CHEM105b - Exam 1 (2008, Spring)c

CHEM105b - Exam 1 (2008, Spring)c - PLEASE PRINT YOUR NAME...

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Unformatted text preview: PLEASE PRINT YOUR NAME IN BLOCK LETTERS Chemistry 105b First letter of last name Spring, 2008 Name: Exam #1 Feb 7, 2008 Last 4 digits of USC I.D.No: Dr. Robert Bau T.A.‘s Name: Lab: (Tu morn. / Tu aft. / Tu eve. / W aft. / W eve. / none) (please circle lab section above) __ _ __r Question Max1mum Score Grader points 1 15 2 16 ' "1 Alwyn w —+———————— L 4 +———— 9 —L-— J— —-—~( III..." 6 r 16 A 4 7 14 I #8 f _ Total 100 I VII _ ___._..1 2 H II III IV V VI VII He 1.00797 4.003 ‘1‘1‘T—j 5‘ 5 ’T a 7 8 ’1 9 1 10 Li Be B C N O F Ne . 0.81 12. 1 14.007 5.999 8.998 . 1 12 15 1 1e 117 7—2-2118 1 Si P S Cl Ar 28.09 30.97 32.06 35.45 39.95 32 33 34 ’1— 35 36 ‘1 Ge As Se Br Kr T 72.59 _‘ 74.92 78.96 __L_ 79.90 83.80 49 50 51 52 53 54 Sn Sb Te I Xe 118.69 121.75 L 127.60 126.90 131.3 82 83 84 85 86 Pb Bi I’o At Rn 207.19 208.98 (209) (210 (222) J 58 59 65 66 ‘1 67 68 e9 1‘ 7o 71 Lanthanides Ce Pr Tb Dy Ho Er Tm ‘ Yb Lu 1 140.12 14091—4 158.93 162.50 164.93 167.26 [68.93 <L130! 174.97 90 91 97 98 99 100 101 102 1 103 ‘5 Actinides Th Pa Bk Cf Es Fm Md No Lw 232.04 231.04 . . (247) (251) (252) (257) (258) (259) (note: a copy of useful equations is given on the last page) I have observed all the rules of academic integrity while taking this exam f signature Question #1 (15 points) A reaction A —> B is a first-order process. The concentration of [A] is 0.258 M at 12:00 noon on Ian. 22, and it decreased to 0.147 M by 12 noon on [an 31. (a) (6 pts) Calculate the rate constant k of this first-order reaction. 4 A 7:“1‘6 1"( 0",“? > 5' “k (7.447!) a ,2, (o.57o)fl ° 251’ : ( 5’79): “4: a_a‘z;d‘1l 74471- fdafa , (b) (3 pts) What is its half-life? 4/ : £3. : 34—5—2:- :: ”~/ dag: A A a . o as a," 1 ‘W‘ (c) (6 pts) What w the concentration of [A] at 12 noon on ]_a_1;1_5? [144.1104 1 ( We M146”!- Arm.) [AL-7 me get 4.5.1.1. a“: [A]: m; at 4. If (6'm=—7447:) ,6» [A] = -14.: + a0)” = (, o-oézs'aq,")(e7d,,,) + 1» (a-zs’JM) :: 4439—1455 .2 ~o~7l7 [A] 5 4.0.90: 0.9-ooM ____.__.—-————-l an 3" MAI-Med 3- Defiu [96‘]: We. “flaw If and [4]: ‘97" 01‘ fl = ,5” (0.1!?) + (o- 0425' da7:’)(74a75) : —(.?:5' + 0.42.! -= ~0~‘71‘7 Question #2 116 points! The initial rate of a reaction P + Q = R was measured for several different starting concentrations of P and Q, with the results given below: Experiment No. [P]0 [Q] 0 Initial Rate 1 0.30 M 0.06 M 0.110 M sec"1 2 0.90 M 0.06 M 0.990 M sec"1 3 0.30 M 0.12 M 0.880 M see"1 (a) (10 pts) Using these data, determine the rate law ( Rate = k[P]n[Q]m ) for the reaction (please show workI or state “by inspectionI comparing lines ## and ##”) MA/Zod 1 ( 12-70; cfi'a'h) '5 an. .- 12¢ /ao-d 2—, "=7- 1 [arhi hf 2;”;5: Rd‘:kfpl[cpj n; awn/arr“;- 11-h» / “he! 3/ fl-‘A‘d Z “‘7) __ h m arm/aw flak (4mm (2.), 0 -HM W 0.97M k[0.?]h[°.0‘]n (2%) = 6%)“; n w— eir-'14”; (“4. CI) Mal (3)) Ov-IIM [((o.3)”(o.cé_)m M: o-FPM /< (0.3)“(a.z2)"‘ (VF): (yz')m/' 4° ":3 A711: flah’kl-PJZ-o]? _________._.—__—__-——-—""__________————-"—“ an (b) (6 pts) Calculate the value (including the correct units) of the rate constant k. Am, af (hfiuf flu... £1p‘h‘un: um .ruffiu. filth: [<[PJLLQ13 k Kate. o-Hon 5““! 1 ______________._ g _. [rTEQJ’ teemi’rmmr —I 3 M’B‘x“ A : 5-7 X10 Question #3 (18 points) The following mechanism has been proposed for a hypothetical reaction: Rate constant of step _ Step 1: A <:> 2 B (fast equilibrium) k1 (forward); l<_1 (backward) Step 2: B + C —> D + E (slow) k2 Step 3: B + E —> F (fast) k3 (a) (2 pts) What is the overall reaction? A + c ——-v D + F (b) (2 pts) What are the intermediates in this reaction? 3 amt E (c) (2 pts) Which is the rate-determining step? [Liar 2 (d) (2 pts) What is the molecularity of the rate—determining step? éa'mo/Aw/ar < 6/: Mo/4¢u/ar:7" = 2) (e) (10 pts) What is the rate law predicted by this mechanism? (Please show work!) flak/«t. :: kl [5.][CJ (f/Ou 5-117) frm fasl' fifki/) k, [A] :— él [(3]!- n B=¢//(fi)fnl k— I 51535-44517 Air 1:45: fl. VII/+4 1—74‘44'0“, /<( ’A (L [cf-e. = /<1_(7<'—) [Ad/[c] Question #4 19 points) The reaction A -> B + C is known to be zero order in A wit 5.0 x 10'2 mol/ L 's at 25°C. An experiment was run at 25° What is the concentration of B after 5.0 x 10'3 sec? h a rate constant of C where [A]0 = 1.0 x 10'3 M. Zara drab—r: flag-4. : [< ': 0,551 “no”, [A] .— ~ké + [AL " —(a_05‘oMr1;’)(a.045—yg> + a.ooloM : ; c.ooo75‘M - a. 06025 4— a.¢°,° [’3] = “nu-“1’ sf [Ajax-«£7 : 0-00/0M __ 0.000755, : OuOOQzSH -—¢+ ~ 53PM .. Question #5 112 points) action has an activation energy of 9.1 k] / mol. As the tempera the rate constant increases by a factor of 6.125. his problem is very similar to #61 of the HW set) A certain re ture is increased from 19°C to a higher temperature, Calculate the higher temperature. (Note: t ME? =f;(—,:--—7—’:> LP 7-, = H’c dud TL 4- fl uufi‘flwfi- k; =(é’lzg)(k/) 7’I Alma/4:, r I [0007' ', ( ’ 75)( J . 5 = “Z” (4 ’2) r"-3197 I<”m.;¢" /6’+273/< l/q' (L. ,,1_ _ (salumagdzy) " - a ll (taco) -3 ’L (f:3/‘F)(l.ill) _ I g —‘$ ._ ,_ 2 - . 5‘ x10 3'4ZS-XIO 1- C?~/)(Iaco) 4 I 5!;K h M f‘ 7;“ ’ 2 r3 " ———-___.__——‘=—7 .7172;qu v- l-és’éxla a, zfz‘c Question #6 [16 points) Consider the reaction: 3 A + B + C —> D + E Whose rate law is: —A[A] / A t = k[A]2[B][C] An experiment is carried out in which [B]0 = [C]0 = 1.50 M and [A]o = 1.50 x 10‘4 M. (a) (10 pts) If, after 135 seconds, [A] = 6.25 x 10—5 M, calculate the value of k. (note: this is very similar to Problem #77 of the homework set) rim" [5]. Mel [c], are 1;“ hay-c Juraer fl}: 1';- q. {J‘Audo‘ fiend ' ord‘v I’ll-‘9‘.“ flail" Z k’[fij‘ “‘Lr‘ k I: k (B]°[c]o / 2’37 = ’6 +z—go l/{-2rm."‘M = #0357“) + 7,.;.x,;"M k’(125‘mcr) = l/{.2szo"'m " l/l-s’oxtng -: (glam/v1". glu7M": 7,3331%" A“ 7 3??fi-/r3;.r..u == éfi‘J M Cm," /<’ 97454—2134" ,3 ._ /< = = - 30.7 M 9:; (b) (6 pts) What is the concentration of [A] after 625 seconds? “I'm? fuudo—mcwd «la-o- ral" [dbl-AZ", _L. [A] H +EAJ° I = (47., warm and 4 (ma) :2 4.317 x:::4 p.144. a,6‘7 XIS’I‘M—‘ : 4.?36xm4 m" ! —5- f»]:-—-————————- : ZNHch 4-7?é¢1o"M" M Question #7 (14 points) A study was made of the effect of the hydroxide concentration on the rate of the reaction I — + CO — ——> 10 ‘ + Cl ‘ in the presence of base: The following data were obtained: [I‘]O [OCl‘]O [OH-k, Initial Rate (mol/L) (mol/L) (mol/L) (mol/L.s) 0.0013 0.012 0.40 9.4x 10-3 AW" 5"“ "‘ "fl ” 0.0026 0.012 0.40 18.7x 10—3 fiffuhd 47 (040‘ maW‘fi‘m 0.0013 0.006 0.40 4.7x 10-3 M [0,7 ' 742m m4...“ 0.0013 0.018 0.40 14.0 x 10—3 - , fl, U 0.0013 0.012 0.20 18.7x10-3 / “7‘7"” V" 7” U 0.0013 0.012 0.80 4.7 x 10—3 ( flu'r ;! hlf' (:q fl-s Val/{aw —3 0.0013 0.018 0.80 7.0x10 “I 7‘“ {4.9,‘ A 40"“ ‘7 . . . - , b, ,‘ (l- “6" Determlne the rate law for thls reactlon ’7‘ " “" ‘7“ l” 4" “.7 /° “' ) (please show work‘ or state "by ins‘gectionl comparing lines ## and ##"1 Z (note: this is very similar to Problem #79 of the homework set) w". —"a—r Ml-n'ad! ([7 A7414: k[IJ[“][”] . ~M__ a. = Cm Chi/~7- it I M“ J m l ’][cc1u] can/am“; ii“: I A.“ '7 ’ a = I #14:: [‘fc : [a 7.. Cmfariu,‘ .21.“: laud 5‘} r: _/ H (N f dual 4 /0u; I'M-flu“ 7- (an “61) In -—3 Cm/odrikf if“: / 44.4 2 (0",0’3) (o'dlz')h (OZ—L4 : we; (a.oozg)"' (0.011)” (o.q-)f [P.7xlo‘3 c4)=<4)"‘,- I. n... d Cmftzfinf L'uu‘ [Md 3 (a,aola)"(o.o,z)”(o_9)’ : 4.493(10'3 [0.6013)m(a.aa()f(o.4)f 4"]on (1)"; (1)“; So n: / IM“ 5. (a.d°1;>m Co.a,t)fl(a'#)f — 7\_"xl°’3 (OM/3)” (0.011)"(a-2.)F /P-7:<Io*’ ouf (2'): fa f: -1 04:11:} /L:ja,_-/7.mr) 1,7 2 __ L; /= (£92)/(/4/a-y) : '1 MIMI“: K442 .7 “IEGH'I—l USEFUL PHYSICAL CONSTANTS AND EQUATIONS Temperature conversion TF = [(Tc x (9/ 5)] + 32 TK = Tc + 273.15 Avogadro’s number NA = 6.022 x 10+23 Atmospheric pressure 1 atm = 760.00 torr = 101.325 kPa Energy units 1 cal = 4.184 J 1 L-atm = 101.325 J Universal gas constant R = 0.08206 L atrn K”1 mol‘1 = 8.314] K“1 mol‘1 Faraday’s constant F = 96,485 C mol-1 Ideal gas law PV = nRT Work w = —FAL P-V work W = —PextAV Heat transfer q = mCsAT 1st law of thermodynamics AB = q + w Enthalpy H = E + PV Arrhenius equation (exponential form) k = A exp( _Ea/ RT) Arrhenius equation (logarithmic form) k2 E 1 1 1n —— =———"— ————— k1 R T1 T2) Quadratic formula x = _ b i (b2 _ 40“: 2a Order Rate law Rate = k Rate = k[A] Rate = MA]2 1 I =— + =—k+ A -——-=k:+—-— Integrated rate law [A] ’0 [No NA] 1 mi 10 [A] [Ah 1 Plot needed to give a straight line [A] versus 1 ln[A] versus t E versus t Relationship of rate constant Slope = _k S‘ope = _k Slope = k to the slope of straight line - = [Mo = 93.2; = _1__ Half-life t1/2 3;— ‘1/2 k ’ W k A10 ...
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CHEM105b - Exam 1 (2008, Spring)c - PLEASE PRINT YOUR NAME...

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