Final '07

Final '07 - CHEM 132 FINAL EXAM K. Lindenberg Winter 2007...

Info iconThis preview shows pages 1–18. 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 DocumentRight 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 DocumentRight 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 DocumentRight Arrow Icon
Background image of page 6
Background image of page 7

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

View Full DocumentRight Arrow Icon
Background image of page 8
Background image of page 9

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

View Full DocumentRight Arrow Icon
Background image of page 10
Background image of page 11

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

View Full DocumentRight Arrow Icon
Background image of page 12
Background image of page 13

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

View Full DocumentRight Arrow Icon
Background image of page 14
Background image of page 15

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

View Full DocumentRight Arrow Icon
Background image of page 16
Background image of page 17

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

View Full DocumentRight Arrow Icon
Background image of page 18
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: CHEM 132 FINAL EXAM K. Lindenberg Winter 2007 NAME if. LIND€NBEKG STUDENTID 1. /37 5. /16 2. /15 6. /17 3. /22 7. /21 4. /12 8. /25 TOTAL FINAL EXAM/ 165 MID/105 + HW/3O + FNL/l65___= _/300 = /100 COURSE LETTER GRADE NOTE: Your REASONING and ANSWERS MUST be clear to me. Guess- work, or apparent guesswork, will earn NO credit. TOTAL NUMBER OF PAGES INCLUDING THIS COVER PAGE: 18. IMPORTANT NOTE: To grade this exam we have to be able to follow your steps, so please be neat and clear. This is especially important for some of the longer problems (e.g. rate law management). We are NOT willing to wade through messy work to follow your reasoning. At the end of the exam there is extra working paper that you can tear off and use for your thinking, no need to hand it in (page count above does not include the blank pages). The actual answers have to be clear and reveal your reasoning. CHEM 132 1. 37 points total; each part 3 points except (g), which counts 4 points The following short questions are of the True/ False or multiple choice type or simply require a short word answer, all concerning reaction kinetics (Chapter 17). Along with making choices you MUST give me a short explanation or an example or a counterex- ample that makes your reasoning clear. A simple T or F or multiple choice answer Will get no points. (a) T or F: Every reaction has an order. «mam ma am WW W W sfuf cm W tux/ms 09 . (b) T or F: Every species that appears in the rate law of a reaction must be a reactant or a product in that reaction. [1:] :thfiWMWMM WW, (0) For the reaction scheme A —> B —> ZC Where the concentrations of any inter- mediates are in a steady state at low concentrations, which of the following is the correct statement? A[A] = A[C]; (ii)2A[A] = A[C]; (iii) —A[A]=A[C]; (iv) —2A[A]=A[C]; (v) —A[A]=2A[C]. F“ W A Mfaliéaw)m C’s aw) 90 Choime a Cc‘l cs rm Lu, CA3. mm m}, flow 14m Poem 11‘ H MOJLW ((1) Explain the word “reversible” in thermodynamics and the same word in re— action kinetics. Yes or No: Are these meanings equivalent? Whether “Yes” or “No”, your explanation must make your selection clear to me. Wmmoalxl ma, ‘ '. Q yé'l’w CS m E § § flow SW 2; E- :3; $- A CHEM 132 PL ‘ ~ q ' ’ (e T or F: The Arrhenius equation holds exactly for ideal gases but not for real .3 O gases or liquids. o k3 (f) T or F: The Arrhenius equation holds exactly for real gases or liquids if the E 3 gas constant R appearing in it is suitably modified to be the appropriate system- dependent constant. 553 El R mméwi‘wom "L‘M‘Ws 17— ° [g a WM WWRW/rwmfi'm i may} . Q? (g) If r = k[A]m[B]” for a reaction, deduce m and n from the following information: when the initial concentration of A is tripled and the initial concentration of B is doubled, the rate is multiplied by 18. When, instead, the initial concentration of \Q A is doubled and the initial concentration of B is tripled, the rate is multiplied - by 12. m wig A: MA]me “4’ M an \811 : )3 13A] L73] ‘4 § § \9n:*§9m EBB 73> 1g:/er9X3-J'm nut/mm (h) T or F: In homogeneous catalysis, the catalyst does not appear in the rate law. 12 mi If g WWWWaWMfin/WW é HWJ/ancmfiswkétmmd'im? (i) T or F: In homogeneous catalysis, doubling the catalyst concentration will not g change the rate. ‘ l 5 $3 E DWWUWWManAv ‘! fl: 3 (j) T or F: The presence of a homogeneous catalyst cannot change the equilibrium constantZafim. o [T3 mama—7U WW e; E S (k) T or F: The presence of a homogeneous catalyst cannot change the equilibrium ft ’ composition of a system. a E amt mmw WW 8 gl— W‘f’w . (l) T or F: The half-life of A in the reaction nA —> Products depends on the initial 2’ concentration of A for a second order reaction, but the time at which 99% of A "jfg is gone no longer depends on the initial concentration. . 15 points total For the dimerization reaction 2A —> A2 of a certain nitrile oxide (compound A) in ethanol solution at 40°C the following data were obtained. [A]/(mmol/dm3) 68.0 50.2 40.3 33.1 28.4 22.3 18.7 14.5 t/min 0 40 80 120 160 240 300 420 Because it would take a fair amount of time on an exam, I will give you additional information that you could easilydeduce graphically from the given data, namely, the time t at which the concentration has fallen to half of its initial value: Hm f; {77L [A]0/(mmol/dm3) 68 60 50 40 30 t/min 114 132 163 198 266 (a) 8 points Use the half-life method to find the reaction order. To do this I want you to use the “graph paper” provided above. I want to see your graph; make sure I understand what you are plotting . Be sure to label your axes clearly, including units! le’w M W W b» 2,095”, (ii-HM) 2‘057— 2.191 96%..Vfll./¢MMJ/M)§ 1,633 [33,2 rm 3966.“)? 12mm Mal—'- sloeei 471% hu/i/‘G’MOMSAW 5W4 mm. 7.15??— 2H1? [.402 L9??- 3,212 W :—I.0«Cl l-’Yl 3. (no—13579 Sammian IIIIIIIIII Ill-lull!- Illlllllll lull-lull- IIIIIZIIII IIIQZIIIII IIHIIIIIII lull-Ill.- null-Ill.- IIIIIIIIII o lo 0 'M 0 300 go o :0 0 t I m m (b) 7 points Find the rate constant for the reaction from the data provided. Since the reaction is 2A —> A2, the rate law could be written as 7" = k[A]" or as r = 2k[A]". Tell me which version you are using when you give me your result, and be sure to include units in your answer. Again, I want you to do this using the graph and I need to understand clearly what you have done. hA]’\/[l-lmrl> W7— l‘lr‘l 218 30.9. 7:52 44.1.9 5’35- 49,0 'if/rmbw O '40 90 \20 140 1‘10 200 wa mm [7-,5'3 CHEM 132 3. 22 points total There is good evidence that the gas—phase decomposition of N205 with overall reaction 2N205 —> 4N02 + 02 occurs by the following multistep mechanism: ka Step(a): N205 :‘ N02+N03 k—a 7% Step(b) : N02 + N03 —> NO + 02 + N02 kc Step(c): NO+N03 —> 2N02 The observed rate of decomposition is d[N205]/dt = —2k[N205]. In this whole problem I encourage you to use the scratch paper provided at the end of the exam for your detailed work (I will not collect this), but show me here enough of what you do to eliminate intermediates as required in each portion of the problem so that I can tell what you have done. Questions start on the next page. CHEM 132 (a) 8 points Apply the steady-state approximation to both intermediates and show that this leads to the observed rate law. Find the expression for the rate coefficient k in terms of the rate constants for the elementary steps in the mechanism. A good fraction of the points will rest on your finding the correct result for the rate coefficient. awful“ (N20!) +184 (N09 (N03) CHEM 132 (b) 8 points Instead of the approximations in (a), apply the rate—determining-step approxima- tion to the N205 mechanism, assuming that speb b is slow compared with steps —a and c. Again, show that this leads to the observed rate law. Fing the expression for the rate coefficient k in terms of the rate constant for the elementary steps in the mechanism. Again, a lot of your points will depend on finding the correct rate coefficient. r‘p Gila? l) (5 ’H“ “m' "Ltd-’W Sic—‘0 M sly) a is m Mbm)m “£2:— : (NOD (N03) fi'a' (N>O§> WWOQWWMJCLWA H” not, ~ A M \; .S‘l'oi‘cWJ/TZL WmUC pct. gs 0, so mam cs insomon -: kbb. (N;0§ «k-» B (c) 6 points Under what condition does the rate law in Part (a) reduce to that in Part (b) of this problem? Does this make sense? In addition to stating the mathematical condition, give me a one—sentence explanation as to why this condition is necessary for the answer to (a) to reduce to the answer to Elli?» mm m Ls opr W a, W ‘ WW 0? in Mtg—WW ’si’cf Wow, 9 76mm CHEM 132 4. 12 points total This is the rare derivation problem, but it is a simple derivation so don’t fret. Consider the elementary reaction M A : 2C. 1% Suppose this system, initially in equilibrium is subjected to a small perturbation. The concentrations [A] and [C] will then deviate from their equilibrium values [A]eq and [C]eq. Initially just after the perturbation is applied the concentrations are [A]0 [C]0 but in time they will relax back to their equilibrium values. Derive the relation [A] - [Aleq = (We - [Alene—“T and give the expression for 7' in terms of the rate coefficients and the equilibrium concentrations. There is more work space on the next page - the rest of this page might not be enough. w;,jq(m+/&L(C)L At M x: (ALT; (A) 5> (c)%-(c3; -—.2>< GUM ; " &:~kg[(fl>€?flj+ ELKCJq-i'ile‘ if? at / =(C); HM (Clef. PW; 10 CHEM 132 MORE WORK SPACE FOR PROBLEM No. 4 Who 0'44 > WV w (W‘W W) 11.5 §$=-—[X¢g 4’ %ib(C3fi—]* : ‘T’ly Jags-L} ah __ —‘ If": * x if M Wt . AX “ q: 1 Q 55 \L =. .— imflr 2‘ ET ’ 0 2: ——‘t/’C ;> X: e’t/t «Wt 11 CHEM 132 5. 16 points total, 8 points each part Consider 35012 (ideal) gas at 300K and 1 atm. Data for this gas: Grot=0.3500K and x. o @m-b2798K. In case you have not written down the quantum energies and degeneracies, W QC; I’m giving you the information: vibrational energy levels in units of K are ism-M = (k + 99m (nondegenerate), With k = 0, 1, 2, - - -, and rotational energies in units of K are 5",“ = GMJU + 1) (degeneracy 2J + 1), With J = 0, 1, 2, - - -. The rotational partition function for 35012 at 300K is equal to 429. (a) What fraction of molecules in the gas has vibrational energy 8mm? What fraction has energy 5mm? ?\l"b '1 l _ Z Lei—148800 , f E ‘5 ‘~ L—e Fnou. 1 vs ,0 gwb -Qvib/T’ ?V\\o (b) What fraction of molecules in the gas has rotational energy 5mm? What fraction has energy grow? €nof)o Ii L—fi ?flar ‘EAQCA’I‘M foo‘l’)\ " 3 e CHEM 132 T E- 6. 17 points total; each part 3 points except (e), which counts 5 points W Here again are T and F and short questions requiring some explanation of your an- ‘ swer to get credit. The questions refer to ideal gases. BEWARE: Remember the i distinction between energy levels and states. ‘2 Q‘_ (a) T or F: For a system in thermodynamic equilibrium, the population of molec- ular energy levels always decreases as the energy of the levels increases. (b) T or F: For a thermodynamic system in equilibrium, molecular states that have the same energy must have the same population. E m Aewwwpw :1: WANG; . (c) T or F: It is impossible for a higher-energy molecular state to have a greater population of molecules than a lower—energy state. SOL.’Y l\ “fl. them 0‘4"wa (d) T or F: For a thermodynamic system in equilibrium, if we set the zero level of molecular energy at the ground state energy, then the fraction of molecules in the ground state equals 1/2. E x Wi’fix lka) i- ng :7,“ ; aw“; ,L W €050. ? '1'" (e) Sketch Cum versus T for a typical ideal diatomic gas, and explain your sketch. , b Nvmsim ’r no+a+~fm + V' “ ‘lfiafl +n 0+5le (M 0 22.53 Vmblmmalfm W 29.1% 40mm 221% 5 g 0? 13 CHEM 132 . 21 points total, 7 points each part AGAIN BEWARE: Remember the distinction between energy levels and states. A certain system is composed of 1 mole of identical, noninteracting, indistinguishable molecules. Each molecules has available to it only three energy levels whose energyes and degeneracies are 51 = 0, 91 = 1; Ez/kB : 100K, 92 = 3; 53/193 = 250K, 93 = 6. (a) Calculate 2 at 200K. 0 ‘I00/200 -250/200 2:- ’x& 4‘ BXQ 4.1x?— 11+ log; + 1,74 l}:- Lf~§3 (b) Calculate the average number of molecules of each energy at 200K. 14 CHEM 132 (c) Calculate the average number of molecules in each energy level in the limit T —> 00. 15 CHEM 132 8. 25 points total In this problem we will deal with the TST approach to the rate coefficient k for the reaction H + D2 —> HD + D. The HD2 activated complex is linear (H—D—D). The rate constant is (ztir/V) Zfot zxiribzil k=N k Th—lex —Aei k T ———-——-— A B p( 0/ B ) (Ztr,H/V)Ze1,H(Ztr,D2/V)Zrot,D2Zvib,D22e1,D2 Here is some data: 0 Boltzmann’s constant k3 = 1.38065 X 10‘16 erg/ K. o The temperature of the system is 600K. 0 For each molecule, the difference between the lowest electronic energy state of the activated complex and those of the two reactants (what the book calls the “classical barrier heigh ”) is 6.68 X 10‘13 erg. The lowest electronic state of H and of the complex are doubly degenerate. That of D2 is nondegenerate. o The vibrational wavenumber of D2 is 3110 cm‘l, corresponding to a vibrational energy of 6.18 X 10‘13 erg (energy = wavenumber th). The HD2 activated complex has four vibrational modes. The vibrational wavenumbers of the relevant ones are 1762 cm‘1 (symmetric stretch, corresponding energy 3.50 X 10‘13 erg) and 694 cm“1 (degenarate bending, corresponding energy 1.38 X 10“13 erg). o I don’t want you to do long calculations that I don’t ask for, so don’t do anything that I am not asking you for! The questions start on the next page. 16 8W23,2M Kaila” 23.g M237 CHEM 132 (a) 9 points Calculate the activation energy factor exp (—Aeé/kBT) . I want a numerical answer. A a: law" _— I —l LAM? +g (351MB 8+|.3Q -4,/8):]xw M [6 411» x600 K ’K/‘ilé 1.3904010“ 2+9” (b) 7 points Calculate the electronic contribution 2:1 __ :2. Zel,HZel,D2 Again, I want a numerical answer. 17 3 M? CHEM 132 (c) 9 points Calculate the vibrational contribution 1/ Zvib _ Zvib,D2 And again, I want a numerical answer. 15.3 may»le WW 8W; ’éiL/z LL23 Xm’ L’s‘AxOJOO emu/«A : L39 = H7» f/ / Ate—r lag " 0'600 r) i 2‘ 2W , 4000,12) ’L” be" \-e 18 ...
View Full Document

Page1 / 18

Final '07 - CHEM 132 FINAL EXAM K. Lindenberg Winter 2007...

This preview shows document pages 1 - 18. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online