CHAPTER 17 CHEMISTRY IN THE ATMOSPHERE

Size: px

Start display at page:

Download "CHAPTER 17 CHEMISTRY IN THE ATMOSPHERE"

Lillian Strickland
6 years ago
Views:

1 CHAPTR 17 CHMISTRY IN TH ATMOSPHR 17.5 or ideal gases, mole fraction is the same as volume fraction. rom Table 17.1 of the text, CO is 0.0% of the composition of dry air, by volume. The value 0.0% means 0.0 volumes (or moles, in this case) out of 100 or 0.0 Χ CO To change to parts per million (ppm), we multiply the mole fraction by one million. (. 10 )( ) 0 ppm 17.6 Using the information in Table 17.1 and Problem 17.5, 0.0 percent of the volume (and therefore the pressure) of dry air is due to CO. The partial pressure of CO is: 1atm PCO Χ CO P ( ) T (. 10 ) 754 mmhg. 10 atm 760 mmhg 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. or further discussion, see Sections 17. and 17. of the text rom Problem 5.10, the total mass of air is 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). 1 1mol 0 Total moles of gases ( g) mol 9.0 g Mass of N (78.0%): g 1 (0.780)( mol) g 1mol kg Mass of O (0.99%): 0.00 g 1 (0.099)( mol) g 1mol kg Mass of CO (0.0%): g 18 (. 10 )( mol).6 10 g 1mol kg The energy of one photon is: J 1 mol J/photon 1mol photons

2 510 CHAPTR 17: CHMISTRY IN TH ATMOSPHR The wavelength can now be calculated. λ 8 hc ( J s)( m/s) m J 60 nm 17.1 Strategy: We are given the wavelength of the emitted photon and asked to calculate its energy. quation (7.) of the text relates the energy and frequency of an electromagnetic wave. hν irst, 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 c m/s 14 ν /s λ m Now, we can calculate the energy difference from the frequency. Δ hν ( J s)( /s) Δ J 17.1 The formula for the volume is 4πr h, where r m and h.0 10 m (or.0 mm). V L 15 4 π ( m) (.0 10 m) m L 1m Recall that at STP, one mole of gas occupies.41 L moles O 1mol ( L) mol O.41 L molecules molecules O ( mol O ) 1mol molecules g O 1kg 1 mass O (kg) ( mol O ). 10 kg O 1 mol O 1000 g 17. The quantity of ozone lost is: (0.06)( kg) kg of O Assuming no further deterioration, the kilograms of O that would have to be manufactured on a daily basis are: kg O 1yr 100 yr 65 days kg/day The standard enthalpy of formation (from Appendix of the text) for ozone: O O Δ f 14. kj/mol H o

3 CHAPTR 17: CHMISTRY IN TH ATMOSPHR 511 The total energy required is: 1molO 14. kj ( g of O ) g O 1 mol O kj 17. The formula for reon-11 is CCl and for reon-1 is C Cl. The equations are: CCl 4 + H CCl + HCl CCl + H C Cl + HCl A catalyst is necessary for both reactions The energy of the photons of UV radiation in the troposphere is insufficient (that is, the wavelength is too long and the frequency is too small) to break the bonds in CCs λ 50 nm m/s 15 ν /s m hν ( J s)( /s) J Converting to units of kj/mol: 479 kj/mol J photons 1 kj 1 photon 1 mol 1000 J Solar radiation preferentially breaks the C Cl bond. There is not enough energy to break the C bond irst, we need to calculate the energy needed to break one bond J 1 mol J/molecule 1mol molecules The longest wavelength required to break this bond is: λ 8 hc ( m/s)( J s) m J 44 nm 44 nm is in the visible region of the electromagnetic spectrum; therefore, C Br will be decomposed in both the troposphere and stratosphere The Lewis structures for chlorine nitrate and chlorine monoxide are: + Cl O N O O Cl O

4 51 CHAPTR 17: CHMISTRY IN TH ATMOSPHR 17.8 The Lewis structure of HCC 1 is: H C C Cl Cl The Lewis structure for C CH is: H C C H Lone pairs on the outer atoms have been omitted The equation is: ZnS + O ZnO + SO 4 1ton molzns 1 ton mol SO ton SO ( ton ZnS) ton ZnS 1 ton mol ZnS 1 ton mol SO tons SO Strategy: Looking at the balanced equation, how do we compare the amounts of CaO and CO? We can compare them based on the mole ratio from the balanced equation. Solution: Because the balanced equation is given in the problem, the mole ratio between CaO and CO is known: 1 mole CaO 1 mole CO. If we convert grams of CaO to moles of CaO, we can use this mole ratio to convert to moles of CO. Once moles of CO are known, we can convert to grams CO. 1 1 mol CaO 1molCO g mass CO ( g CaO) g CaO 1 mol CaO 1 mol CO g CO kg CO Total amount of heat absorbed is: J ( mol) K J K mol The heat of fusion of ice in units of J/kg is: J 1 mol 1000 g J/kg 1mol 18.0g 1kg The amount of ice melted by the temperature rise: 1kg ( J) J kg kj 17.4 thane and propane are greenhouse gases. They would contribute to global warming.

5 CHAPTR 17: CHMISTRY IN TH ATMOSPHR mol S 1molSO 7 (.1 10 g). 10 mol SO g S 1 mol S 7 nrt (. 10 mol)(0.081 L atm/mol K)(7 K) 8 V L P 1atm Recall that ppm means the number of parts of substance per 1,000,000 parts. We can calculate the partial pressure of SO in the troposphere molecules of SO SO 1 atm atm 6 10 parts of air P Next, we need to set up the equilibrium constant expression to calculate the concentration of H + in the rainwater. rom the concentration of H +, we can calculate the ph. SO + H O H + + HSO quilibrium: atm x x K + [H ][HSO ] PSO x x x M [H + ] ph log( ) (a) Since this is an elementary reaction, the rate law is: Rate k[no] [O ] (b) Since [O ] is very large compared to [NO], then the reaction is a pseudo second-order reaction and the rate law can be simplified to: Rate k'[no] where k' k[o ] (c) Since for a second-order reaction then, t 1 1 k[a] 0 t 1 t 1 1 [(A ) ] 0 [(A ) ] 01

6 514 CHAPTR 17: CHMISTRY IN TH ATMOSPHR min 10 ppm ppm t 1 Solving, the new half life is: t min You could also solve for k using the half-life and concentration ( ppm). Then substitute k and the new concentration (10 ppm) into the half-life equation to solve for the new half-life. Try it Strategy: This problem gives the volume, temperature, and pressure of PAN. Is the gas undergoing a change in any of its properties? What equation should we use to solve for moles of PAN? Once we have determined moles of PAN, we can convert to molarity and use the first-order rate law to solve for rate. Solution: Because no changes in gas properties occur, we can use the ideal gas equation to calculate the moles of PAN ppm by volume means: VPAN VT 0.55 L L Rearranging quation (5.8) of the text, at STP, the number of moles of PAN in 1.0 L of air is: n 0.55 L (1 atm) 1.0 L 6 PV 1 10 L mol RT (0.081 L atm/k mol)(7 K) Since the decomposition follows first-order kinetics, we can write: rate k[pan] mol 11 rate ( /s) M/s 1.0 L The volume a gas occupies is directly proportional to the number of moles of gas. Therefore, 0.4 ppm by volume can also be expressed as a mole fraction. Χ n 0.4 O O n 6 total 1 10 The partial pressure of ozone can be calculated from the mole fraction and the total pressure. 1atm PO Χ O P T (4. 10 )(748 mmhg) ( mmhg) atm 760 mmhg Substitute into the ideal gas equation to calculate moles of ozone. no P 7 O V ( atm)(1 L) mol RT (0.081 L atm/mol K)(9 K)

7 CHAPTR 17: CHMISTRY IN TH ATMOSPHR 515 Number of O molecules: O molecules ( mol O ) 1molO O molecules The Gobi desert lacks the primary pollutants (nitric oxide, carbon monoxide, hydrocarbons) to have photochemical smog. The primary pollutants are present both in New York City and in Boston. However, the sunlight that is required for the conversion of the primary pollutants to the secondary pollutants associated with smog is more likely in a July afternoon than one in January. Therefore, answer (b) is correct The room volume is: 17.6 m 8.80 m.64 m m Since 1 m 1 10 L, then the volume of the container is L. The quantity, ppm is: mole fraction of CO The pressure of the CO(atm) is: PCO 1atm Χ COPT ( )(756 mmhg) atm 760 mmhg The moles of CO is: n 5 PV ( atm)( L) RT (0.081 L atm/k mol)( 9 K) 1.5 mol The mass of CO in the room is: mass 8.01 g CO 1.5 mol 1molCO 78 g CO Strategy: After writing a balanced equation, how do we compare the amounts of CaCO and CO? We can compare them based on the mole ratio from the balanced equation. Once we have moles of CO, we can then calculate moles of air using the ideal gas equation. rom the moles of CO and the moles of air, we can calculate the percentage of CO in the air. Solution: irst, we need to write a balanced equation. CO + Ca(OH) CaCO + H O The mole ratio between CaCO and CO is: 1 mole CaCO 1 mole CO. If we convert grams of CaCO to moles of CaCO, we can use this mole ratio to convert to moles of CO. Once moles of CO are known, we can convert to grams CO. Moles of CO reacted: 1 mol CaCO 1molCO 0.06 g CaCO.6 10 mol CO g CaCO 1 mol CaCO

8 516 CHAPTR 17: CHMISTRY IN TH ATMOSPHR The total number of moles of air can be calculated using the ideal gas equation. n 1atm 747 mmhg (5.0 L) PV 760 mmhg RT (0.081 L atm/mol K)(91 K) 0.1 mol air The percentage by volume of CO in air is: VCO Vair n CO.6 10 mol 100% 100% 100% 0.1% nair 0.1 mol The chapter sections where these gases are discussed are: O : Section 17.7 SO : Section 17.6 NO : Sections 17.5, 17.7 Rn: Section 17.8 PAN: Section 17.7 CO: Sections 17.5, 17.7, An increase in temperature has shifted the system to the right; the equilibrium constant has increased with an increase in temperature. If we think of heat as a reactant (endothermic) heat + N + O NO based on Le Châtelier's principle, adding heat would indeed shift the system to the right. Therefore, the reaction is endothermic (a) rom the balanced equation: K c [O ][HbCO] [CO][HbO ] (b) Using the information provided: [O ][HbCO] [ ][HbCO] 1 [CO][HbO ] [ ][HbO ] 6 Solving, the ratio of HbCO to HbO is: [HbCO] [HbO ] 6 (1)( ) ( ) The concentration of O could be monitored. ormation of CO must deplete O (a) N O + O NO NO + O O + NO Overall: N O + O + O O + NO (b) Compounds with a permanent dipole moment such as N O are more effective greenhouse gases than nonpolar species such as CO (Section 17.5 of the text).

9 CHAPTR 17: CHMISTRY IN TH ATMOSPHR 517 (c) The moles of adipic acid are: g 1 mol adipic acid 10 (. 10 kg adipic acid) mol adipic acid 1 kg g adipic acid The number of moles of adipic acid is given as being equivalent to the moles of N O produced, and from the overall balanced equation, one mole of N O will react with two moles of O. Thus, mol adipic acid mol N O which reacts with mol O In Problem 17.6, we determined the partial pressure of CO in dry air to be. 10 atm. Using Henry s law, we can calculate the concentration of CO in water. c kp [CO ] (0.0 mol/l atm )(. 10 atm) mol/l We assume that all of the dissolved CO is converted to H CO, thus giving us mol/l of H CO. H CO is a weak acid. Setup the equilibrium of this acid in water and solve for [H + ]. The equilibrium expression is: H CO H + + HCO Initial (M): Change (M): x +x +x quilibrium (M): ( ) x x x + K (from Table 15.5) [H ][HCO ] x [H 5 CO ] ( ) x Solving the quadratic equation: x M [H + ] ph log( ) irst we calculate the number of Rn atoms. Volume of basement (14 m 10 m.0 m) m L nair nrn 5 PV (1.0 atm)(4. 10 L) mol air RT (0.081 L atm/mol K)(7 K) 6 PRn mmhg 4 5 ( ) ( mol).0 10 mol Rn Pair 760 mmhg Number of Rn atoms at the beginning: Rn atoms (.0 10 mol Rn) 1molRn Rn atoms 0.69 k.8 d d

10 518 CHAPTR 17: CHMISTRY IN TH ATMOSPHR rom quation (1.) of the text: [A] t ln [A] 0 kt x 1 ln (0.18 d )(1 d) x Rn atoms Strategy: rom ΔH f o values given in Appendix of the text, we can calculate ΔH for the reaction NO NO + O Then, we can calculate Δ from ΔH. The Δ calculated will have units of kj/mol. If we can convert this energy to units of J/molecule, we can calculate the wavelength required to decompose NO. Solution: We use the ΔH f o values in Appendix and quation (6.18) of the text. o o o Δ Hrxn nδhf (products) mδhf (reactants) Consider reaction (1): o o o Δ H Δ Hf(NO) + ΔHf(O) ΔHf(NO ) ΔH (1)(90.4 kj/mol) + (1)(49.4 kj/mol) (1)(.85 kj/mol) ΔH 06.0 kj/mol rom quation (6.10) of the text, Δ ΔH RTΔn Δ ( J/mol) (8.14 J/mol K)(98 K)(1) Δ J/mol This is the energy needed to dissociate 1 mole of NO. We need the energy required to dissociate one molecule of NO J 1molNO J/molecule 1molNO molecules NO The longest wavelength that can dissociate NO is: 8 hc ( J s)( m/s) λ m 94 nm J (a) Its small concentration is the result of its high reactivity. (b) OH has a great tendency to abstract an H atom from another compound because of the large energy of the O H bond (see Table 9.4 of the text). (c) NO + OH HNO

11 CHAPTR 17: CHMISTRY IN TH ATMOSPHR 519 (d) OH + SO HSO HSO + O + H O H SO 4 + HO This reaction has a high activation energy The blackened bucket has a large deposit of elemental carbon. When heated over the burner, it forms poisonous carbon monoxide. C + CO CO A smaller amount of CO is also formed as follows: C + O CO The size of tree rings can be related to CO content, where the number of rings indicates the age of the tree. The amount of CO in ice can be directly measured from portions of polar ice in different layers obtained by drilling. The age of CO can be determined by radiocarbon dating and other methods The use of the aerosol can liberate CC s that destroy the ozone layer Cl + O ClO ΔH ΣB(reactants) ΣB(products) ΔH (1)(4.7 kj/mol) + (1)(498.7 kj/mol) ()(06 kj/mol) ΔH 9 kj/mol o o o Δ H ΔHf(ClO) ΔHf(Cl ) ΔHf(O ) 9 kj/mol ΔHf o (ClO) 0 0 Δ H f o 9 kj/mol (ClO) 165 kj/mol There is one C Br bond per CH Br molecule. The energy needed to break one C Br bond is: 9 10 J 1 mol J 1mol molecules Using quation (7.) of the text, we can now calculate the wavelength associated with this energy. λ hc 8 hc ( J s)( m/s) λ m 409 nm J This wavelength is in the visible region of the spectrum and is available in the troposphere. Thus, photolysis of the C Br bond will occur.

12 50 CHAPTR 17: CHMISTRY IN TH ATMOSPHR 17.8 In one second, the energy absorbed by CO is 6.7 J. If we can calculate the energy of one photon of light with a wavelength of 1499 nm, we can then calculate the number of photons absorbed per second. The energy of one photon with a wavelength of 1499 nm is: 8 hc ( J s)( m/s) J λ m The number of photons absorbed by CO per second is: 1 photon 6.7 J J photons 17.8 (a) The reactions representing the formation of acid rain (H SO 4 (aq)) and the damage that acid rain causes to marble (CaCO ) statues are: SO (g) + O (g) SO (g) SO (g) + H O(l) H SO 4 (aq) CaCO (s) + H SO 4 (aq) CaSO 4 (s) + H O(l) + CO (g) irst, we convert the mass of SO to moles of SO. Next, we convert to moles of H SO 4 that are produced (0% of SO is converted to H SO 4 ). Then, we convert to the moles of CaCO damaged per statue (5% of 1000 lb statue is damaged). And finally, we can calculate the number of marble statues that are damaged lb 45.6 g 1molSO 11 (50 10 tons SO ) mol SO 1 ton 1 lb g SO 11 1molHSO4 11 (0.0) ( mol SO ) mol HSO4 1molSO The moles of CaCO damaged per stature are: 45.6 g 1 mol CaCO (0.05) (1000 lb CaCO ) 6.6 mol CaCO /statue 1 lb g CaCO The number of statues damaged by moles of H SO 4 is: 11 1molCaCO 1 statue ( mol HSO 4) 1 mol HSO4 6.6 mol CaCO statues Of course we don t have marble statues around. This figure just shows that any outdoor objects/statues made of marble are susceptible to attack by acid rain. (b) The other product in the above reaction is CO (g), which is a greenhouse gas that contributes to global warming.

13 CHAPTR 17: CHMISTRY IN TH ATMOSPHR (a) We use quation (1.14) of the text. k1 a T1 T ln k R TT s a K 98 K ln s 8.14 J/mol K ( K)(98 K) a J/mol 6.6 kj/mol (b) The unit for the rate constant indicates that the reaction is first-order. The half-life is: s t. 10 s 8 min k Actually, this question has two parts: (1) How can we determine the temperature and () how do we know the time period. (1) Hydrogen has two stable isotopes: 1 H (99.985%) and H or deuterium (D) (0.015%). Oxygen has three stable isotopes. The two major ones are: 16 O (99.759%) and 18 O (0.04%). Therefore, water molecules are made up of these different combinations. H O molecules containing heavier isotopes have greater mass and require more energy to evaporate. During cold periods, water that evaporates contains fewer 18 O and D than during warm ones. As the moist air is transported toward the North and South Poles and cools, the water that condenses and eventually freezes contains fewer heavy isotopes. The reverse holds for water evaporated during warm periods. By comparing the isotope ratio (using a mass spectrometer) with that of average ocean water (for which the temperature is known), it is possible to estimate the temperature during the period when evaporation-condensation occurred. () The formation of ices cores occurs at a steady rate. Therefore, knowing the rate of formation and the depth from which the ice cores are extracted, we can estimate the time of deposit. urthermore, from the CO trapped in the ice core, scientists can apply carbon-14 dating (and other radioactive dating techniques) to determine the age of the ice core In order to end up with the desired equation, we keep the second equation as written, but we must reverse the first equation and multiply by two. S(s) + O (g) SO (g) K SO (g) S(s) + O (g) ' 1 K1 ( K1) 1 SO (g) + O (g) SO (g) K K ( K ) K K ( ) 5 ( K1) (4. 10 ) K Thus, the reaction favors the formation of SO. But, this reaction has a high activation energy and requires a catalyst to promote it. 1

The following gas laws describes an ideal gas, where

The following gas laws describes an ideal gas, where Alief ISD Chemistry STAAR Review Reporting Category 4: Gases and Thermochemistry C.9.A Describe and calculate the relations between volume, pressure, number of moles, and temperature for an ideal gas as