Solutions to 2007 Prelim - Page 1 of 8 Cornell University...

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Unformatted text preview: Page 1 of 8 Cornell University Chem 208- Spring 2007 Prelim 1 Name: SOLUT'DU S CU ID: Last First M.l. Lab: TA Name: Day Time Instructions: Answer all questions in the spaces provided. You must show all work for full credit. The last page of this exam consists of an equation sheet, which may be detached if desired. There are a total of 8 pages in this exam. Graphing calculators and calculators that allow alphanumeric input are not allowed. You have 90 minutes to complete this exam. There are 5 questions; the points for each is indicated below. 1. [25 2. l20 3. {20 4. /25 5. [10 Total I] 00 Page 2 of 8 1. (25) The rate law for the reactiOn described by: 2H: (s) + 2N0 (g) —+ N2(s) + 2H20 (g) d[N2] 2 was measured to be: dt = kobsngllNO] A proposed sequence of elementary steps is: k NOtg)+N0(g)l::1> Nzozlg) —1 k Hztg)+Nzoztg)iN20(g)+Hzotg) k H2(g)+N20(g)—3> N2(g)+ Hzolg) a) Identify any reaction interrnediate(s). N202 and N2??? a“: recto—lion in‘l-ermedtavks b) What is the overall order of the reaction? 3 c) In this problem, yOu will determine the rate law expression using the above elementary steps. The first step is to write down the rate law expression for production of N2 based on % elementary reaction. sedge: Magma cl) Derive an expression for [N20] using the steady state method. I: U10] = COflS+an+ cl [u zo] k 05 e " a: “LMLOJ‘ kai Hailed]:— 0 [kilo]; kgEsziM;OZ:]_ k [no 3 trawl] " 3k; ‘ Page 3 of 8 e) Derive an expression for [N202] using the steady state method. all: o ,2??? = k, [M0] u In, furor} szusoflt H; = O E M1023 {— to - karma} 2 - ELMO}; [Niel]: M. is. Home? 1) Determine the rate law expression. Rain = ammo] : %[#ajszulo‘j E3 $5435 “@072: ken/Jug" k‘l*k2[”zj - it, + k2[”e] g) Does your result in f) agree with the measured overall rate law expression? If not, under what conditions would agreement be observed? Express kobs in terms of the elementary rate constants. No, 1+ doe,an agree. However :FZ k_, >7 sz "3’ 3+ does ' . 7. agree . @145: kill: : klk; ["23 [no] kt, + O I:~1 kobs z % FF kid)” szH‘j h) Can the overall order observed for this reaction ever differ from your result in part b? Under what conditions might this occur, and what is the observed overall order? Yes at r-.<< ma, sew 2- L / r Alisa]: hF/wal k,[~03 “a at m The absemed Drama order is new 2. Page 4 of 8 2. (20) Phosgene (CIZCO) is a highly toxic gas produced by the following reaction: C12 (8) + CO (e) —> C12C0(g) The method of initial rates was used to determine the order of the reaction. Rate = k[C12]"[CO]"'[C12CO]° Ext. Cl M CO M C1_CO M Rate Mfs v- 1 0.0205 0.0620 0.0152 3.06 x 10' 2 0.0310 0.2040 0.0215 1.33 x 10" M3 0.0515 0.0620 0.0152 1.22 x 10" w- 4 0.0205 0.2040 0.0152 1.01 x 10" a) Determine the order of the reaction with respect to C12, C0, and CIZCO, 3113.. find n, m, and p. 312-. O-Zo‘i0>m= IONIC)” E- 0.0210 "5( 2022:): limo) 5" 0-090 3-06Kloq' RH’ 0.02.05 010:5? /.om~* 3.20m : 57.30 (/5/ "5 (24/) = flié EU 2% fwd/fl Li’é 35-: (0.0533 0, "ZZXlO-F‘ P: 1.. RI 020203 * 3.06K'0-L 2,51“ = 3.9? n1032_51:|03339 n- JA——' 3’5 =15 - foiZcS! b) Determine the rate constant, k. USE: any experimen‘ha‘ fun: g M o 3,0;x/0‘Z 3’1 = #50. 020514] [0.090117 [ 2”] : m z e; -/ k [QQOSIQ'Ebacwfl /58 M 5‘ Page 5 of 8 3. (20) Given the following standard enthalpy changes: N0(g) + N02(g) —> N203(g) AH" = 39.8 H 2N0(g) + 02(g) —> 2N02(g) AH” = —114.2 k] N0(g) + N02(g) + 02(g) —> N205(g) AH" = -1125 k] N204(g) —> 2N02(g) AH” = 57.2 H NgOstg) —* N205(s) AH“ = -54.1 k] a} Calculate the standard enthalpy change for: N203(g) + N205(S) -> 2N204lg) N203 f?) 4‘ Wm “102.61 Alf: Hagar #2050, __;. Mao; (9) by”: +5-9Jk3- “‘9 +K/0d(?)+06 Afiz-F/(fo—j a) LING; *9 2 14,0“9; a #- = —2 fizz/:7): 4mg)" Agog?) f/Vzoja-Jn) 214070) 5H: -—2.2.Zted” b) The standard enthalpy of formation (AI—I01“) of N205(s) is -42.8 lemol. Calculate the standard enthalpy offormation of N203(g). You may use any information on this page. .-’ - a '— MG) *We) —-9 “2.03 (7) A H -— —3 9. 2 e.) WM) W)+0L(9’ A30: 1" 5'25‘3- IVZQIG) .-—) Alf: +9‘fi/k3— 44056) -—'> A403(7)+02_(?3 4w: meter 0 k..._,_._._./ ¢¢=—‘/z.?k%,( (/mtgbflo [24039,] + 0 —/m€f-423%} /2g,9kT I me origin/2,039,] = (dos—— 42.8957: 37-Ma— AH1=EU203GJJ 5 WOW”: Page 6 of 8 4. (25) You buy coffee (200.0 ml) served in a styrofoam cup. You are up late working in a chemistry lab= and want to drink the coffee right away, but at 950°C, it is too hot to drink. You decide that the fastest way to cool the coffee is by dropping 40.0 grams of dry ice, CO; (s), initially at 480°C, into the coffee. Note that CO; does not exist as a liquid at 1 atm; the solid sublimes directly to a gas with an enthalpy of sublimation (AHsub) of 8.76 kamol at 480°C. Assume that the final temperature of the C02(g) is the same as that of the coffee. You can neglect all heat losses and the solubility of CO; in coffee. The molar mass of C02 is 44.0 g/mol. The specific heat capacity, 3 = 0.850 JIg°C for cofig); a) Calculate the quantity of heat (qsub) required to sublime 40.0 grams of dry ice to form gaseous CO; at -78.0“C. I CO _ mo 35 2.’ __._9——l_lq‘05/m'DI : as. b = (o, cqumoleS)(5’,'7G-Eg: = 1% k3— Q. IMO-3 b) Write an expression for the quantity of heat (qgas) required to increase the temperature of the gaseous \5 0K CO; from -?8.0°C to the final temperature of the coffee. . Kelvin i5 €236 = m S CST’ g OK. = Gama-saan—we --=- 0.03% 561-43730) . c) Calculate the final temperature of the coffee in 0C. c ea: ms AT 0 Cl “H: =(zoa,oj)(q.is 373%) (TF- 95.0 c) 2. 0.334, KT (Tp~‘?5.0) fist“: + class 1 cl Cage-c: O 7% in??? 0-03‘1(-r,; 09.0)}:3‘4 o. 3 3L (-1:C «6190) IJ= O 7. amt/3.03%}: + 2 .cs + 0.3301 —— 79.42 = O 0, T --— : =. : 8’3}: O 370; em/ 0 )7} g??? rTith d) Calculate the value of AU for the C02, assuming an atmospheric pressure of 1 atm. (760 mmHg). ‘5 ince w = — Pat AVrRT/Ssn : ‘63 "* “7m: K)(273./r7a/)K(0. 90m) -— —2 econ; L24.th Au: 01+ My Clcof chiefly: — .. _ can : gagkj+ 0.0390179.» 73.0 995.37) hr = 3.3“ Page 7 of 8 e) You taste the coffee, and find that carbonated coffee isn’t very appealing. So you buy another 200.0 ml cup of 95”C coffee, and add 40.0 grams of regular (water) ice, initially at —15°C. The final temperature of the coffee is 65°C after the ice melts. Calculate AU for the 40.0 grams of added material. ‘1 added 1-‘1ca‘GPee. = 3‘3?“ “SM {20°00@1837?°03(-30°C)(‘L—’u—) :—25 ET 1000?” l t. k flaw +23 7 mg, ,= 3m z. 25 w L/UZO since Alf-:0 ‘1 1) Explain the origin of the difference in effectiveness of cooling coffee using 38°C dry ice vs. -15°C water ice. “5"‘3 Hos “is @"'$°C ATM'F‘F3’300C mica #aukHF Since we used eqwtl masses all“ ice ice our Compar'isoA mosTl" be on a Per 63mm bas‘rs (no+ r mole. , The “l: me'l' ice anal lt u:cl waler‘lnauq CD Smog], 8:420 (51 > 8:0; 9‘: much attire“, SPECE'FFC hexljy lelg aka? rote. 31cc is alMOS'l' 'huaee. as is no-l rue-ml:j (as-co C9 Per 5mm} mo M. hek‘l' F5 Used in me. l‘l‘inj ice 'l'lum Sublimrn C043) Hip : n:— fillgfiaul: 222ml” "9 Ally”=(L.olk3‘/m.12(2,zzm.t)=13.01: .5”. m 90; n: Jag”! ‘1 ’01q0m0l83”) ourub=(3-75 has/ml) 0301'“):- 90 5. (10) It takes onger to cook f0 0% £111in altitude because water boils at a lower temperature. For example, a 8000 ft. elevation, water boils at 92°C. The major reaction when eggs are boiled is the unfolding of the protein, which has an activation energy of about 45 kJ/mol. ifa “3-minute egg” refers to the consistency of an egg Obtained by boiling an egg for 3.00 minutes at sea level, how long would it take to make a “3-minute egg” at 8000 ft? This is not a trick question; the correct answer is not “3 minutes". 0 " H—Ed- fl fi3l00C=373Kl klzfle hell A4: T2: 92°C: 3¢$K kg fie—EdRTL k2— p‘ 6*Eae/RTL efigé'l‘éfi-E‘LL .._—-——-- .— r a T : 1- K A3 E /‘3 I -LthloT/m’? I h 7 * small/mm 3'73'K‘ 36%] «03:8 .—_» 63 ...
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This note was uploaded on 03/02/2008 for the course CHEM 2080 taught by Professor Davis,f during the Spring '07 term at Cornell University (Engineering School).

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Solutions to 2007 Prelim - Page 1 of 8 Cornell University...

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