This preview shows pages 1–18. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full 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:/er9X3J'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 halflife 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 halflife 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”, (iiHM) 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
Illlull!
Illlllllll
lulllull
IIIIIZIIII
IIIQZIIIII
IIHIIIIIII
lullIll.
nullIll.
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]’\/[llmrl> 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 steadystate 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—determiningstep 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+ ELKCJqi'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
JagsL} 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 @mb2798K. 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 ismM =
(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 higherenergy 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
 Winter '08
 Lindenberg

Click to edit the document details