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1032 exam2 practice key

1032 exam2 practice key - March 2009 1 Nam" PRACTICE...

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Unformatted text preview: March, 2009 1 Nam "@ ' PRACTICE EXAM II 1. The following plot shows the change in concentration of nitrosyl bromide, NOBI‘, with time, for the reaction: 2 NOBr(g) —> 2 NO(g) + Br2(g). Note that [NOBr]., = 1.60 x 10*3 moi/L. 11:1r l I «I H I“ I I i % I II ‘I‘ III I 0.8 I‘ I III III I I! "III I I. I .!II | [near] x 10-3 (Ill I . II II OZIIIIIIIIII IIIIIIIIIIIIiIlIIIIII-In mm {IIIIIII 16 (a) Explain how you can tell from the plot that the reaction is 2nd order in [NOBr]. (“12. INNLI— 11w M.— na’t mtbawb . \Jrhk (l.g_,o-BI':IMC« ,- lot-bk (wed-ow.) 7.1%; EWaCI'Hflm) 2 LI“ dw- (b) What other graph could you plot to prove that this reaction is 2nd order in [NOBr]? What would you plot on each of the axes, and what would the graph look like? I [[0081—3 i : tin-4.. (c) Use the above plot to determine the approximate instantaneous rate of decomposition of NOBr at t = 1.00 min. Show your work. BIN/U ~W "Io—Itha- Wnpt 1'60 Muh- SL-pn—o)‘. 43.14.: :-p€t(_.mIJ-=L- «Datum _..4'3 z-II- Juana: — o- 00"” )1“ = Le Lexie I‘M/m,- 2—; (0 " 1"?)M (d) Is the initial rate of decomposition of NOBr greater or less than the rate at t = 1.00 min? M 'tn'd'e‘tuk nah. [dc 0-0» we.) L. 3mm New... 1am aka. a): boon-.4)». 9.1... u¥+M W ”I: o-oo mi“ w LI'L. wt... 4v... 4w.)- «Jr boo (wick. . _\\ March, 2009 2 NaC EM :14 W25 ) (e) What is the rate of formation of Br; at t = 1.00 min? (W': -—_L£L +L¢l§fi§ l Dev- 3 2. b-L at Ma. WM? BIL -_ 98v.) 2 1/1 ch'ijgtod' :1 '2. -2. no,“ [vs/W (bk. (1) Calculate the rate constant, and include the units. JLl :: _.l——- Dims l” 1% ’Ur.‘ =- l'0° min. (I. k-[e‘\° ‘91.:- bet)”: \‘63‘1 5‘7m k: l ., .sJ—flw—a—w» = 625 W‘nc.“ 421; $99).) @013 Fat.) Ct-‘nodi‘Al --—'—'——__‘— (g) Calculate the concentration of the nitrosyl bromide after 20.0 mins. _\__ -4. 2. La: \Noflt lvofll). '1— '- _,_L——- 1'. 62.5- 25. c _ . ...S" imp-it. vaxto") * U‘C‘EAE — '7 82 no In 2. The reaction: 2 NO + 02 r) 2 NO; was found to be 2nd order in [NO] and 1St order in [02]. Use this information to complete the following data table. Show or explain your work below the table. Experiment [Nolinitial (M) [02]initial (M) Initial Rate (thr) 1 1.30111102 1.10x10_2 3.21x10’3 2 1.3u0x10’2 2.20x10—2 (MC-Ln. n63 3 0-12.50 mil 1.10::10‘2 1.2mm—2 4 1.00 x10“2 C“ [-043 Me"- 1.73 )1110’3 (as) Cm?“ up: \ “1.2. ; [pa-)mh.k_’ L02.) b¢%_ Luv Ln... ten. “Mu (Y‘HX—w-ILI-u— 1M- OL _'_ +1“, (“1M “4‘..." M‘}. d..§\._: 2x3-2tz\G-,= 6-‘1—13“ U3) Cmnflfch l ocJ. 3 r ‘92.) (MIN , (mtg, WM (t-lfixto'gzmw,’ : L6) - 5.14.. Law "HA-I— Mbh L. ZNLW ‘m NO r 3-- Lh-L— MLWGJ): lxi’jdxlbrlz Z'écxto”zm (c1 Un— f-npk :L 4n. at... L : La. rats-L : mums” ‘ __ “Th“: MAU‘ [NOB (0:3 (Kantian-(axle. J Mu... uh.— L400 CA; 73 '1 at“ {9‘3 t“ ”1’? Le- " l9») = ”A“ 7‘ ___\_‘__"_L.°___.-—————- :1 bond“ LL30]?- G.1Tgi?)(l-oo xi 0" 1J1- March, 2009 3 Name JWiKLS 3. In the catalytic decomposition of hydrogen peroxide, the following reactions occur: step 1: 2 H202(aq) + Mn02(s) —> H200) + HzMnO(aq) + 2 02(aq) kl = 1.78 x 10‘2 Mzs“ step 2: 02(aq) + HzMnO(aq) —> Mn02(s) + H200) k2 = 6.26 x 10'5 M's“ (a) What is the overall balanced equation? 2 H101 w #4. 2_ 62.0 (n + O ,_ (“Ll (b) Which is the reaction intermediate and which is the catalyst? In W~EL = B 2.0“" o H.) W = New: (C) Is this an example of homogeneous of heterogeneous catalysis? ‘1 , 3 _ ((1) Which step is the rate-determining step, and what is its molecularity? 9kg 7. L. an. @155 r mum-g =- 7. (e) Write the rate law for the rate-determining step. (It; ;-_ kLiQAi‘é'I-W 0X (1) If step 1 is endothermic, step 2 exothermic, and the overall reaction is exothermic, sketch one diagram (not two separate ones) of energy vs. reaction progress for this sequence of reactions. Label the forward activation energies for each step on your diagram. (g) At a temperature of 25.0 °C, step 2 has an activation energy of 40.0 kJ/mol. If we wish to increase the rate of the reaction for step 2 by a factor of 10, what temperature would be required? til—'- "—'- ta; :0 t .1 . k1: tor-ls “COL)= -Leo.uxto [:L __\_ E“; 145.01“)! SA“ Ll 9'3H-r TL 1“: R: 8‘3“? Vol-k -—lvt-'78’5 like.“ 2 %_ .... 23351 IKE-3 L TL: 1. Tl: 2H8 J— : 2-917xto" le'ZLi-S’K TL TL=1§°L March, 2009 4 Name:§ Eggsmms 2 (h) If we consider just the second step on its own, predict the following: (i) what happens to k; if the temperature is increased? L QM (ii) what happens to Ea; if the temperature is decreased? i“ o ct—lgcdc. (iii) what happens to the rate of this step of more catalyst is used? och. inn-mm (iv) what happens to the frequency factor if the collision frequency (2) increases? PS {nu-M. . (a) What will be the partial pressures of ammonia and hydrogen sulfide that the young woman determines, if the temperature is 25.0 °C and K0 = 4.2 x 10’2 for the reaction: NH4HS(s) = NH3(g) + st(g) AH=+17kJ/mol I 0 Sb (hm O'UDATM “This is a lovely old song that tell f mag worn-n who leaves her carcass; C- "' x "" K 4‘. K + mines off “31mg:- She arrives at her M woman p as lid NI-LHS‘ - shank cont-hin‘lIJOEIIIiI-lzfsgmmonimuldl E "‘ 0 $0 + 2‘- K mum-mmamd mudhydlmuflldawhn \(Q - Q, X? s) equibrhmhmdnd." '- W’ “I ah "'5 2g: 0 SD+x)L>¢3 fpu,..b-Sb+|-t9 X24055" —-1.5"—0 (“3:5341“. g 5:. le' 3 “News... mu - m... (b) The next day, the young woman decides that she must modify the reaction so as to increase the equilibrium partial pressures of ammonia and hydrogen sulfide. (i) Would increasing the temperature help? Explain. YES 51mm racer?» 1. Mime (an 4r) 5. T1“, WT start -—> 5.. (ma T in“? (ii) Would transferring the mixture to a smaller flask help? Explain. No huML,sWé—w s. iuhnLiHyW—szn. (iii) Would adding more NH4HS(s) help? Explain. No News '... .c mu.) m’r~ Lem k4. ewe}... (iv) Would adding a catalyst help? Explain. No Cuhnkbf'k— «Lb arm-Iced. m‘uz. 4.. Mews; tbs-M—imhnfi Mi”- LflJthHux... WWW, M-L—h‘u— pm WW3? “Lat-“65L, March, 2009 5 Name: JUL/6X13 (c) In another experiment, the young woman discovers that the equilibrium partial pressures of ammonia and hydrogen sulfide are 4.00 atm and 6.25 atm, respectively. Then, all of a sudden, the partial pressure of the ammonia falls to 2.00 atm. Complete the following diagram to show how the partial pressures change while equilibrium is being re-established. (This is similar to the example shown in Fig. 14.10, p. 647 in your text). cfih’b. - _ . CLS. “I“: L ll M ((1) Calculate the new equilibrium partial pressures of hydrogen sulfide and ammonia. NHEWSU‘: :3- W7 (31 +- H15 (5) '3: - Lpoo 6-35 1.6““ I -- 2-015 6'15- C— .-)<. +X Jr! + E you“: 64.3 + x- k‘, a ZS": Q-aorXXé 25+ x) 9114- R-lfx -\"-‘5' :- 0 fun: ‘— l-osa-t-fh 2 E-tht'm. x = \-3\ x a --1.s7. has 1 6-2.“ t-m - '1'“ «Fw- T WM 5. If the equilibrium constant for: CD 4N0 + 02 =2 2N203 is K1 and that for: @ N203 + 02 = N205 is K2 What is the equilibrium constant for: 4 NO + 3 02 = 2 N205 in terms of K1 and K2? (9 u-No + on. 2-? awe; K. _._'.:~.. '1- @"l’b7— Zita}: 4* 2%. *‘ lid-mg: [Kbi *- Lfi-MO ’h" 201 :2, lNzor kn T. Kt kkz-y'l. ...
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