F04 MT2 - CHEMISTRY 110A INSTR: R.N. SCHWARTZ MIDTERM #2 ....

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Unformatted text preview: CHEMISTRY 110A INSTR: R.N. SCHWARTZ MIDTERM #2 . Wednesday, November 24, 2004 NAME: v SID # 8 (last) (first) Instructions: 1. This exam consists of 3 questions. Before starting, write your name on every page. 2. Indicate answers, including units, and show 4 your method of calculation and reasoning. 3. No credit will be giyen for an answer alone or for an illegible answer. ' 4. Use the conventions for significant figures and units. 5. . Ifyou 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. N 0 books or notes. 7. Useful Constants and equations are on last pages; ' you can carefully detach these pages. MAKE SURE YOUR EXAM HAS 9 NUMBERED PAGES. . CHEM 110A MIDTERM #2 24 November 2004 NAME: @4— 1. (35 points) Phosgene COCIZ (g) undergoes the following reaction C0C12 (g) —> C0(g) + C12(g) Given the following information: K; = 8.0x10‘9 at 100 °C and AS§73K = 125.5 JK‘lmorl. ‘ a. Calculate the equilibrium value of the extent of reaction fieq for phosgene at 100 °C and 2 bar total pressure. Assume initially 1.00 mol of. COCIZ (g) in the reaction vessel; b. Calculate AH €73 K for the reaction. c. If the reaction, originally in equilibrium at 100 0C and 2 bar total pressure, is suddenly perturbed by increasing the temperature (total pressure held fixed) to say 400 °C , which direction will the reaction equilibrium shift in order to relieve the temperature increase? Explain your answer. a. W moot-J? act; '_*°r~=¢oa got/ate,» Mo MV—f , M 221'.) 5%?) am 7‘? f ‘ p W0"? {ow-f W24?) “00 ~ +5 { ._.. f l to in o \\i \Q 3 1 N CHEM 110A MIDTERM #2 24 November 2004 NAME: W ’ ‘ W 0N go 3:5.»— Zo mm M $110 0 _ 1a ‘ I : “fa.% :- 70X10-9X 3.4—5” — 940x”)? . ~5’ A. o ,, '0 9 , o ‘ 0' Aé,=—fl7&u>4 ., #3735 A4é731(5-/€X3?3z€xflw¢ W0 4%:3’8: ~£3MQ§£¢X39WXAOflD >05?) = 7‘ 5% év/M/ '0 0 o 0" i 0 pt “Tag/z, “ W932 ' 3gg£§m Mr‘ 4”? Mgr) 0 a . 44393;: z 4%” 39%4559326 = val/J WM I M 2 *0 _.o M“€’§é= 13%;” ’ 452:. «2&0 K- W €72. CHEM 110A MIDTERM #2 24 November 2004 NAME: ’ 2. (35 points) Consider the phase diagram for single component system shown below: P/bar _! | l l . 100 400 800 1200 1600 T /K where Solid ,6 and Solid 0: refer to solid phasesa and [3 . a. What are the values of f (degrees of freedom) at the points 1, 2, 3, and 4 indicated in the phase diagram above. b. Sketch and label a plot of ,u vs T for P = 80 bar. c. 1 Sketch and label a plot of ,u vs P for T = 850‘ K. d. -A What can you say about the density p“ of the solid phase 0: Compared to the density of the liquid phase p1. Explain your reasoning. 4- V4” 4 v W tie. (:6 22w 222% '7 ” M ,l: c7»; = 379 lkjfigj .fl; “(7,1 :5 M7533 MW ##sz “viii flflzfeafzf CHEM 110A MID'I‘ERM #2 24 November 2004 72—79le CHEM 110A MIDTERM #2 24 November. 2004 NAME: 3. (30 points). A gas obeys the equation of state: PVm' =0 RT + bP , where b is a constant. If the gas expands from VmJ to Vm’z reversibly at constant T, derive expressions for AA,” , AS," and AGm. Express final answers in terms of R, T, VmJ , and Vm,2 . Show all steps to receive credit. 0) dflw *é'Mcéz'wfl/g” , m 72%) «L729 444": w}; Afigwjf I, I 4-6 2 L~ é) CHEM 110A MIDTERM #2 .ma mp Mfi‘Mm-fi :- Mf. 6 _ h] g . ' (2.1-?) 4/»: £7/&1[fl!3]+/5 .. A } I g [W W A) 3'Sfl7’+%g/P affirm ?:kfaijdp$&d/IZ:A “In > W JP: - 27 (556); ‘Mdgm=%aéa= -fir fifiji ,f m In’ :9”? flaw“ [Mr/90+ J“ dvémw W , UP” V l /‘//m =flén== ‘flr r— e (W55 A b “7' , v “5 T “W Ml ' ’ “M 6"”), ~ "9' END OFTHEEND "'1" my) my kdp~ g’ibfifi pf and...» 24 N Ovember 2004 $4,151) My?» ”' g/I’Zr” 14/ 4%,: W ’%J Jay » * V “Um”, 3 “M70 kW‘ 45,», -’ not}: 4) gfwg ,2 4%,: Mmr‘AMfi AAM‘ Mafia?) Some useful equations and values of fundamental constants R = 8.31451 JK’ mor’ = 0.0820568 L atm K’ mor’ = 0.0831451 L bar K1 mor’, 1 L bar = 100 J, 1 L atm, = 101.325 J, NA = 6.02214x 1023 particles-moi], For f(x)=x“: d%x=axa_1, jx”dx=x"+1/n+1, Ix—Idx=1nx, dgnx=d%, dx —{1 ax+ xdx = '2 na+ + a I — )In( L), {WM fl [1( fix) (OH/M], dU=dq+dw, dw=—Pex,dV, H=U+PV, G=U+PV—TS=H—TS=A+PV, dG = —SdT+VdP, A = U—TS, dA = —SdT—PdV , CV = (OW/51"» ’ CP = (ow/07L, (w/W)T = T (61751")V ‘P ’ (fay/31>» = V"T (61751")15’ dS = %—dT + (fiP/fl)V dV, dS = gde — (UV/07)}, dP TJ-Vi Wacl = (l/avJJW/acl’ We»); =“(%L (%)Z(%L(%c>y =4, WW = (378% +(%L(%)w' 2 PV =nRT, [P+" %2](V—nb)=nRT, For an ideal monatomic gas: CV’m = %R , C15,,” = %R , T V. R/CM PW =°°nstanta 7 = CP,m /CV,m’ 2(4) , 7 For an ideal diatomic gas: CV”, = %R , C15,," = ER , For an ideal gas Cam — C11,," = R , d6 = —SdT + VdP + Z ,uidni , dA = -SdT — PdV + Z ,ul-dni , dy = 4de + deP i i ,ul- = yf(T)+RT1n-I:’—, 0 —> ZViAi , AH; =ZviAfHfii, AG;w =ZviAng,,-, i i i P0 P- " d1n1<° AH” AS" = v. 0 -,K0 = "‘1 AGO =—RT1nK° P = , T l m,T,l P P0 a T Pa dT RT2 g5] _ 8,63 —S,% _ Asa,” dT gégistence Vm —V$ AVg’fl curve Gibbs Phase Rule: f = c - p + 2 ...
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F04 MT2 - CHEMISTRY 110A INSTR: R.N. SCHWARTZ MIDTERM #2 ....

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