{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

chem1B-Spring07-mt1-Saykelley-exam

chem1B-Spring07-mt1-Saykelley-exam - Chemistry 13 Exam Name...

Info iconThis preview shows pages 1–6. Sign up to view the full content.

View Full Document Right Arrow Icon
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Background image of page 2
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Background image of page 4
Background image of page 5

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Background image of page 6
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Chemistry 13, Exam Name- February 8, 2007 Professor R.J. Saykally TA A6 A La; TOTAL EXAM SCORE (100) moi; Rules: Work all problems to 2 significant figures No lecture notes or books permitted N 0 word processing calculators Time: 90 minutes Show all work to get partial credit Periodic Table, Tables of Physical Constzmts, and Conversion Factors included Chemistry 13 8’07, Exam I Na‘ 1 1. (15 points) Given the elementary reaction, determine the time required for the concentration of A to decrease from 0.10 mol-L’1 to 0.080 Incl-L", given that k = 0.015 L-mol“1nin’l for the rate law expressed in terms of the loss of A. Z M” (New 2A—~>B+C /_ -am ., .4 7f KLAJ -1. , .1. my my” .J—\ :t I ' .L. [Ht [A39 1 MW fl 0"” )LL 67mm V : ' I K o, 015 L/ne/Am l9 2. (10 points) Three mechanisms for the reaction N02(g) + C0(g) —-> C02(g) + NO(g) have been proposed: a) Step 1 N02 + CO —> COZ + NO b) Step 1 N02 + N02 -> NO,+ NO3 (slow) Step 2 N03 + CO —> N02 + CO2 (fast) c) Step 1 N02 + N 02 —> NO + NO3 and its reverse (both fast, equilibrium) Step 2 N03 + CO —> N02 + CO 2 (slow) Which mechanism agrees with the following rate law: rule = k[NOZ]2? Explain your reasoning. may a W 3‘4 M’ [Alger [VOL 4) dz —. «Blanca e» 0‘09 M?” «W! 1:) Q : kimagf 4% \0 Li) )1" [Qg‘difal aim/135M” / yCm'iCMfl a ijfrm; W1 M ”f as“! __’~ “L 1 Chemistry 1B S’07, Exam I Name -_ 2 3. (10 points) The rate constant for the decomposition of N205 at 45°C is k = 5.1 X 104 5—1. The activation energy for the reaction is 103 kJ-mol—l. Determine the value of the rate constant at 50°C. ’_’____________, dTI: £57, —_.7 slfld [omen .____———/ M : 3' J,Ia*‘1$-‘I £1, -— :07 :u/M T1 —v5,,~c = 361 w. - e 13/,“— »b‘/fl7'. K1 2 1 KI : A Q ‘— L _ -t’u/m’. K‘ Q‘E‘fllr' KL 'A€ E die/gr + «An K). ‘ Kl e 2 I ‘3 R, WK 3”“) -w (~10: no 31?. 506/: 33 + '0’ ”a /" 1 / ~; 5.1 PIE: Q A : 9.39 rlo'” s" f...— 4. (10 points) The combustion of octane is expressed by the themochemical equation 25 CisHmU) + 7 02(g) —> 8 C02lg) + 9 H200) AH" = —5470 k] How much heat will be evolved from the combustion of 1.0 gal of gasoline (assumed to be exclusively octane)? The density of octane is 0.70 g-mL‘l. I?! “17"" x 3,735sz 122301: $37,613,512" Malachi/V y sq‘JOIUW x 33,.) [e mL— [H.201 M Igté’rfiwldzwww Chemistry 1B 8’07, Exam I Nam-i 3 5. (5 points each) Consider the collision theory result for the bimolccular reaction of K with Br2 at 273 K. 3) Calculate the reduced molar mass (in kg). (m \L MAR/”m; (saunaU‘l/‘lfll”? ’ 63%,,th M : ../ J’— m ; [hm “SW-11"“ WWW ”“3 n, L b) Calculate the average relative speed (8RT/Itu)”2. 7 __ 8(5.3ms)(313) 0) Calculate A (A = 0 CE! No) given that the collision cross section is 3.0 x 10'19 m2. Chemistry 1B S’07, Exam I Name ._i 4 d) The measured value of A is 1.2 X 108 and E. is 180 kJ/mol. Compute the steric factor for the reaction. " Vow Mac! 755 H01 v“, 6. (15 points) The following results were obtained for the rate of the iodine clock reaction in a lecture demonstration [t a time for blue color to appear]: 6I‘(aq) + BrO3(aq) + 6H+(aq) -> 312 + Br“ + 3H20 I — 1“ M “f t(secL T(°K) ;) 33 280 2) 12 355 Calculate the activation energy for this reaction. Chemistry 13 8’07, Exam I Name 1 5 7. (10+5+5 points) The mechanism for the gas phase decomposition of NOZCI is: k, NOZCI <_> NO2 + C1 k4 NOZCI + C1 42“» No2 + C12 A. By making a steady-state approximation for [C1], express the rate of appearance of C12 in tenns of the concentrations of N OZCI and N02. {2 : KZ [Mic/j [4/] 3,214 $1710 : Kfl/WLCI] - KHCMflEcQ + KzUMUJCC‘j 0 2 Kilt/01m] + [at] (- Lima) + KZLNoch) -K‘CNUthj "4 [<4] :7 [CO : LCM/16:] i _ W Rib/act] K4 Elk/0’13 IL, [1002,] - lid)“; (,1 ] J2 2 “Eat/VOLCIJ K‘EWC’] _ K—c [Natl ‘ {CZ [P4160 w ngCNaz] ‘KLUWLCO B. Graph the concentration of C1 vs. time. 3‘: K2 [Nazq E2 H3 ...
View Full Document

{[ snackBarMessage ]}