Chapter17ISM

Chapter17ISM - CHAPTER 17 CHEMISTRY IN THE ATMOSPHERE...

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CHAPTER 17 CHEMISTRY IN THE ATMOSPHERE Problem Categories Biological : 17.65, 17.69. Conceptual : 17.60, 17.68, 17.76, 17.78, 17.79, 17.85. Descriptive : 17.23, 17.67, 17.70, 17.71a, 17.75, 17.77, 17.87. Environmental : 17.7, 17.11, 17.21, 17.22, 17.24, 17.25, 17.26, 17.41, 17.50, 17.58, 17.59, 17.66, 17.73, 17.74, 17.79. Industrial : 17.39, 17.40, 17.49, 17.71. Organic : 17.42. Difficulty Level Easy : 17.5, 17.6, 17.12, 17.24, 17.26, 17.27, 17.28, 17.40, 17.60, 17.68, 17.69, 17.79. Medium : 17.7, 17.8, 17.11, 17.21, 17.22, 17.23, 17.25, 17.39, 17.42, 17.49, 17.67, 17.70, 17.76, 17.77, 17.80, 17.81, 17.84, 17.86, 17.87. Difficult : 17.41, 17.50, 17.57, 17.58, 17.59, 17.65, 17.66, 17.71, 17.72, 17.73, 17.74, 17.75, 17.78, 17.82, 17.83, 17.85. 17.5 For ideal gases, mole fraction is the same as volume fraction. From Table 17.1 of the text, CO 2 is 0.033% of the composition of dry air, by volume. The value 0.033% means 0.033 volumes (or moles, in this case) out of 100 or 2 CO 0.033 100 == 4 3.3 10 Χ × To change to parts per million (ppm), we multiply the mole fraction by one million. (3.3 × 10 4 )(1 × 10 6 ) = 330 ppm 17.6 Using the information in Table 17.1 and Problem 17.5, 0.033 percent of the volume (and therefore the pressure) of dry air is due to CO 2 . The partial pressure of CO 2 is: () 2 4 CO T 1atm (3.3 10 ) 754 mmHg 760 mmHg × × = 2 4 CO 3.3 10 atm P P × 17.7 In the stratosphere, the air temperature rises with altitude. This warming effect is the result of exothermic reactions triggered by UV radiation from the sun. For further discussion, see Sections 17.2 and 17.3 of the text. 17.8 From Problem 5.102, the total mass of air is 5.25 × 10 18 kg. Table 17.1 lists the composition of air by volume. Under the same conditions of P and T , V α n (Avogadro’s law). 21 20 1mol Total moles of gases (5.25 10 g) 1.81 10 mol 29.0 g × = × Mass of N 2 (78.03%): 20 21 28.02 g (0.7803)(1.81 10 mol) 3.96 10 g ×× = × = 18 3.96 10 kg × Mass of O 2 (20.99%): 20 21 32.00 g (0.2099)(1.81 10 mol) 1.22 10 g = × = 18 1.22 10 kg ×
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CHAPTER 17: CHEMISTRY IN THE ATMOSPHERE 523 Mass of CO 2 (0.033%): 42 0 1 8 44.01 g (3.3 10 )(1.81 10 mol) 2.63 10 g 1mol ×× × = × = 15 2.63 10 kg × 17.11 The energy of one photon is: 3 19 23 460 10 J 1 mol 7.64 10 J/photon 6.022 10 photons × ×= × × The wavelength can now be calculated. 34 8 7 19 (6.63 10 J s)(3.00 10 m/s) 2.60 10 m 7.64 10 J ×⋅× == = × = × 260 nm hc E λ 17.12 Strategy: We are given the wavelength of the emitted photon and asked to calculate its energy. Equation (7.2) of the text relates the energy and frequency of an electromagnetic wave. E = h ν First, we calculate the frequency from the wavelength, then we can calculate the energy difference between the two levels. Solution: Calculate the frequency from the wavelength. 8 14 9 3.00 10 m/s 5.38 10 /s 558 10 m × ν= = = × λ × c Now, we can calculate the energy difference from the frequency. Δ E = h ν = (6.63 × 10 34 J s)(5.38 × 10 14 /s) Δ E = 3.57 × 10 19 J 17.21 The formula for the volume is 4 π r 2 h , where r = 6.371 × 10 6 m and h = 3.0 × 10 3 m (or 3.0 mm). 62 3 1 23 1 5 3 1000 L 4 (6.371 10 m) (3.0 10 m) 1.5 10 m 1.5 10 L 1m × × = × × V Recall that at STP, one mole of gas occupies 22.41 L. 15 13 33 moles O (1.5 10 L) 6.7 10 mol O 22.41 L × = × 23 13 3 6.022 10 molecules (6.7 10 mol O ) × × = 37 3 molecules O 4.0 10 molecules × 13 3 3 3 48.00 g O 1kg (6.7 10 mol O ) 1 mol O 1000 g × × = 12 3 3 mass O (kg) 3.2 10 kg O × 17.22 The quantity of ozone lost is: (0.06)(3.2 × 10 12 kg) = 1.9 × 10 11 kg of O 3
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CHAPTER 17: CHEMISTRY IN THE ATMOSPHERE 524 Assuming no further deterioration, the kilograms of O 3 that would have to be manufactured on a daily basis are: 11 3 1.9 10 kg O 1yr 100 yr 365 days × ×= 6 5.2 10 kg/day ×
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Chapter17ISM - CHAPTER 17 CHEMISTRY IN THE ATMOSPHERE...

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