Solution manual of environmental chemistry a global perspective 4th ch05

Discuss the relationship between the production of nitrogen oxide species and the formation of ‘acid rain’ and the role played by carbon monoxide, methane, and the hydroxyl radical.. The

PROBLEMS/SOLUTIONS

1 The tropospheric processes discussed in this chapter cannot be considered as independent reactions

Discuss the relationship between the production of nitrogen oxide species and the formation of ‘acid rain’ and the role played by carbon monoxide, methane, and the hydroxyl radical

Solution

NOx species are formed primarily during combustion processes and are emitted into the atmosphere NO is oxidized to NO2 by species such as HOO•, ROO• and O3 The production of HNO3 is a result of the reaction of HO• with NO2 in the presence of a third body (M)

HO• + NO2 + M → HNO3 + M

As a preventive measure, catalytic converters remove NO from exhaust gases by reaction with CO and

CH4

CH4 + 4NO → 2N2 + CO2 + 2H2O 2CO + 2NO → N2 + 2CO2

CO and CH4 can also react with HO• to produce further reactive species

CO + HO• → CO2 + H•

H• + O2 + M → HOO• + M

CH4 + HO• → CH3• + H2O

CH3• + O2 + M → CH3OO• + M Both of these processes enhance the reactions that form NO2 and at the same time regenerate the HO• and RO• radicals

2 In a particular 3000 km2 region of southern Sweden, the annual rainfall averages 850 mm, its mean pH

is 4.27, and 66% of the hydrogen ion is associated with sulfuric acid with the remaining 34% derived from nitric acid Calculate whether soils of this region are subject to excessive sulfate loading if the only source of sulfate is rainfall and if the recommended maximum is set at 20 kg SO42- ha-1

Solution

The following data has been provided:

area = 3000 km2 = 3 x 105 ha = 3 x 109 m2 (this number will not be required)

1.0 ha = 100 m x 100 m = 10 000 m2 (the calculation will be done on a per ha basis)

rainfall = 850 mm (= 0.850 m)

pH = 4.27

hydrogen ion source is 66% as H2SO4 and 34% as HNO3

Is there excessive sulphate (SO42-) loading compared to a maximum 20 kg SO42- ha-1 ?

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The H+ is derived from both H2SO4 and HNO3 in the following ways:

HSO4- ↔ H+ + SO4- (pKa = 1.89)

H2SO4 has the potential to provide 2 moles of H+, while

HNO3 → H+ + NO3-

can provide only 1 mole of H+

The second proton on H2SO4is not strong (i.e Ka = 1.3 x 10-2), but deprotonation of HSO4- will occur and contribute a second H+ at the pH of 4.27, which is higher than the pKa (1.89) value

Calculation of: volume of rainfall per ha = 10 000 m2 x 0.850 m = 8500 m3 = 8.5 x 106 L

Amount of acid (H+) per L, pH = 4.27, therefore, in 1L there is 5.37 x 10-5 mol of H+

Amount of acid attributed to H2SO4 : (assuming 1 L for the calculation)

0.66 x 5.37 x 10-5 mol = 3.54 x 10-5 mol of H+ from H2SO4

divide by 2] gives: 1.77 x 10-5 mol of SO42- (in 1L)

The total amount of sulfate loading to 1 ha in 1 year is calculated by:

1.77 x 10-5 mol L-1 (of SO4-) x 8.5 x 106 L = 150 mol (of SO4-)

150 mol (of SO4-) x 96 g mol-1 = 1.44 x 104 g (of SO4-) or 14.4 kg (of SO4-)

Based on this calculation, the soil is not subject to excessive sulphate loading

3 The mean monthly pH values of rainfall in Guiyang city in Guizhou province in southern China in 1984

were as follows:

Jan 3.9, 4.0, 3.8, 4.1, 4.0, 4.5, 4.5, 4.1, 3.7, 3.8, 3.7, 3.4 Dec For the same year, the measurements at Luizhang, an adjacent rural area were:

Jan 4.3, 4.4, 4.4, 4.2, 4.5, 4.9, 4.9, 4.6, 4.8, 4.3, 5.4, 5.4 Dec (a) Calculate the mean monthly pH of the rain at the two locations

(b) What is the ratio of the mean hydrogen ion activity at the two sites?

(c) What is likely to be the most important anion in the rainfall at the urban site?

Refer to Dianwu, Z and X Jiling, Acidification in southwestern China, In Acidification in tropical

countries, eds Rodhe, H., and R Herrera, John Wiley and Sons, Chichester; 1988

Solution

a) The average pH of rainfall for Guiyang city: 3.9

The average pH of rainfall for Luizhang is: 4.5

Note: the simple average of the reported pH values gives an incorrect result of 4.0 and 4.7 respectively

Because of the logarithmic nature of the values involved, you must first convert each pH to the hydronium ion concentration, then sum and average, and finally convert back to pH

b) Ratio (Guiyang city/Luizhang) of the mean hydrogen ion activities

1.39 x 10-4 M

Ratio = - = 4.8

2.91 x 10-5 M

The mean hydrogen ion activity of Guiyang city is 4.8 times as great as Luizhang

c) The most important anion in the rainfall for Guiyang city could be either sulfate (in China, coal is a common fuel used in industry and for domestic purposes) or nitrate (from a variety of internal engine combustion sources)

4 The ionic composition (in units of mg m–3) of an atmospheric aerosol in a tropical rain forest is

SO4-, 207; NO3-, 18; NH4, 385; K+, 180; Na+, 247;

The pH of the aerosol is 5.22

Use these data to calculate the total positive and negative charge ‘concentration’ (mol m-3) in the aerosol and suggest reasons that might account for any discrepancy in anionic and cationic charge

Solution

In solving this question, we will begin by first assuming that the concentration is the aqueous concentration

of the ions in the aerosol

Summary of data: Anions

SO4- 207 mg m-3 M.W = 96 g mol-1

NO3- 18 = 62

Cations

NH4+ 385 = 18

K+ 180 = 39

Na+ 247 = 23

(pH = 5.22) H+ ? = 1

Assuming a volume of 1 m3, the amount in moles of each is calculated by:

SO4- 207 mg ÷ 1 x 103 mg g-1 = 2.07 x 10-1 g

2.07 x 10-1 g ÷ 96 g mol-1 = 2.156 x 10-3 mol of SO42-

The amount of charge is two times the number of moles to give:

4.3 x 10-3 equivalents of charge from SO4

-The others are similarly calculated and are provided below (eq m-3 of charge);

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SO4-, NO3-, NH4, K+, Na+, H+ (calculated from pH, see below)

4.3 x 10-3 2.9 x 10-4 2.1 x 10-2 4.6 x 10-3 1.1 x 10-2 6.0 x 10-3

The H+ is calculated from pH, where [H+] = 10-pH = 6.0x10-6 mol L-1

Then multiply by 1000 L (1 m3) to obtain 6.0x10-3 mol H+ (or equivalence of charge)

The total anion (negative charge) concentration is 4.6 x 10-3 mol m-3

The total cation (positive charge) concentration is 4.3 x 10-2 mol m-3

An excess of positive charge exists This type of charge balance is typical for experimental measurements involving precipitation chemistry A large number of other cations and anions could also be present in small amounts Other common ions that could improve the calculation include Ca2+, Mg2+, Cl- and organic matter (in such forms as humic or fulvic acids)

5 A report on rainfall composition in Egypt* gives results obtained at several sites One of these is for average rainfall in Cairo where the pH was 7.24:

SO4-, 12.23; NO3-, 5.99; NH4, 1.85; Cl-, 2.9; Ca2+, 5.3;

Concentrations are in mg L-1 After doing a charge balance calculation, explain how you would interpret the results

*Daifullah, A.A.M., and A.A Shakour, Chemical composition of rainwater in Egypt,

AJEAM-RAGEE, 6 (2003) 32-43

Solution

Calculation of concentration of charges (in mol L-1) are done as shown in Problem 4

SO4- NO3- NH4 Cl- Ca2+ H+

2.5 x 10-4 9.7 x 10-5 1.0 x 10-4 8.2 x 10-5 2.7 x 10-4 5.8 x 10-8 mol L-1

The total anion (negative charge) concentration is 4.3 x 10-4 mol L-1

The total cation (positive charge) concentration is 3.7 x 10-4 mol L-1

There is a significant amount of excess negative charge, indicating that there may be another major cation missing from this analysis Sodium is usually a large contributor to the charge balance and could account for the missing positive charge

6 The tropospheric mixing ratios of carbon monoxide are higher in the Northern Hemisphere than in the

Southern Hemisphere However, it has been observed that there was a general global decline in carbon monoxide concentration everywhere in the 1990s Two reasons have been suggested for this—one is the eruption of Mount Pinatubo and the other is the occurrence of several relatively dry years in the tropics Comment on these two possibilities in terms of tropospheric and stratospheric processes

Solution

The atmospheric level of carbon monoxide in the Northern Hemisphere is greater than in the Southern Hemisphere likely due to the greater populations and the larger amount of industrial and urban development there – leading to more emissions, largely from combustion of fossil fuels

Carbon monoxide removal from the atmosphere is accomplished by reaction with the hydroxyl radical (Chapter 2, Reaction 2.25) and uptake by biological activity, resulting in an atmospheric residence time of between 2-4 months Near the end of Section 5.3 Volcanoes - the 1991 eruption of Mount Pinatubo we indicated that there may have been substantial stratospheric ozone depletion associated with the Mount Pinatubo eruption Ozone destruction results from gases such as methyl chloride being injected into the stratosphere where they could be photolyzed to produce radicals that would take part in catalytic ozone-destroying cycles This would enhance the flux of UV radiation to the troposphere, and increase the rate of production of the hydroxyl radical, leading to a general enhancement of carbon monoxide oxidation However, there are other features of volcanic eruptions that could produce an opposite effect Solid (sulphate) particles injected into the stratosphere reduce the solar flux, causing less hydroxyl radical to be produced This would indirectly slow down the rate of carbon monoxide oxidation Furthermore, carbon monoxide is itself one of the gases released from volcanoes

Similarly the issue of dry years in the tropics could affect carbon monoxide levels in either direction You will see in the next chapter (Reaction 6.7) that carbon monoxide is a product of oxidation of isoprene (and other exudates) from growing trees, and the levels of isoprene production could be smaller in dry years On the other hand, a significant contribution of carbon monoxide comes from forest and grass fires, and one would expect that there would be more fires when there is less rain

7 The name reactive nitrogen has been recommended as a new short term for an extensive group of

atmospheric nitrogen compounds, and is given by:

NOy= NO + NO2 + NO3 + 2N2O5 + HONO + HNO2+ HNO3 + aerosol nitrate + PAN + organic nitrates

Discuss situations of atmospheric chemistry that we have described in this and earlier chapters where use of this term could be employed

Solution

The term reactive nitrogen could be well applied to situations that arise in smog and acid precipitation (day and nighttime) chemistry As well, some of these species are involved with ozone chemistry as reservoir species

8 Nitrogen oxide is formed at night by dissociation of dinitrogen pentoxide (the reverse of reaction 5.9)

For the first-order reaction

3 2 5

the rate constant is 3.14 × 10-2 s-1 at 25C, and 6.88 × 10-1 s–1 at 55C Calculate the half-life of this molecule at the two temperatures For a concentration of N2O5 of 3.6 ppbv, calculate the length of time it could take for the concentration to be reduced to 1.0 ppbv at a constant temperature of 25C

Calculate the Arrhenius parameters, A and Ea, for the reaction

Solution

Half-life = ln 2 / k at 25ºC (298 K) t½ = 22.1 s

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at 55ºC (328 K) t½ = 1.0 s The change in concentration from 3.6 to 1.0 ppbv is just short of two half-lives Expect the time to be just under 44s

use: ln[A] = ln[A]o – kt (solve for t)

ln 1.0 = ln 3.6 – 3.14 x 10-2 t

t = 40.8 s

Arrhenius Equation

k = Ae-Ea/RT or ln k = (– Ea / R) x (1/T) + ln A

y = m x x + b

From the plot of ‘ln k vs 1/T’, a straight line with the slope of ‘– Ea / R’ is obtained From two rates and two temperatures given in the question, Ea can be solved using the following equation to obtain the slope (ln k2 – ln k1) / (1/T2 – 1/T1) = – Ea / R

3.086 / (– 0.000307) = – Ea / 8.314

Ea = 83.6 kJ mol-1

The pre-exponential factor (A) can now be calculated for either temperature and is 1.4 x 1013 s-1

The following two problems use concepts that are developed in Chapter 11.

9 Assuming an atmospheric pressure of 83 kPa, an atmospheric mixing ratio of 1.5 ppbv for hydrogen

peroxide, and a value of KH(H2O2) = 7.0 × 10–1 mol L–1 Pa–1, calculate its solubility in the cloud water droplets Will this concentration depend on pH over the range 5 to 8?

Solution

Assume a temperature of 288 K, and a volume of 1 m3 for this problem The total number of moles in the gas phase is therefore given by n(total) = PV / RT

n(total) = 83 000 Pa x 1 m3 / 8.314 J mol-1 K-1 x 288 K = 34.66 mol 1.5 ppbv (H2O2) needs to be converted to Pa;

moles (H2O2)

- x 109 = 1.5 ppbv moles (H2O2) = 5.2 x 10-8

34.66

using P = nRT/V P (Pa) = 5.2 x 10-8 mol x 8.314 J mol-1 K-1 x 288 K / 1 m3

= 1.25 x 10-4 Pa (H2O2)

Using Equation 11.2 [H2O2] = 7.0 x 10-1 mol L-1 Pa-1 x 1.25 x 10-4 Pa (Note: from Chapter 11) [H2O2] = 8.7 x 10-5 mol L-1 (solubility in the cloud droplets)

The Ka of H2O2 is 2.2 x 10-12 (pKa = 11.7), so it is not expected that the concentration of H2O2 will depend

on the pH over the range of 5-8

10 Using the values for constants provided in reactions 5.25–5.27, calculate the solubility of sulfur dioxide

in water at pH = 9.0

Solution

Assume an atmospheric concentration of 10 ppbv (see paragraph following Reaction 5.27) at Pº and 298 K

n(total) = 101 325 Pa x 1 m3 / 8.314 J mol-1 K-1 x 298 K = 40.90 mol

10 ppbv (SO2) needs to be converted to Pa;

moles (SO2)

- x 109 = 10 ppbv

40.90

moles (SO2) = 4.1 x 10-7

using P = nRT/V P (Pa) = 4.1 x 10-7 mol x 8.314 J mol-1 K-1 x 298 K / 1 m3

P(SO2) = 1.01 x 10-3 Pa

Using Equation 11.2 [SO2] = 1.81 x 10-5 mol L-1 Pa-1 x 1.01 x 10-3 Pa (Note: from Chapter 11) [SO2] = 1.83 x 10-8 mol L-1

Ka1 = 1.72 x 10-2 (pKa1 = 1.76) Ka2 = 6.43 x 10-8 (pKa2 = 7.19)

SO2 (aq) + 2H2O ↔ HSO3- (aq) + H3O+ (aq) Ka1

HSO3- (aq) + H2O ↔ SO32- (aq) + H3O+ (aq) Ka2

The total solubility of SO2 will be the sum of the three aqueous species

s = [SO2] + [HSO3-] + [SO32-]

From Ka1 [HSO3-] = 1.72 x 10-2 x 1.83 x 10-8 mol L-1 / 1 x 10-9

= 0.31 mol L-1

From Ka2 [SO32-] = 6.43 x 10-8 x 0.31 mol L-1 / 1 x 10-9

[SO32-] = 19.9 mol L-1

s = ~0 + 0.3 + 19.9 = 20.2 mol L-1

The total solubility of SO2 at pH 9.0 is 20.2 mol L-1

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11 Considering Example 5.4 shown in the text, use the second-order rate constant for SO2 oxidation by hydroxyl radical and determine the half-life of SO2 (Assume a constant concentration of hydroxyl radical to determine the pseudo first-order rate constant.)

Solution

SO2 OH M HOSO M ( 5 21 )

2 





The reaction as shown is third-order but in the lower troposphere where the concentration of the ‘third body’ M—mostly dinitrogen and dioxygen—is large, it becomes pseudo second-order

The pseudo second-order rate constant given in Example 5.4 is 1.2x10-12 cm3 molec–1 s–1 If we also assume that the hydroxyl radical concentration is constant, then the reaction can further be considered as a pseudo first-order process

Pseudo second-order:

Rate = 1.2x10-12 cm3 molec–1 s–1 x [•OH][SO2]

Assume the concentration of •OH is 1.7x106 molec cm-3

Pseudo first-order:

Rate = 1.2x10-12 cm3 molec–1 s–1 x 1.7x106 molec cm-3[SO2]

Rate = 2.04x10-6 s-1 x [SO2]

The pseudo first-order rate constant is calculated to be 2.04x10-6 s-1

The half-life for the pseudo first-order reaction is determined by:

t1/2 = ln 2 / k

t1/2 = 0.693 / 2.04x10-6 s-1

t1/2 = 339706 s ( ÷ 60 s/min)

t1/2 = 5661.7 min ( ÷ 60 min/h)

t1/2 = 94.4 h ( ÷ 24 h/day)

t1/2 = 3.9 days

The half-life of SO2 is approximately 3.9 days

12 The United States Department of Energy has set a goal within their Innovation for Existing Plants

program to limit the amount of NOx released from their current fleet of fossil fuel-fired electricity generating plants These now have a capacity of 320 GW The goal is to emit less than 0.10 pounds of

NOx per MMBtu of fuel consumed Assume that these facilities operate at 30% efficiency, and over the year at 70% capacity, calculate the number of tonnes of NOx that will be emitted annually Table 8.6 will provide useful information for solving this problem

Solution

From Table 8.6 coal averages 2.5 x 1010 J per tonne 320 GW is equal to 3.2 x 1011 W (giga = 109 see Appendix C.2) One Watt is equal to one joule per second So in one year (which contains 60 x 60 x 24 x

365 = 3.1536 x 107 s) there is the potential for generating 1.01 x 1019 J of energy Since the plants operate at 30% efficiency, the amount of fuel needed to meet demand will be increased by a factor of 3.33 However, the running at 70% capacity would decrease the amount of energy generated to 7.1 x 1018 J annually This would require the consumption of about 9.4 x108 tonnes of coal

If there is 0.0454 kg of NOx released for every MMBtu (1012 Btu), which is equal to 1.05 x 1015 J, the amount of NOx released from 7.1 x 1018 J is 6762 x 0.0454 = 307 kg or about 0.3 tonnes annually

13 Using the SONOX process, how much limestone would be required each year in order to effect

quantitative removal of sulfur dioxide from a power plant whose daily consumption of coal (1.5% sulfur) is 6000 t?

Solution

The SONOX process is described in Section 5.7, and reactions 5.45 and 5.46 indicate that the molar ratio of limestone (CaCO3) to sulfur dioxide (SO2) in these reactions is 1:1

6000 t coal at 1.5% S = 90 t S d-1

90 t S = 90 000 kg = 90 000 000 g S (32.066 g mol-1) Moles of S = 2.807 x 106 = moles of SO2 = moles of CaCO3 (100.089 g mol-1)

Mass of CaCO3 = 2.809 x 108 g = 281 t d-1

On a yearly basis (365 d y-1) the amount of CaCO3 required would be approximately 103 000 t for stiochiometric removal However, due to inefficiencies in the process, it may be necessary to add up to 3 times this amount to ensure efficient removal of SO2

14 In the United States and other parts of the world, control of both sulfur and nitrogen oxide emissions

in the past decades have had a significant impact on their concentrations in precipitation and the

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amounts deposited into water and soil Consult the maps found through Additional Resources 4 Note the regions of major depositions of these compounds What are the most important sources? Explain what legislative initiatives are likely to have had the greatest effect on the reductions observed Comparing nitrate and ammonia, what are the reasons for the geographical difference in their distribution? What is the likely reason for the increase in ammonia levels during the time when amounts of the other compounds are decreasing?

For classroom discussion, no solution provided What relationship to the ‘developed world’ is there?

What can we do to change or alter the observed patterns?

15 Extra problem (Problem 6 from 2 nd edition) The estimated atmospheric carbon dioxide concentration

in the Northern Hemisphere in 1950 was 310 ppmv It may be predicted with some certainty that the

concentration in 2010 will be 390 ppmv Calculate the pH of pure rain which would be in equilibrium with the

carbon dioxide in each of the two years cited, and comment on the contribution which carbon dioxide makes towards precipitation acidity

Solution

Assume total pressure is Pº = 101 325 Pa and temperature is 298 K

The mixing ratios given (in units of ppmv) are also a mole fraction (x 106), and therefore, the partial pressure of carbon dioxide is given by:

310 ppmv CO2 = 310 x 10-6 x 101 325 Pa = 31.4 Pa

Henry’s Law constant for CO2 is 3.3 x 10-7 mol L-1 Pa-1 (Table 11.1, p 237)

Using Henry’s Law : (see p 237, Chapter 11)

[G]l = KHPg

[CO2](aq) = 3.3 x 10-7 mol L-1 Pa-1 x 31.4 Pa = 1.04 x 10-5 mol L-1

H2O (l) + CO2 (aq) → H2CO3 (aq)

H2CO3 (aq) ↔ H+ (aq) + HCO3

(aq) (see p 241 for complete description of carbon dioxide in water)

[HCO3-][H+]

Ka = - = 4.5 x 10-7

[CO2]

Assume [CO2] = 1.04 x 10-5 mol L-1, and that [HCO3-] = [H+]

Solve for [H+], [H+] = 2.16 x 10-6 pH = 5.67

Repeat the calculation for 370 ppmv: result is pH = 5.63