Meteo 532 Atmospheric Chemistry

Problem Set #1. Solutions.

Problem 1.1. The current mixing ratio of methane (CH₄) is ~1760 parts per billion by volume (ppbv) and is increasing at a rate of about 8ppbvper year. Its main sinks are reaction with the hydroxyl radical (OH) in the atmosphere (k_OH= 3x10^-9s^-1), and removal by soils (k_soils=2x10^-10s^-1).

a. What are the lifetimes (in years) of methane due to the two different processes?

b. What is the total methane lifetime (in years)?

c. Which process – reaction with OH or soils – is the most important in determining methane’s lifetime?

a. t_OH = (3x10^-9)^-1 = 3.3x10⁸ s = 10.5 yrs; t_soils = (2x10^-10)^-1 = 5x10⁹ s = 159 yrs.

b. The total lifetime is t_total = (1/t_OH + 1/t_soils)^-1 = (3x10^-9 + 2x10^-10)^-1 = 3.12x10⁸ s = 9.95 yrs.

c. OH is more important that soils in determining methane’s lifetime.

Problem 1.2. On the same typical State College summer’s day, plumes from power plants in the Ohio River Valley can contain as much as 10 ppbv of sulfur dioxide (SO₂).

1. What is the number concentration of SO₂?
2. What is the mass concentration of SO₂?
3. If all of this SO₂is converted over to sulfuric acid,H₂SO₄, what is the resulting mass per unit volume of the sulfuric acid?

solution:

1. In State College in summertime, a typical daytime temperature is 303 K and pressure is 970 hPA. Thus, the air number density is [M] = 2.69x10¹⁹ (970/1013)(273/303) = 2.3x10¹⁹ molecules cm^-3. To get from ppbv to SO₂ concentration, we must do the following:

[SO₂] = 10 x 10^-9 x [M] = 10^-8 x 2.3x10¹⁹ = 2.3x10¹¹ molecules cm^-3

2. To get the mass, we must multiply by the molecular mass.

m_SO2 = (0.032 + 2 x 0.016)/6.02x10²³ = 1.06x10^-25 kg molecule^-1

So the total mass is: M_SO2 = 2.3x10¹¹x1.06x10^-25 = 2.4x10^-14 kg cm^-3 = 24 mg m^-3.

3. If it is all converted over to H₂SO₄, then the total mass associated with each S atom is not 0.064 kg/mole, as for SO₂, but is (0.032 + 4x0.016 + 2x0.001) = 0.098 kg/mole. Thus, the total sulfuric acid mass is M_H2SO4 = (0.098/6.02x10²³)x2.3x10¹¹ = 3.7x10^-14 kg cm^-3 = 37 mg m^-3.

Problem 1.3. The mean global stratospheric ozone column Is about 300 DU.

1. What is the column density in molecules m^-2?

2. What is the total number of ozone molecules in Earth’s atmosphere?

3. What is the total mass of ozone in Earth’s atmosphere?

4. What percentage the ozone is lost each year in the Antarctic ozone hole?

1. 300 DU means 3 mm of pure ozone at STP = 2.69x10²⁵ molecules m^-3 x 0.003 m. The column density = 8.1x10²² molecules m^-2.

2. Earth’s radius is about 6400 km. Thus the total surface area is about 4π (R_E)² = 4π(6400x1000)² = 5.15x10¹⁴ m². Multiplying this surface area by the column density gives the approximate number of molecules in Earth’s atmosphere: N_O3 = 5.15x10¹⁴ x 8.1x10²² = 4.2x10³⁷ molecules = 6.9x10¹³ moles

3. Each mole of ozone has a mass of 0.048 kg. Multiplying by the total number of moles gives:

M_O3 = 3.3x10¹² kg.

4. The ozone hole’s size is roughly 20x10⁶ km², where we have defined the size as the area which has experienced about 150 DU of loss since the 1970’s. Thus, the total amount of ozone lost is the molecular column density equivalent to 150 DU times the surface area of the ozone hole. Note that 150 DU is ½ of the 300 DU in 1.

Lost ozone = 4.05x10²² molecules m^-2 x 2x10¹³ m² = 8.1x10³⁵ molecules = 1.3x10¹² moles.

Compared to the total in 2., fraction = 8.1x10³⁵/4.2x10³⁷ = 0.019, about 2%.