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Unformatted text preview: —'——ﬁ Chapter 14: Chemical Kinetics 2. In each case, we draw the tangent line to the plotted curve.
A[H202]_1.7 M —0.6 M _ = F9.2x10"‘ M s"
400 3—1600 5 (a) The slope of the line is A[i—12()2]
Ar Reaction rate =  = 9.2x10" M s" 0)) Read the value where the horizontal line [HgOgl = 0.50 M, intersects the curve
2 21§0 s or = 36 minutes I
I
l
l
l
l
l MA] (0474 M—0.485 M) _ 3. Ratc=— =— _1.0x10'3 Ms"
' Ar 32.45—71.55
A A — .
4 (a) Rate = —L = — 0'1498 0 1565M = 0.006”.Ir M min"1
1.00 mm — 0.00 mm
Rate: _ A[A] = _0.1433 M—0.t493 M =0m65 M M4 .
A: 2.00 min—1.00 mm
(b) The rates are not equal because, in all except Zeroorder reactions, the rate depends
on the concentration of reactant. And, of course, as the reaction proceeds, reactant is
consumed and its concentration decreases, the rate of the reaction decreases.
; (a) [A] = [AL +A[A] = 0.533 M —0.013 M 2 0.575 M
(b) A[A] = 0.565 M —0.533 M = 0.023 M
A; _—_ A[A]_§_ = ﬂ: 1_[) mm A[A] ~2.2><10'2 Mjrnin
time = t + A: = (4.40+1.0)n11n :54 min 6. Initial concentrations are [ch1.]:0.105 M and [Czof']=0.300 M. The initial rate of the reaction is 7.1x10‘5 M min”. Recall that the reaction is: 2 HgC12(aq) +C203‘(aq) 4, 2 Cl'(aq) + 2 C02(g) + Hg2C12(aq).
The rate of reaction equals the rate of disappearance oszoi' . Then, after 1 hour,
assuming that the rate is the same as the initial rate, (a) 3"  '
[chn]: 0.105 M— 7.1x10’5 waxl hxéomm = 0.096 M
L 5 Into! CEO‘t 1h
(13) [clef]: 0.300 M—[Tlx 10'5 mi x1 h >< 60mm] =0.296 M
Lmin 1b 450 Chapter 14: Chemical Kinetics —A A A c  7. (a) Rate: [ ]=—[—=l.76><10‘5 Ms"    A: 2A: A C ;
% = 2><l.'l6><10*5 M s" = 3.52x10'5 Mfs _;
Aw A[C] . . . . (b) ——~ = —— = —l.76x 10 M 3 Assume this rate IS constant.
A: 2a: 
[A]=0.3530 M+[—1.76x10“ M s"><1.001m'uix160.S ]=0.357 M
min
A A
(c) [ ]= —1.76><10" M s"
A: _
H
At: A[A] =0.3500M—0.53580M=4‘5X102 s I
—l.76x10‘5 W5 —1.'}'6x10 Ws 5.?XIO" mall[202 x 1 mo] 02
l L soln s 2 mol H202 An 0
s. (a) it 1] = 1.00 L solnx = 2.9x104 mol Ozfs o (h) M 2]=2.9x10“‘ “10102 x 60.5 =1.'l><10'1 molOzfmin .. g
A: 3 1mm § 2 . (c) av[oz] z 13x10: 1110102 x 22,414 mL 02 at 3'1? = 3.8x10 02 at STP _
A: min 1 mo] 02 mm . ‘ ill“ a Notice that, for every 1000 mmHg drop in the pressure of A(g), there will be a corresponding
2000 mmHg rise in the pressure of B(g) plus a 1000 mmHg rise in the pressure of C(g). (a) We set up the calculation with three lines of information be10w the balanced equation: (I)
the initial conditions, (2) the changes that occur, which are related to each other by reaction
stoichiometry, and (3) the ﬁnal conditions, which simply are initial conditions + changes. Ms) A 23(g) + C(g) Initial 1000. mmHg 0. mmHg 0. 1111an Changes «4000. mmHg +2000. Iang +1000. mmIIg _ Final 0. l'l'll'an 2000. MmIIg 1000. mmHg Total final pressure = 0. mmIIg + 2000. mmHg + 1000. mmHg 2 3000. nuan
0:) Me) —> 213(3) + an Initial 1000. mian 0. mmHg 0. mmHg Changes —200. mmHg +400. mmHg +200. mint1g Final 800 mmIIg 400. mmHg 200. mmHg Total pressure=800. mmHg+400. mmIIg +200. mmIIg: 1400. ang 451 ﬂﬁ Chapter 14: Chemical Kinetics 10. (a) We will use the ideal gas law to determine NZO5 pressure
O
1.00 gxl_m‘3$_t x0.08’206 L 3”“ x(273 + 65) K
nRT 108.0g mol K 760 mmHg
PENZOS} = —— = x
V 15 L 1 atm
(l1) After 2.38 min, one half—life passes. The initial pressure of NEO5 decreases by half to 6.5
mmHg.
(c) From the balanced chemical equation, the reaction of 2 mol N105(g) produces 4 mol
N02(g) and l molOAg) . That is, the consumption of 2 mo] of reactant gas produces 5 mol of product gas. When measured at the same temperature and conﬁned to the same
volume. pressures will behave as amounts: the reaction of 2 mmHg of reactant produces 5 111lan of product. = 13 mmHg 5 mmHgtpmducn
pmml =13 N205 (initially) — 6.5 H‘IHIHg N105 (mt) mml‘lgtmcwnllxm]
Method of Initial Rates
u. (a) From Expt. 1 to Expt. 3, [A] is doubled, while [B] remains ﬁxed. This causes the
. 10“ "
rate to increases by a factor of M = 2.01 = 2
3.35x 10 M s Thus, the reaction is ﬁrstorder with respect to A.
From Expt. 1 to Expt. 2, [B] doubles, while [A] remains ﬁxed. This causes the rate —3 fl
to increases by a factor of = 4.03 z 4
3.35 x 10 M s I Thus. the reaction is secondorder with respect to B.
(b) Overall reaction order : order with respect to A + order with respect to B = l + 2 = 3 The reaction is third—order overall.
(c) Rate = 3.35x10" M s“ = k(0.185 M)(0.133 in)2 —4 —1
k4 3.35x10 Ms =0_102M_25_l " (0.185 mm. 133 M)2
i 
l
= (13—65 + 16) mmHg = 23 mmHg
I 12. From Expt. 1 and Expt. 2 we see that [B] remains ﬁxed while [A] triples. As a result. the initial rate increases from 4.2 ><10'3 Wmin to 1.3 ><10’2 Mfmin, that is, the initial reaction
rate triples. Therefore, the reaction is ﬁrstorder in [A]. Between Expt. 2 and Expt. 3, we
see that [A] doubles, which would double the rate, and [B] doubles. As a consequence, the
initial rate goes from 1.3)(10'2 Mfmin to 5.2 X 10‘1 Wmin, that is, the rate quadruples.
Since an additional doubling of the rate is due to the change in [B], the reaction is ﬁrst—
order in [B]. Now we determine the value of the rate constant. Rate 52X 107: M; mi“
R :k A 1 B I ' = =—‘H—'——‘
ate [ ][ ] [A13] 3.00Mx3.00M The rate law is Rate = (5.8 x 103 L mol—1min" )[A]'[s]i = 5.8 ><10'3 L madman" 452 Chapter 14: Chemical Kinetics 13. From Experiment 1 to 2, [NO] remains constant while [C12] is doubled. At the same time the
initial rate of reaction is found to double. Thus, the reaction is ﬁrstorder with respect to
[C12]. From experiment 1 to 3, [C12] remains constant, while [N0] is doubled, resulting in a
quadrupling of the initial rate of reaction. Thus, the reaction must be second—order in [NO].
Overall the reaction is thirdorder: Rate = k [NO]2[C12]. The rate constant may be calculated
from any one of the experiments. Using data from exp. 1, Rate 2.27x10‘5 M s" k = w=w
[NO]2[C12] (0.0125 M)2(0.0255M) = 5.70 Mis"l 14. (a) From Expt. 1 to Expt. 2, [B] remains constant at 1.40 M and [C] remains constant at
1.00 M, but [A] is halved(x0.50). At the same time the rate is halved(x0.50) . Thus, the reaction is ﬁrstorder with respect to A, since 0.50r = 0.50 when x = 1.
From Expt. 2 to Expt_ 3, [A] remains constant at 0.70 M and [C] remains constant at 1.00 M, but [B] is halved(x0.50], from 1.40 M to 0.70 M. At the same time, the rate
is quartered (x025) . Thus, the reaction is secondorder with respect to B, since
0.50? = 0.25 when )2 = 2. From Expt. 1 to Expt. 4, [A] remains constant at 1.40 M and [B] remains constant at
1.40 M, but [C] is halved (x050), from 1.00 M to 0.50 M. At the same time, the rate is increased by a factor of 2.0. 1
[mtg =16 rate3 =16><Zrate2 = 4 rate2 =4x31Eratel = 2Xrate1] Thus, the order of the reaction with respect to C is —l , since 0.52 = 2.0 when z = *1. l 2 WI
('3) rates = k(0.70 M)1 (0.70 M)2 (0.50 M)" = [{1'4: M] (M: M] [1‘02 M] 1121 2
=k§ (1.40 M)‘ g (1.40 Mn" (1.00 M)‘l = meg] = ratel[%] = ﬁratel This is based on rate, = k (1.40 M)‘ (1.40 M)2 (1.00 M)‘l FirstOrder Reactions g (a) TRUE The rate of the reaction does decrease as more and more of B and C are formed.
but not because more and more of B and C are formed. Rather the rate decreases because the concentration of A must decrease to form more and more of B and C. (b) FALSE The time required for one half of substance A to reactmthe halflife—is
independent of the quantity of “A” present. 453 a Chapter 14: Chemical Kinetics 16. (a) FALSE For firstorder reactions, a plot of In [A] or log [A] vs. time yields a straight
line. A graph of [A] vs. time yields a curved line.
(b) TRUE The rate of formation of “C” is related to the rate of disappearance of “A” by
the stoichiometry of the reaction. LL. (3) Since the halflife is 180 s, after 900 3 five halflives have elapsed, and the original
quantity of A has been cut in half five times. final quantity of A = (0.5)5 x initial quantity of A = 0.03125x initial quantity of A
About 3.13% of the original quantity of A remains unreacted after 900 s. or
More generally, we would calculate the value of the rate constant, k. using
k = .192 _ 0593 = 000335 5—1 Now ln(% unreacted) : 10 = 0.00385 s‘*x(900s) = —3.46§ rm _ 130 s
(% unreacted) = 0.0313 % (b) Rate = k[A] = 0.00335 5" x050 M = 0.00193 Mrs 18. (a) We note that the ﬁnal concentration is oneeighth of the initial concentration. Thus,
three halflives have elapsed, the first to reduce [A] from 0.800 M to 0.400 M. the
second to reduce [A] from 0.400 M to 0.200 M, and the third half—life to reduce [A]
from 0.200 M to 0.100 M. From this we see 3mm = 54 min, therefore, tug = 18 min.
To reduce the concentration even further requires two additional halflives: the ﬁrst
of these to lower [A] from 0.100 M to 0.050 M and the second of them to reduce [A]
from 0.050 M to 0.025 M. These two halflives equal 2x18 min = 36 rnin . Thus, at 54 min +36 min = 90 min from the start of the reaction, [A] = 0.025 M. (b) We determine the rate constant for this ﬁrstorder reaction:
k _ 0.693 _ 0.693 in, ' 18 rnin
Rate = k[A]l = 0.039 min“ x 0.025 M = 9.3 x10"1 Mmun = 0.039 rnin'1 Then we determine the rate: 19. (a) The mass of A has decreased to one fourth of its original value, from 1.60 g to . l 1 .
0.40 g. Since 1 = ~2— x i , we see that two half—lives have elapsed. 2
Thus, 2mm = 38 min, or rm :19 min.
. A
(b) k : 0.6931“:m = 0 69,3 = 0.036 min"l ln[—]’— s —kt = —0.036 rnin'l x60 min = #22
19 min [AL A
iA—{f e‘” =01; or [A1=[AL e‘“ =l.60gAXO.ll:0.l§gA A
20. (a) in i 1 = —kr : 1n 0'6” M = 4.256 k = —_0—'25§—= 0.0160 min"
[A10 0.816 M 16.0 min
{b} 1m 2 ﬂ — mm = 43.3 min I: _ 0.0160 min—l 454 Chapter 14: Chemical Kinetics (c) We need to solve the integrated rate equation to ﬁnd the elapsed time. A
mu=uhﬂn023SM =—1.245=u0,0160minIxt t_ —l.24S
[AL 0.816 M _W = (d) In [A] =—kt becomes [A] =e'“ which in turn becomes [Aio [Ale [A]: [n]D e‘" = 0.816 M e:t[:u[—0.0160min_I x25 bx 60’1"“ 1h = 0.816x0.0907 = 0.074 M 21. We determine the value of the ﬁrstorder rate constant and from that we can calculate the
halflife. If the reactant is 99% decomposed in 137 min, then only 1% (0.010) of the initial
concentration remains. A
[ ]‘ = —kr = In 0010 = 4.61 = —k x137min k = 46]: = 0.0336 min—1
[A]o 1.000 137 mm 0.0693 _ 0.693 In t = 20.6 min 22. If 99% of the radioactivity of 32F is lost, 1% (0.010) of that radioactivity remains. First we compute the value of the rate constant from the halflife. k = 0:593 = = 0.0435 11"
[1‘2 '
Then we use the integrated rate equation to determine the elapsed time.
A A
ln[ 1‘ = —kr t = —lln[ 1‘ = —m———L———_!—ln (1010 = 95 days
[A]o 1: [AL 0.0435 d 1.000
35
W g; (a) In [A] = ln(0.35) = —kr = (—4.31 x 10“3 min“): t: 218 min.
0 Note: We did not need to know the initial concentration of acetoacetic acid to
answer the question. (b) Let’s assume that the reaction takes placed in a 1.00 L container. 1 mo! acetoacetic acid 10.0 g acetoacetic acid X = 0.09795 mol acetoacetic acid. 102.090 g aoetoacetic acid After 575 min. (— 4 half lives, hence, we expect ~ 6.25% remains), use integrated
form of the rate law to ﬁnd [AL = 575 min. ln[ :2} J = —kr = (4.81 x 10‘3 mm“‘)(575 min) = —2.76§ [A]! : e'm‘j = 0.06293 (~ 6.3% remains) —[ﬂ~——
[A]. 0.09795M 455 ' = 0.063 [A]l : 6.2 XIO'3 moles. Lilith" an...“ In Chapter 14: Chemical Kinetics (a) (b) (a) (b) (c) [Mmcled = [A]o — [A][ = (0.098 — 6.2 x 10'3) moles = 0.092 moles acetoacetic acid. The
stoichiometry is such that for every mole of acetoacetic acid consumed. one mole of
C03 forms. Hence, we need to determine the volume of 0.0913 moles CO; at 24.5 °C (297.6; K) and 748 ton (0.984 atm).
Latm RT 0091311101 0.08206 K 1 297.65 K
v = " _ "1° = 2.3 L C02
P 0.934 atrn
A
lngz— [:11] g =—3.47=“6.2X10_43II, IZﬂrj= S
[A]0 80.0 g 6.2 x10 5 We substituted masses for concentrations, because the same substance (with the same
molar mass) is present initially at time t, and because it is a closed system. 1 mol N205 1 mol o2 0 t0 =77.5 N o x————x———=0.359 r0010
am "n 2 g 2 5 103.0 g N205 2 mol N205 2
RT 0.359 mo] 02 x0_08206 L x(45+273) K
v: " : n“: m =956LO2
P 745 mmHgX———
760 1111an If the reaction is ﬁrstorder, we will obtain the same value of the rate constant from
several sets of data. A
ln[ 1 = —t: = mm: —k><100 s =—0.1ss, k = 0‘188 =1.33x10—3 5"
[AL 0.600 M 100 s
A
lug: —k1 = In 0344 M =  x300 s=—0.556, k = 0556 :1.85><10'3 5'1
[AL 0.600 M 300 s
A
ln[ 1 = —k1 = In 0285 M = —kx400 s = 41744, I: = 0344 =1.36><10‘3 s“1
[AL 0.600M 400 s
A
ln£—= —k: = ln 0'198 M = w1005.00 s = —1.109, I: = L109 =1.85x10‘3 5"
[AL 0.600 M 600 s The virtual constancy of the rate constant throughout the time of the reaction conﬁrms that the reaction is ﬁrstorder.
For this part, we assume that the rate constant equals the average of the values obtained in part (a).
k =1.88+l.85:l.8.6+1.85><10_3 54:1.86x104 3] We use the integrated firstorder rate equation:
[A1750 = [ ALepoz): 0.600 M exp(—1.86x10'3 5“ X750 3) .[ ALSO = 0.600 M a” = 0.143 M 456 \ Chapter 14: Chemical Kinetics (CH3)EO(g) > CHAg) + H2(g) + C0(g) Initial 312 mmIIg 0 l'l'll'an 0 mmIIg 0 mnng
Changes —109 mmHg +109 rang +109 mmHg +l09 mmHg
Final 203 mmHg 109 mmHg 109 1711an 109 mmHg
Plow] : PDME + Pmlham +Rlydmgen +Pco = 203 mmHg+109 mmHgtIOQ I‘ang +109 mlan = 530. mmHg Reactions of Various Orders E (a) Set 11 is data from a zeroorder reaction. We know this because the rate of set II is
constant. 0.25 W25 5 = 0.010 M s‘l . Zero—order reactions have constant rates of reaction. (1)) A firstworder reaction has a constant halflife. In set I, the ﬁrst halflife is slightly less than
75 sec, since the concentration decreases by slightly more than half (from 1.00 M to 0.47
M) in 75 5. Again, from 75 s to 150 s the concentration decreases from 0.47 M to 0.22 M,
again by slightly more than half, in a time of 75 3. Finally, two halfilives should see the
concentration decrease to onefourth of its initial value. This, in fact, is what we see.
From 100 s to 250 sec, 150 s of elapsed time, the concentration decreases from 0.37 M to
0.08 M. i.e.. to slightly less than onefourth of its initial value Notice that we cannot
make the same statement of constancy of halflife for set III. The ﬁrst halflife is 100 s.
but it takes more than 150 s (from 100 s to 250 s) for [A] to again decrease by half. (e) For a secondorder reaction, 1! [A]l — I! [A]0 = kt . For the initial 100 s in set 111, we have l — l =1.0Lmol“=k100s, lc=0.010Lmol"s'1
0.50 M 1.00 M
For the initial 200 s, we have I  1 = 2.0 L moi" = k 200 s, k = 0.010 L mol" 3'1
0.33 M 1.00 M Since we obtain the same value of the rate constant using the equation for second
order kinetics. set [11 must be second—order. 28. For a zeroorder reaction (set II), the slope equals the rate constant:
k = —a[A]lot=1.00 W100 s =0.0100 W3 29. Set I is the data for a firstorder reaction; we can analyze those items of data to determine the
halflife. In the first 75 s, the concentration decreases by a bit more than half. This implies a half
life slightly less than 75 s, perhaps 70 s. This is consistent with the other time periods noted in the
answer to Review Question 18 (b) and also to the fact that in the 150 s interval from 50 s to 200 s,
the concentration decreases from 0.61 M to 0.14 M, which is a bit more than a factoroffour
decrease. The factor—offour decrease. to onefourth of the initial value, is what we would expect
for two successive halflives. We can determine the halflife more accurately, by obtaining a value of it from the relation ln([A1f[AL): —kt followed by rm = 0.693; k For instance,
ln(0.78l1.00)= —k (25 s); k = 9.93 x103 s'l. Thus. to; = 0.693l9.9gl_ x103 5" = 70 s. 458 Chapter 14: Chemical Kinetics 30. We can determine an approximate initial rate by using data from the first 25 s. A A _
Rate = —£I—] = —w = 0.0080 M s“ 25 5—05 3. The approximate rate at 75 s can be taken as the rate over the time period from 50 s to 100 s.
A[A] _ _ 0.00 M — 0.50 M Rt =— =0.010M ‘1
(a) 36" At 100 5—503 5
AA . ﬂ . _
(b) Ratel =__[_]=_w=0_0043 M 5‘
AI IOOsSOs
A  .
(.3) Rate!“ = 9U = _w = 00034 M s"
A: lOOs—50s Alternatively we can use {A} at 75 s (the values given in the table) in the relationship Rate = k[A]"' ,
where m=0 , 1, or 2. (a) Raten = 0.010 M 3" x(0.25 mollL)D = 0.010 M s" (b) Since rm = 705, l: 2 0.693 I 705 '= 0.00993"
RateI = 0.0099 s" x (0.47 rnolt'L)I = 0.0047 Ms" (c) Ratem = 0.010 L mol'1 3" ><(0.57 rims/L)2 = 00032 M 5‘1 32. We can combine the approximate rates from Question 31, with the fact that 10 s have
elapsed, and the concentration at 100 s. (3) [AL 2 0.00 M There is no reactant left after 100 s.
(b) [a]l = [A]m —(10 sx rate) = 0.37 M —(10 sx0.0047 M s“) = 0.32 M (c) [a]In = [ALm ~(10 sx rate): 0.50 M —(10 sx0.0032 M s") = 0.47 M 33. Substitute the given values into the rate equation to obtain the rate of reaction.
Rate = k[A]2[B]" = (0.0103 M'lmin")(0.116 M)” (3.33 M)D = 1.39 x 10“ Mi min 34. (a) A ﬁrstorder reaction has a constant halflife. Thus, half of the initial concentration
remains after 30.0 minutes. and at the end of another halflife—60.0 minutes total—
half of the concentration present at 30.0 minutes will have reacted: the concentration
has decreased to onequarter of its initial value. Or, we could say that the reaction is
75% complete after two halflives—60.0 minutes. {h\ A 79rn_nrdar Martiniu nmnaar‘n nt 4 nnnnan... —nl— 411...... :c a... _.....: .. ,. 1. :nm _ _ __,1 __ Chapter l4: Chemical Kinetics (d) We use the same equation as in part (b), but solve for: , rather than k. 1 1 1 1 1 1 .
t: _ ﬁ_ = M— = 219 min
I: [111 [A]o 0.137 Lind—{mind 00250 M 0.100 M (e) We use the same equation as in part (b), but solve for: , rather than k. 1 1 1 1 1 1 .
I:__ ___M =_M_l ‘ _! _—#——— =136m1n
k [A [A 013?me min 0.0350M 0.100M 37. (a) In the first 22 s, [A] decreases from 0.715 M to 0.605 M. that is, A[A] = —0.1 10 M.
The average rate over these 22 s is then determined as follows. —A A .
Rate: [ 1=M=5.0X10'3Ms
Ar 22 s
In the ﬁrst 74 s, A[A] = 0.345 M ~0.'115 M, and the rate is determined.
#A A .
Rate: [ 1: W=50x104MIs
T4 s
Finally, in the ﬁrst 132 s, A[A] = 0.055 M —0.715M. and the rate is determined as
—A A
follows. Rate = if ] = M = 5.0x10'3Mt's. Since the rate is constant for this
3 reaction, it must be zeroorder in [A]. (b) The halflife of this reaction is the time needed for one half of the initial [A] to react. Thus, MA]: 0715 M+2=0.358M and 1m =—°'—35§—M——=72 s. 5.0)<10_3 MIS 38. (a) We can either graph ll[C “1116] vs. time and obtain a straight line, or we can determine the secondorder rate constant from several data points. Then, if k indeed
is a constant, the reaction is demonstrated to be secondorder. We shall use the
second technique in this case. First we do a bit of algebra. 1 1 _, les 1 =
FETAL—k I[[A], [AL] k
h 1 1
'12.13 min [0.0144 M _ 0.0169 M = 0.843 L mol'lmjn" k 1
1
k=—~—_— “Law—i— =0.a'151.1111;114:1111:1
2455 mm 0.0124 M 0.0169 M l k = ———] = 0.892 L mol"min"'
42.50 rmn 0.0103 M 0.0169 M l
k = = 0.870 1.. moi—Imm—l
68.05 nun 0.00845 M 0.0169 M The fact that each calculation generates similar values for the rate constant indicates
that the reaction is secondorder. Chapter 14: Chemical Kinetics _. The halflife of the reaction depends on the concentration of “A” and, thus, this reaction
cannot be ﬁrst—order. For a second—order reaction, the half‘life varies inversely with the reaction rate: rm = lt[k[ AL) ork = If (rl,2[ AL). Let us attempt to verify the second
order nature of this reaction by seeing if the rate constant is ﬁxed. k = ———1——— = 0.020 L mol'lmin"
1.00 MXSO min I: = = 0...
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 Spring '08
 Davis
 Chemistry, Reaction, Chemical reaction, Rate equation

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