Final W 08 A - NAME This examination is composed of six...

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Unformatted text preview: NAME: This examination is composed of six problems. Doall parts of all the problems. You have three hours to complete the exam. You may use your textbook, three pages of notes (front ‘ Final Exarninatibn,~~ '7 awakens; léggMa‘rvéh; 8:00 CHZOB-JZ; Winter.» 200 and back) and a calculator of your choice in working the exam. Problem 1: Problem 2: Problem 3: Problem 4: Problem—5+v —» Problem 6: TOTAL: 47 O of 30 points I b of 353 points of 30 points 2g of 35 points 13’?) of 35 points / of 35 oints 35 P F [5% ' AH° and AS° values are temperature independent. PbEOwSW. -; P9?“ 50$". A C7345 t ‘56?qu 2 —2q.q%-‘7 - " f " {4%25 K35“ ., _ 4673:, a — Lit-$25" "‘12{20?>) in (39’1“?) 2 330102673 :7 z5P 7 5XP [ ""6 :. (a A'LW {0‘00333C2qg5 _ X "'0‘9 1x07 1‘ DH." \ Q l - \nKs, : \m \gzc‘ + 2a2> -\- ~ ~-- M..-- 3 <5 {2. 25C€5 3’3> O ‘ g & a \ —\ ‘3 fl ,. 592% ‘ ”“4 (F‘ol ma; 3 ‘v bers’so-f“ ”AH; (FbSqug\ Ly “3m. 2 . WW" " :2) — PI «aoalz7 ~(—aIQ.qH-‘) 2.18 q7 k9 ( \n 5‘ 3g; 11 \ [3‘4” _+ 8-5:? I ‘ \ M“ *-~~—~—~—---»-—~~~~-~~-~-»~»~»—w- - N __ _ _ :. *5 0 mecca; 29:8 3:3,} 7 72 —>,L3‘3 " "qu” 0 {GENE->73 7" 2; x \0—5:5 '-— CPbWfiOqI-r] 1 h a g+craH mg wi’rh o . O& M _\ D " Yxk“ a ( 0‘? /’¥\)(O.oi+’fi} (an) (M) ( P523 1 bfimg (507%))” AG?>.aa(WQ«rs>) ~ ALL/’3 —— («31341§3 Na 50;; ”W! w." figpt MWfivMfi & 0.0M“? «1 z ngxxo’g M24001v~sa€>ww $0 1M7. x1“ 3 WW~T-w «N- NW 0|?) )r £0210 MERGE/L77 ‘ " 7 10(20kaer H. 2. (35 points) Quim'ne (C20H2402N2) is a base that iofnzes in water in two stages -With5 :7 - . , _Kb1— _ 3. 31 x 10-6 and Kb'z- _ 1. 35 x. 1o-}.9_.1::A solution of 2. 000 g of quinine in 1Q0;_ ' ‘ ‘ mL of water is titrated With an HCl solution that is 0.09000 M in water. Determine. A the pH of the solution after adding the following volumes of the HCl titrant: (i) 0 mL, ' ‘ 7 (ii) 68.6 mL, (iii) 100 111L135 y'z C20"24-02 “2 V “Ci C3015; 1:“2 * #20 :5:— (20 475 02 r12 +0H l‘b, a 11110 (L011); 02 N +1120 :1 Cw +123. 02ng -1 OH” #1:; 1 NWT.) .3 -114." r y ' . 1 N H 0 N .»-... :17; VOW“ 7C; AaCj/Q) 11.4 I), 2i ,Jfgl 0 0010 ”0’ . ._ «1.- - 2 a l :14 :fl’yl/Ih' H(i ’1'" ‘— 0.; L, 3 #11,-1 -zorzpac’z N2 z a C6 M 7 (1190on1 7 // / ’5’ M f ; / / (on C 0112.0 ”2407 ”2" 7 W w ”19% 5.1013qu :5 1*.» ml EU 2 0.0919179» 15 '7. .:b\ if V7903 (Z,OH}QU:'?1)? 2 0. #VfiDl Hf? 3' OIUDOJVLA' H’H" ? O a {“019 1.13:; 14:53? (202—? 240215 LVN-.245 {2302 1...,” 2 [mafia 0.1050179! 1 9.000114 ) ‘? S g: + \N 3 (30.: points) A samp 6 ~ c661ed such that it ‘ -‘ > «become CeH6(s) at: tin 'd- ‘200 K It 1s then heated such that it becomes CsH6(l ) , at 1 atm and 290 K. Finally, it is heated further to become CsHs (g) at 2 atm and 450 , K. From data in your text determine the entropy change of the benzene due to this process. You may assume that 06H6(g) is an ideal gas and that any relevant heat capacities are independent of temperature. (buy—F 151m!» 3 p 0+ 561 1 x , Uzi H. , .. /‘)-C f 7 K :(> 4 ,/Q:' (if "ff/"d”, ’ Luz/‘q C : r, r .1 f '; 5, a b . ) .-‘/(,.:’ If,‘ r .. , A‘ 2, WW .5 “Nb e1] ”‘4 M“ i U) P? ’5’ ”' ’2) 6‘0““ “1 '—»7-‘;(49H7v(1\‘=7>’cu‘rwn (1.! kew- :57 HOOK. 3“,, *1: - n fi _, . *v a. «‘3 , v 7:! w ”met 2 I’m 2.73 r at” 27—3 in 'rp Hal “ 7‘ I ~ - , 6 /0 b5] \Cpir) ”2—”; [{fiIM-l Cwl‘( \Z‘; ('5‘ .> 4 QCMC / "IJ’I‘ (“57’3“ 57"“; Zeok : as. i! (— M221 (m, H 11m}, ..0. a J/ rm" , -, :55 ’39 6 H K 55 C 35: 7:4. ’ : Md ‘1.” ,: " (pith if: (Q‘s: 4‘1 : 31:). V ugok 234,:65W“ Lil—3 30f" N ‘» r .1 (:5 ~ "0": ~~2¥ e7 42 84+ '8» oz * c“ 7 ‘2 IQ>KO(LV(.1’5—_’ J ‘- 9; w. ‘e ”c.1{r3rv‘\i [" {(3.23}! ( Val-(.1 A: ~ -: ,, "3,.‘3‘ ’ i r" 2755'. ” ??\2:1-~H,( LLJ) H" L; (713 3 (F / " 551' N 55 n i’ infisi- (Mi/19% 1’ "NJ“? .1 ‘ f7 _‘ ; V 1' i! ,; j? 5‘1v11":: 5‘3} "‘_:\_'{" IE ‘ ‘S;7(g w“: (d) For this process calculate (a) (10 points) AU, (b) (5 points) AH, (c) (10 points) 1;, and (d) (10 points) 11). You may assume that the gas behaves as an ideal gas. To} a Q. [BUFV‘CVT CP 1 26.423hzmo‘», (‘12-2’3‘41‘8'313 PO1QIJ « [Fur-10’ CV sz—l ATe‘in .3 . 1a+w\;(Z.OOLL *1 ‘ fl—Vr‘ '2 ( __~._____ > E '5. (04 K n ya , {1.7%2.3,J&gwx \ / T -2 ’10?" fl“! 2 SLX 2 f ___________ '7111K (1-76yooazoe) AH 1 5m 3‘ - (I if ’1 w “ 1'78M°‘(7"I¥’-J3Fm0#‘ ( r711 ': M“) r—/‘—-‘ ~._‘N~"“ . k AH2177.7a 3f w=~1>e~xww “sz L—~\a‘w~o\,€151-”2'DOV\ " ‘O'SL‘W ”7— W '3 -0 5L arty/)(mhglgfi 7— '50 (ab 3 $5,: 127.05J+so.ee_32 177,717-3’MM‘TV W7"P€y-\- [>v 2 —p¢v : 1-:M+M‘1L2.SL*210L 7 “OSLGH , V14 1,]: (“9.93aw\1( gear 325‘} _~ _- 1, _._._._.__1.._.‘ .-- ; m; C 50 bk”; 3 / ,-_,_,,____. 1 -~. , >>77.;,_157_ The decomposition of ethaneigas (C2115) at 700°C in a constant volume" container of '7 , .2, volume V proceeds according to the rate law _ 07min] : k d7 [CZHG] where n is 1 or 2. Two experiments are performed at 700°C in which initial rates are measured. With an initial concentration of 021-16 equal to 0.001 mol/L the rate is measured to be 3 X 10‘6 mol/L—s. When the initial concentration is equal to 0.00176 mol/L the rate is measured as 5.29 X 10‘6 mol/L—s. (a) (15 points) Determine k and n for the reaction. I (b) (20 points) In a third experiment an initial 0.1 g of C2H6 in the container of volume V decomposes at 700°. Determine the mass of 02H6 present after the reaction has proceeded for 1 hour. p _ (7\/é2 Kl ’/ N94: (7) 19:291.] O«OO\MOl/L 79-00177» " '"l, ( ’{9 I. Page 5x4; Wei/Ln SvLfixlD~W°//L"5 mm ! Z HUI ”63.1”“ 7mg 2 562*}531» A“ l ”a“! [C +1 ‘7) ,1...“ ‘7 A «2' OJ: Va. V) 7 K “(2077311, 0 5i2fixm’b ( 7°. @3777.» ‘ A‘ > — IV \ 3,11024'25’ :2 ‘ U'Ovl J l ra‘,..9 z TgiclH-bj 3x(o~b x K ( o.OO\W‘OLIL ) ,_..—«~—r “~""‘"‘"‘" 1 K = 0.003 2 gayeo’jt'lvmm lg I771r 5M5; it?” [Mr jewdwawaz‘g _\_ \ g f,— 7I 002% m 776777.7ch // I 2' y ’0‘ .3 {Nil EC gigv'fffhal‘”: i _ .., ._ 1......__._ ‘;777 ,,,, , 7,, ,, , ,, , , 7,7,, , , , ,, , , ,7, ,7 ,7 , 7,, 7* Mm. \ 7' l I 6. (35 points) The amino acid lysine is known to contain 49. 29% C, 9. 65% H, 19.16% N, and 21. 89% O by mass. A solution of 50 mg of lysine in 1. 500 g of biphenyl (an organic solvent) decreases the fieezing point from that of pure biphenyl by 1. 83°C. Determine the molecular formula of lysine. [K f, the freezing point depression constant for biphenyl is 8.00 K—kg/mol.] -2 AT; -.- “1182-01.: :» -2z_r ( 5X” j \ 1 ’M £9 ”9:23:17 In; / AA — E E’KH/P'lq / ' 95>; ( w _ j '2 gab 1 0! o /5‘xr05/ 7 J/ml M .1»: 117's 2 .2; 4m. 2 1. 1.007s g.-- \. ~—~ 1' ———r- 1: new m " V5 1 :2 ’ .0"... .€ 2 3'2» ./ f 1’ ~. n 45" v“ s ‘7=""-‘l \. . _, a 1...,» : r...”- — .e . - - - J -« “1!.5'" "’ 1'3.- ) m .1 Vt 1 2 g 11- I}, / 0 L119, 3 5 . <1, \ ,u - «M 124 ‘MUBTE’ -' ”—90 4 ’1: E’ / ' ( 11-251 11 T) J ' Lu 3 ‘ M 5"— / manlgwrfi - “15W 1 I}! a I 2 I! 7; Z land. ’ / Y1 r if if W 04.261 Ll 2 f3 / 1h 14 z 9/5 a ; (.2 ”The WLOLthlftj: {Dymuvla by 1,1191“? 1“,: mt/ W..._._._______.—J ...
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