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Ch6[1] - CHAPTER The Greenhouse Effect Problem 6-1 Since...

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Unformatted text preview: CHAPTER The Greenhouse Effect Problem 6-1 Since the rate ofenergy release is proportional to the fourth power of the Kelvin temperature, and since 0°C = 27.3 K and 17°C = 290 K, the ratio of the rates of release is (273 / Z9O)4, the value of which is 0.785. The Kelvin temperature T at which the rate of release is twice that at 0°C would be the solution to the equation ("I/Z73)4 = 2.0 Raising both sides of the equation to the power of 1/4 gives T/273 = 1.189 So T = 325 K, which is equivalent to 52°C. Problem 6~2 H2 and ClZ will not absorb IR light because there cannot be a change in the dipole moment during the vibration; the dipole moment must always be zero for such symmetrical molecules. All the others will absorb IR light because, during at least some of their vibrations, the dipole moment will change. Note that (d), ozone, has a bent structure in which the terminal oxygen atoms are not equivalent to the central one, so the oxygenaoxygen bonds can be polar and the molecule has a dipole moment. Problem 6~3 The diatomic molecules that could absorb IR light are CO and NO (see Problem 6—2). Since they do not absorb much light in the thermal IR region, this implies that their stretching frequencies must lie outside the thermal IR region. 53 54 Chapter 6 Problem 6-4 Using the formula relating energy to wavelength in Chapter 1, that is E : 119627/7L will produce E in k] mol'l ifl is expressed in nm. Since 1 micrometer = 1000 nm it follows that 15.0 um = 15,000 nm 4.26 um = 4260 nm E = 119627/15000 = 8.01<Jmol’1 = 8000] mol—1 E = 119627/4260 = 28.1 kJ mol‘1 = 28,100] mol’1 Since each mole contains 6.02 X 10” molecules, the energies per molecule are: 1 mole 8000 J mol‘1 X ————————73———~— = 1.3 X 10—20 J molecule-l 6.02 X 10“ molecules 1 mole W = 4.67 X 10—20 J molecule’l - X mo ecu es 28,100 J 11101"1 X For the reaction CO2 —) CO + 0 thus AH = AHMCO) + AHMO) — AHl-(COZ) = —110.5 + 249.2 — (—3935) = +5322 1<J mol‘1 Thus the 15 Jim light supplies only 8.0 532.2 of the energy required to dissociate C02, whereas the 4.26 pm light supplies = 0015 (Le, 1.5%) & = 0053 (Le, 5.3%) 532.2 of the energy required. Problem 6-5 From the. balanced equation it follows that 1 mole ofCOl is released per 1 mole ofCaCO3 heated. From molar masses of 44.01 and 100.09 g for CO2 and CaCO3 respectively, it follows that 0.440 g of C02 is released per gram of CaCO3 heated; thus 0440 tonnes of CO2 are released per tonne of CaCQ heated. The Greenhouse Effect 55 Since 1 mole ofC is contained in 1 mole of C02, and because their molar masses are 12.01 and 44.01 respectively, then for each 44.0] grams of carbon dioxide entering the armosphere, 12.01 g of carbon enter it; by proportion, then for each gram of C02, the mass of carbon that enters is (12.01/44.01) = 0.27 grams. Problem 6-6 The ppm scale gives us concentrations on a moles/moles or a volume/volume basis. Because most data in the problem are masses, presumably the moles definition is the most appropriate to use here. So first we convert all masses to moles: First convert mass of carbon into moles of it: 4.7 Gt = 4.9 X 109 tonnes X 1000 kg/ tonne X1000 g/kg = 4.7x 1015gofC Since carbon’s molar mass is 12.01, moles C = 4.7 X 1015 g C/ 12.01 g C mole"1 = 3.9 X 1014 moles of C, which is also the change in the number of moles of CO2. Since air’s molar mass is 29.0 and its total atmospheric mass is 5.1 x 1021 grams, we can compute the number of moles of air in the atmosphere: 5.1 X 1011 g / 29.0 g mole'I = 1.76 X 1020 moles air The increase in ppm is the ratio of the increase in number of moles of CO2 to the total number of mil— lions of moles ofair: 3.9 X 1014 moles col/1‘76 ><10l4 million moles ofair = 2.2 ppm Thus the annual increase in CO2 concentration is about 2.2 ppm. To obtain the total mass of carbon, we first use the CO2 concentration to find moles of it: 382 ppm means 382 moles C02/ 1 million moles air X 1.76 X 1014 moles air = 6.72 X 1016 moles CO2 Since this is also the number of moles of carbon, we can use its molar mass to determine its total atv mospheric mass: 6.72 X 1016 moles C X 12.01 g C / mole C = 8.07 X 1017g C Since 1 Gt = 1015 grams, the total carbon content is 807 Gt, in good agreement with the value of ~ 800 Gt in Figure 610. Problem 6-7 The residence time T, concentration C, and rate R of concentration increase are related by the formula T=C/R 56 Chapter 6 The total amount A of the gas is proportional to C by the same factor that the rate of addition M in mass units is proportional to the rate of increase of concentration, so we can rewrite the equation as T = A / M Solving for A we obtain A = T M =50 y X 2.0 kg / y = 1.0 X108 kg Thus the total amount of it in the atmosphere is 1 X 108 kg. Problem 6-8 From the concentration C and the residence time T, the rate R of its addition can be calculated: R : C/T : 7.0uggmr‘l/14y = 05 “g galril yil If we multiply rhis result by the total mass of the atmosphere, the rate of addition is converted from a pervgramrofiair basis to the atmosphere as a whole: 0.5ug gmr’l y‘1 X 5.1 X 1031ng = 2.5 x 1021 pg y’1 = 2.5 X IOlsgy‘l We had no need to use the molar mass data for the gas or for air. Problem 6-9 Since the reaction is: CH4 + OH —) CH3 + HZO it follows that its rate law is: rate = k [CH4] [OH] Substituting for k and [OH] from data in the problem rate = 3.6 X 10"”cm3molecule’l s’1 X 8.7 X 105 molecule ch [CH4] 2 3.13 X 10’9 [CH4] in units of s—1 Substituting for the current methane concentration, then rate = 3.13 X 10’9 s’1 X 1.74 ppm = 5.45 X 10—9 ppm/second Since 1 ppm CH4 = 1 mole CH4/ 106 moles air = 16.05 5; CH4/ 10(’ moles air thus, rate = 5.45 X 10‘9 X 16.05 gCH4 sec—l / 106 moles air = 8.75 X 10‘14 g CH4 / sec/ mole air The Greenhouse Effect 57 We now need to multiply this rate by the number of seconds in a year 60 sec 60 min 24 hr 365 days X X X — = 3.15 x 107 ~ 1 min 1 hr 1 day 1 year sec/yr and the number of moles of air in the atmosphere to obtain the yearly mass of methane destroyed. Now from data, mass of atmosphere = 5.1 X 1021 g Average molar mass ofair = 29.0 g / mole so moles of air in atmosphere = 5.1 X 1021 g / 29.0 g mole’1 = 1.76 X 1020 moles . 1 “4 f H . 7. M X M x 1.76 X 1020 moles air 1 sec X 1 moles air 1 year Thus, CH4 destroyed = = 4.85 x 1014g01cr14 = 485Tg ofCH4 since I Tg = 1012grams. Problem 6-10 The formula for the hydrate is CH4 - 6HZO. We can deduce the proportion of CH4 in one mole of the compound by comparing the molar mass of CH4 to that of the whole compound: molar mass of CH4 molar mass of CH4 ' 6HZO = 16.05 g/124.17 g = 0.1293 fraction of mass that is CH4 = Thus, in 1000 g of the hydrate, the mass of methane is 0.1293 X1000g =129g Problem 6—11 No, they will have no sink because they possess no C ——H or multiple bonds that the OH radical can attack—and start their tropospheric oxidation—nor are they likely to be soluble in water, or to absorb visible or UVaA light because they Contain no multiple bonds. They would act as greenhouse gases because C —- F stretch and PCP bending vibrations fall in the thermal IR region. For CHJF and CZHSF, attack by OH would begin their oxidation and clean them out from the air in a finite time—though while still present, perhaps for decades, they would act as greenhouse gases. 58 Chapter 6 Green Chemistry Problems 1. (a) (b) 2. Alternative reaction conditions for green chemistry. 1. Prevention of waste. 3. Wherever practicable, synthetic methodologies should he designed to use and generate substances that possess little or no toxicity to human health and the environment. 4. Preserving efficacy of function while reducing environmental damage. 5. The use of auxiliary substances (e.g., solvents, separation agents, etc.) should be made unnecessary whenever possible and innocuous when used. 6. Energy requirements should be recognized for their environmental and economic im— pacts and should be minimized. Synthetic methods should he conducted at ambient temperature and pressure . . The rinse step is no longer necessary, thus eliminating the need for millions of liters ofhighly purified water, the energy required to produce this water, and the associated wastewater. This also reduces the fossil fuels required to produce the highly purified water and the accompanying formation of carbon dioxide. Eliminates or reduces the need for (and waste from) hazardous and toxic chemicals, such as strong acids or bases, or organic solvents in the photoresist removal step. This also results in enhanced worker safety. . Eliminates the need for alcohols used for drying after the aqueous rinse step. The carbon dioxide is recovered after each use and reused. . The only waste left after evaporation (and recovery) of the carbon dioxide is the spent photoresist, which is unregulated. Additional Problems 1. (a) This problem is rather subtle. In particular, since 502 and N02 are nonlinear molecules, they each have a dipole moment, the magnitudes of which would change even during the symmetric stretch vibration. Thus, all three vibrational modes—symmetric and antisym— metric stretch and the bond angle bending—are IR active. The $0, symmetric stretch would probably be the most important, since its wavelength is Close to the “window" region where few other absorptions are active. Both gases have short atmospheric lifetime (~ days), so their concentrations will not build up by accumulation, and without this feature their ability to cause much global warming is slight. The Greenhouse Effect 59 (a) If it had the linear NON structure, then like COX, the symmetric stretch vibration would be unable to absorb 1R. 1n the NNO structure, there is a dipole moment to the molecule and it can change during the symmetrical NNO stretch; thus all 3 vibrations will absorb IR. Thus, because we know NNO has 3 IRrabsorbing vibrations, its structure must be NNO and not NON. (b) No, because the zero dipole moment associated with the tetrahedral structure does not change during such a vibration. From the increase in the CO2 concentration, we can calculate the total amount of the gas that remained in the air. Recall that ppm can be interpreted on a moles of CO2 per million moles of air basis; thus the increase in concentration is equivalent to 11.1 moles of CO7 for each million moles of air. The total number of moles of air in the atmosphere can be calculated from the at- mospheric mass divided by its average molar mass: Moles of air in atmosphere = 5.1 x 102' g / 29.0 g / mole = 1.76 x 1020 moles Thus the total increase in CO2 in air is (11.1 moles C02/ 106 moles air) x 1.76 x1020 moles air = 1.95 X 1015 moles CO2 Since the molar mass of CO2 equals 44.01 g, the increase in the mass of CO2 in the air is 1.95 x 1015 moles COZ X 44.01 g / mole COZ = 8.6 x1016 g. Now the total amount of COZ emitted into the air was 178 Gt = 178 x 1015 g, so the fraction of CO2 that remained in the air was 8.6 X 1016 g/ 178 x 1015 g = 0.48, i.e., 48%. If methane was growing at an annual rate of 0.6%, then because the total is 5000 Tg, the rate of input must have exceeded that of output by 0.006 X 5000 Tg = 30 Tg/year. Thus the rate of input must slow by 30 Tg/year to stabilize it. Since its loss rate was about 530 Tg/year, the pro~ duction rate must have been 530 + 30 = 560 Tg/year, of which tw0athirds or 373 Tg/year was anthropogenic in origin. Thus the percent cutback required in anthropogenic methane is: (30/373) x 100% = 8% This problem can be solved numerically by evaluating F for 0.001 and double that value, and for 2 and its double, and then analyzing how F is changed in the two cases. Since loge (1 — F) = —ch thus for ch = 0.001, loge (1 — F) = —0.001 50 (1 _ 1:) = C—OOO! (1 — F) = 0.9990 so F = 0.0010 Similarly for ch = 0.002, (1 ~ F) = 0.9980 soF = 0.0020 60 Chapter 6 Thus for ch near zero, F doubles with a doubling of the concentration C. For ch = 2, it follows F = 0.865 and for ch = 4, F = 0.982 Thus doubling ch in this region does not nearly double F Sincet1 = 15°C, thus T1 = 273 + 15 = 288 K, and similarly T2 = 291 K. Since the units of K involve joules, the correct units for AH are also in joules: AH = 44,000] mole’1 Thus ln(P_,/P,) = (—44,000/8.3) (1/291 — 1/288) 2 0.190 and hence (Pg/P1) = 1.21 Thus the vapor pressure of water increases by 21% if the liquids temperature is raised from 18°C to 21°C. However, if the average air/surface temperature is raised to 18°C, the 1R absorbed by water may decrease by less than 21% because absorption is linearly related to concentration only at very low concentrations, and because air is not saturated 10000 by water vapor and the relative humidity at 21°C maybe less than that at 18°C. Although CO2 emissions would decrease, the lifetime of much of the previous emissions is so long that the overall level of COZ would not be much affected, and so the heating by it would remain about the same. However, the sulfate aerosol lifetime is very short, so its concentration in air would be greatly reduced by a sharp reduction in 802 emissions, thereby canceling its cooling effect. Thus the immediate effect on Climate would probably be to significantly increase air temperature! Both CH4 and CHFzCl contain 5 atoms, and thus have the same number of vibrational modes. However, in the case of CH4, there is only one type of bond, CrH, so the various types of stretching modes are similar. This is also true for bending modes, all of which involve C~H bonds only. Due to the high symmetry of the molecule, some of the vibrational modes are 1R in~ active because they do not involve an overall change in dipole moment. This is in contrast to the case of CHFZCl, which contains three types of bonds: C—H, CF and C«Cl, all of which have very different bond strengths and atom masses, and thus absorb in very different regions of the IR spectrum. This is also true of the various bending modes, which can involve CHF, CHCl, CFCl, and CClZ. The much lower symmetry of CHFZCl as compared to CH4 also means that more vibrational modes are 1R active. Finally, the CF and C—Cl bonds in CHFZCl are more polar than the CIH bonds in CH4, so that vibrations involving these bonds are much more effi' cient at absorbing 1R because the bond dipoles are larger. 10. The Greenhouse Effect Combustion reaction: C8H18 (1) + 12.5 02 (g) ——-> 8 co2 (g) + 9 1120(1) thus moles co2 / moles C81118 = 8:1 moles C8ng = 1.0 L x 1000 mL/L x 0.702 g/mL/ 114.26 g mol’1 = 6.1411161 moles co2 = 8 x 6.14 mol = 49.2 mol VCOZ = nRT / P = 49.2 mol x 0.082 L atm K-l mol—I x 293 1</ 1 atm = 1180 L Since 1 gallon = 3.785 L, 1 180 L X 3.785 = 4470 L COz/gallon midvsizc sedan: 100 miles / 33 miles per gallon = 3.03 gallon X 4470 L COZ / gallon = 13,500 L CO2 produced SUV: 100 miles/19 miles per gallon = 5.26 gallon X 4470 L COZ / gallon = 23,500 L C02 produced 1.77 ppm CH4 = 1.77 mol CH4/ 106 mol atmosphere total mol atmosphere = 5.1 ><1018 kg X1000 g/kg / 29 g mol—1 = 1.76 ><10ZO mol moles CH4 in atmosphere = 1.76 X 1020 mol X 1.77 mol CH4/ 106 mol atmosphere = 3.11><10l4 mol .‘.mass CH4 in atmosphere = 3.11 X 1014mol>< 16.0 g/mol = 5.0 X 10'5 g 6] ...
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