This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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
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 $110 0 _
1a ‘ I : “fa.% : 70X109X 3.4—5” — 940x”)?
. ~5’
A. o ,, '0 9 , o ‘ 0'
Aé,=—ﬂ7&u>4 ., #3735 A4é731(5/€X3?3z€xﬂw¢
W0 4%:3’8: ~£3MQ§£¢X39WXAOﬂD >05?)
= 7‘ 5% év/M/ '0 0 o 0" i 0 pt
“Tag/z, “ W932 ' 3gg£§m Mr‘ 4”? Mgr)
0 a .
44393;: z 4%” 39%4559326 = val/J WM I M 2 *0 _.o
M“€’§é= 13%;” ’ 452:. «2&0
K W €72. CHEM 110A MIDTERM #2 24 November 2004 NAME: ’ 2. (35 points) Consider the phase diagram for single component system shown below: P/bar _!  l l .
100 400 800 1200 1600 T /K where Solid ,6 and Solid 0: refer to solid phasesa and [3 . a. What are the values of f (degrees of freedom) at the points 1, 2, 3, and 4 indicated in the phase
diagram above. b. Sketch and label a plot of ,u vs T for P = 80 bar. c. 1 Sketch and label a plot of ,u vs P for T = 850‘ K. d. A What can you say about the density p“ of the solid phase 0: Compared to the density of the
liquid phase p1. Explain your reasoning. 4 V4” 4 v W tie. (:6 22w
222% '7 ” M ,l: c7»; = 379 lkjﬁgj .ﬂ; “(7,1 :5 M7533
MW ##sz “viii ﬂﬂzfeafzf CHEM 110A MID'I‘ERM #2 24 November 2004 72—79le CHEM 110A MIDTERM #2 24 November. 2004 NAME: 3. (30 points). A gas obeys the equation of state: PVm' =0 RT + bP , where b is a constant. If the
gas expands from VmJ to Vm’z reversibly at constant T, derive expressions for AA,” , AS," and AGm. Express ﬁnal answers in terms of R, T, VmJ , and Vm,2 . Show all steps to receive credit. 0) dﬂw *é'Mcéz'wﬂ/g” , m 72%) «L729
444": w}; Aﬁgwjf I, I 46
2 L~ é) CHEM 110A MIDTERM #2 .ma mp Mﬁ‘Mmﬁ : Mf. 6 _ h]
g . ' (2.1?) 4/»: £7/&1[ﬂ!3]+/5 .. A } I
g [W W A) 3'Sﬂ7’+%g/P afﬁrm ?:kfaijdp$&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 $4,151)
My?» ”' g/I’Zr” 14/
4%,: W ’%J
Jay » *
V
“Um”, 3 “M70 kW‘ 45,», ’ not}:
4) gfwg ,2 4%,: Mmr‘AMﬁ AAM‘ Maﬁa?) Some useful equations and values of fundamental constants R = 8.31451 JK’ mor’ = 0.0820568 L atm K’ mor’ = 0.0831451 L bar K1 mor’, 1 L bar = 100 J, 1 L atm, = 101.325 J, NA = 6.02214x 1023 particlesmoi], For f(x)=x“: d%x=axa_1, jx”dx=x"+1/n+1, Ix—Idx=1nx, dgnx=d%, dx —{1 ax+ xdx = '2 na+ + a
I — )In( L), {WM ﬂ [1( ﬁx) (OH/M], dU=dq+dw, dw=—Pex,dV, H=U+PV, G=U+PV—TS=H—TS=A+PV, dG = —SdT+VdP, A = U—TS, dA = —SdT—PdV ,
CV = (OW/51"» ’ CP = (ow/07L, (w/W)T = T (61751")V ‘P ’ (fay/31>» = V"T (61751")15’ dS = %—dT + (ﬁP/ﬂ)V dV, dS = gde — (UV/07)}, dP TJVi Wacl = (l/avJJW/acl’ We»); =“(%L
(%)Z(%L(%c>y =4, WW = (378% +(%L(%)w' 2
PV =nRT, [P+" %2](V—nb)=nRT, For an ideal monatomic gas: CV’m = %R , C15,,” = %R , T V. R/CM
PW =°°nstanta 7 = CP,m /CV,m’ 2(4) , 7
For an ideal diatomic gas: CV”, = %R , C15,," = ER , For an ideal gas Cam — C11,," = R , d6 = —SdT + VdP + Z ,uidni , dA = SdT — PdV + Z ,uldni , dy = 4de + deP
i i ,ul = yf(T)+RT1nI:’—, 0 —> ZViAi , AH; =ZviAfHﬁi, AG;w =ZviAng,,,
i i i P0
P " d1n1<° AH”
AS" = v. 0 ,K0 = "‘1 AGO =—RT1nK° P = ,
T l m,T,l P P0 a T Pa dT RT2
g5] _ 8,63 —S,% _ Asa,”
dT gégistence Vm —V$ AVg’ﬂ curve Gibbs Phase Rule: f = c  p + 2 ...
View
Full
Document
This note was uploaded on 04/03/2008 for the course CHEM 110A taught by Professor Schwartz during the Fall '06 term at UCLA.
 Fall '06
 Schwartz

Click to edit the document details