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final + solutions - CHEMISTRY 110A INSTR R.N SCHWARTZ FINAL...

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Unformatted text preview: CHEMISTRY 110A INSTR: R.N. SCHWARTZ FINAL EXAM Friday, March 21, 2003 NAME: . SID # (last) (first) INSTRUCTIONS 1. This exam consists of 8 questions. Before starting, write your name on every page. 2. Indicate answers, including units, and show your method of calculation and reasoning. _ 3. No credit will be given for an answer alone or for an illegible answer. ' ’ 4. Use the conventions for significant figures and units. I _ 5. If you run out of space working a problem, use the back of that page and indicate on the, front that you have done so. 6. You may use a calculator. No books Course Grade: ___—_-__—-——-——- MAKE SURE YOUR EXAM HAS 17 NUMBERED PAGES! CHEM 110A FINAL EXAM 21 March 2003 NANIE: ’ 1.'_ "(25 points) Shown below is the phase diagram for the'carbon system. he , log [Pilatrm]. T/(K) :1) Describe the changes in the carbon system for i) isobarically heating from point a}, ii) isothermally compressing from point a, iii) isobarically heating from point b, and iv) isothennally V decreasing the pressure form point c. CHEM 110A FINAL EXAM 21 March 2003 4 NAME: lb) Identify in phase diagram by circles any triple points and describe What they involve. ' W = c) Draw and label a plot of ,u ’versus T for isobaric heating thrOugh point b. .. .J‘W‘ WWWYW v . (EMF: -é CHEMiioA FINAL V . ' p 21 March 2003 NAME: '2. (25 points) An approximate relation Ween the enthalpy Of vaporization and'the normal boiling point of a liquid (at V1 aim) is given by ’Trouton’s rule: Amp-Ho = chp, where the- constant c is equal to 88 JK’mol‘]. Benzene obeys Trouton's rule; and its normal boiling point is 801°C. "51) Derive an equation‘for the vapOr pressure of benzene as a fimction of temperature V T. b) Find the vapor pressure of benzene at 25°C. c) Find the boiling point of benzene at 620 torr. ' - * _ _ a» Pa 9 T:- L .er O __ A 5:: A VL“.J' {aw " Ag?” .\ T7“ We LR? T1 affine) Ho V i 4 ' ' -: $.61T ) a”) (8* V CHEM 110A FINAL EXAM I 21 March 2003 NAME:.I ’ V I ‘ 72F": {53/9,}? : 35’33/C J g: NOV/AM J’WP‘: fgg‘wwt > . Vagflmfi \ , 2%2/5 $339 7&4 ' “9,73!” M : a 5/5635» x {04325 W an”; ‘ f fle— = O 274% W , . I 72 8W7va x 353.3 2 ___ 310404 will ‘ ‘ - a 93+”? . My a 3,4 A xijogzw W5 KM » g FWQ CHEM 1'10A FINAL 21 March 2003' NAME: 3. ' (30 points) Given the following data fora solution of liquids A in B at 25°C: IPA/tort x A Pnltorr ' ’ 353.4 ‘ 1.00 0.0 V . 0,0 . 290.0 0.80 ' 80.0. 0'1 0 ' - , 230.0 0.60 135.0 34/0 165.0 0.40 185.0 .3. .0 a f 92.0 ‘ 0.20 230.0 .950 0.0 0.0 . 280.4 a) Calculate the activity coefficients of A at xA = 0.80 and 0.40 using the Raoult’s Law basis. ‘ ' ' -» A ,_ ,, V: ___ 9’4] 3’5 M34 1 69:353.¢;/;*Jg7—0m-4é42 (6’, 2.? 7404* 3 090,1 $7.07? ’22 53.!” 941/: aguxS . .V‘W CHEM 110A EXAM * 21 March 2003 NW: b) calculate: GE and SE for themixiure at XA‘= (assume nww _= 1.000 ). , 46 (WW 22+ _ ; 1: Jaw/{x {ts/M? fix-"329‘7-"(0149W-“W 4 ~ ' ' . 9» O-VD'ZV‘HAOO ' ’ . =‘ 944,2 ' I _>,___ _' __ I .’ , '5. 863’; . _ . -_ .='W\é (“v/1%”? 7L.Z6’Z’“bg):' 7" CHEM 110A FINAL 21 March 2003 NAME: 4. (30 points) One mole of an ideal gas initially at a volume of 5.0 L, pressure P], and temperature 298 K experiences the following reversible changes}. ,Step A: Isothermal compression to one half the volume, the new volume and'pressure being V; = V1/2 , and P2. StepB: Cooling at constant volur‘ne, until the pressure is returnai to the original value of P], the final temperature being T2. . ‘ . a) Calculate P5P), and T2. Also calculate q, w,.AU, AH, AS, and AG for. Steps A and B separately. If a r ' quantity can not be calculated because of insufficient information please explain ’b) Are the ' magnitudes (without regard for sign) of AU, q, and w- for Step C greater than, less than, or equal to the g a values of these quantities for the sum of Steps A and B? - ' ' ‘ i ' - _ ~ - -LLQK‘ '_ r ' ' ’ ' P Var: /ms<s’53*¢mm s‘sz’m swam ' ' - r ; , 22M“ «WU “Ii: @7’ ; mzi' ~ ~ F ' ,1 I ‘ I ,5 ’ I " F I, w‘jwi / _ I y H g n V I w — if. t ’z— ' WT! .—. #7 . . fl Wxg;31#¢?x;o_f" (,4 i... t \ «4):: - 5 w J “r , 7 m {57‘ a 7‘“ x“ 2.35"“ now)» ,- ,.%W,()(ff/¢l,$g{imj -' awn-7.171 T CHEM 110A FINAL EXAM 21 Mgrch 2003 NAME: wafipflm/J) Aliza; MM Ayzf-ruj :5“) A“) 14/4 :0. a“: 9% r' Prcb’ 4D '. ;' ' ' ‘ ‘ ’ ) ds~ 4,2,2”,— (§;_w = (saw, M: (Mk V ,, ‘ ’4' V ‘ ' REA; V '" *M“~3“’”EM d" V, $4.: firm! )1 293K} . . ; ' 'vA-é-‘J-szyf" . . - w) d”.- @VdF,‘ [gafl 4u=gy¢r= 6,, z 7;) 44’ ‘ 0 4m (Wavy =- {WK 3"‘3’WfimY (“W ‘ Live/8153.35? ' "V bur) 42¢ .:: .1? g M 049-: O ‘ 3?: 43¢..— —- 5750034 7 > L . w EL 1 4.9 a . 6V) 'dH‘: deT Alf- .: fl (9",mq €/¢?-29§>K AH= «304%.1-0’ ’7 M» 9m "(1)."; ' VI " ‘ wow Azzfiflm/Msznasgfgjw Wk skier, W CHEM 110A FINAL EXAM I 121 March 2003‘ NAME: '5; (35 points) a) I ' Calculate the ionic strength of an aqueous solution containing 0.010 mol kg” of the FeCI; and 0.006 mol kg‘of CdClz. Assume that bothvare strong electrolytes. = ifibrfizrfi f Jeri" + é" ' a+= /x 0.010) b.” = u 0.00!» > b -= 3,3 0.02’0-1- 11:01:95 I 0 - a ;: 0,030 f 0:,_0IL:0006/1 .‘ .65 ' V .7 ‘ _ b) Find expressions for the mean ionic activity in terms of the molality b for solutions containing the strong electrolytes: i) KCI and ii) 1712604); - _ _ our-“9w Vew— ' '7 Uwfb/b" :éb/b" ‘ WM ' , $034500} bi egg/6.549.: gaze/47240: mflfflfi giggly/om a .. I e) Calculate the mean ionic molality bi for a 0.03 mol kg" aqueous solution of the strong electrolyte CeCl3. I Ali I/J , I ,. QC? ‘ ‘ bizé/li) 'J 36:44:)? ‘Aa‘vl‘ .‘ AQ:y\/+r¥\2,_>n 13+ " 4%. Who ‘= 3 4%? 10 »- CHEM'IIOA FINAL EXAM ' " 21 March 2003' NAME: d) Given theexperimental data below for CaCl; (in water at 250C and 1 am): Mean Ionic Coefficient ‘ . b/b" ' yi ‘ 0.001 - ' 0.888 ' 0.01 0.729 0.1 _ 0.517 - 0.5 0.444 1 * 0.496 5 v , ' 5.91 10 .43.1 I i) ‘ Calculate the mean ionic activity; and the solute activity Iatb/b" = 1.0. V 44:50:19; ,5 bL—(va¢_).l:/bn _¢’%,b/b _., 5162+: Obe 41? a a A 2:: 0.4610X'4587il' ' .W': 3 '=~_ 0.184 _ _ I amwmj AL .—-— 00294 V i ii) How do these experimental values compare with that predicted by the Debye- ' Huckcl limiting law? ‘Is this what you would expect? Explain. ' ' ‘ UL fl ' .‘_ 0.612% {QEJI ‘ ‘Jf vb)b0”.=—i, 3W 3W6 . 1: gibwgz)‘+b&-<-4fl ; 56ng =+Vbla=4x451 .J x a -i ,. i; + 1X30 3 ' “armsz -_-. —o,sz>2l25x(-4)la "9 -— J.7§32—. ' ’ 49.052. a 2/7: =. W ‘ 2' 0.01%4 z a+ = = 0. 9“}?le L’- ’/3: , "‘ ~— -#- 115:, r - V 2‘7 5/59 =/_.4&%5n~ CHEM 110A FINAL EXAM 21 March 2003 NAME: 6. ' (30 points) Shown below is a plot of the vapor pressure'versus composition for a 2'-component liquid system made up of molecules A and B at T = 298 K. ' a) In the figure graphically display Henry’s Law and Radult’s Law for both components. To get full credit you must label these features clearly. 12 CHEM 110A FINAL EXAM 21 March 2003 NAME: b) What are thc‘values of Henry's constants for. both_cori1ponéntS? 1' ‘ éq?‘°[0/;{)V‘ “My 45 W ' c)' Find the mole fraction of each component in the ilopor phase in equilibrium with the liquid solution * t of composition 0.60 inole fraction B. ' ' : i z / 75 * J V y 2' 0,0“. M :0")! g 8. ‘ (7% 7M7” .¢5 "’* gazsgw ' PW 96%:(mw i * v -Wm;w/s)-.‘ VP: - «A ’. 564m» _. 7*, 2/1 7.97M 1 2f, pfim.._ flaw. j _, CHEM 110A FINAL EXAM ' 21 March 2003 NAME: 7. (25 points) Solid AX decomposes‘accordingtq the qquation - 1 ~ Mm é’rfis) +5X2(g),- The equilibrium vapor pressures of X 2(g) over AX“) is 3x10’4 bar at 650 K and 63:10"3 Bar at ' - > 850 K Determine A‘,H°, A,G°, and A,S° ai 750 K. Ex (WKKbSDK ,i-‘gésow a 623K) =—- 244% éJ/mw-L W CHEM 110A FINAL EXAM ' 21 March 2003' NAME: 059K, Arh'gp/g ‘7‘35éffw W .fi’” =x<9om7 * r ".“‘£’7’%% (:7auz):—gawgwxm~ ; * X{+3x/4&évd:} V CHEM 110A EXAM _ 21LMarch 2003 NAME: ' 8. (25 points) a) When a small charge dQ is motled through, electric petential difference ¢ , the Work done on the charge is given by dw : ¢dQ ., i-) Write down the eombined 1" and 2"“ law equation for the I .intemal energy Uin differential form and ii) Show that , ’ . ' _16 CHEM 110A FINAL EXAM 21 Marchzom‘ NAME: LP [ vii/4L m aw emw 61nd: alnal 00112 b) ‘ Given that p,- = + RT Ina; , show that [ V j . Blnxl ‘17 Lg), Some useful equations and values of fundamental constants are given below: latm = 101.325 kPa,1bar =105 Pa, lPa = Jm‘3,1atm = 760 torr= 760 mm Hg, R = 8.31447 JK’mor’ = 0.0820574 L atm K1 mor’ = 0.0831447 L bar K’mor’, 1L bar = 100 J, For f(x)=xa: d%x=axa—l, Ixndx=xn+1/n+1, Ix—ldx=1nx, I dx = (thfiww [3), a and ,6 are constants , dfnx = dX/x , [fix = 23031081095 s ax+,6 @fiHW%@@@H@%I dcz acy @xo'xz’ azayoyx ’drz ([email protected] dw=—pexth, w— Ipexth,dS=dqrev,dS2qu, H=U+pV, path T G=H—TS, A=U—TS, dU=TdS—pdV, dH=TdS+Vdp, dG=—SdT+Vdp, dA=—SdT—pdV, [p+1'2—“](V—nb)=nRT, dS=CngTT+(a%T)VdV, ds=CpiTT-—(5%T)pdp, CV=(0”U/5T)V, R/Cym _ o'H _l_ W _ 1 W _ j 7- Cp—( /0'T)p:0!—-V(fl] ,KT-——I;[E T, T. — V ,pV —const, . . . _ 5 p17 C Foranmonatomzc1dealgas.CV,m-— R,Cp,m=ER,Cp’m—CKM=R,Z=__,,/= %V’ I7 5U = Q? _ 5H _ _ 5V _ _ 2—1 (/o’VV)T T( /0’T)V 1” ( /@)T—V T( /5T)p,f-¢PJ"¢-g[_p P, 0 dG=—SdT+Vdp+Z,ul-dn,-,G=Znipi, (“AG /T%T] =—AH7RT2,Gm=G/n=,u, i i p V. Pi,eq I 6111K A H0 "i =no,i +Viga ArGo =_RTgnKp’ KP =H[ 0 J ,( p 51' = r RT2 , i P P For a general chemical equation: 0 = Zvl-Ji , [3,60 = ——RT€nK , where K = H(ai’eq )V" and i i ideal real a,- =1 for pure solids or pure liquids, ,ul- = ,ul-o + RT lnxl- , ,ui = ,u," +RT£nai; al- = yixl- , 'd * I GE =Z"i(#leal *fll gal), indfli = 0, indlnh = 0, Pi = VixiPi = yipTotal =7ixikia i i i inin = O; = , Yi = partial molar quantity, Y any extensive property , i i T ,p,n,-,,- — / 1/ — — 1/ a: =(ai+ar )1 V = aware, at =ribi, n =W73 )“hbi = (W! ) V, where b+ = v+b , b- = v_b , Where it is understood that b = b/ b0 and v = v+ + v_ , 10g107: =—0-509|Z+Z—|11/2’ I=%Zbizi2 , i A S A H = 0:63 _ , At aphaSe transitiona <——) ,6 (Tandp constant): AaHflS = fl ’ dT (19,6 Ade/3V Taefl denp=fififi F=C_p+2. For a (—) ,6 corresponding to sublimation or vaporization: 2 RT ...
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This test prep was uploaded on 04/06/2008 for the course CHEM 110A taught by Professor Schwartz during the Winter '06 term at UCLA.

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final + solutions - CHEMISTRY 110A INSTR R.N SCHWARTZ FINAL...

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