sample1_fall98_key

sample1_fall98_key - Chemistry 105B Hour Examination 1 Fall...

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Unformatted text preview: Chemistry 105B Hour Examination 1 Fall 1998 You must Show all your working on the long answer problems to gain full credit. Useful Physical Constants and Equations TC =TK—273.15 —1 —1 R=8.314 JK moi Zeroth Order reaction: [A] = [A10 ' kt half life t1/2 = [A]0/ (2 k) First Order reaction: In [A] = 1n [A]0 — kt halflife t1/2 = 0.693/k Second Order reaction: 1/ [A] = 1/[A]0 + kt half life t1/2 = l / (k [A]()) Arrhenius Equation: k=Aexp(- EA/ RT) 1n(k) = ln(A) — %[% j Multiple Choice and short answer questions : (You need not show your working to receive full credit) Question 1) If the reaction : 2N205 (616])9 4N02 (061) +02 (8) is first order and has a half life of 35 seconds at 110° C, what is the rate constant for the reaction, if [ N205 10 = 0.1 molL_1? (a) 35 moi"1 L s"1 0020 5'1 (c) 0.020 mol‘1 L s'1 (d) 0.29 s"1 (e) 0.29 mol_1 L s-1 (5 points) Question 2) A catalyst (a) speeds up a reaction by increasing the temperature (b) speeds up a reaction by increasing the reaction exothermicity (c) has no effect on the rate of reaction speeds up the reaction by changing the activation energy (e) none of the above. (5 points) Question 3) The mechanism for the following reaction has 3 elementary steps: C4H9Br + 2H20 a C4H90H + H30+ + Br- Mechanism: C4H9Br —> C4H9+ + Br' (slow) C4H9+ + 2H20 9 C4H90H2++ H20 (fast) C4H90H2+ + H20 —> C4H90H + H3O+ (fast) (a) Which step is rate determining? What is the molecularity of that step? @ Sum STeP (STE? M) ‘5 RATE W WMWWG ® 0M1 Mougww? QTeP (b) Identify the intermediate(s) in the mechanism given above (if any). @ Call; (c) Write down the overall rate law for the production of C4H90H from C4H9Br based on the proposed mechanism. Fate; “AtCthlBrjl: RECqHSBV] (3 At (10 points) Long Questions (Show all your working for full credit): Question 4) The reaction N0 (:4) + 03 (g) -> N02 (g) + 02 (g) obeys the following rate law: -A[NO]/ At =(0.8mol'1Ls—1)[O3]2 (a) In a given experiment, how long would it take for [ NO] to be reduced to one half of its initial value? SECOND 06mm OVBQALL, IN 03 9 E71 ’1 thcglo , 1.15 moL L“ s" I: 0330 Dawn}: m m \kaA Cmng [03] Four possible mechanisms for the decomposition are: I II NO +03 9 N02+02 O3 9 0+02 NO + O 9 N02 III IV NO +03 9 N04 203 9 O6 O3+NO4 9 N02+202+O NO+O6 9 N02+O3+02 N0+2~Ozfl’ Va #79154 +NOL+ 102‘”) %03+ ND —r 94 —-J/ 064 N024.- 94+ 02 N°+103~>~ r2107}. 1014—0 / (b) Which mechanisms are consistent with the overall balanced chemical equation? Show your reasoning. SWMMLA5 M10 WI, 111;) m we WWW an. 6‘35»; m: @ flfl/t MI” QWUUII Zawgli‘m Question 4 continued) (c) If the first step is assumed to be rate-determining for each suggested mechanism above, which is the most—likely mechanism? Why? OAVJWMMJ 13E}?- mmn Ram Law u (WWW; 53 Pm 89 um RD S 1 (d5 [5 radii h‘Cog—ACN 0] vaDW j: (As is grstfle‘o rfijf, :- h'l: UALMD\W ‘3 3 1?: mi; ‘6 (vs\ Si}? O/Ws fl: L: (mde vaH (NW rat’a \cWo 0M 6mg kalmmé OVW Whoa mmmm TI Mos): hub (20 points) Question 5) At elevated temperatures, nitrous oxide decomposes according to the equation 2N20 (g) 9 2N2 (g) + 02 (2;) Given the following data, plot the appropriate graphs using the sheet of graph paper on the next page, to determine the rate law for the reaction. Give both the differential and integrated rate law in your answer. Time (min) 180 [N20] . 0.204 0.190 0.166 TO M iSl mm‘ !o\& (AENZOI \lefswr [we @ “HM/its ST‘QWG'l/(T LINE (W AM? (Day) To m {a M W (3M J, Wm 4M; [ix/Lo”) ‘ Tm ix nol’ Tented/('1‘ UMP, AT LEAST ONE (am anon) PLOT o~ alum WPR ($81me -1.4 \ S’TRA \thT U NF 6:. 1.6 E. E -1.8 o 50 100 150 200 time (min) 0 Sim: ” “03 : —2.275x:o~5 Test for First Order Kinetics ’ 80 "WV! o 50 100 150 200 time (min) Test for Second order kinetics (b) What is the rate constant for the reaction? (Make sure to include the units in your answer) W stag #) V1 @679 ® SlDPQZ—Z.27S Xt‘O—g MiG—J <2 $5 $59 (20 points) Question 6) The first order rate constant for a reaction increases from 10.0 s'1 to 100 s'1 when the temperature is increased from 300.0 K to 400.0 K. (a) What is the activation energy for the reaction? 93133 566‘ W M‘ debt A Whom: giuaiiom 8.2.3 x10" K" 21-131) k3 moL“ [v 395 £3 (b) What would the rate constant be at room temperature (200°C)? at 20.0%, (293 K) _. - _ 1302 (mm ‘Jw‘mot' 2.303 \ < EA) <_ unfit») £7 EA—( ) ) 50W MA *6 Qndrmowbvt (A R Z. (/1 _ 23/000 (mmtzea) PMSiCaL (c) Give two/reasons why the reaction rate constant for a chemical reaction increases when the temperature is raised. 1;} m Mmbqug Miami Minions Madam 06 W i 5 vi Omar J baa/ML Juim LAW an/WW 5"» m M1 \D’U ow} \\mz, u‘gyewu w‘bL (20 points) to ,_ Hm: eta Fae (c) Mm prvHQ/m Question 7) Given the initial rate data below, we may consider the rate law for the reaction 412e,2+ + 02 +4H+ 9 4Fe3+ +2H20 The initial rate, -A [02] / A t , was determined for the following concentrations of each of the reactants in 4 experiments: [FezJ']0 [02]0 [H+]0 Initial rate EXperiment mol L-1 mol L'1 mol L'1 11101 L-l 5-1 1 1x10_3 1x106 0.1 5 x10-4 2 12x10_3 2x 10'3 0.1 8x10-3 3 2x10-3 1x10.3 0.2 8 x10—3 4 2 x 10'3 2 x 10'3 0.2 1.6 x 10‘2 Gwemg RATE um ; R CF22“? [0233 CW]? (a) What is the order of this reaction with respect to Fe2+? mm?) _ A CW3] Wadi : : My raki ” I S 0-9 2/ EWOJJW‘COJT 7“ x Z 2 2/4 / l l list/5 remit h WU! (IO-Ki CH"? 7C6? 2:1, 30 23C: 3 oval mom“ (3 THIRD ORDKQ CF€2+] 10 Question 7 continued) (b) Determine the order of the reaction with respect to 02 Fair} \ ,R/EIXIO’SJE E leO'ljj 8MO‘3 ' w, v w] 3' C, A new pm”? ' S'x ID'Ll LA. ‘3 \U/ N N l! a: {1/ M L; \\ v.) Remy (r (FIRST ORDPQ, Lari; 2:02] . . . + (0) Determine the order of reactlon With respect to H Chuck. Me’s law) 4L ‘lfi extmd (DrdJU with Cut]- wamfifl “£12? % l.é><lo‘1_2 W1 l4 (32*qu EZWBCOJ’Ji : Z : 3x103— @ :> (Z :1] Rudd,“ a; FlRS’F ORDER NRTCHfl ((1) Write down the overall differential rate law for the reaction, based on your answers to parts (a) — (c)- ll (e) Calculate the value of the rate constant, giving appropriate units. 0&3 (lat—a mefl lug:— 3 \ \ Q 3. w "i raiz : 5x ID'L' : k (W031 “‘04) <0 l} G) ' / («on mm (mm) W9 L, :> R: YXIO‘ 9 W '“ $— X10 U” (20 points) 12 ...
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sample1_fall98_key - Chemistry 105B Hour Examination 1 Fall...

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