‐10 s-­‐1

τOH 0.21 yr 0.35 yr 0.12 yr

kOH [OH] = (1.49 × 10-­‐13) (106) = 1.49 × 10-­‐7 s-­‐1 kOH [OH] = (9.12 × 10-­‐14) (106) = 9.12 × 10-­‐8 s-­‐1 kOH [OH] = (2.76 × 10-­‐13) (106) = 2.76 × 10-­‐7 s-­‐1

For loss by photolysis: CHBr3: CH2Br2: C2H5Br:

τp 0.023 yr 2.64 yr 151 yr

The overall tropospheric lifetime for each is computed from:

1 1 1

= + τ total τ OH τ p

The results for τtotal are:

CHBr3

0.021 yr

CH2Br2

0.31 yr

C2H5Br

8.33 yr

Note: 31,536,000 seconds/year

b. Assume

Q = 1.5 Gmol Br yr-­‐1 emission flux into the troposphere τ = 0.021 yr tropospheric lifetime kTS = 0.063 yr -­‐1 flux of air from troposphere to stratosphere

At steady state,

Q – (τ-­‐1 + kTS) c = 0

We want to calculate c. Note that Q = 0.5 x 109 mol CHBr3 yr-­‐1. Solving for c,

C = 0.5 x 109 / (47.7 x 8.14 x 1024) = 1.3 x 10-­‐18 mol cm-­‐3 = 7.83 x 105 molecules cm-­‐3

The mole fraction of CHBr3 is 7.83 x 105/2.7 x 10 19 = 2.9 x 1014 = 0.029 ppt.

This is to be compared with the surface measurements of about 1 ppt. Owing to the relatively short lifetime of CHBr3 in the troposphere, its concentration is not uniform over the entire troposphere. With its source at the surface, the mixing ratio is expected to be highest at the surface.

c. The annual flux of CHBr3 into the stratosphere is

(0,063 yr-­‐1) (1.3 x 10-­‐18 mol cm-­‐3) (8.14 x 1024 cm3) (3 Br atoms)

= 2 x 106 mol Br yr-­‐1

2.9 Reaction with OH is the main sink of CH3CCl3 in the troposphere: CH3CCl3 + OH -------> CH2CCl3 + H2O with the reaction rate constant k = 6.11 × 10-15 cm3 molec-1 s-1 at 273 K. The dynamic behavior of CH3CCl3 due to reaction with OH is given by d[CH3 CCl3 ] =k CH3 CCl3 OH dt

(1)

Assuming constant OH radical concentration, Equation (1) becomes d[CH3 CCl3 ] =k' CH3 CCl3 dt

(2)

where k’ = k[OH]. In this way, the time-dependent concentration of CH3CCl3 follows the first order decay [CH3 CCl3 ] = [CH3 CCl3 ]0 exp -k't

(3)

Plot [CH3CCl3] vs. [t] and apply exponential regression to the data: 8

11

x 10

10

8

7

6

3

[CH CCl ] (molecule cm−3)

9

3

5

4

3

2

1

0

0.5

1

1.5

2

time (s)

2.5

3

3.5

4 8

x 10

We get k’ = 5.698 × 10-9 s-1. Therefore, [OH] = k’/k = 9.3 × 105 molecule cm-3 Or plot log([CH3CCl3]/ [CH3CCl3]0) vs. [t] and apply linear regression to the data:

0

−0.5

log(C/C0)

−1

−1.5

−2

−2.5

0

0.5

1

1.5

2

time (h)

2.5

3

3.5

4 8

x 10

We get k’ = 5.698 × 10-9 s-1. Therefore, [OH] = k’/k = 9.3 × 105 molecule cm-3

2.10 a. The mixing ratios of water vapor in the two seasons are:

3.6 ppm JFM 18 ppm JJA

b. We assume that methane is well mixed in the troposphere owing to its long lifetime. The difference in transport between water and methane can be inferred from the difference in their mixing ratios. The present-­‐day mixing ratio of methane is 1803 ppb and its preindustrial value is 700 ppb. (check these) We assume that the mixing ratios of water vapor in the two seasons are unchanged from pre-­‐ industrial time. Methane carries 4 H atoms versus water with 2 H atoms. In JFM,

Water vapor 7200 ppb H Methane 7212 ppb H (present day) 2800 ppb H (pre-­‐industrial)

In JJA, Water vapor 36000 ppb H Methane 7212 ppb H (present day) 2800 ppb H (pre-­‐industrial)

We conclude that methane competes with water vapor as a source of H to the stratosphere in present day, whereas in pre-­‐industrial time, water vapor was a more dominant source.

ATMOSPHERIC CHEMISTRY AND PHYSICS: FROM AIR POLLUTION TO CLIMATE CHANGE Third Edition John H. Seinfeld California Institute of Technology Spyros N. Pandis University of Patras and Carnegie Mellon University John Wiley 2015

1.4

CH3SCH3 concentrations range from 250 to 500 ng m-3 (0.25 to 0.50 μg m-3).

ξ CH SCH (ppm) = 3

3

=

RT pM CH 3SCH 3

[CH3SCH 3 ] (μg m −3 )

(8.314) (298) (0.25) = 9.86 × 10-5 ppm

(1.0133 × 10 ) 5

= 0.0986 ppb At 0.5 μg m-3, ξ CH 3SCH 3 = 0.197 ppb

1.5 ⎛ 3.7879 L ⎞ ⎛ 1000 cm 3 ⎞ ⎛ 0.85 g ⎞ ⎟⎟ ⎜ ⎟⎟ ⎜⎜ 500 gallons ⎜⎜ = 1,609,857.5 g Gasoline 3 ⎟ L ⎝ gallon ⎠ ⎝ ⎠ ⎝ cm ⎠

C: 12.01 g/mol H: 1.01 g/mol O: 16.00 g/mol gasoline (C8H18) = [8(12.01) + 18(1.01)] g/mol = 114.26 g/mol CO2: (12.01 + 2·16.00) g/mol = 44.01 g/mol ⎞ ⎛ 44.01 g CO 2 ⎞ ⎛ 1 kg ⎞ ⎟⎟ ⎜⎜ ⎟⎟ ⎟⎟ ⎜⎜ 1 mol gasoline 1 mol CO 1000 g ⎠ ⎠⎝ ⎠⎝ 2 ⎠⎝ ⎞⎛

⎛

(1,609,857.5 g )⎜⎜ 1 mol gasoline ⎟⎟ ⎜⎜ 8 mol CO 2 ⎝

= 4960 kg CO2

114.26 g

1.6 Mass of the atmosphere = 5.148 × 1018 kg = 5.148 × 1021 g Moles of the atmosphere = 5.148/29 × 1021 = 1.77 × 1020 mol air Moles of 1 ppm CO2 = (10-6) (1.77 × 1020) = 1.77 × 1014 mol CO2 Mass of 1 ppm CO2 = (1.77 × 1014 mol CO2) (44 g/mol) = 7.8 × 1015 Mass of carbon = (7.8 × 1015) (12/44) = 2.1 × 1015 g = 2.1 Pg C Current mass of C in atmosphere = (2.1) (400) = 840 Pg C

1.7

The scale heights for the atmospheres of Venus and Mars are: Venus Mars

15.85 km 10.89 km

Although the temperature on Mars is colder than that of Earth and the molecular weight of its atmosphere is slightly higher than that of our atmosphere, the scale height on Mars is larger than Earth’s owing to the much smaller gravity on Mars.

Chapter 2

2.1 The mean lifetime of CH4 in the atmosphere is estimated as the ratio of the amount of CH4 in the atmosphere (Tg) to its source or sink strength (~ 600 Tg yr-1).

To determine the abundance of CH4 we need to convert a mixing ratio of 1745 ppb into total atmospheric mass. The total number of moles of air in the atmosphere = 1.62 × 1020. So Moles of CH4 = (1745 × 10-9) (1.62 × 1020) = 2.83 × 1014 Mass of CH4 = (2.83 × 1014 moles) (16 g mol-1) = 45.3 × 1014 g = 4530 Tg Mean atmospheric lifetime of CH4 = 4530/600 ≅ 7.6 years For CO: Mass of CO = (100 × 10-9) (1.62 × 1020) (28) = 45.4 × 1013 g = 454 Tg Mean atmospheric lifetime of CO = 454/2780 = 0.16 year ≅ 2 months Thus the explanation lies in the much longer atmospheric lifetime of CH4.

2.2 z (km)

ξ O (ppm)

cO3 (molec cm-3)

cair (molec cm-3)

cair (ppm)

5 10 20 25 30 35 40 50

0.0326 0.0581 1.0283 5.7632 8.8874 9.6667 6.0214 4.6858

5.00 × 1011 5.00 × 1011 1.90 × 1012 4.80 × 1012 3.40 × 1012 1.70 × 1012 5.00 × 1011 1.00 × 1011

1.53024 × 1011 8.59253 × 1018 1.84755 × 1018 8.32862 × 1017 3.8256 × 1017 1.75861 × 1017 8.30369 × 1016 2.13411 × 1016

736 413.5 88.91 40.08 18.41 8.463 3.996 1.027

3

Mixing ratio is cO3 /cair , and cair is not a constant with altitude. Absolute O3 concentration peaks

at about 25 km; O3 mixing ratio peaks at about 35 km.

2.3 Total amount of N2O in the atmosphere = (315 × 10-9) (1.62 × 1020 moles) (44 g mol-1) = 2245 Tg N2O = 2245 (28/44) = 1429 Tg N Mean N2O lifetime = 1429 Tg N/12.6 Tg N yr-1 = 113 years.

2.4

nO3 = (3 × 1012 molecules cm-3) (15 × 105 cm) = 4.5 × 1018 molecules cm-2 Then

h = (3.72 × 10-20) (4.5 × 1018) = 0.167 cm = 1.67 mm = 167 DU

2.5

2.6

a.

4.2 × 1015 and 2.5 × 1015

b.

5.3 months

9.0 months

5.0 months

6.6 months

1 ppb CH4 = 2.78 Tg CH4 46 Tg y-1/2.78 Tg ppb-1 = 16.5 ppb y-1

2.7

Two compartment model for CH3Br in the atmosphere.

mSH

kOH

kT

kT

mNH

kOH

S

The mass balances on CH3Br for the two hemispheres are:

dmNH = S + kT mSH − kT mNH − kOH mNH dt dmSH = kT mNH − kT mSH − kOH mSH dt

Assume steady state:

dmNH dmSH

= 0 = 0 dt dt

Obtain

O = = S + kT mSH − kT mNH − kOH mNH

kT mNH − kT mSH − kOH mSH O =

a) To get mNH /mSH consider (2),

kT ( mNH − mSH ) = kOH mSH

(1)

(2)

mNH k = 1+ OH mSH kT

= 1+ b)

0.667 = 1.74 0.09

The observed mNH /mSH ratio is 1.3. This suggests that there are sources of CH3Br in the SH, possibly from the ocean.

2.8

a.

For the 3 species at 273 K:

CHBr3: CH2Br2: C2H5Br:

j = 1.4 × 10-­‐6 s-­‐1 j = 1.2 × 10-­‐8 s-­‐1 j = 2.1 × 10-­‐10 s-­‐1

τOH 0.21 yr 0.35 yr 0.12 yr

kOH [OH] = (1.49 × 10-­‐13) (106) = 1.49 × 10-­‐7 s-­‐1 kOH [OH] = (9.12 × 10-­‐14) (106) = 9.12 × 10-­‐8 s-­‐1 kOH [OH] = (2.76 × 10-­‐13) (106) = 2.76 × 10-­‐7 s-­‐1

For loss by photolysis: CHBr3: CH2Br2: C2H5Br:

τp 0.023 yr 2.64 yr 151 yr

The overall tropospheric lifetime for each is computed from:

1 1 1

= + τ total τ OH τ p

The results for τtotal are:

CHBr3

0.021 yr

CH2Br2

0.31 yr

C2H5Br

8.33 yr

Note: 31,536,000 seconds/year

b. Assume

Q = 1.5 Gmol Br yr-­‐1 emission flux into the troposphere τ = 0.021 yr tropospheric lifetime kTS = 0.063 yr -­‐1 flux of air from troposphere to stratosphere

At steady state,

Q – (τ-­‐1 + kTS) c = 0

We want to calculate c. Note that Q = 0.5 x 109 mol CHBr3 yr-­‐1. Solving for c,

C = 0.5 x 109 / (47.7 x 8.14 x 1024) = 1.3 x 10-­‐18 mol cm-­‐3 = 7.83 x 105 molecules cm-­‐3

The mole fraction of CHBr3 is 7.83 x 105/2.7 x 10 19 = 2.9 x 1014 = 0.029 ppt.

This is to be compared with the surface measurements of about 1 ppt. Owing to the relatively short lifetime of CHBr3 in the troposphere, its concentration is not uniform over the entire troposphere. With its source at the surface, the mixing ratio is expected to be highest at the surface.

c. The annual flux of CHBr3 into the stratosphere is

(0,063 yr-­‐1) (1.3 x 10-­‐18 mol cm-­‐3) (8.14 x 1024 cm3) (3 Br atoms)

= 2 x 106 mol Br yr-­‐1

2.9 Reaction with OH is the main sink of CH3CCl3 in the troposphere: CH3CCl3 + OH -------> CH2CCl3 + H2O with the reaction rate constant k = 6.11 × 10-15 cm3 molec-1 s-1 at 273 K. The dynamic behavior of CH3CCl3 due to reaction with OH is given by d[CH3 CCl3 ] =k CH3 CCl3 OH dt

(1)

Assuming constant OH radical concentration, Equation (1) becomes d[CH3 CCl3 ] =k' CH3 CCl3 dt

(2)

where k’ = k[OH]. In this way, the time-dependent concentration of CH3CCl3 follows the first order decay [CH3 CCl3 ] = [CH3 CCl3 ]0 exp -k't

(3)

Plot [CH3CCl3] vs. [t] and apply exponential regression to the data: 8

11

x 10

10

8

7

6

3

[CH CCl ] (molecule cm−3)

9

3

5

4

3

2

1

0

0.5

1

1.5

2

time (s)

2.5

3

3.5

4 8

x 10

We get k’ = 5.698 × 10-9 s-1. Therefore, [OH] = k’/k = 9.3 × 105 molecule cm-3 Or plot log([CH3CCl3]/ [CH3CCl3]0) vs. [t] and apply linear regression to the data:

0

−0.5

log(C/C0)

−1

−1.5

−2

−2.5

0

0.5

1

1.5

2

time (h)

2.5

3

3.5

4 8

x 10

We get k’ = 5.698 × 10-9 s-1. Therefore, [OH] = k’/k = 9.3 × 105 molecule cm-3

2.10 a. The mixing ratios of water vapor in the two seasons are:

3.6 ppm JFM 18 ppm JJA

b. We assume that methane is well mixed in the troposphere owing to its long lifetime. The difference in transport between water and methane can be inferred from the difference in their mixing ratios. The present-­‐day mixing ratio of methane is 1803 ppb and its preindustrial value is 700 ppb. (check these) We assume that the mixing ratios of water vapor in the two seasons are unchanged from pre-­‐ industrial time. Methane carries 4 H atoms versus water with 2 H atoms. In JFM,

Water vapor 7200 ppb H Methane 7212 ppb H (present day) 2800 ppb H (pre-­‐industrial)

In JJA, Water vapor 36000 ppb H Methane 7212 ppb H (present day) 2800 ppb H (pre-­‐industrial)

We conclude that methane competes with water vapor as a source of H to the stratosphere in present day, whereas in pre-­‐industrial time, water vapor was a more dominant source.

Chapter 3

3.1

τ =

1 k[OH]

kOH + CHF2 Cl = 1.7 × 10-15 cm3 molecule-1s-1 @ 250K kOH + CH 2 ClCF3 = 6.9 × 10-15 cm3 molecule-1s-1 @ 250K

τ CHF Cl 2

τ CH ClCF 2

3

1 1.7 × 10−15 106

=

(

=

1 6.9 × 10−15 106

(

)( ) = 18.7 yr )( ) = 4.4 yr

(Note: 3.1536 × 107 s yr-1 ≅ π × 107 s yr-1)

3.2 OH + NO2 + M → HNO3 + M 2 ⎡ ⎛ ko (T ) [M ] ⎞⎤ ⎫⎪ ⎧ ⎡ k o (T ) [M ]⎤ ⎟⎟⎥ ⎬ 1 + ⎢log10 ⎜⎜ k (T, p) = ⎢ ⎥ 0.6 ⎨⎩ k T oo ⎠⎦ ⎪⎭ ⎝ ⎣ ⎢1 + k o (T ) [M] ⎥ ⎢ koo (T ) ⎥⎦ ⎣

From Table B.2:

ko (T) = 2.0 × 10-30 (T/300)-3.0 k ∞ (T) = 2.5 × 10-11

@ T = 300K,

ko

= 2.0 × 10-30 cm6 molecule-2 s-1

k∞

= 2.5 × 10-11 cm3 molecule-1 s-1

[M ] = pAν RT

=

(6.022 × 10 ) p (82.056 cm atm K ) (300 K )

=

2.45 × 1019 p

23

3

p, atm

-1

molecules cm-3

[M], 1019 molecules cm-3

0.05

0.123

0.1

0.245

0.2

0.49

0.5

1.22

0.75

1.84

1.0

2.45

(p in atm)

−1

⎧ 2⎫ −19 ⎡ 2.0 × 10−30 [M ] ⎤ ⎨ 1 + log10 0.8 × 10 [M ] ⎬ k(300, p) = ⎢ 0 .

6 −19 ⎩ ⎭ [M ]⎥⎦ ⎣1 + 0.8 × 10

[

p, atm

k(300),p) cm3 molecule-1 s-1

0.05

0.17 × 10-11

0.1

0.29 × 10-11

0.2

0.46 × 10-11

0.5

0.74 × 10-11

1.0

1.03 × 10-11

(

)]

−1

3.3 [M] = 2.5 × 1019 molecules cm-3 [O2] = 0.21 [M]

τ0 = =

1 k[O 2 ][M] 1

(4.8 × 10 ) (0.21) (2.5 × 10 ) − 33

= 1.6 × 10−6 s

19 2

3.4

Assume rCHF3 = 2.1 × 10-8 cm rOH

πd 2

= 2.0 × 10-8 cm

= π (1.4 × 10-10 m + 2.1 × 10-10 m)2 = 38.5 × 10-20 m2

mOH =

17 g/mol 6.022 × 1023 molecule/mol

= 2.82 × 10-26 kg molecule-1 70 g/mol 6.022 × 1023 molecule/mol

mCHF3 =

= 11.6 × 10-26 kg molecule-1 µ

=

mOH mCHF3 mOH + mCHF3

=

(2.82 × 10 )(11.6 × 10 ) −26

−26

(2.82 + 11.6) × 10− 26

= 2.27 × 10-26 kg molecule-1 Then 1

A = πd

2

⎛ 8kBT ⎞ 2 ⎜⎜ ⎟⎟ ⎝ πμ ⎠ -20

= (38.5 × 10

⎛ 8 × 1.381 × 10− 23 JK -1 × 298K ⎞ ⎟⎟ m ) ⎜⎜ 3.14 × 2.27 × 10− 26 ⎝ ⎠ 2

= (38.5 × 10-20 m2) (680 m s-1) = 2.62 × 10-16 m3 molecule-1 s-1 = 2.62 × 10-10 cm3 molecule-1 s-1 The fraction of collisions that lead to reaction =

2.8 × 10−16 2.62 × 10−10

= 1.07 × 10-6

1

2

3.5 Because the unpaired electrons are on the O atoms, the O atoms want to meet as HOOCl. This transition state falls apart relatively easily to HOO + Cl. In order for the transition state to fall apart as HCl + O2, the H and Cl atoms which are on opposite ends of the transition state must come together. This rearrangement is less favored, and this is reflected in a lower A factor.

3.6 v N 2O5 = 1.45 × 104 (T/108)½ cm s-1 Ap = π D p2 N a

γ = 0.1

3.10 The ratio τ 0 −15 / τ 0 for the species given in the problem are shown in the figure below. Although the absolute rate constants vary by more than four orders of magnitude, the predicted ratios of removal times vary by a factor of two.

Halocarbons (CH3Cl and CHF2Cl) show the largest ratios. Hence, while instantaneous rates of reaction may vary significantly with height in the atmosphere, the integrated rate constants are closer.

A doubling of a twenty minute lifetime is not very significant whereas doubling a 25 or 50 year lifetime is very significant in absolute terms, but in relative terms, the calculated changes are not very great. The variation in τ 0 −15 / τ 0 with absolute integrated rate constant is shown in the figure. The relationship is quite linear on a semilog plot, which suggests that this simple correlation could be used to estimate lifetimes for a given reaction (OH) and assumed scale height.

Chapter 4 4.1 λmax = 173 nm 305 nm 1120 nm

4.2

3.0 × 10-25

W m-2

3.0 × 10-21 2.1 × 10-44 6.0 × 10-55 1.4 × 10-19 0.28 1.2

4.3

Altitude where τ = 1 (@θ = 0°) Schumann-Runge Continuum

140 km

Schumann-Runge Bands

85 km

Herzberg Continuum

30 km

4.4

−τ

F (0, λ ) / FTOA (λ ) =e O3

(λ )

ln(F (0, λ ) / FTOA (λ )) = − τ O3 (λ )

= −mσ O 3 (λ ) Therefore

∫ n (z )dz 0

∫ n (z )dz = − mσ ln⎜⎜⎝ F 0

O3

⎛ F (O ) ⎞ ⎟⎟ TOA ⎠

1

∞

∞

O3

O3

(molecules cm-2)

m = (cos 45°)-1 = (0.707)-1 = 1.414 σ O 3 = 3.2 × 10-19 cm2 F(0)/FTOA = 0.026 ∞

18

∫ n (z )dz = 8.1 × 10 molecules cm 0

O3

= 300 DU

-2

F (z ) = FTOA exp(−τ ( z ))

4.5

∞

τ ( z ) = m ∫ σ n(z )dz z

m=1

m = sec θ0

∞ − z ʹ′ / H ⎛ ⎞ F ( z ) = FTOA exp⎜ − σ nO 2 (0) ∫ e dz ʹ′ ⎟ z ⎝ ⎠

(

)

= FTOA exp − σ nO 2 (0)H exp (− z / H )

σnO (0) H = (10 −24 cm 2 ) (0.21) (2.6 ×1019 cm −3 ) (7.4 ×105 cm) 2

= 4.0

F ( z ) = FTOA exp (− 4.0 exp (− z / H )) At the surface, F(0) = 0.018 FTOA so O2 absorbs nearly all the incoming radiation of λ = 240 nm by the time the light reaches the surface.

If the sun is not overhead,

⎛ 4.0 ⎞ F ( z ) = FTOA exp ⎜⎜ − exp (− z / H )⎟⎟ ⎝ cos θ 0 ⎠ The larger θ0, the more light is absorbed because the pathlength is longer.

4.6 Photolysis rate is of the form

[

J (z ) = α e− z / H exp − βe− z / H ln J (z ) = ln α −

]

z − βe − z / H H

ln x is monotonic, so f (x ) and ln f (x ) have maxima and minima at the same value of x.

Thus the extrema occur at

d ln J (z ) 1 β −z / H =− =0 + e dz H H or

β e− z / H = 1

4.7 Photolysis rate of O2 (z) ≅ [O2 ( z )]∫

λ max, λ min

σ O (z, T ) FTOA (λ ) exp (− τ (z, λ )) dλ 2

Assume: 1. O2 is the only absorber 2. σ O 2 independent of T 3.

[O2 ] decreases with height with scale factor H

4. Assume solar zenith angle = 0°.

τ (z, λ ) = σ O (λ ) [O2 ]surface He -z/H 2

[

Photolysis rate = [O2 ]surface e-z/H σ O2 FTOA exp − σ O2 [O2 ] surface He− z / H

]

Photolysis rate is a maximum at τ = 1

σ O [O2 ] surface He -z/H = 1

(4.57)

2

[

]

Altitude of maximum: z = H ln σ O2 (0.21 n air (0)) H H ≅ 7 km nair (0 ) = 2.6 × 1019 molecules cm-3

σ (cm 2 )

z (@τ = 1) (km)

Schumann-Runge continuum

10-17

125

Schumann-Runge bands

10-20

77

Herzberg continuum

10-23

29

4.8 Photodissociation rate of O2 = φ O2 × Photolysis rate

{

= 0.21nair (0) e− z / H σ O FTOA φ O2 exp − mσ O2 0.21nair (0)He− z / H 2

}

This plot will look like:

Run out of O2 molecles as z → ∞ ; e-z/H → 0

z

Maximum at τ = 1

Run out of photons; as z → 0, e- τ → 0

4.9 From Figure 4.15, φ = 0.5 and λ = 400 nm.

At 20 km, photolysis rate coefficients at the two solar zenith angles from Figure 4.16 are: θo = 0°

j

NO 2

= 1.5 × 10-2 s-1 = 0.8 × 10-2 s-1

= 85°

The lifetimes are just j −1 : NO 2

67 s @ 0° 125 s @ 85°

Chapter 5

5.1 At 230 K k O + O 3 = 1.03 × 10-15 cm3 molecule-1 s-1 Relative loss rates =

[

kO (1 D) + O O (1 D ) 3

kO + O 3 [O]

=

]

0.078

( )

Despite the fact that the rate of the O 1 D reaction is 5 orders of magnitude faster than the O reaction, the O3 loss rate with O is about 13 times faster, since the concentration of O is so much larger.

5.2

τO =

0.21 k2 [M ] k4 jO 3 [O3 ]

k2

= 6.0 × 10-34 (T/300)-2.4 cm6 molecule -2 s-1

k4

= 8.0 × 10-12 exp (-2060/T) cm3 molecule -1 s-1

x

2

z, km

T, K

jO 3 s -1

[M] molecules cm-3

20

217

7 × 10-4

1.4 × 1018

45

265

6 × 10-3

3.6 × 1016

a

Figure 2.6

[O3]a

k2

k4

τO

2 × 1012

13.08 × 10-34

6 × 10-16

6.4 × 108 s

0.4 × 1012

8.1 × 10-34

3.4 × 10-15

2.7 × 104 s

x

5.3 The lifetime of a species can be estimated from

τ=

Concentration Rate of removal

The problem asks to estimate the lifetime of N2O at 30 km using Figure 5.6. Panel (a) of Figure 5.6 gives N2O mixing ratio over the tropics. At 30 km, the concentration of N2O is [N2O] = ξ N 2 O [M ] = (200 × 10-9) (3.1 × 1017) = 6.2 × 1010 molecules cm-3 The loss rate of N2O is given in Panel (b) of Figure 5.6. Since the concentration profile in Panel (a) is that over the tropics, we select the loss rate from Panel (b) as close to 0° latitude as possible. At 30 km, that is 22 × 1012 molecules cm-3 s-1.

Thus,

τN O = 2

6.2 × 1010 = 2.8 × 10− 3 s 12 22 × 10

5.4 The relative rates of O3 destruction by the chlorine and Chapman cycles is given by the ratio, kClO + O [ClO] kO + O 3 [O3 ]

kClO + O = 3.0 × 10-11 exp (70/T) kO + O 3 = 8.0 × 10-12 exp (-2060/T) At 20 km (T = 220 K), kClO + O kO + O 3

=

4.12 × 10−11 = 0.59 × 105 7 × 10 −16

Thus, the ratio of [ClO] / [O3] at which the two rates are equal is 1.7 × 10-5.

If ξ O 3 = 4 ppm, then [O3] = (4 × 10-6) (1.4 × 1018) = 5.6 × 1012 molecules cm-3, and [ClO] = (1.7 × 10-5) (5.6 × 1012) = 9.5 × 107 molecules cm-3

5.5 ClO + ClO + M → ClOOCl + M

a.

ClOOCl + hν → ClOO + Cl ClOO → Cl + O2 2(Cl + O3 → ClO + O 2 ) Net : 2O3 → 3O 2 b.

The rate-limiting step in the above cycle is: ClO + ClO + M → ClOOCl + M For each cycle, two ozone molecules are lost, so

d [O3 ] 2 = − 2 kClO + ClO + M [ClO] dt Now, total ozone loss = 1.2 ppm × [M ] @ 220 K and 60 mbar. d [O3 ] = 2.61 × 1012 molecules cm-3 dt = 31 days × 8 hours × 3600 s hr −1 = 892,800 s Thus, d [O3 ] = − 2.92 × 106 molecules cm-3 s-1 dt Now need to calculate rate coefficient (see Table B.2) kClO + ClO + M = 1.74 × 10-13 cm-3 molecule-1 s-1 Solve for [ClO] :

[ClO] = 2.9 × 109 molecules cm-3 Convert to mixing ratio:

1.33 ppb

5.6

We seek to compute the ratio of the two rates at 20 km and 35 km: f =

kOH + ClO [OH] [ClO] kCl + CH 4 [Cl] [CH 4 ]

where the rate in the numerator is that of the HCl-forming channel. Concentrations of OH and CH4 are given, so we need to compute the ratio [Cl] / [ClO]. This ratio is given by (5.30) based on the steady state within the ClOx family, k [O] + k ClO + NO [ NO] [Cl] = ClO + O [ClO] kCl + O 3 [O3 ]

The steady state concentration of atomic oxygen at any altitute is given by (5.7), jO 3 [O] = [O3 ] 0.21 k2 [M]2 Using values of T and [M] at the two altitudes, we find

20 km

35 km

[O]

1.5 × 106

1.2 × 107 molecule cm-3

[Cl] / [ClO]

4.5 × 10-4

4.6 × 10-3

Now consider the desired ratio f at the two altitudes:

f

20 km

35 km

0.05

0.3

Thus, the minor channel is 30% of the major channel at 35 km.

5.7 ClO + NO2 + M → ClONO2 + M ClONO2 + hν → ClO + NO2 At steady state, j ClONO 2 [ClONO2] = k ClO + NO 2 + M [ClO] [NO2] Where k ClO + NO 2 + M is the rate coefficient given in Table B.2, for which ko (T) = 1.8 × 10-31 (T/300)-3.4 koo (T) = 1.5 × 10-11 (T/300)-1.9 To evaluate the steady state expression, we need estimates of [NO2] at 27 and 37 km. From Figure 5.9, the mixing ratios of NO2 can be estimated as 4 ppb at 27 km and 8 ppb at 37 km, respectively.

The following parameters are needed: z, km

T, K

[M] molecules cm-3

[NO2]

k ClO + NO 2 + M

j ClONO 2 s −1

cm3 molecule-1 s-1

27

224

4.7 × 1017

1.9 × 109

2.67 × 10-13

10-4

37

243

1.0 × 1017

8.0 × 108

4.11 × 10-14

5 × 10-4

We compare the ratio:

[ClONO2 ] = kClO + NO + M [ NO2 ] [ClO] jClONO 2

2

from steady state and Figure 5.14. z, km

[ClONO2]

[ClO]

(Fig. 5.14)

[ClONO2 ] [ClO]

[ClONO2 ] [ClO]

(Fig. 5.14)

(Steady state)

27

~ 1 ppb

~ 0.2 ppb

~5

5.1

37

~ 0.2 ppb

~ 0.6 ppb

~ 0.33

0.07

5.8(a) Steady-­‐state HOx partitioning is governed by

0 = − k1 [OH ] [O3 ] + k2 [ HO 2 ] [O3 ] + k3 [ HO 2 ] [ NO]

so that

[OH ] = [ HO2 ] (k2 [O3 ] + k3 [ NO]) / k1 [O3 ]

(1)

Ozone loss is given by

d [O3 ] = − k1 [OH ] [O3 ] − k2 [ HO 2 ] [O3 ]

(2) dt

Using (1) in (2),

d [O3 ] = − 2k2 [ HO 2 ] [O3 ] − k3 [ HO 2 ] [ NO]

(3) dt

(b) Equation (3) expresses the rate of O3 loss from the HOx cycle reactions 1-­‐3. Now we need to consider the effect of NO2 produced in reaction 3. If the result is no net change in odd oxygen, then equation (3) is correct as it stands. If subsequent NO2 loss leads to consumption of an odd oxygen, then every time reaction 3 happens, an additional O3 is removed. Remember that odd oxygen involves steady-­‐state partitioning between O and O3, and since that partitioning is > 99% O3, consuming an odd oxygen essentially means consuming an O3 molecule. In this case, the term

– k3 [HO2][NO]

should be added to equation (3). On the other hand, if each NO2 loss produces an odd oxygen then the term

+ k3 [HO2][NO] needs to be added to equation (3).

If NO2 loss produces O3, then the two terms cancel and

d[O3 ] = − 2k2 [HO 2 ][O3 ] dt

To determine the effect of NO2, we want to identify the dominant NO2 reaction in the NOx family. The applicable NO2 reactions are:

a)

loss of odd oxygen NO 2 + O → NO + O 2

(4)

b)

c) d)

NO 2 + hν → NO + O

creation of odd oxygen

NO 2 + O3 1 → NO3 + (O 2 ) 2 → NO + O 2

loss of odd oxygen

 → NO3 (O 2 ) → NO 2 + O

null cycle

Reaction (d) produces no effect on either odd oxygen or NO2, so we can ignore it.

Let us compare the fractional rates of NO2 loss, (1/[NO2]) d[NO2] / dt), among (a), (b), and (c):

a) ka [O]

jNO2 b)

c)

the smaller of kc1 [O3] or kc2 [NO3] / [NO2] since the slower step is rate-­‐ limiting

Let us estimate

[O] = 108 molecules cm-­‐3

[O3] = 5 ×1012 molecules cm-­‐3

@ T = 240 K ka = 1.1 ×10-­‐11 cm3 molecule-­‐1 s-­‐1

jNO2 = 10 -2 s-1 kc1 = 4.4 ×10-­‐18 cm3 molecule-­‐1 s-­‐1

Using these, the time constants of reactions (a) – (c) are:

a) ~ 1000 s

b) ~ 100 s

c) ~ 4.5 × 105 s

The shortest time constant is for reaction (b), so reaction (b) is dominant. Most of the NO2 produced in reaction 3 is photolyzed to NO, producing another odd oxygen along the way. Thus, the appropriate rate expression for O3 loss is equation (4).

5.9 The first-order rate constant for collisions is 1

1 ⎛⎜ 8kT ⎞⎟ 4 ⎜⎝ π mN 2 O 5 ⎟⎠

2

Ap

The number of collisions per 12 h is then Z collisions/night. If one is interested in the number in the number of collisions that lead to reaction, then this must be multiplied by the collision efficiency, γ.

5.10 a. At night, some of the NOx is converted to N2O5 via: NO2 + O3 → NO3 + O2 NO3 + NO2 + M → N2O5 + M N2O5 can be photolyzed back to NOx or it can hydrolyze on the surface of stratospheric aerosol: s N2O5 + H2O ⎯ 2 HNO3 ⎯→

Increased aerosol surface area leads to enhancement of the heterogeneous hydrolysis of N2O5 and therefore less back reaction of N2O5 to NOx. b. At 28 km, the loss of odd oxygen is determined almost entirely by the NOx cycle (e.g.

Figure 5.28): O + NO2 → NO + O2 NO + O3 → NO2 + O2 As a result of the decreased NOx, O3 loss slows at this altitude and O3 increases. c. In the lower stratosphere, catalytic O3 destruction results from the HOx cycle, O3 + HO2 → OH + 2O2 OH + O3 → HO2 + O2 Because HO2 increases with decreasing NOx, the strength of the HOx cycle increases.

Also, with decreased NOx, the ClOx cycle is strengthened because formation of ClONO2 decreases.

5.11

Interconversion between HO2 and OH is rapid. Assume HO2 is in steady state, thus

−R2 − R3 − R6 + R1 = 0 This yields

[OH ] = k2 [O3 ] + k3a [ClO] + k3b [ BrO] + k6 [ NO] k1 [O3 ] [ HO2 ] k2 = 1.024 ×10 −15 cm 3 molec-1 s-1 k1 = 2.15 ×10

−14

k6 = 1.12 ×10 −12

[ NO] = 7.5 ×108 molec cm -3 [O3 ] = 1 to 5 ×1012 molec cm -3

Neglecting reactions 3a and 3b, we can compute the ratio {OH]/[HO2] as a function of [O3]: [O3] (1012 molec cm–3) 1 2 3 4 5

[OH]/[HO2] 0.439 0.243 0.178 0.146 0.126

5.13 Column integrated SO2 = 3 × 1016 molec cm–2. Thus column integrated S is also 3 × 1016 S atoms cm–2. If all this S is in the form of H2SO4, column integrated H2SO4 = 3 × 1016 molec cm–2 = 4.88 × 1016 g(H2SO4) cm–2.

For 75% H2SO4/25% H2O aerosol, density = 1.252 g cm–3. Volume of one 0.1µm diameter particle is π V = Dp3 6 = 0.52 ×10 −21 m 3 Mass of one particle = 0.66 ×10 −15 g Number of particles in the entire atmospheric column is ( 4.88 ×10−6 ) / 0.75

(5.246 ×10 ) / 8 −15

= 9.92 ×10 9 particles cm −2

Column-integrated aerosol surface area 2 = ( 9.92 ×10 9 ) ( 4π ) ( 0.1 / 2 ) = 31.2 ×10 7 µm 2 cm −2

Surface area concentration =

31.2 ×10 7 µm 2 cm −2 5 ×10 5 cm

= 624 µm 2 cm −3

Chapter 6

6.1 a. Compute the number of OH produced per O(1D) generated.

This is given by (6.4),

ε OH =

2k4 [H 2O] k3 [M ]

@ 240 K

ε OH =

4.2 × 10−5

@ 298 K

ε OH =

0.13

The difference in H2O content is the reason for the large difference. b. [O(1D)] is given by (6.2).

[O(1D)], molecules cm-3

= 0.69 lower stratosphere = 0.01 boundary layer

POH =

d [OH] is given by (6.3). dt

POH, molecules cm-3 s-1

= 2.5 × 103 lower stratosphere = 2.2 × 106 boundary layer

6.2

Ozone Production Efficiency (OPE) given by (6.22).

OPE =

kHO 2 + NO

[HO2 ] [NO] [OH] [NO2 ]

kOH + NO 2

[HO2 ] ≅ kOH + CO [CO] [OH] kHO + NO [NO] 2

jNO [NO] = [NO2 ] k NO + O3 [O3 ] 2

Thus, OPE =

kOH + CO jNO 2 [CO] kOH + NO 2 k NO + O 3 [O3 ][NO]

[CO] = (200 × 10-9) (2.5 × 1019) = 500 × 1010 molecules cm-3 [O3] = ( 50 × 10-9) (2.5 × 1019) = 125 × 1010 [NO] = ( 40 × 10-9) (2.5 × 1019) = 100 × 1010

jNO 2

= 0.015 s-1

kOH + CO = 2.4 × 10-13 cm3 molecule-1 s-1 kOH + NO 2 = 1.1 × 10-11 k NO + O 3 = 1.9 × 10-14 [OPE] = 0.069

6.3

[NO] / [NO2] is given by (6.6), jNO [NO] = [NO2 ] k NO + O [O3 ] 2

3

Assume [O3] = (100 × 10-9) (2.5 × 1019) = 250 × 1010 molecules cm-3.

Peak noontime jNO 2 = 0.015 s-1. jNO 2 varies over 12 hours as a parabolic function of time with a maximum at 6 hours (noon) of 0.015 s-1, jNO 2 =

4ht (b − t ) b2

where h = 0.015 s-1, b = 12 hours, and t is time in hours. k NO + O 3 = 1.9 × 10-14 cm3 molecule-1 s-1.

6.4 a. The dynamic behavior of DMS due to reaction with OH is given by d[DMS] =k DMS OH dt

(1)

Assuming constant OH radical concentration, Equation (1) becomes d[DMS] =k' DMS dt

(2)

where k’ = k[OH]. In this way, the time-dependent concentration of DMS follows the first order decay [DMS] = [DMS]0 exp -k't

(3)

The fraction of DMS left after 12-hour OH oxidation is f=

[DMS] = exp -k't = exp -5.0 × 10-12 × 2.0 × 106 × 12 × 3600 =0.65 [DMS]0

The fraction of DMS oxidized over a 12-h day is 35%. b. The dynamic behavior of DMS due to reaction with NO3 is given by d[DMS] =k DMS NO3 dt

(1)

Assuming constant NO3 radical concentration, Equation (1) becomes d[DMS] =k' DMS dt

(2)

where k’ = k[NO3 ]. In this way, the time-dependent concentration of DMS follows the first order decay [DMS] = [DMS]0 exp -k't The fraction of DMS left after 12-hour NO3 oxidation is f=

[DMS] = exp -k't = exp -1.0 × 10-12 × 5.0 × 108 × 12 × 3600 = ~ 0 [DMS]0

The fraction of DMS oxidized during a 12-hour nighttime is ~ 100%.

(3)

6.5 NO2 + O3 → NO3 + O2

τ NO = 2

1 k [O3 ]

k = 1.2 × 10-13 exp (-2450/T) cm3 molecule-1 s-1 [O3]

τ NO , s

2.4 × 10-17

125 × 1010

0.33 × 105

1.3 × 1019

0.83 × 10-17

65 × 1010

1.85 × 105

0.65 × 1019

0.2 × 10-17

33 × 1010

15 × 105

z, km

T, K

[M]

0

288

2.5 × 1019

5

256

10

223

k

2

6.6 The theoretical maximum yield of O3 from CH4 + OH reaction would occur if NO concentration was sufficiently high that peroxy radicals CH3O2 and HO2 react exclusively with NO and all the formaldehyde formed photolyzes by the HO2 radical path. Reactions involved are: CH4 + OH

O2

CH3O2 + NO

O2

2O2

HCHO + hv CO + OH

CH3O2 + H2O

O2

HCHO + HO2 + NO2 CO + 2HO2

CO2 + HO2

4(HO2 + NO → OH + NO2) 5(NO2 + hv Net:

O2

NO + O3)

CH4 + 10O2 → CO2 + H2O + 2OH + 5O3

At the surface, p = 1 atm and T = 298 K. Thus the maximum O3 production rate from CH4 photochemistry would be PO3 = 5kCH4 +OH OH [CH4 ] = 5 × 6.3 × 10-15 * 1 × 106 * 2.46 × 1019 *1.8 × 10-6 = 1.40 × 106 molecules cm-3 s-1 = 3.5 × 10-3 g m-3 yr-1 The average height of the troposphere is ~ 12 km. Let us first assume that the air density and temperature do not change with altitude. The total volume of the troposphere would be 6.13 × 1018 m3. Thus, the overall production rate of O3 would be 21455 Tg/year.

The above estimation obviously overestimated P(O3) since both air density and temperature are decreasing along with increasing altitude:

M =

PT0 z T0 M 0 = ex p M0 P0 T H T0 - Γ*z

where P is the pressure at altitude z, P0 is the surface pressure, T is the temperature at altitude z, T0 is the surface temperature, Γ is the lapse rate, and [M]0 = 2.46 × 1019 molecules cm-3 s-1 .

As we know, the volume element in spherical coordinates is given by dV= r2 sin∅ dr d∅ dθ Thus the overall production rate can be expressed as 2π

P O3 =

π

dθ 0

sin∅d∅ 0

R2 R1

5kCH4 +OH M 0 [OH] M 0 [CH4 ] r2 exp(-

≈ π kCH4 +OH M 0 [OH] M 0 [CH4 ]{-H(H2 + 2Hr + r2 ) exp(-

2(r-R1 ) T0 )( ) dr H T0 - Γ(r-R1 )

2(r-R1 ) !! )} |! !

H

where [OH] is the mixing ratio of OH radical (4.06 × 10-­‐14 ) calculated based on its surface concentration, i.e., 1 × 106 molecules cm-3 s-1 , [CH4] is the mixing ratio of methane (1.8 ppm), H is the scale height (8 km), r is the distance from the center of the earth, R1 is the radius of the earth (6371 km), and R2 is the overall height of the troposphere (6383 km). Plugging these numbers into the above equation, we get the ozone production rate of 3390 Tg/year.

6.9 a. Reactions involved in the production of HOx from acetone photolysis CH3C(O)CH3 + hv ! ! ! ! ! ! ! ! ! ! ! !

(CH3C(O)O2 + NO

CH3O2 + CH3C(O)O2

O2

CH3O2 + CO2 + NO2)

(CH3C(O)O2 + NO2 ⇔ CH3C(O)O2NO2) (CH3O2 + NO (HCHO + hv (CO + OH

O2

O2

O2

HCHO + HO2 + NO2) !

HO2 + CO + H2) !

CO2 + HO2)

(HO2 + NO → OH + NO2)

!" !

2O2

(NO2 + hv

O2

NO + O3)

Net: CH3C(O)CH3 + hv + ! NO !

13O2 ! !

!

CH3C(O)O2NO2 + 2CO2 + H2 + 6O3 + 3OH

b.

Reactions involved in the production of HOx from O(1D) + H2O (1) O3 + hv = O(1D) + O2 (2) O(1D) + M = O + M (3) O(1D) + H2O = 2OH Using pseudo steady state for O(1D), we get

!

O(1 D ] =

jO3 →O1 D [O3 ] kO1 D+M M + kO1 D+H2 O H2 O

=

3 × 10-5 *1.1 × 1019 * 50 × 10-9 3 × 10-11 * 1.1 × 1019 + 2 × 10-10 * 1.1 × 1019 * 50 × 10-6

= 0.05 molecule cm-3 The production rate of HOx from O(1D) + H2O: P(HOx)O1 D+H O = 2 kO1 D+H2 O H2 O O(1 D ] 2

= 2 * 2 × 10-10 * 1.1 × 1019 * 50 × 10-6 * 0.05 = 1.1 × 104 molecules cm-3 s-1 The production rate of HOx from acetone photolysis: P(HOx)CH C(O)CH +hv = 3 jCH C(O)CH CH3 C(O)CH3 3

3

3

3

The required mixing ratio of acetone: CH3 C(O)CH3 =

P(HOx)O1 D+H O 2

3 jCH C(O)CH 3

=

3

1.1 × 104 3 * 5 × 10-7 * 6.8 × 1018

= 0.66 ppb

CH3OOH + OH

CH3O2 + H2O CH2OOH + H2O HCHO + OH

CH2OOH + OH

O2 HCHO + HO2 CH3O ⎯⎯→

CH3C(O) chemistry (PAN formation)-O2 CH3C(O) ⎯⎯→ CH3C(O)OO CH3C(O)O + NO2 CH3C(O)OO + NO O2 CH3C(O)O ⎯⎯→ CH3O2 + CO2 CH3C(O)O2 + NO2 ←⎯→ CH3C(O)O2NO2 (PAN) M

b) Assuming that the NOx concentration is sufficiently high that HOx-HOx termination reactions can be neglected, reduce the mechanism in (a). (On the time scale of the laboratory reactor, oxidation of CO can be neglected.)

Neglect HOx-HOx termination reactions (HO2 + HO2 and CH3O2 + HO2) and CO oxidation, the mechanism in 3(a) can be reduced to: HCHO and CH3CHO photolysis and OH reactions--

HCHO + hν ⎯2O ⎯→ ⎯2 2 HO2 + CO H2 + CO O2 HO2 + CO + H2O HCHO + OH ⎯⎯→

CH3CHO + hν ⎯2O ⎯→ ⎯2 CH3O2 + HO2 + CO O2 CH3CHO + OH ⎯⎯→ CH3C(O) + H2O

NOx photochemistry-O2 NO2 + hν ←⎯→ NO + O3

HOx chemistry--

HO2 + NO

NO2 + OH

OH + NO2 ⎯⎯→ HNO3 M

CH3O2 chemistry--

CH3O2 + NO

NO2 + CH3O

CH3O ⎯⎯→ HCHO + HO2 O2

CH3C(O) chemistry (PAN formation)-O2 CH3C(O) ⎯⎯→ CH3C(O)OO CH3C(O)O + NO2 CH3C(O)OO + NO O2 CH3C(O)O ⎯⎯→ CH3O2 + CO2

CH3C(O)O2 + NO2 ←⎯→ CH3C(O)O2NO2 (PAN) M Assume that the reactions of PAN formation and PAN thermal decomposition, i.e., c) Write an expression for the rate of O3 formation in this system, PO3, for the mechanism of (b). Making any reasonable assumptions, express PO3 in terms of the concentrations of stable compounds in the system.

CH3C(O)O2NO2 CH3C(O)OO + NO2 ←⎯→ M are at equilibrium, and therefore the concentration of CH3C(O)OO and PAN are in steady-state.

Also assume steady-state for NOx (NO, and NO2) and HOx (OH, HO2 and CH3O2) species.

The rate of O3 formation, PO3, can be expressed as the sum of the rates of the reactions in the 3(b) that cycle NO back to NO2 without consuming any O3 molecule, which include NO2 + OH HO2 + NO CH3O2 + NO NO2 + CH3O CH3C(O)O + NO2 CH3C(O)O2 + NO and thus the expression for PO3 is PO3 = kHO2+NO[HO2][NO] + kCH3O2+NO [CH3O2][NO] + kCHC(O)O2+NO [CH3C(O)O2][NO]

d) Give an expression for the ozone production efficiency in this system.

The ozone production efficiency (OPE) is defined as the ratio of the rate of ozone formation over the rate of NOx consumption. Because the reaction of PAN formation is assumed at equilibrium with PAN decomposition in 3(c), the only reaction that consumes NOx in the mechanism in 3(b) is OH + NO2. Therefore the OPE can be written as OPE =

(k

HO 2 + NO

)

[HO 2 ][NO] + k CH 3O 2 + NO [CH 3 O 2 ][NO] + k CHC(O)O2 + NO [CH 3 C(O)O 2 ][NO] k OH + NO2 [OH][NO 2 ]

e) If this experiment were repeated at a temperature of, say, 20 oC higher, how do you expect PO3 to change? Why? At higher temperatures, the rate PAN decomposition increases due to a higher rate coefficient, so less CH3C(O)O2 is tied up in the reservoir of PAN. This is the largest effect of temperature on ozone production, since the rate coefficients of the other reactions are less sensitive to temperature variation. Therefore, from the expression of PO3 derived in 3(c), raising the temperature by 20 oC is expected to increase PO3 because of the higher CH3C(O)O2 concentration in the system.

More generally, if the concentrations of NO and NO2 are sufficiently high, then their concentrations can be assumed to be constant at [NO]o and [NO2]o.

In the second part of the problem we consider the experiment conducted also with [NO]o = 0. With no NO in the system, reactions 3 and 4 do not occur. In both cases, with and without initial NO, PAN formation occurs by reaction 1; what is different are the concentrations of the CH3C(O)O2 ⋅ radicals in the two cases. When NO = 0, the steady state concentration of this radical is given by [CH3C(O)O2 ⋅ ]ss =

2 j BA [CH 3 C(O)C(O)CH 3 ] k1 [ NO 2 ]

[NO] o = 0

In this case the rate of PAN formation is then just equal to twice the rate of biacetyl photoloysis, d [PAN] = 2 j BA [CH 3 C(O)C(O)CH 3 ] dt

[NO] o = 0

When NO is initially present, the steady-state concentration of CH3C(O)O2 ⋅ radicals is given by [CH3C(O)O2 ⋅ ]ss =

2 j BA [CH 3 C(O)C(O)CH 3 ] k1 [ NO 2 ] + k 3 [ NO]

[NO] o ≠ 0

and the rate of PAN formation is ⎧ 2 j [CH 3 C(O)C(O)CH 3 ] ⎫ d [PAN] = k1 [NO 2 ]⎨ BA ⎬ dt k1 [ NO 2 ] + k 3 [NO] ⎭ ⎩

=

=

2 j BA [CH 3 C(O)C(O)CH 3 ] k [ NO] 1+ 3 k1 [ NO 2 ] 2 j BA [CH 3 C(O)C(O)CH 3 ] 1 [ NO] 1+ α 1 [ NO 2 ]

[NO] o ≠ 0

The rate of biacetyl photolysis is the same regardless of whether NO is present or not.

Thus, we can divide the rate of PAN formation with [NO] o = 0 by that when [NO] o ≠ 0, d [PAN]NO = 0 d [PAN]

=1+

1 [ NO] α [ NO 2 ]

Evaluating the initial rate of formation in the two cases we get ⎞ [ NO]o ⎛ [PAN(t )]NO = 0 α = ⎜⎜ − 1⎟ ⎟ [ NO ] 2 o ⎝ [PAN(t )] ⎠ −1

k HO2 +O3 [ HO2 ][O3 ]

→ [NO] = fNO [M] = → fNO =

k HO2 +O3 f O3 k HO2 + NO

k HO2 + NO [ HO2 ] =

=

k HO2 +O3 f O3 [ M ] k HO2 + NO

1.9 × 10 −15 × 20 × 10 −9 = 4.69×10-12 = 4.69 ppt −12 8.1 × 10

Assume O3 is at photostationary state, so we have that b) To what level of NOx does the NO level determined in (a) correspond under noontime conditions? Assume jNO2 = 0.015 s-1.

[O3] =

j NO2 [ NO2 ] k NO + O3 [ NO ]

→ [NO2] = fNO2 [M] = → f NO2 =

k NO + O3 [ NO ][O3 ]

k NO +O3 f NO f O3 [ M ] j NO2

j NO2 =

1.9 × 10

−14

=

k NO + O3 f NO f O3 [ M ] 2 j NO2

× 4.69 × 10 −12 × 20 × 10 −9 × 2.46 × 1019 = 2.92×10-12 0.015

= 2.92 ppt → f NOx = f NO + f NO2 = 4.69 + 2.92 = 7.61 ppt c) A competition also exists between NO and HO2 for the CH3O2 radical, and the path depends on the concentrations of HO2 and NO. Assume that the local OH concentration is 106 molecules cm-3. Compute the local HO2 concentration assuming that reaction with CO is the main sink of OH and HO2-HO2 self-reaction is the main sink of HO2.

Assume a CO mixing ratio of 100 ppb.

Assume steady-state of HO2, so the source of HO2, the reaction of OH + CO, is balanced by the sink of HO2, here only the HO2-HO2 slef-reaction: k OH +CO [OH ][CO ] = 2k HO2 + HO2 [ HO2 ] 2

→ [HO2] = f HO2 [ M ] =

[OH ] f CO [ M ] k OH +CO [OH ][CO ] k = OH +CO 2k HO2 + HO2 2k HO2 + HO2 2.4 × 10 −13 × 10 6 × 100 × 10 −9 = 1.30×10-11 = 13.0 ppt −12 19 2 × 2.9 × 10 × 2.46 × 10

→ f HO2 =

k OH +CO [OH ] f CO = 2k HO2 + HO2 [ M ]

Note:

The HO2+HO2 reaction has two channels: H2O2 + O2

(a)

ka = 1.7×10-12

(cm3/molecule s)

M ⎯⎯→ H2O2 + O2

(b)

kb = 4.9×10-32×[M]

(cm3/molecule s)

HO2 + HO2

So the overall rate constant for the HO2+HO2 reaction is the sum of the rate constants of the two channels,

k HO2 + HO2 = ka+kb = 1.7×10-12 + 4.9×10-32×2.46×1019 = 2.9×10-12 (cm3/molecule s)

d) Determine the NO mixing ratio for the conditions of 2(c) at which the reaction of CH3O2 with NO just equals that with HO2. To what NOx level does this NO value correspond? k CH 3O2 + NO [CH 3O2 ][ NO] = k CH 3O2 + HO2 [CH 3O2 ][ HO2 ]

→ [NO] = fNO [M] =

→ fNO =

k CH 3O2 + HO2 [ HO2 ] k CH 3O2 + NO

k CH 3O2 + HO2 f HO2

→ f NO2 =

k CH 3O2 + NO

=

k CH 3O2 + HO2 f HO2 [ M ] k CH 3O2 + NO

5.2 × 10 −12 × 1.30 × 10 −11 = 8.75×10-12 = 8.75 ppt −12 7.7 × 10

k NO +O3 f NO f O3 [ M ] j NO2

=

1.9 × 10 −14 × 8.75 × 10 −12 × 20 × 10 −9 × 2.46 × 1019 = 0.015

= 5.43×10-12 = 5.45 ppt → f NOx = f NO + f NO2 = 8.75 + 5.45 = 14.2 ppt

e) It is desired to prepare a plot of the rates of both O3 production, PO3, and loss, LO3, (molecules cm-3s-1) as a function of NO. Consider NO mixing ratios from 1 to 100 ppt. Ozone production occurs as a result of HO2 and CH3O2 reacting with NO, whereas O3 loss occurs by photolysis and reaction with HO2. Plot PO3 and LO3 as a function of NO. Recall that the results of Admission Ticket #15 can be used to estimate the photolysis term in LO3. For this, assume 50% RH. Assume a local O3 mixing ratio of 20 ppb. Note any assumptions you make in estimating [CH3O2].

PO3 = k HO2 + NO [ HO2 ][ NO ] + k CH 3O2 + NO [CH 3O2 ][ NO ]

(

)

= k HO2 + NO f HO2 + k CH 3O2 + NO f CH 3O2 f NO [ M ] 2

(

)

(

= 8.1 × 10 −12 × 1.3 × 10 −11 + 7.7 × 10 −12 f CH 3O2 × f NO × 2.46 × 1019

)

2

(1)

LO3 = k O (1 D ) + H O [O(1 D)][ H 2 O] + k HO2 +O3 [ HO2 ][O3 ] 2

= k O (1 D )+ H O [O(1 D)] f H 2O [ M ] + k HO2 +O3 f HO2 f O3 [ M ] 2 2

(

)

2

= 2.2 × 10 −10 [O(1 D )] f H 2O × 2.46 × 1019 + 1.9 × 10 −15 × 1.3 × 10 −11 × 20 × 10 −9 × 2.46 × 1019 (2)

Assume fHO2 the same as in 2(c): f HO2 =

k OH +CO [OH ] f CO = 2k HO2 + HO2 [ M ]

2.4 × 10 −13 × 10 6 × 100 × 10 −9 = 1.30×10-11 −12 19 2 × 2.9 × 10 × 2.46 × 10

Assume steady-state of CH3O2: k CH 4 +OH [CH 4 ][OH ] = k CH 3O2 + NO [CH 3 O2 ] + [ NO] + k CH 3O2 + HO2 [CH 3O2 ][ HO2 ] → [CH3O2] = f CH 3O2 [ M ] =

(k

k CH 4 +OH f CH 4 [ M ][OH ] CH 3O2

)

f NO + k CH 3O2 + HO2 f HO2 [ M ]

→ f CH 3O2 =

(k

k CH 4 +OH f CH 4 [OH ] CH 3O2

(assume fCH4 = 1700ppb)

)

f NO + k CH 3O2 + HO2 f HO2 [ M ]

6.3 × 10 −15 × 1700 × 10 −9 × 10 6 = 7.7 × 10 −12 × f NO + 5.2 × 10 −12 × 1.30 × 10 −11 × 2.46 × 1019

(

)

(3)

Derive fH2O and [O(1D)] as in Ticket # 15 at T = 298 K and RH = 50 % (use jO3 = 7×10-6s-1): a = (1-373.15/T) = (1-373.15/298), fH2O = RH × PH2O,S / Ps = RH × exp(13.3185a-1.976a2-0.6445a3-0.1299a4) = 0.015 [O(1 D)] =

jO3 [O3 ] k O (1 D ) + M [ M ] + k O (1 D ) + H O [ H 2 O] 2

=

=

jO3 f O3 k O (1 D ) + M + k O (1 D ) + H O f H 2O 2

7 × 10 −6 × 20 × 10 −9 = 4.32×10-3 (molecule/cm3). −11 −10 2.9 × 10 + ×2.2 × 10 × 0.015

Substitute Eq.(3) into Eq.(1) and derive PO3 as a function of fNO. Use the above fH2O and [O(1D)] values to calculate LO3. The figure below is the plot of PO3 and LO3 at fNO from 1 to 100 ppt. PO3 increases with increasing NO level while LO3 is at constant of 6.6×105 molecule/cm3s.

The two curves cross each other at fNO = 8.4 ppt.

We can judge whether a region is O3-producing or O3-destroying by comparing the local f) Based on the NO mixing ratio at which PO3 = LO3, would you judge the following regions of the troposphere to be O3-producing or O3-destroying?

Remote Pacific Ocean marine boundary layer Downtown Los Angeles Rural Southeastern U.S. [NO] with the value of [NO] at the cross point of PO3 and LO3, which is around fNO = 8.4 ppt as derived in 2 (d). (Assume the regions have the same local [O3], [OH], [CO] and [CH4] levels and the same jO3 as in our calculations). If the NO mixing ratio is higher than 8.4 ppt, the region is O3-producing since PO3 is larger than LO3 at the right hand side of the cross point, and contrarily if fNO is lower than 8.4 ppt, the region is O3-destroying.

Table 2.7 in Seinfeld and Pandis provides typical NOx mixing ratios at different regions, which can be converted to NO mixing ratio by assuming photostationary state: −14 −9 19 ⎛ k NO +O3 f O3 [ M ] ⎞ ⎟ f NO = ⎛⎜1 + 1.9 × 10 × 20 × 10 × 2.46 × 10 ⎞⎟ f NO f NOx = f NO + f NO2 = ⎜1 + ⎜ ⎟ ⎜ ⎟ j NO2 0.015 ⎝ ⎠ ⎝ ⎠

→ fNO = 0.6 fNOx Regions

fNOx (ppt)

fNO (ppt)

O3-producing/destroying

Remote Pacific Ocean MBL

20-40

12-24

slightly O3-producing

4

4

Downtown Los Angeles

10 -10

6

6

10 -10

Rural Southeastern U.S.

200-104

100-104

highly O3-producing moderately O3-producing

Note: 1. The NO level at the cross point varies with local conditions. For example, a higher jO3 value can shift the cross point to higher fNO to around 50 ppt. In this case the region of remote Pacific Ocean can become ozone-destroying.

So the remote regions is actually at a place where the ozone production and loss are about at balance.

2. In 2 (c) we assumed that the HO2+HO2 reaction is the main sink for HO2 and ignored the other HO2 loss reaction to derive the HO2 level. This is only true under low NOx condition. In regions with high NOx level, such as LA downtown, the HO2+NO can become the dominant HO2 sink, and the way to evaluate local HO2 level has to be modified. The rigorous solution is to consider the full balance of HO2, i.e.

k OH +CO [OH ][CO ] = 2k HO2 + HO2 [ HO2 ] 2 + k HO2 + NO [ HO2 ][ NO ] + k CH 3O2 + HO2 [CH 3 O2 ][ HO2 ] + k HO2 +O3 [ HO2 ][O3 ] Substitute the steady-state approximation of CH3O2 from 2 (d) into the above equation and solve for fHO2 as a function of fNO. Use this solution of fHO2 in calculating PO3 and LO3, the plot shows that the two curve cross at a slightly higher NO level (around 12 ppt), and the magnitudes of both PO3 and LO3 are smaller than the results from using just low-NOx limit for HO2.

6.13 The system is

NO 2 + hν 1 → NO + O

O + O 2 + M 2 → O3 + M

k2 = 6.09 × 10 −34 cm 6 molecule-2 s−1

NO + O3 + M 3 → NO 2 + O 2

k3 = 1.9 × 10 −14 cm 3 molecule−1 s−1

Given:

[NO2] = 10 ppb, [NO] = O

Solve

jNO2 = 1.5 × 10 −2 s−1

d [ NO 2 ] = − jNO2 [ NO 2 ] + k3 [ NO] [O3 ]

dt

d [O3 ] = k2 [O] [O 2 ] [ M ] − k3 [ NO] [O3 ]

dt

Assume pseudo-­‐steady state for [O]:

[O] = jNO2 [ NO2 ] / k2 [O2 ] [ M ]

[M] = 2.5 × 1019 molecules cm-­‐3

[NO2]O = 2.5 × 1011 molecules cm-­‐3

Solve the two differential equations using MATLAB. The steady-­‐state concentrations achieved are

[NO2]ss = 4.91 × 1010 molecules cm-­‐3

[NO]ss = 1.97 × 1011

[O3]ss = 1.97 × 1011

The time to achieve the photostationary state is ~ 5 min.

6.14B

VOC

A

OH, O2 HO OO

R1 O2

B

NO +

HO

NO 2

O frag.

O

CARB

HCHO +

HO +

HO 2

OO

R2O2

OH

NO, O2

D

OH, O2

HO

OO

O

C

isom., + O2

R3O

O

+

HO 2

+

OH

2

DIHOCARB

NO O

O +

NO 2

isom., + O2

E

OH

O

R4O2

OO NO OH

O

+

O

NO 2

frag., + O2 O OO

isom.

OH

O

R5O2

HODICARB OH

NO O

O O2NOO

OH

O

OH

O +

NO 2

frag., + O2

H

O

PAN

O

HO

G

NO 2

F

O

NO 2

A: OH can abstract an H atom or add to the double bond. Addition is much faster than abstraction in general. OH will preferentially add to the ionically favored location.

B: Peroxy radical can react with NO, HO2, or RO2 (often peroxy radical). In a polluted environment, NO is sufficiently high so that the peroxy radical mostly reacts with NO. C: Alkoxy radicals can react with O2, decompose, or isomerize. In the absence of H atom on the alpha position of an alkoxy radical, reaction with O2 cannot occur. Decomposition is likely to be favored as HCHO is a stable leaving group. Isomerization would result in a compound with two alcohol groups and one carbonyl group. Its decomposition is unlikely as there is no good leaving group.

D: According to Structure-Reactivity Relationships, a carbonyl group facilities the OH abstraction of a H-atom located two carbons away from the carbonyl group. We also know that abstraction from secondary carbon is much faster than abstraction from primary carbon.

E: This alkoxy radical is likely to undergo isomerization, as isomerization tends to be faster than decomposition (when there is no good leaving group) and reaction with O2.

F: This case is different since both decomposition and isomerization are favorable.

Decomposition: because it leads to an acylperoxy radical which is a good leaving group.

Isomerization: because the H-atom on the alpha position of the alcohol is very reactive and leads to a very stable alkyl radical.

G: Acyl peroxy radical can react with NO2 to form peroxyacyl nitrate.

H: CO2 is a very good leaving group so decomposition is likely to be favored.

b.

The fraction of 2-methyl 1-pentene that reacts with O3: fO = 3

=

kO3 [O3 ] kO3 O3 + kOH [OH] 1.60 × 10-17 * 3.0 × 1012

1.60 × 10-17 * 3.0 × 1012 + 6.07 × 10-11 * 5.0 × 106

= 0.14

c.

At night, NO3 becomes the primary oxidant for 2-methyl 1-pentene. O3 can oxidize the double bond. Ozonolysis of alkene can lead to the formation of OH. Therefore it is possible that OH plays a role in the oxidation of 2-methyl 1-pentene even at night.

d.

The key process that results in the formation of ozone is the reaction of HO2 or RO2 with NO. Each time HO2 + NO reaction occurs, an additional O3 molecule is produced as the resulting NO2 molecule photolyzes. In other words, the yield of O3 from an HO2 radical is 1. Each time RO2 + NO reaction occurs, an additional O3 molecule, as well as an HO2 radical (via the fragmentation of RO radical), is produced. So the yield of O3 from a RO2 radical is 2. Similarly, the yield of O3 from a peroxy acetyl radical (R(O)O2) is 3, since one O3 molecule and one RO2 radical are produced from each R(O)O2 + NO reaction. Thus the overall production rate of O3 is given by: PO3 = kHO2+NO HO2 NO + 2kRO2+NO RO2 NO + 3kR(O)O2+NO R(O)O2 NO Under high-NOx conditions, the removal rate of NOx is given by: LNOx = kOH+NO2 OH NO2 Thus, the ozone production efficiency (OPE) can be expressed as : OPE =

kHO2+NO HO2 NO + 2kRO2+NO RO2 NO + 3kR(O)O2+NO R(O)O2 NO PO 3 = LNOx kOH+NO2 OH NO2

OPE depends on four ratios: [HO2]/[OH], [RO2]/[OH], [R(O)O2]/[OH], and [NO]/[ NO2].

We can obtain expressions for these ratios using steady state assumptions for the HOx balance. But complicated calculations are needed for the 2-methyl 1-pentene photooxidation system. Here we will express OPE in terms of the OH reactivity ratio of VOCs to NO2.

Let us derive a simplified scheme, focusing on the yield of HO2 or RO2, from the detailed 2-methyl 1-pentene photooxidation mechanism under high-NOx conditions (see the Table below). Let us take the first three reactions as an example. Assuming that reaction (1) is the rate-limiting step. Reactions (1-3) provide a complete description of an O3 formation cycle, see the net reaction (1.1). We obtain PO3 = 5 2 kOH+VOC VOC OH Note that VOC here only denotes 2-methyl 1-pentene. Further photooxidation of the 2methyl 1-pentene products, including pentanone (CARB), acetaldehyde (CH3CHO), and formaldehyde (HCHO), also generate O3 (see net reaction 1.2, 1.3, and 1.4). Thus, the overall O3 production rate is give by PO3 = 5 2 kOH+VOC VOC OH + 5 2 kOH+CARB CARB OH + 3 kOH+CH3CHO CH3 CHO OH + kOH+HCHO HCHO OH Note that the photolysis and OH oxidation of HCHO both give one HO2 radical. In addition, HCHO photolysis rate and HCHO + OH reaction rate are in the same order of magnitude (~ 105 s-1). So we will express the contribution of HCHO to O3 production by its reaction with OH. The yields of pentanone (CARB), acetaldehyde (CH3CHO), and formaldehyde (HCHO) from the photooxidation of 2-methylpentene are 1/2, 1/8, and 3/4, respectively. Thus, The ozone production rate can be simplified as PO3 = 39 8 VOC OH ( 5 2 k

OH+VOC

+ 5 4 kOH+CARB + 3 8 k

OH+CH3CHO

+ 3 4 k

OH+HCHO

OPE, which depends on the OH reactivity ratio of VOCs and NO2, is given by OPE =

PO 3 VOC (20 kOH+VOC + 10 kOH+CARB + 3 kOH+CH3CHO + 6 kOH+HCHO ) = LNOx 8 kOH+NO2 NO2

)

≈

(1)

VOC + OH

(2)

R1O2 + NO

(3)

! !

O2

at 298 K

!

!

R1O2 (a)

O2 ! !

(R2O2 + NO

20 VOC NO2

!

HCHO + ! CARB + ! HO2 + ! R2O2 + NO2 (b)

O2

DIHOCARB + HO2 + NO2)

HO2 + NO = OH + NO2 !

(NO2 + hv !

(1.1) Net: VOC + hv

O2

O2 ! !

NO + O3) !

O2

(4)

!

(5)

!

isom.

(6)

!

isom. & frag. !

!

(R5O2 + NO2 ↔ PAN) (d)

(7)

!

(CARB + OH

(R3O2 + NO !

(R4O2 + NO ! !

R4O2 + NO2)

frag., +O2

(9)

!

O2

! !

(NO2 + hv ! !

(10)

!

(11)

!

(12)

!

R6O2 + NO2)

CH3CHO + HCHO + HO2 + NO2)

(HO2 + NO = OH + NO2)

!

(1.2) Net:

O2

NO + O3)

!

!

CARB + ! OH + ! NO !

(CH3CHO + OH !

O2

(CH3O2 + NO ! ! !

O2

O2 ! !

O2

CH3O2 + CO2 + NO2)

HCHO + HO2 + NO2)

(HO2 + NO = OH + NO2)

(NO2 + hv !

O2

!

NO + O3)

!

!

!

HODICARB + ! PAN + ! CH3CHO + ! HCHO + ! O3

CH3C(O)OO)

(CH3C(O)OO + NO !

!

!

! R5O2 + ! HODICARB + NO2) (c)

!

(R6O2 + NO !

!

R3O2) (b)

(8)

(R5O2 + NO !

!

HCHO + ! CARB + ! DIHOCARB + ! O3

(1.3) Net:

!

O2 !

(13)

!

O2

(14)

!

O2

CH3CHO !

(HCHO + hv !

(HCHO + OH ! ! ! !

!

!

!

HO2 + CO + ! H2) (e) HO2 + CO + H2O) (f)

(HO2 + NO = OH + NO2)

(NO2 + hv !

(1.4) Net:

!

! CO2 + ! HCHO + ! O3

O2

HCHO + hv !

NO + O3) O2 !

!

!

!

!

! CO2 + !" H2 + ! OH + ! H2O + ! O3

(a): Abbreviations of compounds can be found in figures in question 3(a). (b): 2-methylpentene oxidation products can undergo further reactions. (c): Assuming the branching ratios of fragmentation and isomerization are 0.5 and 0.5, respectively. (d): Assuming the branching ratios of R5O2 reacting with NO and NO2 are 0.5 and 0.5, respectively. (e): The overall molar yield of HCHO from 2-methylpentene is ¾. The photolysis rate of HCHO is ~ 105 s1

. The HCHO + OH reaction rate is also ~ 105 s-1 at OH concentration of ~ 106 molecules cm-3. So here we

assume half HCHO undergoes photolysis and half HCHO undergoes OH oxidation. We also assume that during photolysis, half HCHO undergoes the radical production pathway. (f): The molar yield of CO is ¾, since CO + OH reaction rate is ~ 10 times slower than those for the OH oxidation of other VOCs, here we ignore its contribution to the O3 production.

Chapter 7

7.1

Preindustrial

8.17 = - log [H+]

[H+] = 10 -8.17 = 6.76 x 10-9 M Present day

8.07 = -log [H+]

[H+] = 10 -8.07 = 8.51 x 10-9 M (8.51 – 6.76)/ 6.76 = 0.26

26% increase in [H+]

7.2

The hydrogen ion concentration in an aqueous solution of CO2 is given by (7.29), 3

!"H + #$ − ( K w + H CO K c1 pCO )!"H + #$ − 2H CO K c1K c2 pCO = 0 2 2 2 2 Given the temperature, which determines the values of K w , H CO2 , K c1, and K c2 and the partial pressure of CO2 ( pCO2 = 350 ×10 −6 atm ), [H+] can be calculated (Fig. P7.2-1). Note that as the temperature varies from 0° to 30°C, the pH increases by 0.12 units.

For ξ CO2 = 700 ppm ( pCO2 = 700 ×10 −6 atm ) the pH = 5.5. Therefore, the "background" precipitation pH will decrease by 0.1 pH units.

Problem 7.5 (a) The concentration of the dissolved H2O2 can be calculated using Eq. 7.107:

H H 2O2 pHo 2O2

10 5 10 −9 [H 2 O 2 (aq )] = = = 80.3 µM 1 + H H 2O2 wL RT 1 + 10 5 10 −7 0.082 298 O3 and NO2 are not that soluble in water so we can use for them the corresponding equation for an open system to get:

[O3 (aq)] = H O3 pOo 3 = 1.1× 10 −2 40 × 10 −9 = 4.4 × 10 −4 µM o [ NO 2 (aq)] = H NO2 p NO = 10 −2 10 −8 = 10 −4 µM 2

Finally nitric acid is extremely soluble at pH=4 and all of it will be dissolved in the aqueous phase. Its concentration then will be given by:

[HNO 3 (aq )] =

o p HNO 3

wL RT

=

5 × 10 −9 = 2230 µM 10 −7 0.082 298

(b) The effective Henry’s law coefficient for pH=4 is given by:

⎡ K K K ⎤ -1 H S* ( IV ) = H SO2 ⎢1 + s+1 + s1 + s22 ⎥ =161 M atm ⎣ [ H ] [ H ] ⎦ Then the corresponding aqueous-phase fraction of SO2 will be equal to:

X

S ( IV ) aq

161× 22.4 × 10 −7 = = 3.6 × 10 −4 −7 1 + 161× 22.4 × 10

and practically all the SO2 will be in the gas phase. (c) As all the SO2 will be in the gas phase we can use the open system equation: [S(IV)]=H*S(IV) pSO2=161 5x10-9 M=0.8 µM (d) For the conditions of the fog the reaction of the dissolved H2O2 with S(IV) dominates over the ozone reaction. Also the rate of the reaction is very high so it will consume the limiting reactant in a few hours. By 6 am practically all the H2O2 will have been consumed and 1 ppb of sulfate will be produced. The corresponding aqueous-phase concentration is:

10 −9 [S(VI)]= −7 = 445 µM 10 22.4

(e) The initial [H+] concentration is 10-4 M=100 µM. The addition of 445 µM of sulfate corresponds to an addition of 890 µM of equivalent acidity. So roughly the [H+] concentration will increase to 1000 µM and the pH will drop by approximately one unit to 3. A more accurate calculation requires knowledge of the complete fog water composition that is not given in this problem.

Problem 7.6 Let us assume that only a small fraction of the available formic acid will be dissolved in the cloud water so that we can treat the system as open. We will check the validity of this assumption later.

For the system here the ion balance is: [HCOO-] + [HCO3-] + 2 [CO32-] + [OH-] = [H+] The pH of the cloud water with just the CO2 will be around 5.5 so taking into account the HCOOH it will be even lower. For these conditions the concentrations of CO32and OH- will be much lower than the other terms in the equation and can be neglected.

Therefore: [HCOO-] + [HCO3-] = [H+] The concentrations of the ions can be connected to the corresponding partial pressures to obtain:

Kf [H ] +

o H HCOOH p HCOOH +

K c1 H CO2 pCO2 = [H + ] + [H ]

The concentration of H+ can be then calculated as: [H+] = (1.8x10-4 3600 10-9 + 4.3x10-7 3.4x10-2 400x10-6)1/2 M or [H+] = 2.56x10-5 M and pH=4.6.

Therefore even 1 ppb of HCOOH is enough to lower the clean cloud pH to 4.6 and make it acidic.

We can estimate the fraction of the HCOOH that will be in the aqueous phase using Eq. 7.9. The effective Henry’s law coefficient for HCOOH for pH of 4.7 is equal to:

K f ⎤ ⎡ ⎡ 1.8 × 10 −4 ⎤ -1 * H HCOOH = H HCOOH ⎢1 + + ⎥ = 3600⎢1 + = 28900 M atm −5 ⎥ ⎣ 2.56 × 10 ⎦ ⎣ [H ] ⎦ Its fraction in the aqueous phase is then from Eq. 7.9 equal to 0.066 (see also Figure 7.3) and therefore our assumption of an open system is quite accurate for this case.

7 .

8

7.9

The results are shown in Figures P7.9-1 and P7.9-2.

7.10

The results are shown in Figures P7.10-1 – P7.10-3.

7.12

See Hoffman and Jacob (1984).

9.3 vt = Dp, µm 0.05 0.1 0.5 1.0

D p2 ρ p gCc

18μ Cc 4.95 2.85 1.326 1.164

vt, cm s-1 9.86 × 10-5 2.27 × 10-4 2.64 × 10-3 9.27 × 10-3

9.4 Calculate the terminal settling velocities of 0.001, 0.1, 1, 10, and 100 μm water droplets in

air for a pressure of 0.1 atm.

The challenge in this problem is the calculation at a pressure that is different than 1 atm. The terminal settling velocity of a spherical particle is given by (9.42): 1 D p ρ p gCc ( D p ) vt = 18 μ 2

There are two quantities in the formula that at least in principle depend on pressure, the slip correction factor Cc through its dependence on the air mean free path and the viscosity of air μ.

The viscosity of gases is primarily a function of temperature and depends very weakly on pressure (for variations of pressure by a factor or 10 or so). We will perform the calculation assuming ambient temperature 25oC and therefore μ=1.8x10-5 kg m-1 s-1.

The air mean free path is inversely proportional to pressure (equation 9.6). As a result the mean free path at a pressure of 0.1 atm will be 650 nm or 0.65 μm. The slip correction factor is given by (9.34) Cc = 1 +

⎛ 1.1D p ⎞⎤ 2λ ⎡ ⎟⎟⎥ ⎢1.257 + 0.4 exp⎜⎜ − D p ⎣⎢ 2 λ ⎝ ⎠⎦⎥

The results of the calculation are given in the following table together with the corresponding Reynolds number. The density of air at p =0.1 atm can be found from the ideal gas law and is equal to 0.118 k g m-3.

Diameter (μm)

Kn

Cc

vt (cm h-1)

Re

0.001

0.0013

2155

0.023

4.3x10-13

0.1

0.13

22.1

2.4

4.4x10-9

1

1.3

2.86

31.1

5.7x10-7

10

13

1.16

1270

2.3x10-4

100

130

1.02

1.1x105

0.2

For all drops the Reynolds number is of order one or smaller and the above calculation is accurate. We do not need to correct these values using the drag coefficient.

9.5

T = 20°C, p = 1 atm µ = 0.000178 g cm-1 s-1 g = 980 cm s-2 ρp = 1 g cm-3

9.6

The stop distance of a particle moving with velocity V is (see 9.102) sp = τ V

where V = 1 m s-1 in this case. The characteristic relaxation time is given by (8.38)

τ =

In this case mp = ρp

m p Cc

3πμD p

3 1 3 kg 1 πD p = 1500 3 π (10 −6 ) m 3 7.85 × 10 −16 kg 6 m 6

Cc = 1.164 (see Table 9.3)

µ = 1.72 × 10-4 poise = 1.72 × 10-5 Pa s (see Table A.7) Dp = 10-6 m

and τ = 5.6 × 10-6 s The stopping distance is then sp = 5.6 × 10-6 × 1 m = 5.6 µm.

9.9

Constants: p = 101325 Pa T = 298 K g = 9.8 m/s2 χ = 1.32 λ = 0.0651 μm μ = 1.8 x 10-5 kg/m/s ρp = 2000 kg/m3

Volume equivalent diameter: 1

⎛ 6 ⎞3 Dve = ⎜ V p ⎟ = 0.159 μm ⎝π ⎠

Settling velocity (with shape factor): Dve ρ p gCc (Dve ) 2

Vt1 =

18χμ

= 2.45x10-6 m/s

where Cc = 1 +

1.1Dve − ⎞ 2λ ⎛ ⎜1.257 + 0.4e 2 λ ⎟ ⎟ Dve ⎜⎝ ⎠

Settling velocity (without shape factor): Dve ρ p gCc (Dve ) 2

Vt 2 =

18μ

= 3.23x10-6 m/s

Error in calculation if shape factor is not included: Vt 2 − Vt1 x 100 % = 32 % Vt1

9.10

Equation of motion for a charged particle in an electric field: mp

dv 3πμDve χ (u − v ) + qE = dt C c ( Dve )

For steady motion in still air (u = 0): 3πμDve χ v e = qE C c ( Dve ) ve = Be E

where Be is the electrical mobility: Be =

qC c ( Dve ) 3πμDve χ

(1)

The electrical mobility can also be expressed in terms of the electrical mobility diameter Dem: Be =

qCc ( Dem ) 3πμDem

Equating (1) and (2), we will get: Dem = Dve χ

C c ( Dem ) C c ( Dve )

(2)

9.11

Electrical Mobility:

Be =

qCc (D p )

3πμD p

=

neC c (D p )

3πμD p

where 1.1D p ⎞ − 2λ ⎛⎜ Cc = 1 + 1.257 + 0.4e 2 λ ⎟ ⎟ D p ⎜⎝ ⎠

To calculate the diameters of particles with different charges but the same electrical mobility: n1eC c (D p ,1 ) n2 eC c (D p , 2 ) = Be = 3πμD p ,1 3πμD p , 2 n1C c (D p ,1 ) D p ,1

=

n2C c (D p , 2 ) D p,2

Constants: λ = 0.0651 μm (at STP) n1 = 1 Dp,1 = 0.1 μm Solve equation (1) numerically for Dp,2: For n2 = 2, Dp,2 = 0.15 μm For n2 = 3, Dp,2 = 0.20 μm For n2 = 4, Dp,2 = 0.24 μm

(1)

One source of discrepancy between the two equations is that in (11.74) i* is assumed to be an integer, and thus the value of i* must be rounded to the closest integer to calculate J. This is not required in (11.47). On the other hand, (11.47) employs the integral

approximation for sufficiently large i*. If i* is small, this is not necessarily an accurate approximation.

Problem 9.12 For a spherical particle its electrical mobility equivalent diameter, Dem, and its physical diameter, Dp, are equal: Dem=Dp

(1)

The classical aerodynamic diameter, Dca of this particle is given by (9.110):

⎛ ρ p ⎞ Dca = D p ⎜ o ⎟ ⎜ ρ ⎟ ⎝ p ⎠

1/ 2

⎛ C c ( D p ) ⎞ ⎜⎜ ⎟⎟ ⎝ C c ( Dca ) ⎠

1/ 2

(2)

Combining (1) and (2) and taking the square: 2 Dca2 = Dem

ρ p Cc ( Dem ) ρ op Cc ( Dca )

(3)

Solving (3) for the particle density one then obtains the expression:

ρp =

Dca2 C c ( Dca ) o ρp 2 C c ( Dem ) Dem

For this problem: Dca=600 nm and therefore Cc(Dca)=1.273 and also Dem=500 nm and as a result Cc(Dem)=1.326.

Substituting these values into (4) we find that:

600 2 1.273 ρp = 1 = 1.38 g cm-3 2 500 1.326

(4)

Problem 9.13 (a) For a non-spherical particle the electrical mobility equivalent diameter and the volume equivalent diameter are connected with (9.114):

Dem = Dve χ

Cc ( Dem ) Cc ( Dve )

(1)

The classical aerodynamic diameter for the same particle is connected to its Stokes diameter with (9.109):

⎛ ρ p ⎞ Dca = DSt ⎜ o ⎟ ⎜ ρ ⎟ ⎝ p ⎠

1/ 2

⎛ C c ( Dst ) ⎞ ⎜⎜ ⎟⎟ C ( D ) ⎝ c ca ⎠

1/ 2

(2)

The Stokes diameter using (9.107) can be written as a function of the volume equivalent diameter as: 1/ 2

⎛ C ( D ) ⎞ DSt = Dve ⎜⎜ c ve ⎟⎟ ⎝ χCc ( DSt ) ⎠

(3)

Replacing (3) into (2) one gets:

1 ρ p C c ( Dve ) χ ρ op C c ( Dca )

Dca = Dve

(4)

Solving (4) for Dve and replacing the result into (1) we find that:

Dem = Dca

ρ op χ 3Cc ( Dem ) 2 Cc ( Dca ) ρ p Cc ( Dve ) 3

(5)

which is the relationship linking the electrical mobility equivalent diameter with the classical aerodynamic diameter. (b) Solving (1) for the electrical mobility equivalent diameter we find that:

Dve = Dem

1 Cc ( Dve ) χ Cc ( Dem )

(6)

The vacuum aerodynamic diameter is linked to the volume equivalent diameter with:

Dva = Dve

1 ρp

χ ρ op

(7)

Combining (6) and (7) we find the expression linking the electrical mobility equivalent diameter with the vacuum aerodynamic diameter:

Dva = Dem

1 ρ p Cc ( Dve ) χ 2 ρ op Cc ( Dem )

(8)

10.3

Total Ammonia = 10 ppb Total Nitrate = 15 ppb (a)

RH = 30%, T = 273–308 K x = NH4NO3 concentration (solid @ RH = 30%) 10 – x = gas phase ammonia 15 – x = gas phase nitric acid

(10 − x ) (15 − x ) = K p (T ) Using (10.91) to determine Kp, T (°C) 0 7 14 21 28 35

K (ppb2) 0.029 0.228 1.616 10.39 61.00 329.51

Solving for x, T (°C) x (ppb) x (µg m–3) 0 9.99 35.68 7 9.95 34.65 14 9.70 32.95 21 8.42 27.92 28 4.23 13.70 35 –5.8* 0 *Temperature is too high for any formation of NH4NO3(s) as one cannot have x < 0. (b)

T = 298 For RH < 61.8%, aerosol NH4NO3 is solid, so Kp = 28.87 ppb2 (15 – x) (10 – x) = 28.87 ppb2 x = 6.57 ppb = 21.5 µg–3

For RH < 61.8%, use Fig. 10.21 with Y = 1 to determine Kp and solve for x as in part (a) K (ppb2) 39 30 22 18 6.4 2.1

RH (%) 65 70 75 80 90 95 (c)

x (ppb) 5.77 6.48 7.18 7.58 8.94 9.61

x (µg m–3) 18.89 21.20 23.51 24.79 29.26 31.44

T = 288 K, RH = 30% Total Ammonia = 0–20 ppb, Total Nitrate = 15 ppb Kp (288) = 2.12 ppb2 (Total Amm – x) (15 – x) = 2.12 ppb2 where x = NH4NO3 concentration Total Ammonia (ppb) 0 4 8 12 16 20

x (ppb) 0 3.81 7.71 11.41 13.96 14.61

x (µg m–3) 0 12.90 26.10 38.59 47.26 49.34

The response of NH4NO3(s) to ammonia is nonlinear as there is a term second order in x. (See Figure P10.3-1.) (d)

TA = total ammonia TN = total nitric acid T = 15 °C, RH = 30%, ⇒ Kp = 2.12 ppb2 2 µg–3 NH4NO3 = 0.59 ppb (TA – 0.59) (TN – 0.59) = 2.12 There are many combinations of TA; TN (TA; TN > .59) which would satisfy this.

(e)

High ammonia, nitric acid concentrations, low temperatures, and high relative humidity favor NH4NO3 formation.

10.4

(a)

Total Ammonia = 10 ppb Total Nitrate = 15 ppb Total Sulfate = 4 µg–3 ~1 ppb (assume constant at all T) Preferred form of sulfate in this case is (NH4)2SO4. 2 ppb of ammonia are needed to neutralize the available sulfate. Hence, 8 ppb of ammonia remain for reaction with HNO3. Repeating the calculation of Problem 10.3

(8 − x ) (15 − x ) = K p (T ) and solving for x = (NH4NO3), T (°C) 0 7 14 21 28 35 (b)

Kp (ppb2) 0.029 0.228 1.616 10.39 61.00 329.51

x (ppb) 7.996 7.968 7.776 6.74 2.94 0

x (µg m–3) 28.56 27.75 26.42 22.35 9.52 0

T = 25 °C (NH4)2SO4 = 1 ppb Available Ammonia = 8 ppb Total Nitrate = 15 ppb For RH < 61.8% (assuming that DRH NH 4NO3 is constant with T), aerosol NH4NO3 is solid so Kp = 28.87 ppb2 (15 – x) (8 – x) = 28.87 and x = 5.09 ppb2 = 16.65 µg–3 NH4NO3 For RH > 61.8% use Fig. 10.21 to determine Y. Based on results in 9.3(b), Y ~ 0.7, let us assume Y ≅ 0.7 for all cases.

(15 − x ) (8 − x ) = K p

K (ppb2) 34 25 21 14 5.8 1.9

RH (%) 65 70 75 80 90 95

x (ppb) 4.70 5.40 5.73 6.38 7.25 7.74

x (µg m–3) 15.37 17.66 18.76 20.86 23.73 25.32

The accuracy of the calculation can be improved by calculating the updated values of Y and iterating. (c) T = 288, RH = 30% Total Ammonia = 2–20 ppb Total Nitrate = 15 ppb Kp (288.15) = 2.12 ppb2 Total Sulfate = 1 ppb 2 ppb ammonia are needed to neutralize sulfate, so the total ammonia remaining is always 2 ppb less. [(Total Ammonia – 2) – x] [15 – x] = 2.12 Total Ammonia (ppb) 2 6 10 14 18 20

x (ppb) 0 3.81 7.71 11.41 13.96 14.41

x (µg m–3) 0 12.90 26.10 38.63 47.26 48.79

Problem 10.7 (a) The corresponding input for E-AIM (Model IV, Comprehensive Calculation) is: Sulfate: 0.42x10-6 moles m-3 Nitrate: 0.32x10-6 moles m-3 Ammonium: 0.56x10-6 moles m-3 Sodium: 0.17x10-6 moles m-3 Chloride: 0.17x10-6 moles m-3 [H+]: 0.6x10-6 moles m-3 Please note that E-AIM requires neutrality of the input composition so the [H+] is calculated to satisfy this input constraint of the model. The results are shown in Table P10.7-1. The total dry PM concentration is 57.5 µg m-3 consisting mainly of sulfates and ammonium. There is also all the available sodium and some nitrate.

Table P10.7-1 Results of simulations Base case -50% sulfate -50% nitric acid -50% ammonia Gas-phase (ppb) HNO3

6.3

2.0

3.1

7.3

HCl

3.7

2.0

3.7

3.9

NH3

0.01

0.2

0.02

0.001

PM (µg m-3) Ammonium

10

9.9

10

5

Sodium

3.9

3.9

3.9

3.9

Bisulfate

12.5

1.0

11.3

23.5

Sulfate

28

19.1

29.1

17.1

Nitrate

2.8

14.5

1.6

0.3

Chloride

0.3

2.9

0.4

0.3

Water

77.2

76.3

75.4

78.1

Total dry mass

57.5

51.3

56.3

50.1

Practically all the chloride from the NaCl is transferred to the gas phase in this system.

Also the nitrate exists mostly in the gas phase as nitric acid, while all the ammonia is in the particulate phase.

(b) Let us assume that the sulfate concentration has been reduced by 50% to 0.21x 10-6 moles m-3. Using the same input for the rest of the species with the exception of [H+] which should be now 0.18x10-6 moles m-3 we get the results shown in Table P10.7-1. The sulfates are reduced by approximately 20 µg m-3 so one would expect a reduction of the dry PM by 35%. However, the reduction is only 6.2 µg m-3 (around 10%). The reason for this nonlinear response is the increase of nitrate from 2.8 to 14.5 µg m-3 and of chloride from 0.3 to 2.9 µg m-3. The change in the partitioning of these two acids offsets the reductions in sulfate. The composition of the PM changes significantly as now the nitrate and chloride are as important as the sulfates. All the ammonia remains in the particulate phase. (c) A 50% reduction is the available total nitric acid has a relatively small effect on the PM concentration and composition (Table P10.7-1). The nitrate is reduced by 1.2 µg m-3 and there is a marginal increase of the chloride by 0.1 µg m-3. The change in the dry PM is only 2%. (d) A 50% reduction in ammonia actually has the biggest effect on the dry PM concentration resulting in a 13% reduction. The ammonium concentration is reduced by 5 µg m-3 and all remains in the particulate phase. There is also a significant reduction of nitrate (2.5 µg m-3).

One of the interesting results of these simulations is the relative stability of aerosol water which varies by less than 2.5% in all cases. Actually in the case with the largest reduction of dry PM mass the water increases due to the increase in bisulfate.

11.4 Ccrit (µg m −3 ) = 0.16 exp (0.1T − 3.5RH − 27.7)

a.

Convert µg H2SO4 m–3 to molecules H2SO4 cm–3: €

273 K

€

b.

298 K

Ccrit × 6.14 ×10 9 = nH 2SO 4 (molecules cm–3) 50% RH

1.2 × 108

100% /RH

2 × 107

50% RH

1.4 × 109

100% RH

2.4 × 108

molecules cm–3

The only case that comes close to ambient H2SO4 concentrations is 273 K and

100% RH.

11.5 2

"→ A2 ⋅ B A2 + B " a.

At steady state d[ A 2 ] 2 € = 0 = 12 k1f [ A1 ] − k1b [ A2 ] − k2 [ A2 ][B] dt

giving 2

k [A ] [A2 ] = 1f 1 k1b + k2 [B] 1 2

€

Nucleation rate

J = k2 [ A2 ][B]

€ b.

€

k2 = 3.4 × 10–10 cm3 molecule–1 s–1

[B] = 2.34 × 109 molecules cm–3 (10 ppt)

€ Calculate

k1b = 104 s–1

2.34 × 1010 molecules cm–3 (100 ppt)

€ €

P=

k2 [B] k1b + k2 [B]

@ [B] = 2.34 × 109 €

@ [B] = 2.34 × 1010

P = 7.96 × 10–5 P = 7.96 × 10–4

c.

Figure 11.17 has the lower limit line

(5 ×10 ) × k [A ] −6

1 2

1f

2

1

which can be compared with the two values of P in part (b).

€

Problem 13.2 Exact solution of the problem requires a numerical solution of the continuous coagulation equation to simulate the evolution of the size distribution of the tobacco smoke particles. Please note that even if the initial particle population is monodisperse, the distribution will start to broaden due to the collisions. However, we can still estimate the approximate change in size of the particles.

Assuming a constant coagulation coefficient the total number concentration of particles will be given by Eqs. 13.66 and 13.67. According to 13.67 the characteristic time for coagulation in this system is:

τc =

2 = 0.083s 24 × 10 −10 1010

The total number concentration after 2 s will be then according to Eq. 13.66

Nf =

1010 = 0.04 × 1010 cm-3 1 + (2 / 0.083)

so it will be reduced by approximately 25 times. We can estimate the new average size of the tobacco smoke particles from the initial and final particle concentrations and the fact the particle mass is conserved during coagulation. If Dp0 and Dpf. Assuming that the particles remain spherical during coagulation:

1 1 N 0 ρ πD 3p 0 = N f ρ D 3pf 6 6 and solving for the final particle diameter assuming that their density does not change:

⎛ N ⎞ D pf = D po ⎜ 0 ⎟ ⎜ N ⎟ ⎝ f ⎠

1/ 3

⎛ 1010 ⎞ ⎟ = 20⎜⎜ 10 ⎟ ⎝ 0.04 × 10 ⎠

1/ 3

≅ 60 nm

so we estimate that their size will increase by approximately a factor of 3. The selfcoagulation rate constant does not change dramatically as the particles grow from 20 to 60 nm (Figure 13.5) so our assumption of a constant coefficient during the growth is reasonable.

However the above result does require some additional representation. The particle size distribution Nk at 2 s is given by Eq. 13.71 is shown in Figure P13.2-1. The particles do not grow uniformly but rather the distribution spreads out. Actually after 2 s the highest concentration of particles is predicted to still have a size of 20 nm (no

collisions). However, there are 10,000 cm-3 with k=69 (approximate diameter 82 nm) and also lower concentrations of even larger particles. The concentration of 60 nm particles is around 55,000 cm-3, so they represent only a few percent of the particle population. The estimated 60 nm size can be viewed as a volume average diameter.

180000 160000 140000 120000

Nk (cm-3)

100000 80000 60000 40000 20000 0 1

6

11

16

21

26

31

36

41

46

51

56

k

Figure P13.2-1

61

66

71

76

81

86

91

96

13.3 We are interested in the lifetimes of particles of diameters 0.01 and 0.1 µm in a typical urban atmosphere as a result of coagulation with other particles. The Brownian coagulation coefficients can be calculated from Table 13.1 or estimated from Figure 13.5.

Typical urban aerosol size distribution parameters are given in Table 8.3.

The 0.01 µm diameter particle lies in Mode I and the 0.1 µm diameter particle lies in Mode III. We can use the following: Mode I II III

N (cm–3) 7100 cm–3 6320 cm–3 960 cm–3

Dpg (µm) 0.0117 0.0373 0.151

Coagulation coefficients can be estimated from Figure 13.5.

Dp1 (µm) 0.01

Dp2 (µm) Mode I II III

K12 (cm3 s–1) ~2 × 10–9 ~9 × 10–9 ~7 × 10–8

0.1

Mode I II III

~2 × 10–8 ~5 × 10–9 ~2 × 10–9

Dp1 (µm) 0.01

Dp2 (µm) Mode I II III

0.1

Mode I II III

Results are: τ = 2/KN0 (s) 1.4 × 105 3.5 × 104 3.0 × 104

τ (h) 39 9.7 8.3

1.4 × 104 6.0 × 104 1.0 × 106

3.9 16.7 278

Chapter 14 14.1 The fraction of the OC emissions from meat cooking that is cholesterol is: 15.3 kg C/day _____________

= 0.011

1400 kg C/day Then the total OC concentration from meat cooking is 14.6 ng/m3 __________ = 1327 ng/m3 = 1.34 µg/m3 0.011 So the fraction of the OC from meat cooking is 1.34 µg/m3 /7.5 µg/m3 = 0.18

14.2

The saturation mass concentration of propionic acid is obtained from the ideal gas law, Co =

po M RT

where M = molecular weight of propionic acid (CH3CH2COOH) = 74 g mol—1 po =vapor pressure of propionic acid = 0.005 atm R = universal gas constant = 8.2 × 10—5 atm m3 mol—1 K—1 T = temperature = 298 K so

C o = 15.14 g m −3 Convert 15.4 g m—3 to ppb: 3

Ceq =

#

3

&

m ( ( 298 K ) (15.4 g/m ) %$8.2 ×10 atm mol K ' −5

(1 atm ) ( 74 g/mol)

= 5.0 ×10 = 5.0 ×10 6 ppb −3

a.

Propionic acid forms an ideal solution with the organic aerosol phase. In this case, Caer will be given by: Caer, i =

$ ai RT !

M i mo # & ΔVOC M VOC " mo RT + M o pio %

where ai = yield of propionic acid = 1 mo = preexisting organic aerosol mass concentration = 100 µg m—3 Mo = average molecular weight for preexisting organic aerosol = 100 g mol—1 We know that: ΔVOC = amount of precursor (1-butene) reacted = 100 ppb MVOC = molecular weight of 1-butene = 56 g mol—1 Mi = molecular weight of propionic acid =- 74 g mol—1

$

atm ⋅ m 3 ' ) ( 298 K ) mol ⋅ K ( 56 g/mol

(1) &8.2 ×10−5 Caer, i =

%

* , / −6 3 ( 74 g/mol) (100 ×10 g/m ) , /(100 ppb) 3' , $ / atm ⋅ m −6 3 −5 ) ( 298 K ) + (100 g/mol) ( 0.005 atm ) / , (100 ×10 g/m ) & 9.2 ×10 mol ⋅ K ( % + .

Caer, i = 6.5 ×10 −4 ppb = 0.65 ppt

This is a negligible amount compared to the 100 µg m—3 of preexisting organic aerosol assumed. b.

Propionic acid forms an ideal solution with the aqueous aerosol phase. For an ideal solution, pi = pio xi , where pi = partial pressure of compound i, pio = vapor pressure of compound i, and xi = mole fraction of compound i in the solution.

Now pi = 100 ppb = 10 × 10—9 atm. Then

pi 100 ×10 −9 atm xi = o = = 2 ×10 −5 pi 0.005 atm Assuming 100 µg m—3 of H2O, we calculate the aerosol concentration of propionic acid.

Let C = aerosol concentration of propionic acid in moles m—3 Then xi = 2 ×10 −5 =

C C + (100 µg/m ) (1 mol /18 g) (1 g /10 6 µg) 3

and C = 1.1 × 10—10 mol m—3 So

Caer =

(1.1×10

−10

$ atm ⋅ m ' mol/m 3 ) &8.2 ×10 −5 ) ( 298 K ) % mol⋅ K ( = 2.7 ×10 −12 = 2.7 ppt 1 atm

and is still quite small.

14.3 First, find ΔVOC* from (14.34),

1 a a = 1o + 2o * ΔVOC C1 C2 where a1 = 0.05; a2 = 0.15; C1o = 0.15 ppb; C2o = 6 ppb. So

ΔVOC * =

1 = 2.79 ppb 0.05 0.15 + 0.15 6

So for ΔVOC < 2.79 ppb, there will be no aerosol phase; both products will exist solely in the gas phase. For ΔVOC > 2.79 ppb, an aerosol phase will exist and the component concentrations will be given by (14.22) and (14.23). First, however, we must calculate the amount of the species in the gas phase, which means we must find x1, which is given by (14.27). Equation (14.27) can be solved for different values of ΔVOC. For example, for ΔVOC = 100 ppb, 1/2

) , o o o o 2 1 # ( a1 + a2 ) ΔVOC − C2 + C1 & 1 +# ( a1 + a2 ) ΔVOC − C2 + C1 & 4a1ΔVOC . x1 = − % (± % ( + o 2$ C2o − C1o C2o − C1o C2 − C1o .' 2 +*$ '

1/2

2 ) , 1 # ( 0.05 + 0.15) (100 ) − 6 + 0.15 & 1 +# ( 0.05 + 0.15) (100 ) − 6 + 0.15 & 4 ( 0.05) (100 ) . x1 = − % ( ± +% ( + 2$ 6 − 0.15 6 − 0.15 6 − 0.15 .' 2 *$ '

x1 = −1.21±1.52 or x1 = 0.31

Thus, x2 = 1− x1 = 0.69 Then, from (14.24) and (14.25),

Cg, 1 = x1C1o = ( 0.31) ( 0.15 ppb) = 0.047 ppb Cg, 2 = x2C2o = ( 0.69 ) ( 6 ppb) = 4.14 ppb Then, from (14.23) and (14.24), assuming M1 ≈ M 2 ≈ M VOC ,

Caer, 1 = a1ΔVOC − Cg, 1 = ( 0.05) (100 ppb) − 0.047 ppb = 4.95 ppb Caer, 2 = a2 ΔVOC − Cg, 2 = ( 0.15) (100 ppb) − 4.14 ppb = 10.86 ppb See Figures P14.3-1 to P14.3-4.

14.4

To simplify the calculations we assume that all compounds have approximately the same molecular weight of 150.

From the mass balances:

Cg, 1 + Caer, 1 = α1ΔVOC

(1)

Cg, 2 + Caer, 2 = α 2 ΔVOC

(2)

Assuming that all three compounds form an ideal solution: Cg, 1 = x1C1o

(3)

Cg, 2 = x2C2o

(4)

Finally the corresponding mole fractions are given by: x1 =

x2 =

Caer, 1

(5)

Caer, 1 + Caer, 2 + m0 Caer, 2

(6)

Caer, 1 + Caer, 2 + m0

where m0 is the concentration of the pre-existing aerosol. These equations form an algebraic system for the six unknowns Caer, 1, Caer, 2, Cg, 1, Cg, 2, x1, x2. Combining them we are left with the following two equations:

α1ΔVOC − x1C1o (α1 + α 2 ) ΔVOC − x1C1o − x2C2o + m0

(7)

α 2 ΔVOC − x2C2o x2 = (α1 + α 2 ) ΔVOC − x1C1o − x2C2o + m0

(8)

x1 =

for the two unknown mole fractions. Assuming an approximate molecular weight of 150 g mole—1, then 1 ppb corresponds to ~ 6 µg m—3 and the two saturation concentrations are C1o = 0.9 µg m—3 and C2o = 36 µg m—3.

Substituting the numerical values and solving (7) for x2 we find that:

x2 = 0.0056ΔVOC + 0.303− 0.025x1 −

0.0014ΔVOC x1

(9)

Equation (8) can be written as:

0.2ΔVOCx2 − 0.9x1 x2 − 36x22 +10x2 − 0.15ΔVOC + 36x2 = 0

(10)

Substituting (8) into (9), we have one equation for one unknown, x1. This equation can be solved numerically for any given value of ΔVOC. The solutions are given in the following table.

ΔVOC

x1

x2

(ppb)

Cg, 1

Cg, 2

Caer, 1

Caer, 1

(µg m—3)

(µg m—3)

(µg m—3)

(µg m—3)

0

0

0

0

0

0

0

1

0.027

0.025

0.02

0.89

0.28

0.01

10

0.183

0.175

0.16

6.31

2.84

2.69

50

0.288

0.517

0.26

18.63

14.7

26.3

100

0.277

0.623

0.25

22.45

29.75

67.55

200

0.266

0.701

0.24

25.2

59.76

154.8

500

0.256

0.69

0.23

24.8

149.77

425.2

1000

0.253

0.695

0.23

25

299.77

875

14.5

Let us solve

(

)

⎡ a + a ΔROG − C2o + C1o ⎤ a ΔROG x12 + ⎢ 1 2 o =0 ⎥ x1 − 1 o o C2 − C1 C2 − C1o ⎢⎣ ⎥⎦

without summing constraints

Caer, 1 ≥ 0 a1ΔROG − x1C1o ≥ 0 At ΔROG * , a1ΔROG * = x1ΔROG * = x1C10 , so x1 =

a1ΔROG * C1o

Substituting into (14.32):

(

)

* o o a12ΔROG *2 ⎡ a1 + a2 ΔROG − C2 + C1 ⎤ a1ΔROG * a1ΔROG * +⎢ − o =0 ⎥ o C1o2 C2o − C1o C2 − C1o ⎢⎣ ⎥⎦ C1

(C − C ) a ΔROG + a (a + a ) ΔROG − a ΔROG C + a ΔROG C − a ΔROG = 0 o 2

o 1

2 1

*2

*2

1

1

C1o2

2

*

1

C1o

o 2

*

o 1

1

C1o

(C − C ) a ΔROG + a (a + a ) ΔROG C − a ΔROG C C = 0 o 2

o 1

2 1

*2

*2

1

1

2

o 1

*

1

o 2

o 1

a12 ΔROG *2C2o − a12 ΔROG *2C1o + a12 ΔROG *C1o + a1a2 ΔROG *2C2oC1o = 0 a1ΔROG *2C2o + a2 ΔROG *2C1o − ΔROG *C2oC1o = 0

(

)

ΔROG * a1C2o + a2C1o = C2oC1o a a 1 = 1o + 2o * ΔROG C1 C2

Or equivalently solve without summing constraints:

Caer, 2 ≥ 0 a2 ΔROG − C2o + x1C2o ≥ 0 At ΔROG * , −a2 ΔROG * + C2o = x1C2o , so x1 =

*

1

C1o

C2o − a2 ΔROG * a1ΔROG * = 1− C2o C2o

Substituting into (14.32):

(

2

)

⎛ a2 ΔROG * ⎞ ⎛ a2 ΔROG * ⎞ ⎡ a1 + a2 ΔROG * − C2o + C1o ⎤ a1ΔROG * =0 ⎥− o ⎜ 1− ⎟ + ⎜ 1− ⎟⎢ o C2o C2o C2o − C1o ⎝ ⎠ ⎝ ⎠ ⎢⎣ ⎥⎦ C2 − C1 o * * * o o 2a2 ΔROG * a22 ΔROG *2 ⎡⎣C2 − a2 ΔROG ⎤⎦ ⎡⎣ a1ΔROG + a2 ΔROG − C2 + C1 ⎤⎦ a1ΔROG * 1− + + − o =0 C2o C2o2 C2 − C1o C2o C2o − C1o

(

1−

2a2 ΔROG * a22 ΔROG *2 a1ΔROG *C2o + a2 ΔROG *C2o − C2o2 + C1oC2o + + C2o C2o2 C2o C2o − C1o

(

)

a1a2 ΔROG − a2 ΔROG + a2 ΔROG C − a2 ΔROG C1o *2

−

)

*2

*2

(

C2o C2o − C1o

(

)

o 2

*2

a1ΔROG * − o =0 C2 − C1o

)

Multiply by C2o2 C2o − C1o :

(

)

(

)

(

)

C2o2 C2o − C1o − 2a2 ΔROG *C2o C2o − C1o + a22 ΔROG *2 C2o − C1o + a1ΔROG *C2o2 + 2a2 ΔROG *C2o2 −C2o3 + C1oC2o2 − a1a2 ΔROG *2C2o − a22 ΔROG *2C2o − a2 ΔROG *2C1oC2o − a1ΔROG *C2o2 = 0

C2o3 − C1oC2o2 − 2a2 ΔROG *C2o2 + 2a2 ΔROG *C1oC2o + a22 ΔROG *2C2o − a22 ΔROG *2C10 + a1ΔROG *C2o2 +2a22 ΔROG *C2o2 − C2o3 + C1oC2o2 − a1a2 ΔROG *2C2o − a22 ΔROG *2C2o − a2 ΔROG *C1oC2o − a1ΔROG *C2o2 = 0

(

)

ΔROG * =

a2C1oC2o a22C1o + a1a2C2o

ΔROG * a22C1o + a1a2C2o = a2C1oC2o

ΔROG * =

(

a2C1oC2o

a2 a2C1o + a1C2o

)

a a 1 = 1o + 2o * ΔROG C1 C2

Actually, if a1ΔROG * = x1*C10 , and a2 ΔROG * − C2o + x1*C2o = 0 , then a1ΔROG * o a2 ΔROG − C + C2 = 0 C1o *

o 2

⎛ a Co ⎞ ΔROG * ⎜ a2 + 1 o2 ⎟ = C2o C1 ⎠ ⎝

1 = ΔROG *

a1C2o C1o a a = 2o + 1o o C2 C2 C1

a2 +

a a 1 = 1o + 2o * ΔROG C1 C2

All of the above development assumes that there is not discontinuity at the point of aerosol phase formation.

14.6 500 µg m-3 of α-pinene oxidized

0.01 bin

0.004 x 500 = 2 µg m-3

2 >> 0.01 all in bin

0.1 bin

0.002 x 500 = 1

3 >> 0.1

1.0 bin

0.05 x 500 = 25

28 >> 1.0 all in bin

10.0 bin

0.09 x 500 = 45

73 >> 10

100 bin

0.12 x 500 = 60

133 vs. 100 (assume this is the 50:50 point) so

all in bin all in bin 50% of

the 60 µg m-3 is in this bin. This gives 103 vs. 100 µg m-3 – a good match 1000 bin

0.18 x 500 = 90

193 vs. 1000 C*/COA ~5 so put (1 + C*/COA) -1 ~ 1/6 in this

bin ( ~ 15 µg m-3). This gives 118 vs. 1000 µg m-3.

Say we stop here. Summarizing, we have: ΔVOC = 500 µg m-3 COA = 118 µg m-3 Y = 118/500 = 0.236

23.6% yield

14.7 The table of overall lifetimes is given below (all units are days) C* nO

0.001

0.01

0.1

1.0

10

100

0

4.3

4.3

2.2

0.58

0.32

0.30

4

4.3

4.3

0.81

0.14

0.10

0.09

8

4.3

4.3

0.67

0.13

0.07

0.07

________________________________________________________________________ For C* = 0.001 and 0.01 µg m-3 virtually all the compound is in the particle phase. The lifetime corresponding to the particle-phase rate constant is 4.3 days. As C* increases, the fraction of the compound in the gas phase increases. The gas-phase OH rate constant is greater than the particle-phase OH rate constant and depends on the molecular composition of the compound, increasing strongly as the number of O atoms in the molecule increases. The faster OH rate constant in the gas phase implies a shorter overall lifetime the greater the fraction of the species in the gas phase. When C* = 100 µg m-3 virtually all the compound is in the gas phase, and its lifetime is the shortest. The effect of number of oxygen atoms is also apparent; the lifetime decreases by about a factor of 4 going from no O atoms to 8 O atoms.

14.8

See Figures P14.8-1 through P14.8-10.

Fig. P14.8-1

Fig. P14.8-2

Fig. P14.8-3

Fig. P14.8-4

Fig. P14.8-5

Fig. P14.8-6

Fig. P14.8-7

Fig. P14.8-8

Fig. P14.8-9

Fig. P14.8-10

14.9 Initially the volume of WSOC in the dry particle = 261 × 10-24 m3.

Case 1. Particle volume = 1591 × 10-24 m3.

Molar concentration of WSOC = 261 × 10-24 g WSOC × (1000 cm3/L)/[1591 × 10-18 cm3 × 200 g mol-1] = 0.82 M Case 2. Droplet volume = 522 × 10-12 cm3.

Molar concentration of WSOC = 261 × 10-24 g WSOC × (1000 cm3/L)/[522 × 10-12 cm3 × 200 g mol-1] = 2.5 10-6 M.

PS.14.10.pdf (5 pages)

Prob. Sol. 14.11 (page 1 of 2)

Prob. Sol. 14.11 (page 2 of 2)

14.12 (four pages) 1. List of Species and Reactions

Table 1. List of Gas Phase Species

Name Molecular Formula Glyoxal CHOCHO Table 2. List of Aqueous Phase Species

Name Molecular Formula Glyoxal (unhydrated) (CHO)2 Glyoxal (monohydrated) CHOCH(OH)2 Glyoxal (dihydrated) (CH(OH)2)2 Total Glyoxal (undhydrated,

monohydrated, dehydrated) Glyoxal Dimer

Glyoxylic Acid HOCCO2H Oxalic Acid HO2CCO2H Hydroxyl Radical OH Ammonium Ion NH4+ Table 3. List of Aqueous Phase Reactions

Reaction Glyoxal (g) ↔ Glyoxal (aq) (CHO)2 + H2O ↔ CHOCH(OH)2 CHOCH(OH)2 + H2O ↔ (CH(OH)2)2 Gly + Gly ↔ Dimer Gly + NH4+ → Products

Abbreviation Gly_gas

Initial Concentration 1 ppb

Abbreviation Gly_unhydr Gly_hydr1 Gly_hydr2 Gly

Initial Concentration

Dimer GlyAc OxlAc OH NH4

3 x 10-­‐12 M 0.01 M

Equilibrium Constant K_hydr = 4.19 x 105 M atm-­‐1 K_hydr1 = 350 K_hydr2 = 207

Rate Constant

Ref (1) (1) (1)

K_dimer = 0.56 M-­‐1

k_NH4 = 1.3 x 10-­‐7 M-­‐1 s-­‐1 (pH 2) k_NH4 = 2.4 x 10-­‐4 M-­‐1 s-­‐1 (pH 5) k_NH4 = 0.43 M-­‐1 s-­‐1 (pH 7) k_OH = 1.1 x 109 M-­‐1 s-­‐1 k_OX = 3.62 x 108 M-­‐1 s-­‐1

(1) (1)

(1,2) Gly + OH → GlyAc

(2) GlyAc + OH → OxlAc 1. Model Description

A program was created to simulate the chemical reaction of glyoxal in the dilute aqueous phase. The model was used to predict the timescale over which glyoxal is consumed as well as the composition of products produced as a function of time. Four main reactions were chosen for this experiment (Table 3) due to their primary importance. In addition, Henry’s Law is assumed to hold for all time. These equilibria (Table 4, E2a-­‐c) determine glyoxal partitioning into the unhydrated, monohydrate, dihydrate forms.

Table 4. Rate Equations and Equilibria

The production of products and depletion of glyoxal were modeled by rate equations (Table 4, R1 -­‐ R4).

E1 E2a

Equation

E2b

E2c E3 R1 R2

R3 R4 G1

This set of differential equations were evolved through time using the built-­‐in MATLAB function ode45, a non-­‐stiff differential equation solver. The constant concentrations of OH and NH4, given in Table 2, are based on typical levels present in the atmosphere. We initialized the model with a gas-­‐phase concentration of glyoxal of 1 ppb.

Depletion of glyoxal from the gas-­‐phase as products are formed in the aqueous-­‐phase was accounted for. A 1m3 volume of air was chosen as a model system with total pressure of 1 atm. Glyoxal gas was set initially at a partial pressure of 10-­‐9 atm. For one cubic meter of air, it was assumed that there are 10-­‐6 cubic meters of liquid water droplets. Aqueous-­‐phase depletion rate of glyoxal (Table 4, R1) in moles L-­‐1 s-­‐1 was converted to gaseous-­‐phase depletion in moles m-­‐3 of air s-­‐1 by multiplying the conversion factor given above (Table 4, G1).

3. Results

Glyoxal depletion in the gas-­‐phase and the production of aqueous-­‐phase products was simulated in three different pH regimes. The resulting concentrations as a function of time are shown in Figure 1. In the low pH regime, over the timescale of an hour, total glyoxal concentration decreases by three orders of magnitude from its initial concentration. Total glyoxal is more rapidly consumed as pH increases. The concentration of glyoxal dimer, in equilibrium with total glyoxal, decreases rapidly as total glyoxal is depleted by reactions with OH and NH4+.

Figure 1. Aqueous-­‐Phase Glyoxal Kinetics Simulation. Concentrations of products produced for different Glyoxalic acid, Spimulations roduced by the reaction of 4a queous lyoxal and oOver H, itncreases to a m pH regimes. for pH 2 and pH levels wgere carried he timescale of aximum an hour. The -­‐4 concentration of 10 M at 200 s, before decreasing as it reacts with OH to produce oxalic acid. Oxalic simulation for pH 7 is done over the timescale of 30 minutes. acid concentration quickly increases to 10-­‐5 M and then slowly increases to 10-­‐4 M over the course \

Glyoxalic acid, produced by the reaction of aqueous glyoxal and OH, increases to a maximum concentration of 10-­‐4 M at 200 s, before decreasing as it reacts with OH to produce oxalic acid. Oxalic acid concentration quickly increases to 10-­‐5 M and then slowly increases to 10-­‐4 M over the course of an hour. Ammonium products are produced from the reaction of glyoxal and ammonium. The kinetics of this reaction are strongly dependent on pH. The concentration of these products spanned 10-­‐10 M (pH 2) to 10-­‐4 M (pH 7).

References

Ervens, B. and Volkamer, R.: Glyoxal processing by aerosol multiphase chemistry: towards a kinetic modeling framework of secondary organic aerosol formation in aqueous particles, Atmos. Chem. Phys., 10, 8219-­‐8244, doi:10.5194/acp-­‐10-­‐8219-­‐2010, 2010.

Tan, Y., Perri, P., et al: Effects of precursor concentration and acidic sulfate in aqueous glyoxal−oh radical oxidation and implications for secondary organic aerosol. Environ. Sci. Technol. 2009, 43 (21), pp 8105– 8112, doi: 10.1021/es901742

Chapter 15

15.1

Rayleigh scattering is the irreducible minimum of light scattering owing to air molecules along an atmospheric sight path. The amount of scattered light varies as λ-4. Equal amounts of light are scattered into the forward and backward hemispheres.

The Rayleigh scattering coefficient at three wavelengths is: ⎛ 293 ⎞ bsg = 25.9 ⎜ ⎟ p ⎝ T ⎠

λ = 450 nm

⎛ 293 ⎞ = 11.4 ⎜ ⎟ p ⎝ T ⎠

λ = 550 nm

⎛ 293 ⎞ = 5.8 ⎜ ⎟ p ⎝ T ⎠

λ = 650 nm

where T is in K and p is in atmospheres. At λ = 550 nm, bsg = 11.4 × 10-6 m-1

at sea level (T = 288 K; p = 1 atm)

= 10.5 × 10-6 m-1

at 1 km altitude (T = 282 K; p = 0.887 atm)

= 9.3 × 10-6 m-1

at 2 km altitude (T = 275 K; p = 0.785 atm)

= 8.6 × 10-6 m-1

at 3 km altitude (T = 269 K; p = 0.692 atm)

15.2

bsp = eSulfate [Sulfate] + eNitrate [Nitrate] + eOC [Organic Carbon]

+ eSoil [Soil] + eCoarse [Coarse] bap = eBC [Black Carbon] bsp = (3) (20) + (3) (5) + (4) (25) = 175 (m2 g-1) (µg m-3) bap = (10) (75) = 75 (m2 g-1) (µg m-3) bext = bsp + bap

= 250 (m2 g-1) (µg m-3) = 250 × 10-6 m-1 xv =

=

3.912 bext 3.912 = 15.6 km 250 × 10 −6

Note that this calculation assumes dry extinction efficiencies. At ambient RH, these extinction efficiencies can be several times their dry values, with a consequent decrease in xv.

15.3

a. Size distributions n(r) for p = 2 and 5 are given in Figures P15.3-1 and P15.3-2, respectively. b. T given in Figures P15.3-3 for β = 2 and P15.3-4 for β = 5. Note that it can be assumed that Qext(r;λ) = 2.

Chapter 16

16.1

Dry air parcel 5 °C at surface lifted adiabatically to 4.5 km. − R/c p

" p% θ=T$ ' # po & Calculate p at z = 4.5 km.

p ( z ) = po e−z/H

Use H = 7.4 km p ( z ) −4.5/7.4 =e = 0.544 po

With R = 8.314 J mol-1 K -1 c p = 1006 J kg-1 K -1 M air = 28.97 g mol-1

R = 0.285 cp Then

278 = T ( 0.544) T = 234 K

−0.285

16.2

dT T = −Γ dz T' If T ' = Toʹ′

dT T = − dz T' Integrating we get T(z) = To exp (− Γz / Toʹ′ )

16.3

⎛ T ʹ′ ⎞ ⎟ T ' = T = To ⎜ ⎜ T ʹ′ ⎟ ⎝ o ⎠ ⎛ 1 ⎞ ⎟ To ⎜ ⎜ T ʹ′ ⎟ ⎝ o ⎠

Γ

Γ

Λ

Λ

= T ʹ′(Λ −Γ ) / Λ

⎛ Toʹ′ − Λz ⎞ ⎟ Λ ⎟ ʹ′ T o ⎝ ⎠

= To (Λ−Γ )/ Λ ⎜ ⎜ Solving for z

1 ⎤ ⎡ Γ − ⎛ Toʹ′ − Γ ⎞ Λ ⎥ 1 ⎢ ⎟ z = Toʹ′ − ⎜ ⎥ ⎜ T ʹ′Λ ⎟ Λ ⎢ o ⎠ ⎝ ⎥ ⎢⎣ ⎦

Solution is not valid for Γ = Λ .

ΓΛ−Γ Λ Λ

16.4

The pressure of the atmosphere is governed by ∂p = −ρ g ∂z

Since p = ρ RT M a , 1 ∂p 1 ∂ρ 1 ∂T = + p ∂z ρ ∂z T ∂z ρ g gM a =− − p RT

Thus,

1 ∂ρ 1 # ∂T gM a & =− % + ( ρ ∂z T $ ∂z R ' The condition for ∂ρ ∂z = 0 is

∂T gM a =− ∂z R = −3.42 ×10 −2  C m −1

16.5 Lawrence (2005) proposed the following simple formula for the dewpoint of an air parcel: Td = T - 0.2 (100 - RH) where Td and T are in oC and RH is in %. Compare the results of this equation with those of (16.63) for T = 15oC and RH varying from 50 to 100%.

According to the Lawrence (2005) formula for T = 15oC Td = 15 - 0.2 (100 - RH)

(1)

The latent heat of evaporation in (16.63) for T =288 K can be found in Appendix 16.1 and is equal to 2.46 kJ g-1 Using T 0 = 288 K and substituting in (16.63):

8.314 × 2882 ⎛ RH ⎞ ⎛ RH ⎞ Td = 15 + ln⎜ ⎟ = 15 + 15.57 ln⎜ ⎟ 2460 × 18 ⎝ 100 ⎠ ⎝ 100 ⎠

(2)

In both equations (1) and (2) Td is in oC and the RH is from 0 to 100. The dew points estimated from these two equations are compared in the table below. For RH above 55% the prediction of the Lawrence (2005) formula is approximately within 5% of the actual value calculated from the more complicated (16.63).

RH 50 55 60 70 80 90 100

Dew Point (oC) (Eq. 1) 5 6 7 9 11 13 15

Dew Point (oC) (Eq. 2) 4.21 5.69 7.05 9.45 11.5 13.4 15

Difference (%) of (Eq. 2) -18.8 -5.4 0.7 4.7 4.6 2.7 0

16.6 Compare the values predicted by Lawrence (2005) hLCL = (20 + 0.2 T) (100 - RH) where T is in oC with (16.61) for T = 20oC and RH = 70%.

From the Lawrence (2005) equation for T = 20oC and RH = 70% hLCL = (20 + 0.2 × 20) (100 – 70) = 720 m Use of (16.61) requires first the calculation of the temperature T L at the lifting condensation level using (16.60). The parameters in this equation have the following values: T0 = 293 K Mw = 18 g mole-1 Ma = 28.8 g mole-1 pa0 = 1 atm κa = 0.286 The water vapor mixing ratio of the air parcel can be calculated from (16.38) for the initial state of the parcel (pa = pa0 = 1 atm). The water vapor saturation partial pressure at 20°C (using the expression in Table 17.1) is 0.023 atm and for 70% RH es = 0.7 × 0.023 = 0.016 atm Therefore from (16.38) wv = (18 × 0.016) / (28.8 × 1) = 0.01 Substituting these values into (16.60) we find that:

" 18 × es (TL ) % TL = 293$ ' # 28.8 × 0.01×1 &

0.286

= 293 ( 62.5es (TL ))

0.286

Solving numerically the above equation using once more the expression in Table 17.2 we find that T L = 12.9oC. Now from (16.61)

hLCL =

20 − 12.9 = 0.727 km = 727 m 9.76

and we can see that the simple Lawrence (2005) expression is accurate within 1% in this case without requiring the solution of a nonlinear algebraic equation.

16.7 1

) g # dTa &, 2 N =+ % + Γd (. '*Ta $ dz 1

2 ) 0.0098 km s−2 # K K &, =+ + 9.77 %−6.5 (. $ 273 K km km '*

= 0.011 s−1 Period = 1/N = 91 s

€

16.8 a.

Parcel is 35°C at 2 km altitude. When the parcel is lowered, its temperature follows the dry adiabatic lapse rate, Γd = 9.8 K km -1 . At the surface, the temperature of the parcel is its potential temperature, θ = 35 o C + (9.8)(2)

= 54.6 o C

€

b.

The parcel is now raised adiabatically up to its lifting condensation level (LCL).

The vapor pressure of water in the parcel decreases at a rate dictated by the scale height € saturation vapor pressure decreases at the more rapid rate for pressure, H, whereas the given by the water vapor scale height, H v . The two curves meet at the LCL, zLCL =

€

H Hv e ln s0 H − H v fv p0

We can assume that H v = H / 5 , so zLCL =

€

H e ln s0 4 fv p0

€ vapor pressure at the surface temperature es0 = saturation p0 = surface pressure € €

€

fv = number fraction of water vapor in the parcel Given the specific humidity of the air parcel, q = 10 g kg–1.

From

€ fv =

q ε − (ε −1)q

we find, using q = 0.01 kg kg–1, € At 54.6°C, es0 = 118 mm Hg, so

€

fv = 0.016 es0 118 = = 0.155 p0 760

€ using H = 8000 m, Then, €

€

0.155 0.016 = 2000 ln 9.6875 = 4540 m

zLCL = 2000 ln

16.9 The RH of the parcel = 4 g kg-1/14 g kg-1 = 0.29 = 29% a. Follow the dry adiabat on Figure 16.10 to 750 hPa. The temperature of the parcel is ~ 6 oC. The potential temperature is 18 oC. b. At 750 hPa, the dry adiabat crosses the 4 g kg-1 mixing ratio line. The parcel is now saturated. Continue to lift the parcel along the moist adiabat up to 500 hPa. At this altitude its T ~ -20 oC.

( 1 7 .

3 1 ) :

17.2

5 g NaCl To = 20 °C vwater = 200 cm3 e = water vapor partial pressure RH: 0.80 =

e es (To )

From Lowe and Ficke correlation in Table 17.1, es (To) = 23.371 mbar. Thus at 80% RH, e = 18.7 mbar.

In solution:

Molar Concentrations:

es, soln = γ w xw eso (To )

! 1 g $! 1 mol $ nw = 200 cm 3 # 3 &# & " cm %" 18 g %

es, soln =

nw γ w eso (To ) nw + ns

Let γ w →1.0 :

es, soln =

11.1 11.1+ 0.0856

nw = 11.1 mol

ns =

5 g NaCl mol 58.44 g

ns = 0.0856 mol

es, soln = 0.992eso (To )

So the water vapor pressure over solution is es, soln ≈ eso (To ) = 23.2 mbar, which means water will evaporate. To bring eso (To ) to18.7 mbar, the NaCl concentration in the water volume must be increased. Assume there is no change in temperature, and salt does not evaporate.

After equilibrium: xwater ⇒ xnew es, soln = xneweso (To )

18.7 mbar = xnew (23.371 mbar) xnew = 0.8

xnew = 0.8 =

nw nw + ns

0.8 ( nw + ns ) = nw nw (1− 0.8) = 0.8 ( ns )

nw =

0.8 ( ns ) 0.2

nw = 0.34 mol vnew =

! 18 g $! cm 3 $ nw M = ( 0.34 mol) # & &# " mol %" 1 g % ρw vnew = 6.16 cm 3

for RH = 0.95 =

pw , p (T0 ) o

xnew =

nw = 0.95 nw + n s

nw (1− 0.95) = 0.95ns

nw = 1.625 mol vnew =

! 18 g $! cm 3 $ nw M = (1.625 mol) # & &# " mol %" 1 g % pw vnew = 29.3 cm 3

17.3

Grandma and Grandpa This is a very good question. You are right, there has been an impressive improvement in the gadgets used by meteorologists, but still we cannot predict rain that well.

Think of clouds as a lot of tiny water drops floating in air. There are probably a couple of million of drops in a volume of two liters (equal to the volume of this soda bottle). For the formation of a rain drop, we need a really big cloud drop that will collide with smaller ones, grow and eventually become a rain drop that will fall from the cloud.

If one out of the million cloud drops is big enough it will become a rain drop. It is like a lottery where sometimes there is the right droplet and then we have a winner (rain formation), sometimes there is no lucky drop (no rain). Think of each small volume of air, as having its own lottery, for the “lucky drop”. If there are enough winners in a given cloud we get rain, if there are only a few we get drizzle, and if there are none we get these clouds that stay around for a while and then disappear without raining.

Even if scientists know that there are clouds and they also have other information about it, they do not know about these one out of a million drops. So the best that they an do is calculate the odds of rain. Meteorologists tell you that there is, for example, a 80% chance of rain. They have estimated based on the radar information that there is a four out of five chance that there are going to be enough “lucky drops” to form rain. These are pretty good odds and several weather predictions translate them into “it will rain tomorrow” instead of giving the odds. What they do not tell us is that there is a one out of five chance that it will not rain.

17.5

How long does it take for the particle to reach a diameter of 5 µm? Assume

α c = αT = 1.0 Dpo = 100 nm of (NH4)2SO4

T∞ = 283 K

S = 1.003 3

4 # 0.10 ×10 −4 & π% ( (1.769 ) 3 $ 2 ' ns = 132 −18 = 7.017 ×10 mol Mw = 18

Moles of (NH4)2SO4

ρw = 1

R = 8.314 × 107 0.167+3.67×10−4 T , ) ! $ 273.15 . ×10 3 ×10 7 lv = +2.5# & " % T .+*

Latent heat of evaporation

2 4 c p = #$4.175 +1.3×10 −5 (T − 308) +1.6 ×10 −8 (T − 308) %& ×10 7 Heat capacity of pure water

σ w = #$0.0761−1.55 ×10 −4 (T − 273)%& ×10 7 ×10 −4 ρa =

Surface tension for pure water

1.0133×10 6 RT

Density of air

M a = 28.9

Molecular weight of air 1.94

0.211 ! T $ D= # & p " 273 %

Diffusivity of water vapor

k = 10 −3 ( 4.39 + 0.071T ) ×10 5

Thermal conductivity of air

Non-continuum corrections for thermal conductivity and diffusivity.

D

Dv ( Dp ) = 1+

2D α c Dpo

2π M w RT

k

ka ( D p ) = 1+

2k αT Dpo ρ a c p

2π M a RT

Differential equation for diameter growth with time,

% " 6ns M w ' 4M wσ w $ − S − exp 3 $ RT ρ D πρ w ( Dpo ) '& w p o dDp 1 # = + dt ( ( +D ρ w RT - + lv ρ w * lv M w −1- po * * 4 pD D M - 4k D T ) RT , v ( p) w, a( p ) ) Solve this equation numerically. For Dp = 5 µm t = 31.45 s.

17.6

Set D = Dv and K = Ka to neglect the non-continuum effect. For Dp = 5 µm t = 17.6 s. In the case where non-continuum effects are ignored, the growth time for the particle to reach 5 µm is considerably reduced (by 56%). The diffusivity of vapor to the particle as well as the thermal conductivity of air are greatly reduced for very small particles. If this is ignored, the rate of growth of the particle is over-predicted.

18.3

Procedure same as Problems 18.1 and 18.2, however, we now have (1) Sunny Day, wind speed = 4 m/s ⇒ Stability Class = B-class

Thus

{

}

σy (x) = exp − 1.634 + 1.035l nx − 0.0096 (l nx ) ; 2

{

σy (x) = exp − 1.999 + 0.8752 l nx − 0.0136 (l nx )

2

}

(2) Inversion Layer H = 300 m (3) q = 1.3889 × 108 μg/s (4) h = 100 m (5) Gaussian Plume Eqn. (total reflection): ⎡1 ∞ 2q ⎛ nπz ⎞ cos⎜ < c(x, y, z ) >= ⎟ ⎢ + nΣ 1/ 2 0 = (2π ) u σ y H ⎣ 2 ⎝ H ⎠

(A) 2

⎛ nπh ⎞ ⎧ ⎛ nπ ⎞ σ ⎫⎤ ⎛⎜ y ⎞⎟ × cos⎜ ⎟exp⎨− ⎜ ⎟ ⎬⎥ exp − 2 ⎝ H ⎠ ⎩ ⎝ H ⎠ 2 ⎭⎦ ⎜⎝ 2σ y ⎟⎠ 2 z

2

Results are presented in Figures P18.3-1 – P18.3-3. Figure P18.3-4 is presented to show the influence of the inversion layer. Various inversion layer heights (H= 150, 300, ∞ m) have been used. This figure shows the lower the inversion height, the higher the downwind concentration.

18.15 Thickness of the Ekman layer = 0.2 u* /f Mean vertical velocity of the cloud = u* Time needed to travel to top of layer = 0.2/f This is the time needed for a cloud to become large enough for crosswind shear effects to become important.

Mean wind velocity = U Distant traveled by cloud = 0.2U/f To estimate this distance for a cloud released in midlatitudes, let U = 20 km hr1. f = 2Ωsin β

β = 30º f=

π 12

h -1

Thus, distance traveled ≅ 15 km.

19.2

Dp = 1 μm

ρ p = 1 g cm-3 desert surface @ 298 K u* = 10 m s-1

neutral stability vd =

ra =

rb =

1 ra + rb + ra rbvs

+ vs

⎛ z⎞ 1 ln ⎜⎜ ⎟⎟ κu* ⎝ zo ⎠

(19.7)

neutral stability 1

2 ⎡ ⎛ St ⎞ ⎤ −γ ⎟⎟ ⎥ R1 3u * ⎢Sc + ⎜⎜ ⎢⎣ ⎝ α + St ⎠ ⎥⎦

(19.27)

where the interception term is neglected over a smooth surface. The terminal settling velocity is vs =

ρ p D p2 gCc 18μ

R1 = exp (− St½)

and

Sc =

(19.18) (19.26)

ν D

For smooth surface, St =

vsu*2 gν

The following parameter values apply (Table 19.2):

γ = 0.54, α = 50, zo = 0.04 m

(19.23)

From Figure 9.6 we estimate vs = 20 cm h-1 = 0.0056 cm s-1 at 298 K for Dp = 1 μm.

From Table 9.5, we find D = 2.74 × 10-7 cm2 s-1 for Dp = 1 μm. The kinematic viscosity of air (see Table A.7) is

ν=

μ ρ

=

1.72 × 10−4 0.0013

g cm −1 s −1 g cm- 3

= 0.13 cm2 s-1 Thus

ν

0.13 cm 2 s −1 Sc = = = 4.74 × 105 −7 2 −1 D 2.74 × 10 cm s

(

)( )(

) = 44 )

0.0056 cm s -1 1000 cm s -1 vsu*2 St = = gν 980 cm s - 2 0.13 cm 2 s -1

(

Let us take z = 10 m in the formula for ra. Then 1 ⎛ 10 ⎞ ln ⎜ ⎟ 3 -1 (0.4) 10 cm s ⎝ 0.04 ⎠

ra =

(

=

)

1 ln (250) 400

= 0.0138 s cm-1

Sc − = (4.74 × 105 )

−0.54

γ

2

2

= 0.00086

⎛ St ⎞ ⎛ 44 ⎞ ⎜⎜ ⎟⎟ = ⎜ ⎟ = 0.22 ⎝ α + St ⎠ ⎝ 94 ⎠ R1 = exp (− St1/2 )

= exp (− 6.63) = 0.0013

2

Then rb =

1 (3000 cm s ) [0.00086 + 0.22] (0.0013) −1

= 0.86 s cm-1 The overall dry deposition velocity is vd =

=

1 ra + rb + ra rbvs

+ vs

1 + 0.0056 0.0138 + 0.86 + (0.0138)(0.86)(0.0056)

= 1.15 cm s-1 We see that the predominant resistance contributions to vd is rb, and that the predominant contribution to rb comes from impaction of particles on the surface.

19.3

Dp = 0.01 – 10.0 μm u* = 1 m s-1

Gases; seasonal category 2 in Table 19.2 (A = 2.0 mm; α = 1.2; γ = 0.54) 2 2 ⎡ ⎛ St ⎞ 1 1 ⎛ Dp ⎞ ⎤ −γ ⎟⎟ + ⎜⎜ ⎟ ⎥ R1 = 3u* ⎢Sc + ⎜⎜ rb 2 ⎝ A ⎟⎠ ⎥ ⎢⎣ ⎝ α + St ⎠ ⎦

R1 = exp (− St½) Sc =

ν D

St =

vsu* gA

From Table 9.5 (ν = 0.13 cm2 s-1, see Problem 19.2 solution) Dp, μm

0.01

0.1

1.0

10.0

D, cm2 s-1

5.24 x 10-4

6.82 x 10-6

2.74 x 10-7

2.38 x 10-8

Sc

248

1.91 x 104

4.74 x 105

5.46 x 106

vs

6.74 x 10-6

8.68 x 10-5

3.5 x 10-3

3.05 x 10-1

St

3.74 x 10-7

4.43 x 10-6

1.79 x 10-4

1.56 x 10-2

R1

1.0

1.0

1.0

1.0

Sc − γ

0.05

0.0049

0.00086

0.0002

~0

~0

~0

0.00017

~0

~0

~0

~0

15

1.47

0.26

0.11

⎛ St ⎞ ⎜⎜ ⎟⎟ ⎝ α + St ⎠

2

⎛D ⎞ 1/2 ⎜⎜ p ⎟⎟ ⎝ A ⎠

2

1 ( cm s -1 ) rb

19.4

Dry deposition of O3 as a function of windspeed (1 – 10 m s-1) at 298 K. O3 is a slightly soluble gas, so we use (19.48), rcw , O 3 =

1 ~ k L , o3 H o 3

For kL we can use either the Liss and Merlivat or Wanninkhof formulas. Let’s use the latter, kL, o3 =

0.31u102 S o132 3600

(cm s-1)

(19.46)

So 3 =

Sc co 2 (293 K ) Sc o 3 (298 K )

To compute Sco 3 we need the kinematic viscosity of water and the molecular diffusivity of O3 in water at 298 K. For water

ν = 0.01 cm2 s-1 Do 3 = 1.75 x 10-5 cm2 s-1 Sc o 3 = ν

Do 3

= 571

Sc co 2 (293 K ) = 660 So 3 = 1.16 ~ H o 3 = (0.0113) (24.44) = 0.28 u10, m s-1

1

5

10

k L , o 3 , cm s -1

9.3 x 10-5

0.0023

9.3 x 10-3

1552

384

rcw , o3 , s cm −1 38402

19.5

Surface resistance for dry deposition of SO2 to a flat deciduous forest at 298 K under midsummer conditions with lush vegetation. rc

=

⎛ 1 1 1 1 ⎜ + + + ⎜r + r rlu rdc + rcl rac + rgs m ⎝ st

⎞ ⎟ ⎟ ⎠

−1

(19.50)

Use Table 19.3. Land use type is (4), and Seasonal category is 1:

rst

=

rj

=

70

rlu

=

2000

rac

=

2000

rgsS

=

500

rcl

=

2000

rj

2 ⎡ ⎞⎤ ⎛ 200 ⎞ ⎛ 400 ⎟⎟⎥ ⎟⎟ ⎜⎜ ⎢1 + ⎜⎜ ⎢⎣ ⎝ G + 0.1 ⎠ ⎝ Ts (40 − Ts ) ⎠⎥⎦

rstSO 2 + rmSO 2 = 1.89 rst +

1 * 3.3 × 10 − 4 H SO 2

* H SO = 105 2

⎛ ⎞ 1 ⎟ = rlu rluSO 2 = rlu ⎜⎜ − 5 5 ⎟ ⎝ (10 ) (10 ) ⎠

rdc

⎛ 1000 ⎞ ⎟ = 100 ⎜⎜1 + G + 10 ⎟⎠ ⎝

(Note: θ = 0°)

Then for SO2: rst

= 70 [1 + 0.04 (1.07)] = 2.99

rst + rm = (1.89) (2.99) + 0.03 = 5.68

(19.51)

(Table 19.4)

rc

rlu

= 2000

rac

=

2000

rdc

= 150

rgs

=

500

rcl

= 2000

=

1 1 1 ⎞ ⎛ 1 + + + ⎜ ⎟ 2000 2150 2500 ⎠ ⎝ 5.68

≅ 5.68 s cm-1

−1

19.6

Neutral conditions ra =

1 ⎛ z ⎞ ln ⎜ ⎟ ku* ⎜⎝ z o ⎟⎠

z = 10 m, zo = 1, k = 0.4, u* = 1 m s-1. ra = 5.76 s m-1 DH2O = 0.239 cm2 s-1, DSO2 = 0.126 cm2 s-1 Sc =

v 0.1 = = 0.79 D 0.126

rb =

5Sc 2/3 = 4.276 s m −1 u*

vd =

1 = 0.1 m s −1 ra + rb

When rc = 0,

If rc = 30 s m-1 vd =

1 = 0.025 m s −1 ra + rb + rc

rc = 200 s m-1 vd = 4.76 × 10-3 m s-1

Problem 20.2 (a) At 298 K the Henry’s law coefficient of HCl is HHCl=2 M atm-1 and also based on Table 7.3 ΔH=-4000 cal mol-1. Using equation (7.5) we find that: HHCl (283 K) = 2.8 M atm-1.

The effective Henry’s law coefficient for HCl is given by:

⎡ 1.74 × 10 6 ⎤ 4.9 × 10 6 * M atm-1 H HCl = H HCl ⎢1 + ⎥ ≈ + + [H ] ⎦ [H ] ⎣

(1)

using the data in Table 7A.1 for the dissociation of HCl in water. * * For pH=2 H HCl =4.9x108 M atm-1, pH=4 H HCl =4.9x1010 M atm-1, and for pH=6 * =4.9x1012 M atm-1.

H HCl

For these very high values of the effective Henry’s law coefficient (see Figure 7.3) all the HCl will be in the aqueous phase for the atmospheric pH range of 2-6. (b) The very high solubility of HCl in water simplifies the situation and Equation (20.25) can be used for the calculation of the scavenging coefficient:

Λ=

6 × 10 −3 p 0 K c = 0.6 h-1 D pU t

(2)

Using the example in page 871 the fraction of the HCl that is removed by the rain is given by:

C HCl = e −0.6×2 = 0.3 o C HCl

(4)

So 70% of the HCl will be removed by the rain. (c) 1 ppb of HCl is equivalent to 1.5 µg m-3. Therefore the removed concentration will be equal to 0.7x1.5x40 = 42 µg m-3. For the 2000 m, the deposited amount will be equal to 2000x42=84 mg m-2. The lake has a surface area of 106 m2 so we will have the deposition of 84x106 mg=84 kg of HCl.

20.4 For a more accurate determination of the scavenging rate of a gas, we need to include in the calculation the full droplet size spectrum. Show that if the droplet size distribution is given by the function N(Dp), the rate of removal of an irreversibly soluble gas is given by

Wbc =

∞

2 p

∫ K πD C N (D )dD 0

c

g

p

p

Using this, calculate the scavenging coefficient for the Marshall-Palmer distribution (17.99) N ( D p ) = 0.08 exp − 41D p p0−0.21

(

)

for p0 = 1 mm hr-1 and 25 mm hr-1.

The rate of transfer of an irreversibly scavenged soluble gas to the surface of a raindrop is given by (20.13): (1)

Wt = K c C g Recall that Wt has units of moles of gas transferred per unit surface area per time.

If the size distribution of droplets is given by N(Dp) then the surface area of droplets per volume of air having diameters in the range Dp to Dp+dDp is given by (see section 8.1.4 and Equation 8.4): π Dp2 N(Dp) dDp The rate of mass transfer per volume of air to droplets having diameters in the range Dp to Dp+dDp is equal to: π Dp2 N(Dp) Kc Cg dDp To calculate the overall rate of mass transfer (removal) to all droplets we need to integrate the above expression from zero to infinity and therefore:

Wbc =

∞

2 p

∫ K πD C N (D )dD 0

c

g

p

(2)

p

If the raindrop spectrum has the Marshall-Palmer distribution:

(

N ( D p ) = 0.08 exp − 41D p p0−0.21 then combining (2) and (3)

)

(3)

Wbc = 0.08 K c πD p2 C g exp(− 41D p p 0−0.21 )dD p

∫

∞

(4)

0

The scavenging coefficient is then Λ=

(

)

∞ Wbc = 0.08 K c πD p2 exp − 41D p p 0−0.21 dD p Cg 0

∫

with Kc given by (20.12).

The integral in Equation (5) can be calculated numerically resulting in: Λ= 0.43 h-1 for p0 = 1 mm hr-1 and Λ= 2.74 h-1 for p0=25 mm hr-1

(5)

Problem 20.8 (a) The number concentration of raindrops, N , is given by: σ

6 × 10 −3 p0 Nσ = π D 3pU t

(1)

therefore the concentration of rainwater, L, below the cloud is equal to:

L = Nσ

10 −3 p0 1 π D 3p = 6 Ut

(2)

For the conditions of the problem we find that L=0.28 g m-3. (b) For pH=4 the reaction of S(IV) with H2O2 dominates the aqueous-phase sulfate production. For 1 ppb of SO2 and 1 ppb of H2O2 the corresponding reaction rate is 300 µM h-1. Therefore for 5 ppb SO2 the rate is 5 times higher and equal to 1500 µM h-1. (c) The droplets move with speed equal to 300 cm s-1 (Table 20.1) and therefore need 0.18 h to cover the 2 km from the cloud base to the ground. Assuming that the SO2 and H2O2 concentrations remain constant and using the reaction rate calculated in (b) the produced S(VI) will be equal to 300x0.18=270 µM. As the initial concentration is 500 µM, the final one will be 770 µM. (d) Assuming that the below-cloud air remains well-mixed during the rain and using the expressions derived in Section 7.4 for the aqueous-phase reaction rates:

dξ SO 2 dt

* = −3.6 × 10 −12 LRTk ʹ′H H 2O2 H S(IV ) ξ SO 2 ξ H 2O2

(3)

where we have assumed that the reactants are always in equilibrium with the falling rainwater. In this case k´ is the effective reaction constant shown in Figure 7.17.

Let us call x the mixing ratio (in ppb) of SO2 (and therefore H2O2 as the reaction is 1:1) that have reacted at a given time. In this case at any time:

ξ SO = (5 − x) and ξ H O = 1 − x

(4)

* A = 3.6 × 10−12 LRTk ʹ′H H 2 O 2 HS(IV )

(5)

2

2

2

Defining as

then the mass balance for SO2 can be written as:

dx = A(5 − x)(1 − x) dt This can be solved by separation of variables giving:

(6)

1 [ln(5 − x) − ln(1 − x)] = At +Q 4

(7)

where Q is the integration constant. For t=0, x=0 therefore Q=(1/4)ln5. Substituting this into (7) and solving for x:

x(t ) = 1 −

4 5e

4 At

−1

(8)

Finally substituting (8) back to (4):

ξ SO (t ) = 2

20e4 At 5e4 At − 1

(9)

(e) An equation similar to (3) can be derived for the hydrogen peroxide mixing ratio:

dξH 2 O 2 dt

* = −3.6 × 10 −12 LRTk ʹ′H H 2 O 2 H S(IV )ξSO 2 ξ H 2 O2

(10)

However, there is no need to solve it as we can calculate the evolution of the hydrogen peroxide concentration combining (8) and (4) to obtain:

ξ H O (t ) = 2

2

4 5e

4 At

−1

(11)

Please note that the other pollutants in the problem (NOx, VOCs and O3) have very low solubility in water and their chemical consumption in the aqueous phase is very slow (the pH is too low for the reaction of O3 with S(IV) to proceed with an appreciable rate). As a result their concentrations will remain constant during the rain event for all practical purposes. Of course, this assumes that there are no sources emitting NOx and VOCs in the area and also no photochemistry. (f) The concentration of sulfate produced in the rainwater as a function of time is equal to the x(t) given by (8). The amount of sulfate deposited will be equal to the sulfate in the cloud that is rained-out plus the amount that is produced in the falling cloud droplets. There is no information about the below-cloud sulfate concentration so we assume that the sulfate washed-out is negligible compared to the other two.

The deposition rate due to rainout, Wrainout, will be constant with time and equal to the rainfall rate times the concentration of sulfate. Considering 1 m2 of land the rainfall rate is 4x10-3 m3 h-1 m-2. The concentration of sulfate is 500x10-6 M=0.5 mol m-3=48 g m-3. Therefore: Wrainout(t)=4x10-3x48=192 mg h-1 m-2

The produced sulfate deposition rate will be equal to the production rate in the 2 km column of air between the cloud base and the ground. Considering that 1 ppb of sulfate corresponds to approximately 4 µg m-3 and using (6) the production rate, RS(IV) will be equal to:

RS(VI) = 4 A(5 − x)(1 − x)

(in µg m-3 h-1)

For the 2 km of air below the cloud the corresponding deposition rate due to reaction is:

Wreaction = 8 A(5 − x)(1 − x) (in mg h-1 m-2)

Problem 20.11 The ion balance for this system is: 2[SO42-]+[HSO4-]+2[SO32-]+[HSO3-]+[HCO3-]+2[CO32-]+[OH-]=[H+] with all the concentrations in M. The concentration of sulfate can be calculated as: [Sulfate (in µM)]= (Sulfate in µg m-3)/ {L (in g m-3) x 10-3 x 98} and the corresponding aqueous-phase concentrations for the conditions of the table are: Table P20.11-1 Sulfate (in µM) L (in g m-3) Sulfate (µg m-3)

0.1

0.5

0.04

4.1

0.82 0.16

0.2

20.2

4.1

0.82

1.0

100.2 20.4

4.1

2.5

The sulfate concentrations for all cases are higher than 0.1 µM so ions with much lower concentrations can be neglected in the ionic balance. We also know that the 340 ppm of CO2 on their own can lower the pH of the cloud water to approximately 5.6.

There are only acids in the system so we do know that the pHs that will be calculated will all be lower than 5.6. Also given that this natural cloud water the pH will be higher than 2. With these constraints it is easy to show that the concentrations of HSO4-, SO32-, CO32-, and OH- will all be much lower than that of sulfate and can be neglected. As a result the ion balance can be simplified to: 2[SO42-]+[HSO3-]+[HCO3-]=[H+] or using the expressions developed in Chapter 6: 2[SO42-] [H+] + HSO2 Ks1 pSO2 + HCO2 Kc1 pCO2 = [H+]2 Using pCO2=330x10-6 atm and pSO2=10-10 atm and solving the corresponding quadratic equation for the [H+] we find the following pH values: Table P20.11-2 Cloud pH L (in g m-3) Sulfate (µg m-3) 0.1 0.5 2.5 0.04

5.1 5.4 5.5

0.2

4.4 5.1 5.4

1.0

3.7 4.4 5.1

These results suggest that the cloud pH in the relatively pristine atmosphere can vary significantly (from 4.5 to5.5 in the above cases) just because of the variability of the of the sulphur cycle along (sulfate levels of 0.04-0.2 µg m-3). See also Charlson and Rodhe (Nature, 295, 683-685, 1982).

20.14 See Oppenheimer (1983b).

Chapter 21

21.1 ug = −

1 ∂p fρ ∂y

f = 2 Ω sin 40° = 9.35 × 10-5 s-1 Pressure and density at 5000 m from Table A.8: p = 54,050 Pa

ρ = 0.74 kg m-3

For ∂p / ∂y − 400 Pa Δp = – 200 × 10-5 Pa m-1 = 5 Δy 2 × 10 m

Then 200 × 10−5 Pa m -1 ug = 9.35 × 10− 5 s −1 0.74 kg m -3

(

)(

)

= 29 m s-1

21.2

τ H 2O =

mH 2 O DH 2 O

DH 2 O = 0.2 cm/day × 4 πR 2 × ρ H 2 O ≅ 1.0 × 1015 kg day-1 e

τH 2O ≅

1.3 × 1016 kg ≅ 13 days 1.0 × 1015 kg/day

This value is reasonably close to that estimated by the total fluxes in Table 21.2 of 10 days.

21.3

Assume Horizontal wind speed

~ 10 m s-1

Horizontal length scale

~ 40,000 km (at equator)

Time scale =

40,000 km = 4 × 106 s = 46 days −1 0.01 km s

If we take a length scale of 20,000 km, reflecting mid-latitudes, then the time scale is 23 days.

21.4

∂⎛ ∂u ⎞ 1 ∂p + fv = 0 ⎜ KM ⎟− ∂z ⎝ ∂z ⎠ ρ ∂x

(A)

∂⎛ ∂v ⎞ 1 ∂p − fu = 0 ⎜ KM ⎟− ∂z ⎝ ∂z ⎠ ρ ∂y

(B)

−

1 ∂p = fuG ρ ∂y

Since x-axis is in direction of geostrophic wind, which is perpendicular to the pressure gradient, ∂p =0 ∂x KM

∂ 2u + fv = 0 ∂z 2

(C)

KM

∂ 2v + f (uG − u ) = 0 ∂z 2

(D)

KM

∂ 4u ∂ 2v + f =0 ∂z 4 ∂z 2

(E)

Take ∂ 2 ∂z 2 of (C),

Combining (D) and (E), 2

∂ 4u ⎛ f ⎞ ⎟ (u − uG ) = 0 +⎜ ∂z 4 ⎜⎝ K M ⎟⎠

21.5

The outer boundary of the planetary boundary layer can be defined by 0 = uG e − az G sin azG or azG = π.

(F)

22.2

Worldwide oil production rate = 2.28 × 1015 gC/yr Worldwide gas production rate = 9.33 × 1014 gC/yr Total = 3.212 x 1015 gC/yr.

22.3 Emissions

= 1.0 kg day-1 × 365 days yr-1 × 7 × 109 = 2555 × 109 kg yr-1 = 2.555 × 1015 g yr-1 = 2.555 Pg CO2 yr-1

Current fossil fuel emissions (Figure 22.5) are ~ 9 Pg C yr-1 = 33 Pg CO2 yr–1.

The CO2 that is exhaled arises from carbon in plants and in animals, of which the latter is originally derived from plants. That carbon was originally removed from the atmosphere by photosynthesis, so the photosynthesis-consumption-exhale cycle is a closed cycle with no net addition of carbon to the atmosphere.

22.4 CH4 + 2O2 → CO2 + H2O MW CH4 = 16

19.9 g CH 4 = 1.24 mol CH4 16 g/mol a.

Complete combustion gives 1.24 mol CO2 (× 44 g/mol) = 54.6 g CO2

b.

44 g Coal to produce 50.3 MJ Coal (0.85 C by mass) 37.4 g C 3.12 mol C

→

CO2 3.12 mol CO2 || 137.3 g CO2

22.5 a.

44 1 ppm CO2 = 2.1 Pg C = 2.1× 12 = 7.7 Pg CO2

10 ppm CO2 = 21 Pg C = 77 Pg CO2 15 mol 77 Pg CO2 × 44 g = 1.75 × 10 mol CO2

b.

5 Gt C yr—1 × 8 yr = 40 Pg C Fraction added to atmosphere

21 Pg C = 52.5% 40 Pg C

22.6 Global Tropospheric Balance on CO Sources (Tg yr–1) CH4 oxidation Biogenic oxidation Biomass burning Industrial

Global 800 380 700 900

NH 400 190 0 900 1490

SH 400 190 700 0 1290

400

267

133

Sinks (Tg yr–1) OH reaction Soil uptake NH Balance:

kSH/NHQSH − kNH/SHQNH + PNH − SNH = 0

SH Balance:

kNH/SHQNH − kSH/NHQSH + PSH − SSH = 0 kSH/NH = kNH/SH = 1 yr −1

OH oxidation sink: SNH = kOH+COQNH

SSH = kOH+COQSH kOH+CO = 4.115 yr −1 (@277 K)

Balances become: QSH − QNH +1223− 4.115QNH = 0

QNH − QSH +1157 − 4.115QSH = 0 Solve to determine QNH and QSH

QSH = 282 Tg

QNH = 294 Tg

Now determine the CO mixing ratios in NH and SH. These are the mole fractions (expressed as mixing ratios) in each hemisphere. The total moles of air in the troposphere can be estimated as the product of the fraction of air in the troposphere and the estimated total moles in the atmosphere.

Total moles of air in the atmosphere (Chapter 1) ≅ 1.62 × 1020 moles Fraction of air in the troposphere ≅ 0.85 Total moles of air in the troposphere = 0.85 × 1.62 × 1020 = 1.38 × 1020

So

Total moles of air in each hemisphere = 0.69 × 1020

Now we calculate the mole fractions of CO in each hemisphere. The total moles of CO in the NH is estimated to be "

%

CO ' = 10.5 ×10 moles CO in NH (294 ×10 g) $# mole 28 g & 12

12

and the mole fraction of CO in the NH is

10.5 ×1012 moles CO = 1.52 ×10 −7 = 0.152 ×10 −6 = 0.152 ppm 6.9 ×1019 moles air = 152 ppb In the SH, the CO mixing ratio is estimated as 146 ppb.

Chapter 23

23.1

Global mean precipitation rate = 1m/year Determine corresponding atmospheric energy input Surface area of earth = 4π Re2 = 4π (6.4 × 106 m)2 = 5.14 × 1014 m2 Mass of water precipitated per year = (5.14 × 1014 m3) (1000 kg/m3) = 5.14 × 1017 kg Heat released upon condensation = (5.14 × 1017 kg) (2.5 × 106 J/kg) = 12.85 × 1023 J where we have used the latent heat of vaporization at 273 K (Table A.9) Now 1W = 1J/s The energy of 12.85 × 1023 J is released over 1 year. The number of seconds per year is 86,400 × 365 = 31,536,000. So the energy released per second is 4.07 × 1016 W. Dividing this by the area of the earth, 4.07 × 1016 W ≅ 80 W 2 m 5.14 × 1014 m 2

23.3 Earth’s albedo = 0.30 Fraction of stratocumulus clouds = 0.20 Average albedo of stratocumulus clouds = 0.60 so 0.30 = 0.20 × 0.60 + 0.80x where x = albedo of non-stratocumulus fraction of Earth. x = 0.225 Doubling of CO2 = 3.7 W m–2. By how much would Earth’s albedo need to change to reflect 3.7 W m–2 back to space?

Current net input and output = 239 W/m2 Fractional change due to 2 × CO2 =

3.7 = 0.0155 239

The overall change in Earth’s albedo needed to offset 2 × CO2 then = 0.0155 × 0.3 = 0.00465

€

This change in albedo is achieved by an increase Δy in the global fractional coverage of stratocumulus clouds.

0.00465 = Δy × 0.60 Δy = 0.00775

Thus, the global coverage would need to increase from 20% to 20.775%, an approximate 5% increase in the global coverage of stratocumulus. €

23.4

Taken together, all of the Earth’s clouds exert a cooling effect on the planet.

Low (stratus) clouds lie within ~2 km of the surface and resemble a blanket covering thousands of kilometers of sky. These clouds block sunlight from reaching the surface, acting to cool the Earth. Their temperature is close to that of the surface, so they radiate at approximately the surface temperature. Thus, the outgoing longwave radiation from such clouds is more or less the same as from the surface itself.

High (cirrus) clouds are thin, wispy ice clouds located high above Earth’s surface.

Cirrus clouds let much of the sunlight pass through them and so do not exert much of an effect in preventing sunlight from reaching the Earth’s surface. On the other hand, these clouds can intercept longwave outgoing radiation from the surface, just as greenhouse gases do, preventing escape of some of that radiation to space. Cirrus clouds thus exert a net warming effect on climate.

23.5

Using the given values of ΔF, Ta, and Ts, and solving for Δε gives Δε = 0.019. At the new value of ε, the new surface temperature is Ts = 289.5 K, an increase of 1.2 K over the present day temperature of 288.3 K. This estimate does not take into account climate feedback processes.

23.6

For solution see:

Stocker, T.F. (2013) The closing door of climate targets, Science 339, 280-282.

23.7

Single layer radiative model See equations (23.3) and (23.4)

Ta = 245 K Te = 255 K 1

$ 2 '4 Ts = & T % 2 − ε )( e

€

0.7

0.8

0.9

Ts (K )

284.1

289.7

296.1

σTs4 ( W/m 2 )

369.1

399.1

435.4

ε

0.7

0.8

0.9

σTs4 − σTe4 ( W/m 2 )

129.3

159.3

195.6

ε

€ σ = 5.67 ×10 −8 Wm −2 K −4 σT 4 =€239.8 W/m 2 e

€

€

23.8

a. Solve (23.18) using an integrating factor to obtain the given equation. b. For ΔF = F ΔT (t ) = ( F α )(1− e−t τ )

€

c. For the given ΔF (t ) €

€

. bτ & t ) −t τ 0 < t ≤ t1 00 ( −1+ e + * α 'τ ΔT (t ) = / 0 bτ &( t1 − e−( t−t1 ) τ + e−t τ )+ t ≥ t 1 01 α ' τ *

As t → ∞

€

ΔT (t ) =

€

€

bt1 α

23.9

ΔT (t ) =

ΔF 1− e−t τ ) ( α

As t → ∞ ΔT = ΔF α

€ Given ΔT = 3 K and ΔF = 3.7 W m –2

€

€ €

Then

α = 1.23 W m –2 K –1 € €

τ=

C 1 GJ m –2 K –1 = α 1.23 W m –2 K –1 = 25.8 years

€

23.10

a.

ΔF = 5.3 ln 1.01 = 0.0527 W m –2 Then

€

ΔF = bt ΔF = 0.0527 W m –2

t = 1 year

b = 0.0527 W m –2 yr –1 b. For the CO2 level to increase from 400 ppm to 560 ppm

€

ΔF = 5.3 ln ( 560 400 ) = 1.78 W m −2

Then

ΔF = bt –2

1.78 W m = 0.0527 W m –2 yr –1 × t t = 33.9 yr

€

24.2

In Problem 24.1 we used the mass scattering efficiency for a dry (NH4)2 SO4 particle of diameter 0.4 µm. Here we are interested in the particle at 80% RH. Figure 24.7 can be used. The abscissa of Figure 24.7 is the dry radius of the particle (three different species), and the ordinate is the mass scattering efficiency of the wet particle per mass of sulfate. The dry radius of interest here is 0.2 µm. There is not a tick mark on the x-axis at 0.2 µm dry radius, so we can either eyeball the 0.2 point or actually calculate the value on the x-axis ourselves. Let us do the latter.

Volume of 0.4 µm diameter particle = 0.033 × 10-18 m3 Density of dry (NH4)2 SO4

= 1130 kg m-3

Mass of (NH4)2 SO4 / particle

= (1130 kg m-3) (0.033 × 10-18 m3) = 37.3 × 10-15 g

Molecular weight of sulfate

= 96 g mol-1

Mol of SO 24 − / particle

= 3.9 × 10-16

This quantifies the value of the abscissa.

At this value, the wet mass scattering efficiency at 80% RH ≅ 15 m2 g-1 SO 24 − Now,

τ = αmH But we need to express m as mass of sulfate only since α has units per mass of sulfate.

For (NH4)2 SO4,

mass of SO 24 mass of (NH 4 ) 2 SO 4

so 1 µg m-3 of (NH4)2 SO4 =

=

96 132

96 µg m-3 of SO 24 − 132

Thus, ⎞ ⎛ 96 μg m -3 ⎟⎟ (104 m) ⎝ 132 ⎠

τ = (15 m2 g-1) ⎜ = 0.11

24.3 ΔF = −

⎡ 1 2 Rs ⎛ 1 ⎞⎤ 2 FoT 2a (1 − Ac) ω β τ ⎢(1 − Rs ) − ⎜ − 1⎟ ⎥ 2 β ⎝ω ⎠⎦ ⎣

From Table 24.2: Fo

=

1370 W m-2

1 − Ac

=

0.4

Ta

=

0.76

1 − Rs

=

0.85

β

=

0.29

Consider 0 ≤ τ ≤ 1.0 at ω = 1.0, 0.9, 0.8, 0.7, 0.5.