390_prelim3_key_07

390_prelim3_key_07 - Chem 390 Physical Chemistry I...

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Unformatted text preview: Chem 390 Physical Chemistry I! Chemistry 390 Prelim 111 'Thursday, April 12, 2007 Name «Va—e Problem 1 [22 Problem 2 /28 Problem 3 _____.,__________../18 Problem 4 / 12 Problem 5 [20 Total / 100 When working problems, yOu should clearly state any assumptions made. The last page of this exam has a list of useful formulas and constants. You may wish to detach the last sheet for easy reference. not permitted. Spring 2007 Calculators that have infrared transmitters or that allow alphanumeric input are lof13 Chem 390 Physical Chemistry II Spring 2007 ____________________._.____._.—————-——————-_— Problem 1 Toluene (A) and benzene (B) form an ideal solution. At 300 K, p; = 4.27 kPa and 393 = 13.73 kPe. A liquid mixture is composed of 3 moles of toluene and 2 moles of benzene. The pressure over the mixture at 300 K is reduced firom an initial large value (>> 335) until the first vapor forms, at pressure p = p'. The pressure is then reduced further until the last drop of liquid disappears, at pressure p = p” . (1) Sketch this process on a pressure/composition diagram. Be sure to label axes carefully, and to indicate pressures 101, pg, 19’ and p”. Cohfi'l' I lolfiml SoluL‘OV‘ 5: {3:1 = O Q 3+2. —‘v-- H r.— 2» u >\ 20f13 Chem 390 Physical Chemistry II Spring 2007 (iii) What is the composition of the first trace of vapor formed? f ‘ LL+ ‘-= Wok“ AFWGQLOW OF A 11/1 Vafo'v {PLIA$(__+ tun (71:,utrlml" AJ/ “(Tho/I MIX'JLL'VL. / w o PM :1 0‘4 XA ,g I: I dt‘ at l 5 (‘4 #A fl!“ Xfi c o = / f; / h4g2! Vafc‘v W I‘K / X :> JA -‘ V ffA 30f13 Chem 390 Physical Chemistry II Spring 2007 (iv) What is the value of the pressure 3)”? L46"; diva}, of (liglv'lo’! 9’4*l50}:}><’.4¥5' e ‘erSS-‘LVL'. I? = F” AP #3: f5” / WOLL flagrjfi'w OF A v’fl: I” - F St, JA '" “(/4 LL; “flak Pcc Liam IDL—uswtw '53: 4of13 : 5.8) KPQ Chem 390 Physicai Chemistry II Spring 2007 Problem 2 (1) Starting with the Gibbs—Duhem equation, derive the general relation 6 6 we) we) 3’1 Tm “‘2 Tm for a binary mixture, where M is the chemical potential of component 3' and 32,- is the corresponding mole fraction. Suppose the chemical potentials in a binary mixture of volatile liquids have the form in = of + RTln 3:1 +0131ch (2a) #2 = a; + RT in :62 + 0532331 (21)) where M; is the chemical potential of pure 3' and a is a constant independent of T and p. Verify iW‘uf... 1P0“ X,:XL. that the chemical potentials (2) satisfy relation 3 e":— L Q, it: L“) On] we @ filoi/VLI —r 142 owl/ml : 0 AS :— r‘t-ioL xJ :> )<l Ojf/‘l + X2 (Ti/“3— :— 0 Al- ConS‘i’ T/ is ‘ or a! J’X’Q/M/ Q :> 2<,(,M;/9Xr)T/}olx. + X2 rQ/“z/Qxl/Lf/u : Sinai 01K; 3 4X2 w__ n _ in! {1:3 , i _ r l i G x; \/” gal/iii} _ x2< girl/Til? From 25/ b (Kg/“I/ \. :: E: + K 9/ [X’UFXIJ] CD \ 9X1/T/F X: QXl ( Xe “‘9/“1/ \ 35—2 Chem 390 Physical Chemistry II Spring 2007 xr<gkygxifi 7' ET 7* 0" "H" “2"”; 9 _ x;(/”’/3XJ_) : R; + a< XL(/~2x;) ‘ wit. ” Ih ) {<1 (9/4/9“ # x3 {abuzz/fl ’ U S) r- ‘ / \ dr\2_./..- Xf-z {‘1 AB Q7. (1) IS‘ no!” 34+I S {:ch {:ow QH wath 0? )(i. I TL»: modal |§ aoPua/U ULHFL\%3'}C‘QI . vcs.‘ oi» fim'l" it“; 01065 Mutt} agtrg‘!’ .712; (ii) Derive an expression for the activity coefficient 71 of component 1 with respect to the Raoult’s law reference state. {Walk/w { h? ’ J Showthat'yl—rlasxlalandasxlao. Fm Raw»; ,_ 0? fl! + aghviij at 3:. fllf "f‘ [H [X.] ‘5' XXI/<3" lefk Jaw Ware-(mac: ) 96.?” RT ,VliQ‘j ® Chem 390 Physical Chemistry il Spring 2007 (iii) Calculate AmixG for 71 moles total of the solution If we set AmixG = AmixGideal + AmixG’ , give an expression for AmixG’ . Derive expressions for AmixS and AmixH for the mixture ’1 29/ JF Awux CT : "I/“I + “z/AL "” "l/Mt “nib/Hz, : VIJCRTIH K! #- 9(X:)<;) +“LCKTLX; + (XXI’KL) :: HKT [Xrlhxi +XLIHXZ] 2. Z. + Ho<[)<,,~<2_+x2_x, f'c in! I :: Améx Q <= nRTLK, lnxl +29 nXéj) 'IL Awix Q I <_ no< [XCXL + «LEA/(J AWIK S I _ 9AM,)< : —nR[XJ Mx, TX; h XL’] “9464 2 AH —-TAS :> Am H r ~+ n o< [xfxm xxx-l] 7of13 Chem 390 Physical Chemistry II Spring 2007 Problem 3 (i) Under a 10 bar pressure of inert gas, liquid water at 298 K has a larger vapor pressure (partial pressure) than it has in the absence of inert gas pressure. [El Explain briefly the thermodynamic reason for this increase. i, L)me fvcc'ir..v€,_ ? >> i’nzo 2 4/ _, J} flHLO > /4 “L0 (#142: .,/1 —fihr~i\7“§ (Lmirai fora-(f of II7UI.0/{ HLO f -+ Hip '3 ‘‘‘‘‘‘ 7' I , I veep Cincinnati *oibninfii of HLO M C2; F vqitw qui" If {N 1—1 9'; HLO if ’—'—‘ {M mm 6.7L) l i I :>- lpaiv («all F'c. 930 Vt» . rap (ii) Calculate the increase in vapor pressure produced by the pressure of the inert atmosphere on the water. (The vapor pressure pfho of pure water at 298 K in the absence of inert gas pressure is 3167133.) MHlo ,—_ 18 j moi-i / YHLD [\7 Cm’g ii '2 O 3 2‘ + Cs 9 1/1 f i IF\\ “‘7‘ ‘1: i; *1 \_,_ H x V“ 0 0 + 7v *1 3‘ \\~7 ‘U- 3 \9 «A—U. \ 0 Chem 390 Physical Chemistry II Spring 2007 Blank page fun: HLo 13 CM?) le—l .— 2? 2? _ j ’3 a} VH2} {2 H10 ‘ PM F1 .4 : 2—723 K : [O Law 3" ,..( H20 :7: lg CAM ma{ 2/ _ - \ FHLO "" 1514,} T ’25 I 90fl3 m kw Chem 390 Physical Chemistry ll Spring 2007 —-—————~———————_flu__—._._—_—_fi___ Problem 4 Oxygen can be removed from a gas stream by passing the gas over hot copper particles causing the reaction 1 2Cu(s) + %02(g) : 0120(5). (3) Given that ATG" : oxygen in {for example) nitrogen when equilibrium is reached at a total pressure p = 1 bar. in» K I r s mewcs N f “I F H E / )ii gait?! “‘85 pie Hroi” Z/PO 9})?ear M “fail-‘ng -VAVQO/FT“ : & Argo : slicer mm“ fl" : foo K _> ': X (OR i 2? : Fi i:'o]_/FD 1 kl -23 = 1773! x to iDO : X )0 —- 2:5 Law r If?) 2 4 10 of 13 —130.75 kJ/mol for reaction (3) at 500 K, calculate the partial pressure of J Chem 390 Physical Chemistry Ii Spring 2007 Problem 5 The following reaction is at equilibrium at T = 1273 K and p = 30.4 bar: COg(g)[43 moi] + C(3)[68 moi] r: 200(g)[210 mol]. (4) (Equilibrium amounts are given in square brackets after each substance.) (i) Calculate the composition of the gas at T = 1273 K and p = 20.3 bar. State all assumptions made. r“ —- I): 30 ’1’ 19“" l XCO : "ca 7; HCO '- 2N3 f MM I “(3.. l 1-": K H + W «5"?— JJAl \ / K ! Co Co ‘ 1 2— 1’.- NAG?ch 2: 0.33 ,. ,9 £0, ELM!“ @ :> : 30 if; KCOL ~ la Kw - .2 gm A a l 20 ) Kl = 203 "'50 : [23 2.": 9 T: - \ (J'XCG/ SOL/Q {W X60 3: —- 1-i- @ ch * 0 A _ 0 mg I) Co). — Ed— 11 of 13 Chem 390 Physical Chemistry II Spring 2007 (ii) With the system in equilibrium as in part (i), several moles of inert N2 gas are pumped into the system at fixed temperature and pressure. After equilibrium is restored, will the partial pressure of CO be higher, lower, or the same as you calculated in part (i)? To obtain full credit you must clearly explain your reasoning. RX" (:02ng +c_(5) a9 2036) Av =+|. 943/ l .. , (2) \ 33"? : M 7} : fiKTleK# 1‘ ng“QF /*-\ = 62); (F/Fo) m _‘ 2. / Q A 5 Q; Co I" Co IL ,1" a : I"I C c [ ll Co A n t; 0 & Mow-l— Jas 2? w {:05 If //" z? a 43C. \0 => (9&4; i) J, ,r 73 i“ ‘ e :> d3 > o (Dc—cw:de :> Marc vom/lUC-ehaA 12 of 13 ...
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390_prelim3_key_07 - Chem 390 Physical Chemistry I...

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