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Unformatted text preview: Michigan State University Fall 2007
CEM 383
Fourth Hour Exam
30 November This exam consists of four problems on the next seven pages. Please examine the booklet to make sure
you have a complete examination. Equations are provided on the last 3 pages. Chemical data are
provided in the problems along with constants that you may need. Answer each question in the space provided, continuing on the reverse side of the same page if more
space is needed. If a question is not clear, insufﬁcient information is given, or there is an apparent error, please notify a member of the instructional staff immediately. Pay attention to units and to signiﬁcant ﬁgures of your numerical answers. Show your reasoning for
all problems onthe exam! Student Name (<9— :1 StudentNumber RecitatiOn Section Number Scoring: 1. (25 points) Name 2. (25 points) Section number
3. (25 points) 4. (25 points) Total (100 pts) 1. (25 points) A galvanic cell is constructed using Br2(g)/Br'(aq) and Al(s)/Al3+(aq) half cells. The standard reduction
potentials for these two cells are Br2(g) + M —> ZBr“ E° =1.09V
Al3+ + 36 —> Al(s) E0 = —1.66V The bromine cell features a Pt electrode with a Br2 (g) stream of 0.5 bar bubbled over it and 0.1M
NaBr dissolved in solution. The Aluminum cell has an Al(s) electrode and is prepared with a 0.5 M
AlCl3 solution. a) Write the half reaction occurring at the anode. A,Q(s\ *4 4113+ + 3a“ b) Write a balanced chemical equation for the oxidation—reduction reaction that runs
spontaneously when these two cells are connected. 2 (NCO9 A2234” +3e“>
3 (5,; (P49 26 ~—> 2.8;?» “JP—W 2mm + 3%,}? —> $14.83* +48}: c) What is the cell potential at 298K when this galvanic cell begins operation? (remember that
RT/F = 0.0257V at 298K) g: :0 ..._ 0.03157Vﬂn M1 goz'ID‘if—Koél/
é /7 V {8:3 (333 = ,2. 731/ E», 2579/  ,oasﬁ U». mead“
(9 5503 2 Qnsx/ . (205 V3 l QXIV E: 257/1/ (1) Given the oxidation—reduction reaction determined in part (b), would you expect E° to increase or
decrease with an increase in operating temperature to 360K? (explain your answer) Ar? (T—m) 0 0
51‘: 52w * “Taft‘
36,0 M 732991,;@
77LWWWAV§%«% W
WW,AV§aW/Luo “3.7;; (We’d—<9»
WW%&’$\W%WW’W +1140 W W ou/ 89 W; 42(s)4£3°24f> ‘ Circle on :1 ea D
AND I E i y “(I e ncr se 2. (25 points) a) A 1.0 L solution of 0.01 M HCN, is titrated with 4.0 M NaOH. What is the pH at the equivalence
point? (Ka = 4.9 x 10'10 for the dissociation of HCN) : 1‘5, ruck eiuiv‘ aka; Wm m Hcm=> cu” 7+.(35ML) 20%“0'5 :
.Oi’X 20"” ‘9'7'20‘751/0’5; 2 {t X1 ‘rQo‘lxw’fx  2041/0'7 = O X 9134,00 — (204 lb 3 X ’0 3 a}! “04 2 [law] 2. WW we} (4.4 m4!) : 3.35 PW [Li/3.353 [OLS’ b) What mass of potassium cyanide, KCN, must be added to a 1.0 L solution of .01 M HCN to
prepare a buffer solution with pH = 8.9? (K, = 4.9 x 10‘10 for the dissociation of HCN, MKCN = 65.1 g/mole) HUG—vi H‘+ +cw‘ ‘ 01m 1*
ﬁsh. ._—— (PH: “PEA + jog, 733me [Hani3
3:9 6’3 +16? [C53
[Hun
10(56443) Bun} .O(M (OHOXMMV 1 [C if] LMDJ M Z [00‘] ‘ Kq= ‘43 “040 @IQ 1 Q‘SW WM “AWL lamJae) HCNEﬂI/n, K4 1‘;
50 SM) 44% M £2} ﬂef/ﬁ;él‘t ﬂessocamlom “to give Clo" +Autuu, Hit?
[1 3. (25 points) a) 1.0 g of N204 (g) is placed in a 1.0 L tank at 298K. At this temperature, some of the N204
(g) dissociates to N02(g) according to the equation N204(g) (Z) 2N02(g).
.(DH'X 2% ”Torr a '0” +X
The equilibrium constant for this reaction is 0.143 at 298K. What is the total pressure after the
system comes to equilibrium? (M N204 = 92g / mole,R = 0.083Lbar / (molK)) 1 .,_ I 4
K: ‘PWL 1 hm. Paw MW Vma‘g 10L.
Rhea “Allow ”M EH hm RT
\/
'7.
= .m ; n ~ __ =
w W (Li, WW
“luluq Mr V 92 WW
.1.
,H3< Lem 1 (2%
03315.5 .QQS’R
_ ,on»
M44 " Emu" “he! ‘31—
>2 “3 Ll L V
. x >
' /,0L.
~5’ 3 = 41"
4.39m ~S‘Xxl0 X = 0.35 La».
LLB +58Mo’3x v (34"0—510 K" ’ﬂﬁlo‘j i rift/0’3) 1— LINN£3410”?
Y X" 3.33H0“3mo€w b) Argo = 57.0kJ / mole for the equilibrium given in part a at 298K. Will the equilibrium
constant increase or decrease if the temperature is lowered to 240K? (Explain your answer!) kUd’ + ”20%“; 1/00.; lL 9——
W l ‘f + [‘66th MN
1,1436 W“ 31‘7”“) +0
Judge [4 , Lou/Ia. ~12; Circle one: Increase
meld SWLL
4. (25 points) K» Jam/M , M W a. In aqueous solution, CD? will hydrolyze water according to the equation cog—(0(1) + H200) ﬁ Hcog<aq)+ 0H'(aq)
.05m ‘1 X x with Kb = 2.09 x 104. What is the pH of 22— L of a 0.05 M solution of NaQCO3? —.——. KB“— EH005’3 [0 W3
[cos—3 2.06110“ 4 Z @0613 L_ 4f i "2
[.03’51/0 f2_&‘,a0 X ‘ K )(1 +1061)“qu ;osrx/o‘5'0 I" "Q .O‘i «Cto‘q t (221‘ 40”“)L'LI (og/rs’x/o’f) L _ OH: — '3 2
“P 1.3(“1103 1.5/0 , “Flak HL'S’O: IUCJFOH— 7 3 SUM/0'3 b) How many moles of concentrated HCl have to be added to the solution from part a to bn'ng the ﬁnal
pH of the solution to 10.0? W H+ +0 3%) W W 80;?» yea; ' A » 11/044 2 q = HJXK/VN
' «ﬁt/o" Play 1037. .H?(.oo(0+sc\~—.04‘1~X
QﬁS’HO's—l—,L(8x= {04’1X
“mm .04: “Mimi/ma x: .Oél‘mdw CEM 383 Eguation Sheet PV= nRT V: V/n
V m7
Ptotalz ZPia Pi=)(iPtotal Z27
i=gases
P= .1" 212, 2Xi=1 Z=1+BP+CP2+DP3+...
(V—b) V i
_ 2 3
Xi= n‘ Z=5Z=1+B[i]+c[i) +D(i] +
ntotal RT V V V
3RT
Ems: 3nRT/2, Crms = 7
— 8RT _ 2RT
c= — cmp — —— nM M Z]: ﬁndzzﬁij Zn: £91)
V 2 V
hi 1: 2%
Z] r2 M1 w = Pex,dV w = nRTln (Vz/VI)
AU= q + w H = U+ PV 8U 8H
CV = — C = .— [aTjw ” (8T )
AU=CV(T2—T1) AH=CP(T2—T1)
Cp—Cv=nR AH=AU+A(PV) AU = qv = 'Cbomb(T2  TI) Pl‘lly = PZVZY
Argo = vaAfﬁoQﬁod)—2v,Af§0(reac)
prod reac
ArHT2 ”= [3an .iA,CP(T2 —Tl)
ArH E 2BE(react) — ZBE(prod)
AS: ‘1'” AS=annﬁ
T V1
T T
AS=C,,ln—2 efﬁciency: 1——1=l—Wl
T1 T2 qz
T1 q]
C.O.P. = =— _ ASW =AS+ASSW 20 TZ—T1 w Aﬁﬂ H
Amix =—R(nAlnXA+nBlnXB) AIMS: T
f
A H T2 C
Amp S = W AS = —PdT
Tb T1 T
__.. __ T2 C
=2vS°(pr0d)—ZvSo(react) AS: J—TVdT
T1
T T
AS=CV1n—2 ArSz=ArSI+ArCPln—2
1 1
AG = AH — TAS AA = AU — TAS dG = SdT + VdP
dA = SdT — PdV _" — iv A 5‘1 prod) — ZvrAfEO(reac)
P AH T P A? T —T
AG=nRT1n —2— P— P=—=1n—— 1n —2 =— 2 1
I: Av 11 I: R TIT;
— — — 8V —
V=n1V1+n2V2 Vi: [— ] pi=G.=[§—q—)
ani TPnJ an! T,P.nj
13th = nRT(X1 ln X1 + X2 1n X2) P1 = X11);
0 P. P. ym y.[i] .=. RTl. .='=.X. =——' .=" ' .=.l=1
u, u, + na, 0, P; y, , a, lbar a, my [11, a,(s) a,( )
P. =KX,. =K'm ATf=Kfm°l ATb=Kbmz 7r=RTMz MsolvRszlb
Kf,b=——————— 1—1 a+av v=v++v
(1000g / kg)A”H
um = (my: + v_/.Lf)+ RT In a; _ = 71’”: m+ = (m:+m1’)“"
A,E= A‘,E° +RTan Aﬁ" = ——RTan
TJ= T T
K1 R TITZ aA+ [)3 2‘2 cC+ dD
K=¢¢ c1202
K = PéPd_ X5 CXg .Pc+d— a— —b =nénf). Rota! Chi—(Pb P P1P; XaXZ W "in? "m; K: K KP _7ci’p P5P; K=K = YEYZJCHDJ" : [C]c[D]d mi; P“P” m; [A]“[B]” ‘ [A]“[B]” Kw = 1014 = KaKb = [H+][0H‘] pH = —log[H+] pOH = —log[OH'] pH+pOH = 14 pK = logK pH = pKa +1ogW—Wﬂ
[aczd]
0 a o ,, —A,E" RT M a ,, A,§"(T — 298)
E = Ecathode _ Eanode E : VF = VF ET = E298 + T
RT aéaf, 0.0257V agag =E"———ln a b
vF aAaB v “A03 ...
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 Fall '07
 MCCRACKEN

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