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F05 MT2 - CHEMISTRY 110A INSTR R.N SCHWARTZ MIDTERM EXAM#2...

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Unformatted text preview: CHEMISTRY 110A INSTR: R.N. SCHWARTZ MIDTERM EXAM #2 Wednesday, November 23, 2005 NAME: SID # $53—— (last) (first) Instructions: 1. This exam consists of 3 questions. Before m 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. 5. If you run out of space working a problem, use the back of that page and indicate on the flow that you have done so 6. You may use a calculator. No books or notes. 7. Useful constants and equations are on last page; may carefully detach this page. MAKE SURE YOUR EXAM HAS 8 NUMBERED PAGES. CHEM 110A MIDTERM #2 23 Novuher 2005 NAME: 1. (40 points) a. Solid 190(5) decomposes when heated according to the equation K20(s)-+2K(s)+%02(g). I Given thatthe pressure OH); is 0.300 mbar at 700 Kand 1.50 bare: 800K, determine Angm , Aagm , and A830“, at 800 K. \ Am # dQ/nke :3. bum, CAT 0le AT > T 3°32 SGML“ H%:*( (%,L\ ; DMWHQJMW (TC) bu —- R [In (w y: 173331 (I. __L\"\ AH?- LE ydrilmmflm. 52¢} (seem) z m. :rzln—rw? (@Eook CHEM '1 10A MIDTERM #2 24 November 20“ NAME: b. Consider the reaction H2(g)+ Mg) ->'2H1(g). One mole of H;(_g) and one mole of 12(g) ate initially placed in a reaction vessel. The temperature is raised to 600 K and at equilibrium KP = 7-8.3. Find the partial pressures of the various Species at 600 Kgiven that the total pressure in the vessel equilibrates to 2.00 bar. CHEM 110A MIDTERM #2 23 November 2005 NAME: 2. (35 points) A strip of rubber is subjected to reversible elongation and reflections with dw = —}d€_, where y is the tension (units: N) and E is the length (m). a. AssumingtheintemalenergyU, tobeaftmctionthelenghé' andtemperaun-e T,showthat dqm hasthefollowingformdqm =[6U] (iii-{[23] +7}d£’. wumfie om (9» fir\dT*(%-u 0“ m dCswu‘KO‘zQ’gT Adf+ (315T A’Qd °“I>W (9;; 0‘“ {(3—2—133‘10‘“ b.13ynotingthat dS'=dq;”, showthat[%—3] 3r+ 3T?) cisadjqkn Tfi’fidW—H'Dgi Jrqd. $1; C\%“w~h [SQ/{1330 ( QR) dS- Whficifi N(T1\di3 (‘;—~ flyefil (CHM—(0U (”3W H3 K51 [T693 (or ”1' .(LJ J. on J.” a = j: «A fliflfififfi 3%: 17% -—L 92. T d T“ B - ( B +K]: [(w. T 0-.- L as 4 P CHEM 110A NHDTERM #2 23 November 2005 NAME: c. Setupthetotaldifierentialoanndshowthat [E] {921] as T” 5T, R= Ur'I'S de: duflSdT—TdS dtb dag-u " XCUL “ 1’ng ‘Xdl cm: 138—de ¢ng—Td& dw— - My. — gm (C29 :" (.QE‘Bt‘S £23.92 ii 9(gp‘x 2 (@3113 dldT an); a1" )1 T- 5 CHEM 110A MIDTERM #2 23 November 2005 NAME: 3. (25 poim:)A Carnot cycle operating in reverse becomes a refrigeration engine (refiigerator). The enthalpy effusion forwaterat 0°Cand 1 barpressme is AH}; = 6.010kI/mol :1. Draw, label and describe a schematic model of the reverse Carnot heat engine. Wincwafi [mo-MB Tm V @Vtai . @JMLALMK ASE?) H d, J «egg 93‘ 6* {I N\ SEW/c: W W {W ”“9 ($139033 5‘ of) ce‘s‘fiuw (“asexvorf 5‘? W W “iv-ht. We" “39‘4”” T (gotuewdlm‘xsb Caed- resa uni-«r h. Calculate the minimmn amount ofwork required to fi'eeze one mole ofwater at 0°Ciffl1e mundings is at 25°C. //// = C6le : E Cd'ed :: TCol‘J m. will“ to (gnaw (gm The 'Tml Wmfie‘sz Q7$\< '- ‘Olql £93K -973K 3 _ 1‘ H: a (bowl! m9 ue%w%m LWMQM A” 0 (o. 91 6 CHEM 110A MIDTERM #2 23 November 2005 NAME: c. How many Joules of energy in the form of heat are given up to the surroundings for each mole of ice produced under these conditions? V0 7 (EM +wa£cfl 66h; " W’Cécoch : SQO‘LFS“ OOIDXEIE ENDOFTHEEND 7 Some useful equations and values of fundamental constants K = 8.31451 JK" mor‘ = 0.0820568 L arm K" mor’ = 0.08314511. bar K" mor’, 1 L bar = 100 J, 1 L aim, = 101.325 J, N, = 6.022142: 1023pamcles-mor’, For f(x)=x‘-’: d%x=ar“’l, Ix”dx=x”+lln+l, Jx'ldx=1nx, dgm=d¥A, I dx (1):»(mfi).Ji=fi—2[m(a+m+(a“ ] ar+fi dU=dq+dw. dw=-Pde, (08:91:- , dU=TdS—PdV, H=U+PV, CV = (av/fly 0? = (071/50 (0%»)? = 05%)., -—P, (5%,), = V~T(a%r),,, dS=C?VdT+(ayfl.)ydi/, dS=%§-dT—(W/fl)PdP, 17 fl =fia=fi’ R/C. _ . V In e=—————q"“ +9“; =—T"°' Tm“ , PV” =constant, 7=CP,m lCV,m’ [%]=[%] , i ‘10:» Thu: j PV=nRT. Foran ideal manatomicgas:CV,m =%R, Cp_,, =%R, 7 . Foranidealdiatamicgaschr’m =§R, Cp’m 25R, Foramdealgas CF," —Cy,,,, = R, d6 = —SdT+VdP+Zp,-dn, , 0 = 2m, , n, =n;’ +V,§, ML,“ = 2mm)?” acgmm =Zv,AG}J, l‘ 1 I I P Mfume" = thsfr.” KP =1_I[ It? I } J , new,” =AG;m +RT mg,” 06,101,“ = ~RTanP, danP _ Aerm dT RT” ...
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