Fourth Hour Exam Key - Michigan State University Fall 2007...

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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 (NCO-9 A2234” +3e“> 3 (5,; (P49 26- ~—> 2.8;?» “JP—W 2mm + 3%,}? —> $14.83* +48}: c) What is the cell potential at 298K when this galvanic cell begins operation? (remember that RT/F = 0.0257V at 298K) g: :0 ..._ 0.03157Vfln M1 goz'ID‘i-f—Koél/ é /7 V {8:3 (333 = ,2. 731/ E», 2579/ - ,oasfi U». mead“ (9 5503 2 Qnsx/ .- (-205 V3 l QXIV E: 2-57/1/ (1) Given the oxidation—reduction reaction determined in part (b), would you expect E° to increase or decrease with an increase in operating temperature to 360K? (explain your answer) Ar? (T—m) 0 0 51‘: 52w * “Taft‘ 36,0 M 732991,;@ 77LWWWAV§%«% W WW,AV§aW/Luo “3.7;; (We’d—<9» WW%&’$\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) 2-0%“0'5 : .Oi’X 20"” ‘9'7'20‘751/0’5; 2 {t X1 ‘rQo‘lxw’fx - 2-041/0'7 = O X 9134,00 —- (204 lb 3 X ’0 3 a}! “0-4 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 [Han-i3 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 lam-Jae) HCNEflI/n, K4 1‘; 50 SM) 44% M £2} flef/fi;él‘t flessocaml-om “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 = 0.083L-bar / (mol-K)) 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 “lulu-q 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!) k-Ud’ + ”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= {04’1-X “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 -.i-A,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—W-l 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—TV-dT 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=Kbm-z 7r=RTM-z 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 = 10-14 = 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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