CHAPTER 5 GASES. Questions

Transcription

1 CHAER 5 GASES Questios 0. Molecules i the codesed phases (liquids ad solids) are very close together. Molecules i the gaseous phase are very far apart. A sample of gas is mostly empty space. herefore, oe would expect mol of H O(g) to occupy a huge volume as compared to mol of H O(l).. he colum of water would have to be.6 times taller tha a colum of mercury. Whe the pressure of the colum of liquid stadig o the surface of the liquid is equal to the pressure of air o the rest of the surface of the liquid, the the height of the colum of liquid is a measure of atmospheric pressure. Because water is.6 times less dese tha mercury, the colum of water must be.6 times loger tha that of mercury to match the force exerted by the colums of liquid stadig o the surface.. A bag of potato chips is a costat-pressure cotaier. he volume of the bag icreases or decreases i order to keep the iteral pressure equal to the exteral (atmospheric) pressure. he volume of the bag icreased because the exteral pressure decreased. his seems reasoable as atmospheric pressure is lower at higher altitudes tha at sea level. We igored (moles) as a possibility because the questio said to cocetrate o exteral coditios. It is possible that a chemical reactio occurred that would icrease the umber of gas molecules iside the bag. his would result i a larger volume for the bag of potato chips. he last factor to cosider is temperature. Durig ski seaso, oe would expect the temperature of Lake ahoe to be colder tha Los Ageles. A decrease i would result i a decrease i the volume of the potato chip bag. his is the exact opposite of what actually happeed, so apparetly the temperature effect is ot domiat.. he versus / plot is icorrect. he plot should be liear with positive slope ad a y- itercept of zero. k, so k(/). his is i the form of the straight-lie equatio y mx + b. he y-axis is pressure, ad the x-axis is /.. he decrease i temperature causes the balloo to cotract ( ad are directly related). Because weather balloos do expad, the effect of the decrease i pressure must be domiat. 5. d (molar mass)/; desity is directly proportioal to the molar mass of a gas. Helium, with the smallest molar mass of all the oble gases, will have the smallest desity. 6. Rigid cotaier: As temperature is icreased, the gas molecules move with a faster average velocity. his results i more frequet ad more forceful collisios, resultig i a icrease i pressure. Desity mass/volume; the moles of gas are costat, ad the volume of the cotaier is costat, so desity must be temperature-idepedet (desity is costat). 7

2 8 CHAER 5 GASES Flexible cotaier: he flexible cotaier is a costat-pressure cotaier. herefore, the iteral pressure will be uaffected by a icrease i temperature. he desity of the gas, however, will be affected because the cotaier volume is affected. As icreases, there is a immediate icrease i iside the cotaier. he cotaier expads its volume to reduce the iteral pressure back to the exteral pressure. We have the same mass of gas i a larger volume. Gas desity will decrease i the flexible cotaier as icreases. 7. No; at ay ozero Kelvi temperature, there is a distributio of kietic eergies. Similarly, there is a distributio of velocities at ay ozero Kelvi temperature. he reaso there is a distributio of kietic eergies at ay specific temperature is because there is a distributio of velocities for ay gas sample at ay specific temperature. 8. a. Cotaiers ii, iv, vi, ad viii have volumes twice those of cotaiers i, iii, v, ad vii. Cotaiers iii, iv, vii, ad viii have twice the umber of molecules preset tha cotaiers i, ii, v, ad vi. he cotaier with the lowest pressure will be the oe that has the fewest moles of gas preset i the largest volume (cotaiers ii ad vi both have the lowest ). he smallest cotaier with the most moles of gas preset will have the highest pressure (cotaiers iii ad vii both have the highest ). All the other cotaiers (i, iv, v, ad viii) will have the same pressure betwee the two extremes. he order is ii vi < i iv v viii < iii vii. b. All have the same average kietic eergy because the temperature is the same i each cotaier. Oly temperature determies the average kietic eergy. c. he least dese gas will be i cotaier ii because it has the fewest of the lighter Ne atoms preset i the largest volume. Cotaier vii has the most dese gas because the largest umber of the heavier Ar atoms are preset i the smallest volume. o figure out the orderig for the other cotaiers, we will calculate the relative desity of each. I the table below, m equals the mass of Ne i cotaier i, equals the volume of cotaier i, ad d equals the desity of the gas i cotaier i. Cotaier mass, volume i ii iii iv v vi vii iii m, m, m, m, m, m, m, m, desity mass volume m m d d m d m d m d m d m d m d From the table, the order of gas desity is ii < i iv vi < iii v viii < vii. d. µ rms (/M) / ; the root mea square velocity oly depeds o the temperature ad the molar mass. Because is costat, the heavier argo molecules will have a slower root mea square velocity tha the eo molecules. he order is v vi vii viii < i ii iii iv. 9. NH (g) N (g) + H (g); as reactats are coverted ito products, we go from moles of gaseous reactats to moles of gaseous products ( mol N + mol H ). Because the moles of gas doubles as reactats are coverted ito products, the volume of the gases will double (at costat ad ).

3 CHAER 5 GASES 9, (costat); pressure is directly related to at costat ad. As the reactio occurs, the moles of gas will double, so the pressure will double. Because o mol of N is produced for every mol of NH reacted, N (/)NH. Owig to the : o mole ratio i the balaced equatio, (/). o H NH Note: total H + N (/) NH + (/) NH NH. As we said earlier, the total pressure will double from the iitial pressure of NH as reactats are completely coverted ito products. 0. Statemets a, c, ad e are true. For statemet b, if temperature is costat, the the average kietic eergy will be costat o matter what the idetity of the gas (KE ave / ). For statemet d, as icreases, the average velocity of the gas particles icreases. Whe gas particles are movig faster, the effect of iterparticle iteractios is miimized. For statemet f, the KM predicts that is directly related to at costat ad. As icreases, the gas molecules move faster, o average, resultig i more frequet ad more forceful collisios. his leads to a icrease i. 0. atm L. he values of a are: H, ; CO,.59; N,.9; CH,.5 mol Because a is a measure of itermolecular attractios, the attractios are greatest for CO.. he va der Waals costat b is a measure of the size of the molecule. hus C H 8 should have the largest value of b because it has the largest molar mass (size).. ; Figure 5.6 is illustratig how well Boyle s law works. Boyle s law studies the pressure-volume relatioship for a gas at costat moles of gas () ad costat temperature (). At costat ad, the product for a ideal gas equals a costat value of, o matter what the pressure of the gas. Figure 5.6 plots the product versus for three differet gases. he ideal value for the product is show with a dotted lie at about a value of. L atm. From the plot, it looks like the plot for Ne is closest to the dotted lie, so we ca coclude that of the three gases i the plot, Ne behaves most ideally. he O plot is also fairly close to the dotted lie, so O also behaves fairly ideally. CO, o the other had, has a plot farthest from the ideal plot; hece CO behaves least ideally.. Dalto s law of partial pressures holds if the total pressure of a mixture of gases depeds oly o the total moles of gas particles preset ad ot o the idetity of the gases i the mixtures. If the total pressure of a mixture of gases were to deped o the idetities of the gases, the each gas would behave differetly at a certai set of coditios, ad determiig the pressure of a mixture of gases would be very difficult. All ideal gases are assumed volumeless ad are assumed to exert o forces amog the idividual gas particles. Oly i this sceario ca Dalto s law of partial pressure hold true for a ideal gas. If gas particles did have a volume ad/or did exert forces amog themselves, the each gas, with its ow idetity ad size, would behave differetly. his is ot observed for ideal gases. o o

4 0 CHAER 5 GASES Exercises ressure 5. a..8 atm 760 mm Hg atm.6 0 mm Hg b..6 0 mm Hg torr mm Hg.6 0 torr c..8 atm.0 0 atm 5 a a d..8 atm.7 psi atm 7 psi 6. a. 00 psi atm.7 psi 50 atm b. 50 atm.0 0 atm 5 a Ma 6 0 a 5 Ma c. 50 atm atm. 0 5 torr cm 0 mm cm 65 mm Hg 65 torr; 65 torr atm atm atm.0 0 atm 5 a a i Hg.5 cm i 0 mm 508 mm Hg 508 torr; 508 torr cm atm atm 9. If the levels of mercury i each arm of the maometer are equal, the the pressure i the flask is equal to atmospheric pressure. Whe they are uequal, the differece i height i millimeters will be equal to the differece i pressure i millimeters of mercury betwee the flask ad the atmosphere. Which level is higher will tell us whether the pressure i the flask is less tha or greater tha atmospheric. a. flask < atm ; flask torr 6 torr atm 0.85 atm 0.85 atm.0 0 atm 5 a a b. flask > atm ; flask 760. torr + 5 torr 975 torr

5 CHAER 5 GASES 975 torr atm.8 atm.8 atm.0 0 atm 5 a a c. flask torr; flask torr 0. a. he pressure is proportioal to the mass of the fluid. he mass is proportioal to the volume of the colum of fluid (or to the height of the colum assumig the area of the colum of fluid is costat). mass d desity volume ; i this case, the volume of silico oil will be the same as the volume of mercury i Exercise 9. d m ; Hg oil ; m d Hg Hg m d oil oil, m oil m Hg d d Hg oil Because is proportioal to the mass of liquid: oil doil.0 Hg Hg (0.0956) Hg dhg.6 his coversio applies oly to the colum of silico oil. flask 760. torr! ( ) torr 760.!. 79 torr 79 torr atm atm; atm.0 0 atm 5 a a Gas Laws flask 760. torr + ( ) torr torr atm.0 0 a 78 torr.0 atm;.0 atm a atm b. If we are measurig the same pressure, the height of the silico oil colum would be times the height of a mercury colum. he advatage of usig a less dese fluid tha mercury is i measurig small pressures. he height differece measured will be larger for the less dese fluid. hus the measuremet will be more precise.. At costat ad, costat, so ; at sea level,.00 atm 760. mm Hg mm.0 L 500. mm Hg.0 L 5

6 CHAER 5 GASES he balloo will burst at this pressure because the volume must expad beyod the.5 L limit of the balloo. Note: o solve this problem, we did ot have to covert the pressure uits ito atm; the uits of mm Hg caceled each other. I geeral, oly covert uits if you have to. Wheever the gas costat R is ot used to solve a problem, pressure ad volume uits must oly be cosistet ad ot ecessarily i uits of atm ad L. he exceptio is temperature, which must always be coverted to the Kelvi scale.. he pressure exerted o the balloo is costat, ad the moles of gas preset is costat. From Charles s law, / / at costat ad ml 00. K 9 ml ( ) K As expected, as temperature decreases, the volume decreases.. At costat ad, Avogadro s law holds ( ). 0. L 0.50 mol, 0.89 mol. L As expected, as icreases, icreases.. As NO is coverted completely ito N O, the moles of gas preset will decrease by oe-half (from the : mole ratio i the balaced equatio). Usig Avogadro s law:, 5.0 ml.5 ml N O (g) will occupy oe-half the origial volume of NO (g). his is expected because the moles of gas preset decrease by oe-half whe NO is coverted ito N O. 5. a., L atm 5.00 atm.00 mol (55 + 7) K.0 L 0.00 atm.00 L b., L atm 55 K.7 0 mol c., R.7 atm 5.0 L L atm.0 mol 678 K 05 C d., L atm.5 L 0.5 mol (7 + 75) K atm

7 CHAER 5 GASES 6. a a atm a atm; K, atm 0.0 L L atm 98 K mol b., L atm 0. mol 0.00 L K 79 atm c L atm mol (+ 7) K atm 55 torr.6 L d. R atm 75 mm Hg. L 760 mm Hg L atm 0.0 mol K 6 C 7. 5 atm 00.0 L L atm (7 + ) K. 0 mol For He:. 0 mol For H :. 0 mol.00 g He mol.06 g He mol. 0 g He. 0 g H 8. R; for a gas at two differet coditios: ; because ad are costat: 500 torr 9. K 758 torr 970 K EC 9., 5 atm ( L) 0.9 mol O L atm 95 K

8 CHAER 5 GASES 50. mol L atm 0.60 g (7 ) K.00 g atm 5.0 L 5. a. ; 75 g Ar mol Ar 9.95 g Ar.8 mol Ar 0.0 atm.50 L 69.6 K R L atm.8 mol L atm.8 mol 55 K b.,.50 L. atm ml.9 g ml mol O.8 0 mol O.00g L atm mol.0 atm 0. K.6 0 L 6 ml 5. For a gas at two coditios: Because is costat:,.50 mol 00. torr 800. torr 98 K K.77 mol Moles of gas added mol For two-coditio problems, uits for ad just eed to be the same uits for both coditios, ot ecessarily atm ad L. he uit coversios from other or uits would cacel whe applied to both coditios. However, temperature always must be coverted to the Kelvi scale. he temperature coversios betwee other uits ad Kelvi will ot cacel each other. 5., is costat. R costat, (.00), psi.000 (7 + 58) K 8 psi.00 (7 + 9) K

9 CHAER 5 GASES At two coditios: idetity of the gas i cotaier B is uimportat as log as we kow the moles of gas preset. ; all gases are assumed to follow the ideal gas law. he B A BB.0 L.0 mol 560. K A B AA.0 L.0 mol 80. K.0 he pressure of the gas i cotaier B is twice the pressure of the gas i cotaier A. 56. rocesses a, c, ad d will all result i a doublig of the pressure. rocess a has the effect of halvig the volume, which would double the pressure (Boyle s law). rocess c doubles the pressure because the absolute temperature is doubled (from 00. K to 00. K). rocess d doubles the pressure because the moles of gas are doubled (8 g N is mol of N ). rocess b wo t double the pressure sice the absolute temperature is ot doubled (0 K to K). 57. a. At costat ad, b.,, 7 K 50. atm 0.0 atm 0.0 atm.0 0 K 8 K 7 K 6.6 atm c. 7 K 5.0 atm 0.0 atm 7 K 58. Because the cotaier is flexible, is assumed costat. he moles of gas preset are also costat., ; sphere / πr r / π(.00 cm), / π(r ) 80. K 6 K 80. K 6 K.9, r (.9) /.09 cm radius of sphere after heatig 59. R costat, 70. torr ml 5 ml ( ) K (7 + 0.) K 5. 0 torr 60., costat, ; moles molar mass mass R

10 6 CHAER 5 GASES (molar mass) (molar mass) mass mass, Mass mass.00 0 g 9 K 650. psi 09 g 99 K 050. psi 6., is costat. R costat,,.00 L 760.torr 0. torr (7 ) K.8 L; L (7 + ) K 6., is costat. costat, R 9 K 5 K.0 0 m m Gas Desity, Molar Mass, ad Reactio Stoichiometry 6. S: 7 K ad.00 atm; at S, the molar volume of a gas is. L. mol O.00 L O. L mol Al 6.98 g Al. g Al mol O mol Al Note: We could also solve this problem usig, where have to memorize. L/mol at S. O /. You do t 6. CO (s) CO (g);.00 g CO mol CO.0 g CO 9.09 At S, the molar volume of a gas is. L NaN (s) Na(s) + N (g) 0 mol CO 0 mol CO. L mol CO.0 L N.00 atm 70.0 L. mol N eeded to fill air bag L atm 7 K mol NaN 65.0 g NaN Mass NaN reacted. mol N 5 g NaN mol N mol NaN 66. Because the solutio is 50.0% H O by mass, the mass of H O decomposed is 5/ 6.5 g.

11 CHAER 5 GASES g H O mol H O.0 g H O mol O 0.99 mol O mol H O L atm 0.99 mol atm 76 torr 00. K.0 L O 67. H.0 atm 800 m 00 cm m L atm 7 K L 000 cm. 0 5 mol. 0 5 mol H is i the balloo. his is 80.% of the total amout of H that had to be geerated: 0.80(total mol H ). 0 5, total mol H mol mol H mol Fe mol H g Fe g Fe mol Fe mol H g S mol S.07 g mol H mol SO H 0.56 mol S g HSO mol H SO 00 g reaget 98 g H SO g of 98% sulfuric acid 0.56 mol S will react with 0.56 mol O to produce 0.56 mol SO. More O is required to covert SO ito SO mol SO mol O mol SO mol O otal mol O reacted mol O L atm 0. mol 5.5 atm 6 K.8 L O 69. Xe(g) + F (g) XeF (s); Xe atm 0.0 L 0.8 mol Xe L atm 67 K We could do the same calculatio for F. However, the oly variable that chaged is the pressure. Because the partial pressure of F is triple that of Xe, mol F (0.8) 0.5

12 8 CHAER 5 GASES mol F. he balaced equatio requires mol of F for every mole of Xe. he actual mole ratio is mol F to mol Xe. Xe is the limitig reaget. 0.8 mol Xe mol XeF 07. g XeF 7.5 g XeF mol Xe mol Xe 70., ad are costat., We will do this limitig-reaget problem usig a alterative method tha the oe described i Chapter. Let's calculate the partial pressure of C H N that ca be produced from each of the startig materials assumig each reactat is limitig. he reactat that produces the smallest amout of product will ru out first ad is the limitig reaget. C H N C H N Ma C H N Ma C H Ma Ma if C H 6 is limitig 6 Ma C H N Ma NH Ma Ma if NH is limitig Ma C H C H N.500 Ma.000 Ma if O is limitig Ma O N C H 6 is limitig. Although more product could be produced from NH ad O, there is oly eough C H 6 to produce Ma of C H N. he partial pressure of C H N i atmospheres after the reactio is: a atm a.9 atm.9 atm 50. L 0. mol C H N L atm 98 K 0. mol 5.06 g mol.6 0 g C H N ca be produced. 7. CH OH + / O CO + H O or CH OH(l) + O (g) CO (g) + H O(g) 50.0 ml g ml mol. mol CH OH(l) available.0 g O.00 atm.8 L L atm 00. K.85 mol O available

13 CHAER 5 GASES 9. mol CH OH mol O mol CH.00 mol O OH.00 mol O is required to react completely with all of the CH OH available. We oly have.85 mol O, so O is limitig..85 mol O mol H mol O O.7 mol H O 7. For ammoia (i miute): NH 90. atm 500. L NH. 0 mol NH L atm 96 K NH NH flows ito the reactor at a rate of. 0 mol/mi. For CO (i miute): CO 5 atm 600. L CO mol CO L atm 96 K CO CO flows ito the reactor at mol/mi. o react completely with. 0 mol NH /mi, we eed:. 0 mol NH mi mol CO mol NH mol CO /mi Because 660 mol CO /mi is preset, ammoia is the limitig reaget.. 0 mol NH mi mol urea mol NH g urea mol urea. 0 g urea/mi 7. a. CH (g) + NH (g) + O (g) HCN(g) + H O(g); balacig H first, the O, gives: CH + NH + O b., ad costat; HCN + H O or CH (g) + NH (g) + O (g), HCN(g) + 6 H O(g) he volumes are all measured at costat ad, so the volumes of gas preset are directly proportioal to the moles of gas preset (Avogadro s law). Because Avogadro s law applies, the balaced reactio gives mole relatioships as well as volume relatioships. herefore, L

14 50 CHAER 5 GASES of CH, L of NH, ad L of O are required by the balaced equatio for the productio of L of HCN. he actual volume ratio is 0.0 L CH to 0.0 L NH to 0.0 L O (or : : ). he volume of O required to react with all of the CH ad NH preset is 0.0 L (/) 0.0 L. Because oly 0.0 L of O is preset, O is the limitig reaget. he volume of HCN produced is: 0.0 L O L HCN L O. L HCN 7. he reactio is : betwee ethee ad hydroge. Because ad are costat, a greater volume of H ad thus more moles of H are flowig ito the reactor. Ethee is the limitig reaget. I miute: CH 5.0 atm 000. L L atm 57 K 5 mol C H reacted 5 mol C H heoretical yield mi ercet yield mol CH mol C H 5.0 kg/mi % 6.0 kg/mi 0.07 g CH mol C H kg 000 g 6.0 kg C H 6 /mi d 75. Molar mass, where d desity of gas i uits of g/l. Molar mass L atm.6 g/l 7. K.000 atm g/mol he gas is diatomic, so the atomic mass 70.9/ 5.7. his is chlorie, ad the idetity of the gas is Cl. mass 76. (molar mass) d, d, volume (molar mass) mass Molar mass mass L atm g 7 K 96.9 g/mol atm (750. torr ) 0.56 L Mass of CHCl ; ; molecular formula is C H Cl.

15 CHAER 5 GASES d UF 6 (molar mass) atm 75 torr L atm 5.0 g/mol K.6 g/l 78. d (molar mass)/; we eed to determie the average molar mass of air. We get this by usig the mole fractio iformatio to determie the weighted value for the molar mass. If we have.000 mol of air: d air artial ressure average molar mass 0.78 mol N.00 atm 9 g/mol L atm 7 K. g/l 8.0 g N mol N + 0. mol O mol Ar.00 g O mol O 9.95 g Ar mol Ar g 79. CO mol L atm 7.8 g.0 g.0 L 00. K. atm With air preset, the partial pressure of CO will still be. atm. he total pressure will be the sum of the partial pressures, total + air. CO total. atm + atm 70 torr atm 80. H.00 g H H H mol H.06 g H 0.96 mol H ; He.00 g He L atm 0.96 mol (7 + 7) K.00 L mol He.00 g He 0.50 mol He. atm He He 6.5 atm; total H + He. atm atm 8. atm 8. reat each gas separately, ad use the relatioship ( ad are costat). For H : 75 torr.00 L.00 L 7 torr

16 5 CHAER 5 GASES For N : 0.00 atm.00 L.00 L atm; atm atm 50.7 torr total torr H N 8. For H : 60. torr.00 L.00 L 0. torr total H +, 0. torr 0. torr 80. torr N N total H For N : 80. torr.00 L.00 L 0 torr 8. ; the total volume is.00 L +.00 L +.00 L.00 L. For He: For Ne: 0.00 atm H 00. torr H.00 L.00 L.00 L.00 L 50.0 torr He 0.00 atm; 0.00 atm H atm 76.0 torr Ne For Ar:.0 ka.00 L.00 L total torr.0 ka;.0 ka atm 0. ka atm 90.0 torr Ar 8. We ca use the ideal gas law to calculate the partial pressure of each gas or to calculate the total pressure. here will be less math if we calculate the total pressure from the ideal gas law. O.5 0 mg O NH molecules NH total N O NH g mol O.7 0 mol O 000 mg.00 g O mol NH molecules NH 8. 0 mol NH mol total L atm 7 K.0 L 6. 0 mol total total. atm N mol N χ N total, χ N ; N. atm. atm 6. 0 mol total O. atm 0.0 atm; NH. atm atm

17 CHAER 5 GASES a. Mole fractio CH χ CH χ O.000! CH total 0.75 atm 0.75 atm atm 0. total b., total c. χ CH 6.6 CH total, CH 0 mol CH 0.5 atm 0.5 L 0.6 mol L atm 8 K χ CH total mol mol CH 6.0 g CH mol CH.06 g CH O mol mol O ; 9.7 mol O 86. If we had 00.0 g of the gas, we would have 50.0 g He ad 50.0 g Xe. χ He He + He Xe.00 g O mol O.0 g O 50.0 g.00 g/mol.5 mol He g 50.0 g mol He mol Xe.00 g/mol. g/mol He χ He total torr 58 torr; Xe torr 87. total + H O,.0 atm H 9.56 H H 0 mol H H + torr atm, H atm 0.0 L mol H L atm 0 K mol Z mol H 65.8 g Z 0.65 g Z mol Z atm 88. o calculate the volume of gas, we ca use total ad total ( total / total ), or we ca use He ad He ( He / He ). Because is ukow, we will use He ad He. He + O H H O.00 atm 760. torr, He +.8 torr 760. torr, He 76 torr He g mol.00 g 0.6 mol He

18 5 CHAER 5 GASES He He L atm 0.6 mol atm 76 torr 98 K.69 L 89. NaClO (s) NaCl(s) + O (g) total + O, O H total O 7 torr 9.8 torr 7 torr O H O O 7 torr L atm atm L (7 + ) K. 0 mol O Mass NaClO decomposed. 0 mol O mol NaClO mol O 06. g NaClO mol NaClO Mass % NaClO 0.58 g % g 0.58 g NaClO atm atm.8 atm is the pressure of the amout of F reacted., ad are costat. Moles F reacted.8 atm Moles Xe reacted. atm costat, or.00; so Xe + F XeF 9. HN (g) N (g) + H (g); at costat ad, is directly proportioal to. I the reactio, we go from moles of gaseous reactats to moles of gaseous products. Because moles doubled, the fial pressure will double ( total 6.0 atm). Similarly, from the : mole ratio betwee HN ad H, the partial pressure of H will be.0/.5 atm. he partial pressure of N will be (/).0 atm.5 atm. his is from the : mole ratio betwee HN ad N. 9. SO (g) + O (g) SO (g); because ad are costat, volume ratios will equal mole ratios ( f / i f / i ). Let x mol SO mol O preset iitially. SO will be limitig because a : SO to O mole ratio is required by the balaced equatio, but oly a : mole ratio is preset. herefore, o SO will be preset after the reactio goes to completio. However, excess O (g) will be preset as well as the SO (g) produced. Mol O reacted x mol SO mol O mol SO x/ mol O Mol O remaiig x mol O iitially x/ mol O reacted x/ mol O

19 CHAER 5 GASES 55 Mol SO produced x mol SO mol SO mol SO x mol SO otal moles gas iitially x mol SO + x mol O x otal moles gas after reactio x/ mol O + x mol SO (.5)x f (.5) x x f i i ; f / i 0.75 : l or : mol (CH) NH g (CH ) N H 60.0 g mol N mol (CH ) N 7.5 mol N produced H L atm 7.5 mol N 50 L 00. K 0.7 atm We could do a similar calculatio for H O ad CO ad the calculate total ( N + HO + CO ). Or we ca recogize that 9 total moles of gaseous products form for every mole of (CH ) N H reacted. his is three times the moles of N produced. herefore, total will be three times larger tha. total N N 0.7 atm. atm. 9. he partial pressure of CO that reacted is torr. hus the umber of moles of CO that react is give by: mol CO Mass % MgO 50. atm.00 L mol CO L atm 9 K mol MgO mol CO. g 00 8.% MgO.85 g 0.g MgO. g MgO mol MgO Kietic Molecular heory ad Real Gases 95. KE avg (/); the average kietic eergy depeds oly o temperature. At each temperature, CH ad N will have the same average KE. For eergy uits of joules (J), use R 8.5 J/KCmol. o determie average KE per molecule, divide the molar KE avg by Avogadro s umber, molecules/mol.

20 56 CHAER 5 GASES 96. Ar At 7 K: KE avg At 56 K: KE avg 8 g 9.95 g/mol 8.5 J 7 K.0 0 J/mol J/molecule 8.5 J 56 K J/mol. 0 0 J/molecule 5.7 mol Ar; χ CH CH CH + Ar CH CH ( + 5.7) CH,.7 (0.50), 0.6 mol CH CH CH CH KE avg for mole of gas KE total ( ) mol / 8.5 J/KCmol 98 K J 60.6 kj / 97. µ rms, M where R 8.5 J ad M molar mass i kg. For CH, M.60 0 kg, ad for N, M.80 0 kg. 8.5 J 7 K For CH at 7 K: µ rms.60 0 kg/mol Similarly, µ rms for CH at 56 K is 9 m/s. / 65 m/s 8.5 J 7 K For N at 7 K: µ rms.80 0 kg/mol Similarly, for N at 56 K, µ rms 697 m/s. / 9 m/s / 98. µ rms ; M M UF6 µ UF 6 UF6 / µ He He M He / M M We wat the root mea square velocities to be equal, ad this occurs whe M M. he ratio of the temperatures is: He UF6 UF6 He MUF M.00 UF6 6 He He He UF6 UF6 He /

21 CHAER 5 GASES 57 he heavier UF 6 molecules would eed a temperature 87.9 times that of the He atoms i order for the root mea square velocities to be equal. 99. a b c d Avg. KE icrease decrease same (KE % ) same Avg. velocity icrease decrease same ( mv KE % ) same Wall coll. freq icrease decrease icrease icrease Average kietic eergy ad average velocity deped o. As icreases, both average kietic eergy ad average velocity icrease. At costat, both average kietic eergy ad average velocity are costat. he collisio frequecy is proportioal to the average velocity (as velocity icreases, it takes less time to move to the ext collisio) ad to the quatity / (as molecules per volume icrease, collisio frequecy icreases). 00.,, ad are all costat, so must be costat. Because we have equal moles of gas i each cotaier, gas B molecules must be heavier tha gas A molecules. a. Both gas samples have the same umber of molecules preset ( is costat). b. Because is costat, KE avg must be the same for both gases [KE avg (/)]. c. he lighter gas A molecules will have the faster average velocity. d. he heavier gas B molecules do collide more forcefully, but gas A molecules, with the faster average velocity, collide more frequetly. he ed result is that is costat betwee the two cotaiers. 0. a. hey will all have the same average kietic eergy because they are all at the same temperature [KE avg (/)]. b. Flask C; H has the smallest molar mass. At costat, the lighter molecules have the faster average velocity. his must be true for the average kietic eergies to be the same. 0. a. All the gases have the same average kietic eergy sice they are all at the same temperature [KE avg (/)]. b. At costat, the lighter the gas molecule, the faster the average velocity [µ avg µ rms (/M) / ]. Xe (. g/mol) < Cl (70.90 g/mol) < O (.00 g/mol) < H (.06 g/mol) slowest fastest c. At costat, the lighter H molecules have a faster average velocity tha the heavier O molecules. As temperature icreases, the average velocity of the gas molecules icreases. Separate samples of H ad O ca oly have the same average velocities if the temperature of the O sample is greater tha the temperature of the H sample.

22 58 CHAER 5 GASES 0. Graham s law of effusio: Rate Rate M M Let Freo- gas ad Freo- gas : / / M,., M M g/mol he molar mass of CF Cl is equal to g/mol, so Freo- is CF Cl. / Rate M.0 ml 0. ; Rate M rate ; mi 7.8 ml rate ; mi M 6.0 g ; M? mol / 6.0 M 0.50, 6.0 (0.50) M, M g mol / Rate M 05., Rate M rate( rate( C C 7 8 O) O) / Rate( Rate( C C 6 8 O) O) / he relative rates of effusio of C 6 O to C 7 O to C 8 O are.0 :.0 :.00. Advatage: CO is't as toxic as CO. Major disadvatages of usig CO istead of CO:. Ca get a mixture of oxyge isotopes i CO.. Some species, for example, C 6 O 8 O ad C 7 O, would effuse (gaseously diffuse) at about the same rate because the masses are about equal. hus some species caot be separated from each other. 06. Rate Rate / M M, where M molar mass; let gas () He ad gas () Cl. Effusio rates i this problem are equal to the volume of gas that effuses per uit time (L/mi). Let t time i the followig expressio..0 L.5 mi.0 L t /, t.5 mi.09, t 9 mi

23 CHAER 5 GASES a L atm mol ( ) K. atm.0000 L b. + a ( b) ; for N : a.9 atm L /mol ad b 0.09 L/mol atm (.0000 L L). L atm.0000 ( atm)( L). L atm. L atm L 0.8 atm atm c. he ideal gas law is high by 0. atm, or %. 08. a L atm mol 98. K. atm L b. + a ( b) ; for N : a.9 atm L /mol ad b 0.09 L/mol atm (0.000 L L). L atm ( atm)(0.000 L L). L atm atm. L atm L c. he results agree to ±0.00 atm (0.08%)..6 atm, atm d. I Exercise 07, the pressure is relatively high, ad there is sigificat disagreemet. I Exercise 08, the pressure is aroud atm, ad both gas laws show better agreemet. he ideal gas law is valid at relatively low pressures.

24 60 CHAER 5 GASES Atmospheric Chemistry 09. χ He from able 5.. He χ He total atm atm atm L atm 98 K. 7 0 mol He/L. 0 L 7 mol L 000 cm mol atoms. 0 atoms He/cm 0. At 5 km, -50 C ad 0. atm. Use sice is costat..0 L.00 atm K 7 L 0. atm 98 K. N (g) + O (g) NO(g), automobile combustio or formed by lightig NO(g) + O (g) NO (g), reactio with atmospheric O NO (g) + H O(l) HNO (aq) + HNO (aq), reactio with atmospheric H O S(s) + O (g) SO (g), combustio of coal SO (g) + O (g) SO (g), reactio with atmospheric O H O(l) + SO (g) H SO (aq), reactio with atmospheric H O. HNO (aq) + CaCO (s) Ca(NO ) (aq) + H O(l) + CO (g) H SO (aq) + CaCO (s) CaSO (aq) + H O(l) + CO (g) Coectig to Biochemistry. atm 5.0 L. 6.6 mol NO L atm 98 K L atm 6.6 mol 98 K 60 L.00 atm As expected, the same quatity of N O at the same temperature occupies a larger volume whe pressure is decreased.

25 CHAER 5 GASES 6.00 atm 6.0 L. a. 0.5 mol air L atm 98 K.97 atm 6.0 L b. 0.8 mol air K 0.96 atm 6.0 L c. 0. mol air K Air is ideed thier at high elevatios. 5. Assumig 00.0 g of cyclopropae: 85.7 g C mol C.0 g 7. mol C. g H mol H.008 g. mol H; he empirical formula for cyclopropae is CH, which has a empirical mass.0 + (.0).0 g/mol. (molar mass) d, molar mass d L atm.88 g / L 7 K.00 atm. g/mol Because./.0.0, the molecular formula for cyclopropae is (CH ) C H (/) (costat); at costat ad, the pressure of a gas is directly proportioal to the moles of gas preset. Because both gases are i the same balloo, ad are ideed costat torr 70. torr O O cyclopropae cyclopropae 7. Because ad are costat, ad are directly proportioal. he balaced equatio requires L of H to react with L of CO ( : volume ratio due to : mole ratio i the balaced equatio). he actual volume ratio preset i miute is 6.0 L/5.0 L 0.60 (0.60 : ). Because the actual volume ratio preset is smaller tha the required volume ratio, H is the limitig reactat. he volume of CH OH produced at S will be oe-half the volume of H reacted due to the : mole ratio i the balaced equatio. I miute, 6.0 L/ 8.00 L CH OH is produced (theoretical yield)..5

26 6 CHAER 5 GASES CH OH.00 atm 8.00 L L atm 7 K 0.57 mol CH OH i miute 0.57 mol CH OH.0 g CH OH. g CH OH (theoretical yield per miute) mol CH OH actual yield 5.0 g ercet yield % yield theoretical yield. g ml juice ml C H5OH 90. ml C H 5 OH preset 00 ml juice 90. ml C H 5 OH 0.79 g C H5OH ml C H OH 5 mol CH5OH mol CO.5 mol CO 6.07 g C H OH mol C H OH 5 5 he CO will occupy ( ) 75 ml ot occupied by the liquid (headspace) L atm.5 mol 98 K CO 90 atm 75 0 L CO Actually, eough CO will dissolve i the wie to lower the pressure of CO to a much more reasoable value. 9. total +, 76 torr.8 torr 70 torr H N HO N atm 0.9 atm N N 0.9 atm.8 0 L.0 0 mol N L atm 98 K Mass of N i compoud.0 0 mol N H 8.0 g N.6 0 g itroge mol Mass % N.6 0 g 0.5 g 00.% N 0..5 mg CO. mg H O.0 mg C.0 mg CO.06 mg H 8.0 mg H O 9. mg C; % C.60 mg H; % H 9. mg 00 6.% C 5.0 mg.60 mg 00.% H 5.0 mg

27 CHAER 5 GASES 6 N 70. atm L N mol N L atm 98 K. 0 - mol N 8.0 g N mol N.98 0 g itroge 9.8 mg itroge Mass % N 9.8 mg % N 65. mg Or we ca get % N by differece: % N (6. +.) 60.8% Out of 00.0 g: 6. g C mol.0 g.7 mol C; g H mol.008 g.0 mol H; g N mol.0 g. mol N; Empirical formula is CH 6 N. Rate Rate / M , M (.07) g/mol Empirical formula mass of CH 6 N hus the molecular formula is also CH 6 N.. a. If we have L of air, the there are.0 0 L of CO. CO χ CO total ; χ CO CO total.0 0 because % ; CO torr 0.9 torr b. CO CO ; assumig.0 m air, m 000 L: 0.9 atm (.0 0 L) CO mol CO L atm 7 K. 0 mol molecules CO molecules i.0 m of air mol

28 6 CHAER 5 GASES molecules m molecules CO c. m 00 cm cm. For bezee: g mol 78. g mol bezee bezee bezee L atm mol 96 K atm 78 torr L Mixig ratio L 8 L ppmv vol. of X 0 Or ppbv total vol L ppbv.00 L mol bezee.00 L L 000 cm molecules mol. 0 molecules bezee/cm For toluee: g C 7 H 8 mol 9. g mol toluee toluee toluee L atm mol 96 K atm 78 torr L Mixig ratio L 8 L ppmv (or.7 ppbv) mol toluee.00 L L 000 cm molecules mol. 0 molecules toluee/cm

29 CHAER 5 GASES 65 Additioal Exercises. a. b. c. R costat cost cost R d. e. R costat f. costat costat R costat / Note: he equatio for a straight lie is y mx + b, where y is the y-axis ad x is the x-axis. Ay equatio that has this form will produce a straight lie with slope equal to m ad a y itercept equal to b. lots b, c, ad e have this straight-lie form.. At costat ad, Avogadro s law applies; that is, equal volumes cotai equal moles of molecules. I terms of balaced equatios, we ca say that mole ratios ad volume ratios betwee the various reactats ad products will be equal to each other. Br + F X; moles of X must cotai moles of Br ad 6 moles of F; X must have the formula BrF for a balaced equatio i HgCi.5 cm i 0 mm cm atm 760 mm.5 cm i L 000 cm 0.77 atmcl Boyle s law: k, where k ; from Example 5., the k values are aroud atm L. Because k, we ca assume that Boyle s data ad the Example 5. data were take at differet temperatures ad/or had differet sample sizes (differet moles).

30 66 CHAER 5 GASES 6. M(s) + x HCl(g) MCl x (s) + x H (g) H 0.95 atm. L L atm 7 K 0.00 mol H Mol Cl i compoud mol HCl 0.00 mol H x x mol Cl mol H 0.00 mol Cl Mol Cl Mol M 0.00 mol Cl mol M.77 g M 5.9 g M he formula of compoud is MCl mol Cl mol M 7. We will apply Boyle s law to solve. costat, Let coditio () correspod to He from the tak that ca be used to fill balloos. We must leave.0 atm of He i the tak, so atm ad 5.0 L. Coditio () will correspod to the filled balloos with.00 atm ad N(.00 L), where N is the umber of filled balloos, each at a volume of.00 L. 99 atm 5.0 L.00 atm N(.00 L), N 9.5; we ca't fill 0.5 of a balloo, so N 9 balloos or, to sigificat figures, 90 balloos. 8. Mol of He removed.00 atm L atm L 98 K mol 5 I the origial flask, mol of He exerted a partial pressure of.960! atm mol atm 98 K L 7.00 ml 9. For O, ad are costat, so. 785 torr.9 L.00 L 76 torr O total +, 785! 76 torr O HO H O 0., ad are costat. or Whe ad are costat, the pressure is directly proportioal to moles of gas preset, ad pressure ratios are idetical to mole ratios.

31 CHAER 5 GASES 67 At 5 C: H (g) + O (g) H O(l), H O(l) is produced at 5 C. he balaced equatio requires mol H for every mol O reacted. he same ratio ( : ) holds true for pressure uits. he actual pressure ratio preset is atm H to atm O, well below the required : ratio. herefore, H is the limitig reaget. he oly gas preset at 5 C after the reactio goes to completio will be the excess O. O (reacted).00 atm H atm O atm H.00 atm O O (excess) O (iitial) O (reacted).00 atm -.00 atm.00 atm O At 5 C: H (g) + O (g) H O(g), H O(g) is produced at 5 C. he major differece i the problem at 5 C versus 5 C is that gaseous water is ow a product, which will icrease the total pressure. H O (produced).00 atm H atm HO atm H.00 atm H O total (excess) + O (produced).00 atm O +.00 atm H O.00 atm O H kg Mo 000 g kg mol Mo.0 0 mol Mo 95.9 g Mo.0 0 mol Mo mol MoO mol Mo 7/ mol O.6 0 mol O mol MoO O O L atm mol.00 atm 90. K L of O L O 00 L air L O. 0 6 L air.0 0 mol Mo mol H mol Mo. 0 mol H H L atm mol.00 atm 90. K L of H. For NH : atm.00 L.00 L 0. atm

32 68 CHAER 5 GASES For O :.00 L.50 atm atm.00 L After the stopcock is opeed, ad will be costat, so %. he balaced equatio requires: O O 5.5 NH NH O he actual ratio preset is: NH atm atm he actual ratio is larger tha the required ratio, so NH i the deomiator is limitig. Because equal moles of NO will be produced as NH that reacts, the partial pressure of NO produced is 0. atm (the same as reacted).. Out of g of compoud there are: NH mol C g C.87 mol C;. 00.0g C.5 mol H g H 7. mol H; g H.5 mol N.5. g N.5 mol N; g N.5 he empirical formula is C H N. / / Rate M M ; let gas () He;.0, M Rate M.00.0 g/mol he empirical formula mass of C H N (.0) + (.0) + (.0).0. So the molecular formula is also C H N.. If Be +, the formula is Be(C 5 H 7 O ) ad molar mass.5 + 5() + () + 6(6) g/mol. If Be +, the formula is Be(C 5 H 7 O ) ad molar mass () + () + (6) 07 g/mol. Data set I (molar mass d/ ad d mass/): molar mass L atm 0.0 g 86 K mass 09 g/mol atm (765. torr ) (.6 0 L)

33 CHAER 5 GASES 69 Data set II: molar mass mass L atm 0. g 90. K 0 g/mol atm (76.6 torr ) (6.0 0 L) hese results are close to the expected value of 07 g/mol for Be(C 5 H 7 O ). hus we coclude from these data that beryllium is a divalet elemet with a atomic weight (mass) of 9.0 g/mol g CO.0 g C.0 g CO g C; % C g g % C g H O.06 g H 8.0 g H O. 0 g H; % H. 0 g 0.0 g % H, N.00 atm L atm L 7 K. 0 mol N. 0 mol N 8.0 g N mol N.5 0 g itroge Mass % N g g 00 7.% N Mass % O 00.00! ( ) 8.% O Out of g of compoud, there are: mol 7.78 g C 6. mol C; 7. g N.0 g mol.0 g 0.50 mol N 0.9 g H mol.008 g 0.8 mol H; 8. g O mol 6.00 g 0.5 mol O Dividig all values by 0.5 gives a empirical formula of C H NO. Molar mass d.0 g L L atm 56 torr atm 00. K 9 g/mol Empirical formula mass of C H NO 95 g/mol; hus the molecular formula is C H N O.

34 70 CHAER 5 GASES 6. At costat, the lighter the gas molecules, the faster the average velocity. herefore, the pressure will icrease iitially because the lighter H molecules will effuse ito cotaier A faster tha air will escape. However, the pressures will evetually equalize oce the gases have had time to mix thoroughly. Challege roblems 7. BaO(s) + CO (g) BaCO (s); CaO(s) + CO (g) CaCO (s) i i iitial moles of CO f f fial moles of CO 750. atm.50 L L atm 0. K 0. atm.50 L L atm 0. K mol CO 0.08 mol CO ! mol CO reacted Because each metal reacts : with CO, the mixture cotais a total of 0.0 mol of BaO ad CaO. he molar masses of BaO ad CaO are 5. ad g/mol, respectively. Let x mass of BaO ad y mass of CaO, so: x + y 5. g ad x 5. y mol or x + (.7)y Solvig by simultaeous equatios: x + (.7)y 6.!x!y!5. (.7)y.9, y.9/ y g CaO ad 5. - y x.5 g BaO Mass % BaO.5 g BaO % BaO; %CaO 00.0! 86.6.% CaO 5. g 8. Cr(s) + HCl(aq) CrCl (aq) + / H (g); Z(s) + HCl(aq) ZCl (aq) + H (g) atm 750. torr 0.5 L Mol H produced L atm (7 + 7) K mol H mol H mol H from Cr reactio + mol H from Z reactio

35 CHAER 5 GASES 7 From the balaced equatio: mol H mol Cr (/) + mol Z Let x mass of Cr ad y mass of Z, the: x + y 0.6 g ad (.5) x y We have two equatios ad two ukows. Solvig by simultaeous equatios: (0.0885)x + (0.050)y (0.050)x (0.050)y (0.055)x, x mass of Cr g y mass of Z 0.6 g 0.57 g 0.05 g Z; mass % Z 0.05 g 0.6 g % Z 9. Assumig.000 L of the hydrocarbo (C x H y ), the the volume of products will be.000 L, ad the mass of products (H O + CO ) will be:.9 g/l.000 L 5.56 g products Mol C x H y C H x y atm.000 L L atm 98 K 0.09 mol.5 atm.000 L Mol products p L atm 75 K 0.96 mol C x H y + oxyge x CO + y/ H O Settig up two equatios: (0.09)x (y/) 0.96 (moles of products) (0.09)x(.0 g/mol) (y/)(8.0 g/mol) 5.56 g (mass of products) Solvig: x ad y 6, so the formula of the hydrocarbo is C H Let x moles SO moles O ad z moles He. MM a., where MM molar mass

36 7 CHAER 5 GASES.9 g/l.000 atm MM L atm 7. K, MM mixture. g/mol Assumig.000 total moles of mixture is preset, the: x + x + z.000 ad: 6.07 g/mol x +.00 g/mol x +.00 g/mol z. g x + z.000 ad (96.07)x + (.00)z. Solvig: x 0. mol ad z 0. mol hus: χ He 0. mol/.000 mol 0. b. SO (g) + O (g) SO (g) Iitially, assume 0. mol SO, 0. mol O, ad 0. mol He. Because SO is limitig, we ed up with 0. mol O, 0. mol SO, ad 0. mol He i the gaseous product mixture. his gives iitial.0000 mol ad fial mol. I a reactio, mass is costat. d mass ad % at costat ad, so d %. iitial.0000 dfial.0000, dfial.9 g/l, d fial.7 g/l d fial iitial. a. he reactio is CH (g) + O (g) CO (g) + H O(g)., CH CH costat, air airair CH he balaced equatio requires mol O for every mole of CH that reacts. For three times as much oxyge, we would eed 6 mol O per mole of CH reacted ( O 6 ). CH Air is % mole percet O, so O (0.) air. herefore, the moles of air we would eed to delivery the excess O are: O (0.) air, 6 CH air air 9 CH, 9 I miute: CH 00. L 9 air CH air CH CH air.50 atm.00 atm L air/mi b. If x mol of CH were reacted, the 6x mol O were added, producig (0.950)x mol CO ad (0.050)x mol of CO. I additio, x mol H O must be produced to balace the hydroges.

37 CHAER 5 GASES 7 CH (g) + O (g) CO (g) + H O(g); CH (g) + / O (g) CO(g) + H O(g) Amout O reacted: (0.950)x mol CO mol O mol CO (.90)x mol O (0.050)x mol CO.5 mol O (0.075)x mol O mol CO Amout of O left i reactio mixture (6.00)x (.90)x (0.075)x (.0)x mol O Amout of N (6.00)x mol O he reactio mixture cotais: 79 mol N mol O (.6)x x mol N (0.950)x mol CO + (0.050)x mol CO + (.0)x mol O + (.00)x mol H O + x mol N (0.)x mol of gas total (0.050) x (0.950) x (.0) x χ CO 0.007; χ CO 0.0; χ O (0.) x (0.) x 0. (0.) x (.00) x x χ HO 0.067; χ N (0.) x 0.77 (0.) x. he reactios are: C(s) + / O (g) CO(g) ad C(s) + O (g) CO (g), (costat) Because the pressure has icreased by 7.0%, the umber of moles of gas has also icreased by 7.0%. fial (.70) iitial.70(5.00) 5.85 mol gas + + O CO CO CO + CO 5.00 (balacig moles of C). Solvig by simultaeous equatios: O + CO + CO 5.85 ( CO + CO 5.00) 0.85 O

38 7 CHAER 5 GASES If all C were coverted to CO, o O would be left. If all C were coverted to CO, we would get 5 mol CO ad.5 mol excess O i the reactio mixture. I the fial mixture, moles of CO equals twice the moles of O preset ( ). CO O.70 mol CO;.70 + CO CO 5.00, O CO.0 mol CO χ.70 CO ;.0 χ CO 0.56; χ O a. olume of hot air: πr π(.50 m) 65. m (Note: Radius diameter/ 5.00/.50 m) 65. m 0 dm m L dm L atm 75 torr L. 0 mol air L atm (7 + 65) K Mass of hot air. 0 mol Air displaced: 9.0 g mol Mass of air displaced.66 0 mol g 75 atm L mol air L atm (7 + ) K 9.0 g mol Lift g g.0 0 g g b. Mass of air displaced is the same, g. Moles of He i balloo will be the same as moles of air displaced,.66 0 mol, because,, ad are the same. Mass of He.66 0 mol.00 g mol.06 0 g Lift g.06 0 g g

39 CHAER 5 GASES 75 c. Hot air:.95 0 mol Air displaced: 60. atm (6.5 0 L) mol air L atm 8 K 9.0 g mol g of hot air 60. atm (6.5 0 L) mol air L atm 9 K g mol g of air displaced mol Lift g g g. a. Whe the balloo is heated, the balloo will expad ( ad remai costat). he mass of the balloo is the same, but the volume icreases, so the desity of the argo i the balloo decreases. Whe the desity is less tha that of air, the balloo will rise. b. Assumig the balloo has o mass, whe the desity of the argo equals the desity of air, the balloo will float i air. Above this temperature, the balloo will rise. MMair d air, where MM air average molar mass of air MM air g/mol g/mol 8.9 g/mol d air d argo.00 atm 8.9 g/mol L atm 98 K.00 atm 9.95 g/mol L atm.8 g/l.8 g/l, K Heat the Ar above K or 0. C, ad the balloo would float. 5. a. Average molar mass of air g/mol g/mol 8.9 g/mol Molar mass of helium.00 g/mol A give volume of air at a give set of coditios has a larger desity tha helium at those coditios due to the larger average molar mass of air. We eed to heat the air to a temperature greater tha 5 C i order to lower the air desity (by drivig air molecules out of the hot air balloo) util the desity is the same as that for helium (at 5 C ad.00 atm).

40 76 CHAER 5 GASES b. o provide the same lift as the helium balloo (assume.00 L), the mass of air i the hot air balloo (.00 L) must be the same as that i the helium balloo. Let MM molar mass: MM CMM d, mass ; solvig: mass He 0.6 g 8.9 g/mol.00 atm.00 L Mass air 0.6 g L atm 50 K (a very high temperature) 6. a + (! b), + a a b b + a a b b At low ad high, the molar volume of a gas will be relatively large. hus the a / ad a b/ terms become egligible at low ad high because is large. Because b is the actual volume of the gas molecules themselves, b << ad the b term will be egligible as compared to. hus. 7. d molar mass(/); at costat ad, the desity of gas is directly proportioal to the molar mass of the gas. hus the molar mass of the gas has a value which is.8 times that of the molar mass of O. Molar mass.8(.00 g/mol). g/mol Because H O is produced whe the ukow biary compoud is combusted, the ukow must cotai hydroge. Let A x H y be the formula for ukow compoud. Mol A x H y 0.0 g A x H y mol A xh y. g 0.6 mol A x H y Mol H 6. g H O mol H O mol H.8 mol H 8.0 g mol H O.8 mol H 0.6 mol A x H y 8 mol H/mol A x H y ; A x H y A x H 8 he mass of the x moles of A i the A x H 8 formula is:. g 8(.008 g) 6. g From the periodic table ad by trial ad error, some possibilities for A x H 8 are ClH 8, F H 8, C H 8, ad Be H 8. C H 8 ad Be H 8 fit the data best, ad because C H 8 (propae) is a kow substace, C H 8 is the best possible idetity from the data i this problem.

41 CHAER 5 GASES a. Iitially N H.00 atm, ad the total pressure is.00 atm ( total N + ). H he total pressure after reactio will also be.00 atm because we have a costat-pressure cotaier. Because ad are costat before the reactio takes place, there must be equal moles of N ad H preset iitially. Let x mol N mol H that are preset iitially. From the balaced equatio, N (g) + H (g) NH (g), H will be limitig because three times as may moles of H are required to react as compared to moles of N. After the reactio occurs, oe of the H remais (it is the limitig reaget). Mol NH produced x mol H mol NH mol H x/ Mol N reacted x mol H mol N mol H x/ Mol N remaiig x mol N preset iitially x/ mol N reacted x/ mol N After the reactio goes to completio, equal moles of N (g) ad NH (g) are preset (x/). Because equal moles are preset, the partial pressure of each gas must be equal ( ). N NH total.00 atm + ; solvig: N NH.00 atm N NH b. % because ad are costat. he moles of gas preset iitially are: + x + x x mol N H After reactio, the moles of gas preset are: + N NH x x + x/ mol after iitial after iitial x/ x he volume of the cotaier will be two-thirds the origial volume, so: Itegrative roblems /(5.0 L) 0.0 L 9. he redox equatio must be balaced. Each uraium atom chages oxidatio sates from + i UO + to +6 i UO + (a loss of two electros for each uraium atom). Each itroge atom chages oxidatio states from +5 i NO to + i NO (a gai of three electros for each itroge atom). o balace the electros trasferred, we eed two N atoms for every three U atoms. he balaced equatio is:

42 78 CHAER 5 GASES NO H + (aq) + NO (aq) + UO + (aq) UO + (aq) + NO(g) + H O(l).5 atm 0.55 L 0.05 mol NO L atm 0 K mol UO 0.05 mol NO 0.0 mol UO + mol NO. g 50. a. 56 ml 09 g HSiCl actual yield of HSiCl ml + HCl 0.0 atm 5.0 L L atm 08 K 5.9 mol HCl 5.9 mol HCl mol HSiCl mol HCl 5.5 g HSiCl 68 g HSiCl mol HSiCl actual yield 09 g ercet yield % theoretical yield 68 g b. 09 g HiSCl mol HSiCl 5.5 g HSiCl mol SiH 0.86 mol SiH mol HSiCl his is the theoretical yield. If the percet yield is 9.%, the the actual yield is: 0.86 mol SiH mol SiH SiH L atm 0.59 mol 08 K 0.0 atm L 907 ml SiH 5. hf,.0 + (9.00) 08.0 g/ml molar mass d 08.0 g/mol.5 atm L atm ( ) K.8 g/l he gas with the lower mass will effuse faster. Molar mass of hf 08.0 g/mol; molar mass of UF (9.00) 95.0 g/mol. herefore, UF will effuse faster. Rate of effusio of UF Rate of effusio of hf molar mass of hf 08.0 g/mol.0 molar mass of UF 95.0 g/mol UF effuses.0 times faster tha hf.

43 CHAER 5 GASES he partial pressures ca be determied by usig the mole fractios. methae total χ methae. atm atm; ethae.. 0. atm Determiig the umber of moles of atural gas combusted: atural gas. atm 5.00 L mol atural gas L atm 9 K methae atural gas χ methae mol mol methae ethae 0.898! mol ethae CH (g) + O (g) CO (g) + H O(l); C H O (g) CO (g) + 6 H O(l) mol H O 8.0 g HO 0.8 mol CH 9.6 g H O mol CH mol H O mol C H 6 6 mol H mol C O 8.0 g HO. g H O H mol H O 6 he total mass of H O produced 9.6 g +. g.7 g H O. Maratho roblem 5. We must determie the idetities of elemet A ad compoud B i order to aswer the questios. Use the first set of data to determie the idetity of elemet A. Mass N g g g N g N mol N 8.0 g N 0.0 mol N L atm 0.0 mol 88 K 0.7 L atm 790. torr Moles of A atm 75 torr 0.7 L mol A L atm (7 + 6) K Mass of A g.0 g A