Earth and Planetary Sciences 5 Final Exam

January 16, 2009 Name:________________________________________________ Teaching Fellow:_______________________________________ INSTRUCTIONS

• PUT YOUR NAME ON EACH PAGE. • Complete the problems directly on the exam. • Extra paper is available if needed. • Please show ALL your work so partial credit can be given! • Neatness is appreciated.

Scoring: Problem Score 1. Multiple Choice __________/22 2. Short questions __________/21 3. CO2-O2 Relationship __________/10 4. Carbon Chemistry in the Oceans __________/10 5. Ocean salinity __________/10 6. Stratospheric Ozone – Chapman Mechanism __________/16 7. Stratospheric Ozone – Radical-Catalyzed Ozone Loss __________/8 8. Tropospheric Chemistry: Ozone Production __________/14 Total __________/111 There are 111 possible points, and the value for each problem indicates the approximate time allotted to solve the problem.

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Section I: Multiple Choices. Circle the best answer (1 pt each)

1. An air parcel has a temperature 20oC and a relative humidity of 50%. What is

the dew point of the air parcel? Use the graph below.

a) 20oC b) 9oC c) 15oC d) 5oC

2. At 4 km, the pressure on Planet X is 600 mbar and the pressure on Planet Y is 400 mbar. Assuming both planets have the same surface pressure, which planet has a smaller scale height? a) Planet X b) Planet Y c) Not enough information to tell

3. From the following box model of the terrestrial carbon cycle, what is the

lifetime of carbon in the soil (i.e., the combination of the box labeled detritus/decomposers and active soil)?

a) 104 yrs

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b) 19 yrs c) 90 yrs d) 52 yrs

4. Which of the following likely has the lowest albedo?

a) a lake b) clouds c) an asphalt parking lot d) a desert

5. If an air parcel has a temperature of 10°C and its dew point is 5°C, at what

altitude will it start to form a cloud (assume the background atmosphere is unstable; the dry adiabatic lapse rate is 10K/km and wet adiabatic lapse rate is 5 K/km)?

a) 0 km b) 0.5 km

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c) 1 km d) 2 km

6. Deep water forms in the following locations:

a) North Atlantic b) North Pacific c) Antarctic d) a and c

7. Off the coast of Peru, you would expect to find __________ levels of nutrients due to __________ except during ____________.

a) low; upwelling; the winter. b) low; upwelling; the summer. c) high; upwelling; an El Niño. d) high; downwelling; an El Niño.

8. During El Nino, the trade winds in the Pacific slacken. Assuming all else on the earth remained the same, the angular momentum of the atmosphere would _______ and that of the solid Earth would _____________. Take angular momentum to be positive for west to east motion.

a) Increase; decrease b) Increase; increase c) Decrease; increase d) Decrease, decrease

9. Which of the following statements about ocean circulation is NOT true? a) Gyres move clockwise in the Northern Hemisphere and counterclockwise in

the Southern Hemisphere b) The Ekman spiral can be explained by the combined effects of wind

stress, Coriolis force and vertical convection. c) In both Hemispheres, upwelling arises as a consequence of divergence of the

horizontal currents.

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d) The Gulf Stream plays an important role of transporting heat from the equatorial region to high latitudes.

10. Which of the following does NOT accurately explain why deep water forms

over the North Atlantic? a) It contains more organic matter than the underlying water b) It is colder than the underlying water c) It is more saline than the underlying water d) It is denser than the underlying water

11. Salinity is greater in the surface Atlantic than in the surface Pacific, because of:

a) The height of North-South mountain chains in North and South America b) The direction of the jet stream c) The direction of the trade winds d) a & b e) a & c f) b & c

12. If 4 units of carbon are incorporated in the organic fraction of typical Redfield marine organisms (C:N:P = 105:15:1), what is the change in the abundance of total inorganic carbon, alkalinity, and carbonate ion (CO32-), respectively? Recall that for every 4 units of carbon incorporated in organic matter, one additional unit of carbon is incorporated as a component of shells.

a) -5, -1.43, and +3.57 b) -5, -1.43 and -3.57 c) -4, -1.43, and +2.57 d) -4, +0.57, and -4.57

13. The following occurs during respiration: a) CO2 is consumed b) O2 is consumed c) Solar energy is consumed d) H2O is consumed

14. What causes the oscillations in CO2 concentrations shown in the Keeling curve below?

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a) The difference in amount of surface area covered by the biosphere in the Southern Hemisphere vs. the Northern Hemisphere

b) The seasonality of photosynthesis vs. decay c) Larger fossil fuel emissions in winter vs. summer d) The time lag for oceans to respond to CO2 disturbances

15. Which of the following contribute to the contemporary rising atmospheric

concentrations of CO2 (as shown by the Keeling curve): a) Rising fossil fuel emissions (6 Gt/yr) into the atmosphere b) The release of CO2 associated to deforestation c) The release of CO2 associated to mid-latitude forest re-growth d) Increase in emissions of CH4 e) All of the above f) a and b g) a, b, and d

16. If the lifetime of HCl were to decrease (perhaps due to an increase in [OH]) how would this affect the rate of chlorine-catalyzed ozone loss?

a) Ozone loss would increase. b) Ozone loss would decrease. c) Ozone loss would remain the same. d) There is not enough information.

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17. Air tends to sink over the cold poles in winter. As a consequence the pressure in

the stratosphere at 10 km above the surface is lower than the pressure at the same altitude in the sub-polar regions. Which figures show possible wind direction in the polar vortices?

1 3

2 4

a) 1 & 3 b) 1 & 4 c) 2 & 3 d) 2 & 4

18. Why does the ozone hole develop in the spring?

a) The catalytic processes that destroy ozone can only take place if there is light to make reactive chlorine.

b) The polar vortex sucks in air with low ozone concentrations from lower latitudes.

c) There is always an ozone hole, but we can only see it if there is sunlight.

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d) There is an increase in water vapor during spring, which acts as a catalyst for ozone depletion

19. Stratospheric ozone is important for life on Earth because a) It provides a cap for the atmosphere, preventing lighter molecules such as N2

and O2 from escaping to space. b) It reflects high energy radiation from the sun back to space. c) It is a primary source of O2 in the atmosphere. d) It absorbs the Sun’s harmful UV radiation.

20. What is the first step in OH production in the troposphere?

a) hv + O2 O + O b) O(1D) + O2 O(3P) + O2 c) hv + O3 O(1D) + O2 d) hv + NO2 O(3P) + NO

21. HOx and NOx catalyze O3 _______________ in the troposphere and O3

_____________ in the stratosphere. a) destruction, destruction b) production, production c) destruction, production d) production, destruction

22. The main source of tropospheric O3 is the cycling of:

a) NOx b) HOx c) Cly d) Bry

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Section II: Short Questions

1. Climate change and albedo feedback (9 points) a) Although 70% of the earth's surface is covered by ocean, its average albedo is 0.33 (about 33% of the incoming solar energy is reflected back to space). Using the table below, reason why the average albedo of the Earth is so high. (3 points)

Surface cover Albedo Fresh snow 0.75 - 0.95 Clouds 0.3 – 0.9 Ice 0.3 – 0.4 Desert 0.15 – 0.45 Grassland 0.1 – 0.3 Ocean 0.05 Forest 0.1

Solution:

While 70% of the Earth's surface is covered by ocean (with a low albedo of 0.05), 30% of the Earth's surface is covered by land surfaces that all have a higher albedo (fresh snow has the highest albedo). Given the known distribution of land and ocean, the Earth's albedo would never reach 0.33 without the influence of clouds. Clouds are very reflective, and they prevent a significant portion of solar radiation from ever reaching the surface of the Earth. For example, even though the ocean doesn’t tend to reflect much incoming radiation, the clouds present over the ocean reflect a portion of this radiation before it comes in contact with the ocean. Actually, approximately 2/3 of the Earth's albedo id due to the consistent presence of cloud reflectance. Thus, the Earth's albedo fluctuates with the level of cloud cover of the Earth.

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Let’s now look at the specific effect of snow cover on the Earth’s albedo. Suppose that in the next twenty years, snow cover decreases by 5%. b) Would the average albedo of the Earth increase or decrease? (2 points) Solution:

The average albedo of the Earth would decrease. Land covered with snow has a high albedo. Replacing snow cover with land surfaces that have a lower albedo would decrease the average albedo of the Earth.

c) If all other factors were held constant, what would happen to the average

temperature? (2 points)

Solution: The temperature of the Earth would increase.

d) Based on your answer in c), do you think the snowmelt will continue? Explain why. (2 points) Solution:

The snowmelt would continue – this is an example of a positive feedback. Earth warms -> snow cover retreats -> albedo decreases -> Earth warms some more -> snow cover retreats some more -> albedo continues to decrease -> etc.

2. Implications of atmospheric stability (6 points) a) For each segment (i.e. A, B, C) on the graph shown below, calculate the lapse

rate of the atmosphere and indicate if the atmosphere is stable or unstable for dry air parcels. Consider a moist adiabatic lapse rate of 3K/km as compared with the dry adiabatic lapse rate. (3 points)

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A- dT/dz = (288-300)/2 = -6 Atmosphere is CONDITIONALLY STABLE

since the lapse rate is less than the dry adiabatic lapse rate, but greater than the smallest moist adiabatic lapse rate (3K/km).

B- dT/dz = (290-288)/2 = 1 Atmosphere is STABLE C- dT/dz = (266-290)/2 = -12 Atmosphere is UNSTABLE b) If the meteorological conditions shown in the figure above persist over an area

with surface air pollution sources, is the region at risk for a severe pollution episode? Explain whether you think the surface air will stagnate or ventilate, and whether this will cause a pollution to accumulate. (3 points)

Yes. There could be a severe pollution episode because the stable air at the surface (region A) will inhibit the dilution of surface emissions with clean air from above. The emissions will remain near the surface where they could harm people, plants, and animals. In addition the inversion (region B) will effectively trap air below 2 km, causing stagnant conditions.

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3. Geostrophic wind (6 points) a) The diagram below shows a hypothetical high–pressure system in the northern hemisphere. On the diagram, draw vectors showing the directions of the following at point A: wind velocity, pressure gradient-force, Coriolis force, and frictional force. How would the direction of the vectors change if this system were in the southern hemisphere? (4 points)

Solution:

b) Would the wind likely higher at point A or point B? Why? (2 points) Wind would be higher at point B; stronger

Section III: Problems

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3. CO2-O2 Relationship (10 points) Measurements of the long-term trend of atmospheric O2 have been used to determine the fate of fossil fuel CO2 in the atmosphere and the relative importance of uptake by the ocean and by the biosphere. Assume that the only fossil fuel used in a particular year was natural gas (CH4), and that observations indicate that CO2 concentrations increased by 2.0 ppm, while O2 decreased by 2.7 ppm during that year. Assume also that 6.36 Gton (1 giga ton = 109 tons) of carbon were emitted as CO2 by burning natural gas in this year.

a) If the combustion of CH4 is given by: aCH4 + bO2 → cH2O + dCO2, balance this reaction and answer how many moles of O2 are consumed from the atmosphere for each mole of CO2 emitted. (3 points)

CH4 + 2O2 → 2H2O + CO2

Two moles of O2 are consumed for each mole of CO2 emitted from CH4

burning. Assume that the 3 primary processes that affect CO2 concentrations are ocean uptake/emission, biospheric uptake/emission, and fuel emission. Note that 2.12 Gton of C emitted as CO2 leads to an increase of 1ppm in atmospheric CO2. Photosynthetic activity consumes 1 molecule of CO2 for every 1.1 molecule of O2 released.

b) Based on the above information, calculate the biosphere contribution for both O2 and CO2 in ppmv during this year. (2 points)

Fuel combustion: 3 ppm CO2 emitted and 6 ppm O2 consumed; Biospheric uptake: 3 ppm CO2 removed and 3.3 ppm O2 released;

c) Was the ocean a sink or a source for CO2 in this year? (2 points) Ocean emission: 2 ppm CO2 emitted

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d) Using the plot provided below, schematically plot the CO2 versus O2

concentrations for observations, fossil fuel emission, ocean, and biosphere during this year. The initial O2 and CO2 concentration is indicated by point A in the figure. Changes in concentrations of O2 and CO2 are expressed in units of parts per million (ppm). (3 points)

Solution:

AB = change due to fuel emission

CO2

O2

CO2

O2

B

C D

A

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BC = change due to biospheric uptake CD = change due to ocean AD = net change Slope of AB = -2 Slope of BC = -1.1 Slope of CD = 0

4. Carbon Chemistry (10 points) The direct injection of carbon dioxide into the ocean has been suggested as an approach to sequestrate some of the CO2 released by human activity in order to limit the rise in atmospheric CO2 levels. Let’s examine the change in the chemistry of ocean water caused by this approach. The chemical equilibrium of inorganic carbon species in ocean water may be described by the following three reactions: (a)-(c). The equilibrium constant (α, K1, K2) for each of the chemical reactions is listed in the table. (a) , α

(b) , K1

(c) , K2

Table: Values of equilibrium constants as a function of temperature

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The net reaction for reactions (a), (b) and (c) is CO2 (g) + CO32- + H2O �2HCO3

-

1) Use K to denote the equilibrium constant for the net reaction. Write down the expression for K in terms of the concentrations of the species. (2 points)

2) Consider a scenario where 8×10-5 mol/L of CO2 is directly injected into the ocean at some depth. It is inevitable that some of the injected carbon escapes the ocean. After the new equilibrium is established, only half of the initial amount of carbon added is absorbed by the water . Suppose that the concentrations of HCO3

- and CO32- are given initially (i.e., before the

injection) as 1.4×10-3 mol/L and 2×10-4 mol/L, respectively. Calculate the concentrations of HCO3

- and CO32- in the new equilibrium. (3 points)

½ of CO2 (4×10-5 mol/L) added is converted to bicarbonate ion in the new equilibrium. According to the net reaction, for every one molecule of CO2 (g) consumed, there are one molecule of CO3

2- consumed from the ocean water and two molecules of HCO3

- added in. Therefore, Concentrations of CO3

2- should be decreased by 4×10-5 mol/L and concentrations of HCO3

- should be increased by 2×4x10-5 mol/L = 8×10-5 mol/L In the new equilibrium: [CO3

2-] = 2×10-4 mol/L - 4×10-5 mol/L = 1.6 ×10-5 mol/L [HCO3

-] = 1.4×10-3 mol/L + 8×10-5 mol/L= 1.48 ×10-3 mol/L

3) Suppose the temperature is fixed at 293 K. Calculate the pH of the water in the initial equilibrium (i.e., before the injection) and in the new equilibrium (i.e., after the injection). Does the pH increase or decrease after the direct carbon injection? (5 points)

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, K2 = 6.52 × 10-10 mol/L at 293K (20oC) Before the new equilibrium is reached: [CO3

2-] = 2×10-4 mol/L and [HCO3-] = 1.4×10-3 mol/L

pH = -log([H+]) = 8.34 After the new equilibrium is reached: [CO3

2-] = 1.6×10-4 mol/L and [HCO3-] = 1.48×10-3 mol/L

pH = -log([H+]) = 8.22 pH has decreased after the direct carbon injection.

5. Ocean Salinity (10 points) Imagine that you have the ability to track a cube of ocean water with sides of 100 m. Initially, the water in the cube has a salinity of 25 parts per thousand (2.5%), and its temperature is 15 °C. Imagine there is no transport of water, salt, or heat into or out of the cube. a) Using the following chart (density unit: g/cm3), estimate approximately the initial density of the cube? (2 points)

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Density ~ 1.018 g/cm3 b) As the cube travels from the south Atlantic to the north Atlantic, a good deal of evaporation occurs (i.e., the cube no longer has sides of 100 m). By 40°N, 6.43×109 moles of water have evaporated. What is the new salinity? Note: You can approximate the molar mass of the cube to be that of water (18 g/mol). (4 points) To accurately calculate the new salinity, it is necessary to know the molar mass of NaCl, which wasn’t included. However, the following method gives a good approximation:

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= 5.6*1010 moles

= 4.96*1010 moles

c) What is the density of the water if the temperature is then 10 °C? (2 points)

Density ~ 1.023 d) As the cube travels further north, its temperature reaches the freezing point. What is the density of the cube at that point? (2 points)

Density ~ 1.024 6. Stratospheric Ozone Chemistry – Chapman Mechanism (16 points total)

(R1) O2 + hv O + O J1@ 40 km = 1×10-10 sec-1 (R2) O + O2 + M O3 + M k2@ 40 km = 6×10-34 cm6 sec-1 (R3) O3 + hv O + O2 J3@ 40 km = 1×10-3 sec-1 (R4) O + O3 O2 + O2 k4@ 40 km = 1×10-13 cm3 sec-1

(a) If we assume that the concentrations of the Ox species ([O] + [O3]) adjust

such that the rates R2 and R3 rapidly equilibrate, i.e., R2 = R3, calculate the value of the ratio of [O]/[O3] at 40 km. Let [M] = 1×1017 cm-3. Recall that the mixing ratio of O2 is 0.21. (4 points)

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R2 = R3 → k2[O][O2][M] = J3[O3] → [O] / [O3] = J3 / k2[O2][M] [O] / [O3] = [1×10-3 sec-1] / { [6×10-34 cm6 sec-1](0.21)[1×1017]2 }

Note: [O2] = (0.21)[M], Oxygen is 21% of air, and [M] is the number density of air at a given altitude with units # / cm3.

[O] / [O3] = 7.9×10-4. There are 0.00079 O’s for every 1 O3. (b) Given your answer in part (a), is the approximation that [Ox] ≈ [O3] valid?

Do you expect this approximation to be better or worse at 25 km? Why? (3 points)

Yes, the approximation is valid @ 40 km. [Ox] = [O3] + [O] and 1 + 0.00079 ≈ 1!

The approximation should be better at 25 km. The concentration of [O] @ 25 km is less than at 40 km, and [O3] is near its maximum.

Furthermore, looking at the expression for the ratio we can see that:

a. at 25 km J3 will be less than at 40 km (UV light gets absorbed as you penetrate deeper into the atmosphere), and

b. the density of air (and consequently O2) increases as we go to a lower altitude.

Decreasing the numerator and increasing the denominator makes the ratio smaller, and the approximation better.

(c) If the photochemical lifetime of [Ox] is sufficiently short, we can use the steady-state assumption to derive an expression for the concentration of ozone, i.e. [O3], in terms of the four rate constants, and the ambient number density [M].

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Write an expression for the rate of production of odd oxygen. (1 point) ROx, PROD = 2 × R1 = 2 × J1[O2]; Two odd-oxygen produced in R1, O + O. Write an expression for the rate of loss of odd oxygen. (1 point) ROx, LOSS = 2 × R4 = 2 × k4[O][O3]; Two odd-oxygen lost in R4, O + O3. Note: No odd-oxygens are produced or lost in reactions R2 & R3. What is true about these rates under photochemical equilibrium (or steady-state) assumption? (1 point) Under the steady-state assumption the rates of Ox production and loss will be equal.

(d) Calculate the lifetime of [Ox] given [O3] = 1.6×1011 cm-3. In general, the assumption of photochemical equilibrium is justified if the chemical time constant is less than a month, i.e., ~106 seconds. Do you think this assumption is valid at 40 km? (4 points) τOx = Reservoir Size / Loss Rate From Reservoir = [Ox] / { 2 × k4[O][O3] } ≈ [O3] / { 2 × k4[O][O3] } = 1 / { 2 × k4[O] } . = 1 / { 2 × [1×10-13 cm3 sec-1][1.6×1011 cm-3][7.9×10-4] } = 3.9×104 sec < 106 seconds. Yes, the assumption is valid at 40 km.

(e) If the photochemical lifetime of Ox is longer than a month, we can expect that, depending on location, transport of air can be either a net sink or source for Ox. Do you expect transport processes to play a larger role in

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determining Ox concentrations at altitudes above or below 40 km? Why? (2 points)

Looking at the expression for the τOx we can see that as [O] decreases, the lifetime of odd-oxygen increases, thus at lower altitudes, e.g. 25 km, we expect that transport processes will play a larger role in determining Ox concentrations.

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7. Stratospheric Ozone Chemistry: Radical-Catalyzed Loss Cycle (8 points)

(1) Cl + O3 → ClO + O2 k1 = 9.5 × 10-12 cm3 molecule-1 s-1 (2) Cl + CH4 → HCl + CH3 k2 = 2.6 × 10-14 cm3 molecule-1 s-1 (3) ClO + O → Cl + O2 k3 = 3.8 × 10-11 cm3 molecule-1 s-1 (4) NO2 + hv → NO + O k4 = 5.0 × 10-3 s-1 (5) ClO + NO → Cl + NO2 k5 = 4.5 × 10-11 cm3 molecule-1 s-1 (6) ClO + HO2 → HOCl + O2 k6 = 2.1 × 10-11 cm3 molecule-1 s-1 (7) ClO + NO2 + M→ ClNO3 + M k7 = 1.3 × 10-13 cm3 molecule-1 s-1 (8) HOCl + hv → OH + Cl k8 = 2.5 × 10-4 s-1 (9) OH + O3 → HO2 + O2 k9 = 2.8 × 10-14 cm3 molecule-1 s-1

An air parcel at 30 km altitude (30oN, equinox) contains the following concentrations: [O3] = 3.0 × 1012 molecules cm-3 [O] = 3.0 × 107 atoms cm-3 [NO] = 7 × 108 molecules cm-3 [NO2] = 2.2 × 109 molecules cm-3 [HO2] = 8.5 × 106 molecules cm-3 [CH4] = 2.8 × 1011 molecules cm-3

a) Consider the above series of reactions, and write expressions for the chemical lifetimes of Cl and ClO. (4 points)

b) Calculate the chemical lifetimes for Cl and ClO. (2 points)

3.5×10-2 s 30s

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c) Identify the major sinks for Cl and ClO, respectively. (2 points)

Ozone (R1) is the dominant sink for Cl. NO (R5) is the dominant sink

for ClO. 8. Tropospheric Chemistry: Ozone Production (14 points total) You are the mayor of Smogville, CA, a city in which ozone levels are consistently above the federal standard of 90 ppb. The EPA has decided to crack down and announces that Smogville will lose federal highway funds unless ozone levels are brought down into compliance with the federal standard. Recall that as part of your re-election campaign you are currently proposing a new highway around the city - largely funded by federal dollars. Since you are anxious to comply with EPA, you hire an environmental consulting firm to help develop a strategy for ozone reduction. The firm provides you with the following diagram, which gives the midday peak concentration of ozone (ppb) in Smogville on a typical summer day as a function of hydrocarbon and NOx emissions.

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(a) Current hydrocarbon emissions in Smogville are 2.0×1011 atoms C cm-2 s-1 and NOx emissions are 8×1011 molecules cm-2 s-1. From the diagram, what midday ozone peak concentration would you expect from these emissions? Is ozone production in Smogville hydrocarbon-limited or NOx-limited? (2 points)

120ppb; hydrocarbon-limited.

To bring Smogville into compliance, you consider the possibility of reducing emissions of either hydrocarbons or NOx (with the other remaining at current levels), or both.

(b) Find the percentage change in peak ozone concentration for the following control strategies:

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(i) Reduce NOx emissions by 50% and maintain current hydrocarbon emissions. (2 points)

(110-120)/120*100% = -8.33% (ii) Reduce hydrocarbon emissions by 50% and maintain current NOx emissions. (2 points)

(90-120)/120*100% = -25%

(c) Based on the answers in (b) above, which emission reduction strategy is optimal for getting Smogville to compliance? (1 point)

Reduce hydrocarbon emissions.

(d) Assuming that it costs the same to reduce emissions of hydrocarbons by 1.0×1011 atoms C cm-2s-1 as it does to reduce NOx by 2.0×1011 molecules cm-2s-1, what would be the cheapest reduction strategy to bring Smogville into compliance? (1 point)

Reduce hydrocarbon emissions.

(e) The consulting firm advises you to increase NOx emissions, by relaxing controls on cars, as being the cheapest way to control ozone levels. Does it make sense, as judged from the diagram alone? By how much would emissions have to increase to reach the 90 ppb O3 standard? (2 points)

Yes. It would require emissions of NOx to double in order to reach the 90ppb O3 standard.

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(f) Briefly explain in chemical terms why this approach might work. Also list a

few shortcomings of this approach. (4 points)

As [NOx] increases, OH will be depleted by NO2 + OH + M HNO3 + M. As OH is the key species to initiate the breakdown of hydrocarbon, hydrocarbon oxidation and thus ozone production is limited as [OH] is reduced. (2 points) NOx is a health hazard that can cause respiratory irritation and diseases. NOx also contributes to the formation of acid rain.