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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 mootJ? act; '_*°r~=¢oa got/ate,» Mo MV—f , M 221'.) 5%?) am 7‘? f ‘ p W0"? {owf
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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
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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
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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 particlesmoi], 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 TJVi 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 ,uldni , dy = 4de + deP
i i ,ul = yf(T)+RT1nI:’—, 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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 Fall '06
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