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Unformatted text preview: 60 E 57. C. CHAPTER 4 SOLUTION STOICHIOMETRY The oxidation state of copper increases by 2 (0 to +2) and the oxidation state of silver decreases
by 1 (+1 to 0). We need 2 Agatoms for every Cu~atom. The balanced equation is: Cu(s) + 2Ag+(aq) > Cu2+(aq) + 2Ag[s) The equation is balanced. Each hydrogen atom gains one electron (+1 * 0) and each zinc atom
loses two electrons (0 > +2). We need 2 Hatoms for every Znatom. This is the ratio in the given equation:
Zn(s) + HZSO4(aq) 4* ZnSOr(aq) + His) Review section 4.1 l of the text for rules on balancing by the halfreaction method. The ﬁrst step
is to separate the reaction into two halfreactions then balance each halfreaction separately. No; —» NO + 2 H20
(3e+4H++No;—>No+2H20)x2 (Cu > Cu2+ + 2 e‘) >< 3 Adding the two balanced halfreactions so electrons cancel: 3Cu>3Cu""+6e‘
6e'+8H*+2N03‘—*2N0+4H20 3 Cu(s) + 3 H”’(aq) + 2 NO3'(aq) 4' 3 Cu1+(aq) + 2 NO(g) + 4 H200) (3503 + 2 Cr“ + 7 H20
6 e‘ +14 H+ + C1307” —+ 2 Cr3+ + 7 H20 (2C1—>c12+2e)x3 Add the two balanced halfreactions with six electrons transferred: 6Cl'>3C12+6e'
6 e'+]4H++Cr2012'e+ 2cﬁ*+ ero 14 irraq) + Cr1012‘(aq)+ 6 Cl'(aq) —> 3 C12(g) + 2 C13+(aq) + 7 H200) Pb —> PbSO‘ rho2 a Pbso,t
Pb + st04 a Pbso.1 + 2 11* rho2 + st0.1 A PbSO4 + 2 H20
Pb + sto. a Pbso4 + 2 E + 2 e' 2 e" + 2 Ir + Pbo2 + H2804 a Pesoe + 2 H20 Add the two halfreactions with two electrons transferred: 2 e' + 2 H’" + rho2 + also, > lasso,1 + 2 H20
Pb + stor » PbSO4 + 2 11* + 2 e' Pb(s) + 2 HZSOdaq) + Pb03[s) + 2 PbSO4(s) + 2 11200) This is the reaction that occurs in an automobile lead storage battery. CHAPTER 4 SOLUTION STOICHIOMETRY 61 d. Mn“ 4 MnO4'
(4H20+Mn2+—+Mn04'+8H*+5 e‘) x 2
NaBiO; —> Bi“ + Na+
5 H‘“ +NaBi03 —> Bi“ + Na* + 3 H20
(26'+6H*+NaBi03 —> Bi3*+Na*+3 H10) x 5 8H30+2Mn2*—>2Mno4'+16H*+10e'
10e+30H++5Na3103+513i3++5Na++15H20 8H20+301I“+2Mn¢‘+5NaBiO3—>2M1104'45Bi3++5Na++15H20+16H+
Simplifying:
14 H*(aq) + 2 Mn“(aq) + 5 NaBi03(s) + 2 Mn04'(aq) + 5 Bi3+(aq) + S Na+(aq) + 7 H200) e. HaAsO4 > AsH3 (Zn —> Zn” + 2 e‘) x 4
H3A504 —> AsH, + 4 H20
8 e'+ 8H++H3As04+ AsH3+4 H20 8 6+ 8 H+ +H3A504 4» A5143 +4 H20
4Zn>4Zn2++8c' 8 HTaq) + H3A504(aq) + 4 Zn(s) * 4 anTaq) + AsII3(g) + 4 H200)
f. A3203 + H3A504
A5203 *> 2 H3A504
(S HZO+A5203+ 2H3ASQ4+4H*+4e') X 3
N03; > NO + 2 H10
41F+N0;>N0+2Hzo
(3 e'+4H++N03‘—>N0+2H10) X 4 12e‘+16H‘+4N03‘>4NO+8H10
15 H20+3Aszop 6H3A504+ 12H"+ 12 e' 7 1420(1) + 4 H+(aq) + 3 1451045) + 4 NO3‘(aq) 4 4 N0(g) + 6 H3A504(aq) (5 6+ 8H++Mn04‘>Mn2+‘+4HZO) x 2 10 Br‘—PSBr2+103‘ g. (2 BrA Br; +2 6') X 5  Mao; ~> Mn2+ + 4 H20
10 6+ 16H++2MnO4'—>2Mn2++8H20
I 16 H+(aq) + 2 Mn04‘(aq) + 10 Br'(aq) —> 5 Br2(1) + 2 Mn1"(aq) + 8 H200) 62 CHAPTER 4 SOLUTION STOICHIONLETRY h. CH3OH a CHQO C130,” a Cr“
(CH30H ~> CH20 + 2 11* + 2 3) x 3 14 H* + Crzoﬁ' er 2 Cr“ + 7 H20
6e‘+14H++ Cr2072'> 2013+ +7 H10 3 CH3OII+ 3 CH20+6 H++6 e‘
6 e' + 14 H” + Orzo,2“ a 2 013* + 7 H20 M
8 H"(aq) + 3 CH30H(aq) + Cr20,1‘(aq)—* 2 C15+(aq) + 3 CH20(aq) + 7 H20(l)
58. Use the same method as with acidic solutions. After the ﬁnal balanced equation,_then convert H’ to OH' as described in section 4.11 of the text. The extra step involves converting IF into H10 by
adding equal moles ofOH' to each side of the reaction. This converts the reaction to a basic solution while keeping it balanced.
a. A] or ARCH); . M110; 4 MnO,
41120+Al—>A1(OIrI),{+4H+ 3 e'+4II++Mn04'+Mn01+2H20 4 H20 + A1 a A1(0H),; + 4 1r + 3 e“ 4H20+Al >Al(OH)4'+4II++3 e'
3 e'+4H"+MnO4‘>Mn02+2H20
____,___—__
2 H100) + A1(s) + MnOﬂaq) > Al(OH);(aq) + Mn02(s)
Since II+ doesn’t appear in the ﬁnal balanced reaction, we are done.
b. Cl2 —> C1' C12 —> 010'
2e‘+C12>201' 2H20+C12—>2010+4H++2e“ 2e‘+C12+2C1‘
2H20+c1,+2c10+4H*+2e 21120+2C12+2C1'+2C10‘+4H+ Now convert to a basic solution. Add 4 OH‘ to both sides of the equation. The 4 011' will react
with the 4 H" on the product side to give 4 H20. After this step, cancel identical species on both
sides (2 H20). Applying these steps gives: 4 0H“ + 2 C12 —> 2 01' + 2 C10“ + 2 H20, which can be further simpliﬁed to:  2 OH‘(aq) + Clz(g) r Cl‘(aq) + ClO'(aq) + H200) N02'>NH3 Alr A10;
6e'+7H*+N02'*NH;+2H20 (2 H10+Al+A101'+4H*+3 e')><2 Common factor is a transfer of 6 e'. 6e‘+7H"+N02'—>NH3+2H20
4H20+2A1+2Al02'+81I*+6e‘ W
OH‘+ 21120 +N01‘+ 2 AlPNH3+2A102‘+H++OH'
Reducing gives: OH‘(aq) + H200) + N02'(aq) + 2 Al(s) » NH3(g) + 2 A102‘(aq) 61. 62. 63. 64. CHAPTER 4 SOLUTION STOICHIOMETRY HNO; *r N02
HNO3 a No1 + H20
(c'+H++HNC)3—>NOZ+HQO) x 2 Mn—PMn2*+2e' N1114Mn2++2€
2e‘+2H*+2HNO3*2N02+2H10 _.
2 H*(aq) + Mn(s) + 2 HNO3(aq) —> Mn2+(aq) + 2 N02(g) + 2 H200) or 4 mag) + Mn(s) + 2 NO3’(aq) + Mn2+(aq) + 2 1002(3) + 2 H200) (HINO3 is a stlfong acid.) (4H30+Mn2++Mn04'+8H*+5 (3') x2 [2e‘+2H"+104'—r103‘+H20) x5 8H20+2Mn2++2Mn04'+16H*+10e'
106+10W+510,{>5103'+5H20 M
3 H200) + 2 Mnl+(aq) + 5 104'CaCi) " 2 Mn04‘(a€D + 5 103139) + 5 H+(aq) HCl and HNO3 are strong acids which completely dissociate to Il+ and the anions in solution. Au+4Cl'—>AuCl;+3e‘ 3e'+4H*+N03'—*NO+2H20 Adding the two balanced half—reactions:
Au(s) + 4 Cl‘(aq) + 4 IY(aq) + N03‘(aq). > AuClﬂaq) + N0(g) + 2 H300) (S‘e'+8H"+MnO4‘>Mnﬂ++4H20)>< 2 5H2C204+10002+10H*+10e'
10e‘+16H++2MnO4'—>2Mn2*+8H20 mzczo4+2coz+2w+2ejxs M
6 H*(aq) + 5 H2C204(aq) + 2 MnOJaq) —> 10 00103) + 2 Mn" aq) + 8 H200) 1 mol H3C204 2 mol MnOd: >< ——— = 4.700 >< 10“ mo] Mn04'
90034 g 5 mol chzo4 0.1053 g ch104 x 4.700x10“1 ano ‘
___m#:n0_2__i x M :15” x 102 Mmod Molarity =
28.97 mL L 3. Fe?" —+ Fe?“ + c’ 5 e‘ + s H“ + Mno; a Mn2+ + 4 H20 The balanced equation is:
3 WW!) + Mn04'(aCI) + 5 FezTaCD —" 5 F83+(3ID + MBZTﬁQ) + 4 H200) 0.0216 mol M1104. 5 11101129,2+
________._.._. x .__._._.__
mol Mn04_ 20.62 x 10'3 L 50111 X = 2.23 X 10“3 mol Fe2+ L soln ,3 3+
M = 4.46 x 102 MFe” Molarity =
50.00 x 103 L 488 CHAPTER 15 CHEMICAL KINETICS b. For the ﬁrst experiment: 12.5 x 1074 mol = k[ 0.080 mol] ( 0.040 1661] ’ k = 3.9 x 10., L “101.15.. L s L L
The other values are: Initial Rate k
(mol L“ s“) (L 11101“ 5")
12.5 >< 10“3 3.9 x 103
6.25 X 10"5 3.9 X 10'3
6.25 X 10'6 3.9 X 10‘3
5.00 X 10'6 ' 3.9 X 10'3
7.00 X 10'5 3.9 X 10'3 km,“ = 3.9 X 10'3 L meal'1 5'1 17. a. Rate = k[NOCl]“; Using experiments two and three: 2.66 x 104 z k(2.0 x 1016)" , 4.01 = 2.0“, n = 2; Rate = k[NOCl]2
6.64 >4103 141.0 ><10“5)r1 3 3 5.98 x 104 molecules = k 3.0 x 1016 molecules
CHI 5 2
, k = 6.6 X 10'” reni3i'niolecules‘ls“l
cm The other three experiments give (6.7, 6.6 and 6.6) x 10'29 cm3 molecules'15‘1, respectively. The mean value for k is 6.6 X 10'29 cm3 molecules'1s". 6.6 ><10'29 cm3 x 1L X 6.022 x 1013molecules 4.0 x 1031. molecules 5 1000 cm 3 mol m0] 5 18. a. Rate = k[Hb]"[C0]"
Comparing the ﬁrst two experiments, [CO] is unchanged, [Hb] doubles, and the rate doubles.
Therefore, x = I and the reaction is ﬁrst order in Hb. Comparing the second and third
experiments, [Hb] is unchanged, [(30) triples and the rate triples. Therefore, y = 1 and the
reaction is ﬁrst order in CO.
b. Rate = k[Hb] [CO]
c. From the ﬁrst experiment: 0.619 ,umol L‘1 s‘1 = k (2.21 jm:noir'L)(1.00 nmoUL), k = 0.280 L tnnol‘l s'1 The second and third experiments give similar k values, so km“ = 0.280 L mnol‘1 s‘l. CHAPTER 15 CHEMICAL KINETICS 489 _________________..__..—————————————~— 1 d. Rate = k[Hb][CO] = 0280 L x M X M = 2.26 mot L1 5'1 ,umol 5 L L 19. a. Rate = k[ClOz]“[OII']Y; From the ﬁrst two experiments:
2.30 x 10" = k(0.100)"(0.100)’ and 5.75 x 10'2 = k(0.0500)"(0.100)” Dividing the two rate laws: 4.00 = M = 2.00“, x = 2
(0.0500)* Comparing the second and third experiments: 2.30 x 101 = k(0.100)(0.100)" and 1.15 x 101 = k(0.100)(o.0500)v (0.100)y
(0.050)y Dividing: 2.00 = = 2.0’, y = l The rate law is: Rate = k[C103]2[OH'] 2.30 X 104111011." 5'1 = k(0.100 molfL)2(0.100 moUL), k = 2.30 X 102 L2 n:Itol‘2 s" = kmall 2 2 2
b, Rate: =o,594m01L.1 mo! 2 s L L 20. Rate = k[NO]“[02]Y; Comparing the ﬁrst two experiments, [01] is unchanged, [N0] is tripled,
and the rate increases by a factor of nine. Therefore, the reaction is second order in NO (32 = 9).
The order of 02 is more difﬁcult to determine. Comparing the sécorid and third experiments; I? 18 2 18
M = W, 1,74 = 0.694 (2.50)); 2,51 = 2502, y =1
1.30 x 10” k (3.00 x 1018mm x 10””)3r Rate = k[NO]2[021; From experiment 1:
2.00 x 101‘5 molecules cm“3 s“ = k (1.00 >< 1013 moleculest'cm’)2 (1.00 x 10“ moleculesfcm’) k = 2.00 X 10'“ cm‘ moleeules‘2 5‘1 = kImam 2.00 x 10'33 ern‘5 x 6.21 x 1013 molecules 2 X 7.36 x 1018 molecules Rate =
3 molecules.2 s cm3 em
= 5.68 x 10” molecules cm'3 s“ 21. Rate = k[N1051"; The rate laws for the ﬁrst two experiments are: 2.26 X 10'3 = k(0.190)" and 3.90 X 10“ = M00750)x 490 CHAPTER 15 CHEMICAL KINETICS X
Dividing: 2.54 = ﬂ = (2.53)*, x = 1; Rate = k[N,O,] (00750)"
—4 —1 —l .
= Rate = X mol L S = x 5.1; km”: X104 Sl
[N205] 0.0750 moUL
Integrated Rate Laws 22. 3. Since the 1! [A] vs time plot was linear, then the reaction is second order in A. The slope of
the If [A] vs. time plot equals the rate constant it. Therefore, the rate law, the integrated rate law and the rate constant value are: Rate = k[A]2; L = kt + _L; k = 3.60 x 102 L mol“ 5"
[A] {A}, I b. The halflife expression for a second order reaction is: tm = k [A]
O For this reaction: tIn = 1 = 9.92 x 103 s
3.60 ><10'2 L 11101 '1 s ‘1 x 2.80 ><10’3 mollL Note: We could have used the integrated rate law to solve for t1,2 where
[A] = (2.80 x 10'3 f2) moUL. c. Since the half—life for a second order reaction depends on concentration, then we must use
the integrated rate law to solve. 1 1 1 =3.60x10'2LXH 1
7.00 x 104114 mol 5 2.30 x 10‘3 M 1.43 x103  357 = 3.60 x102 t, t= 2.93 x 10‘s 23'. 3. Since the ln[A] vs time plot was linear, then the reaction is ﬁrst order in A. The slope of the
ln[A] vs time plot equals —k. Therefore, the rate law, the integrated rate law and the rate constant value are:
Rate = k[A]; ln[A] = kt + ln[A]o; k= 2.97 X 10‘2 min" b. The halflife expression for a ﬁrst order rate law is: tm ___ E 2 0.6931 , tm = 0.6931 = 23.3 min
k k 2.97 x 10'2 min1 c. 2.50 X 10'3 M is US of the original amount of A present initially, so the reaction is 87.5%
complete. When a ﬁrst order reaction is 87.5% complete (or 12.5% remains), then the
reaction has gone through 3 halflives: 100% a 50.0% —> 25.0% ~> 12.5%; t=3 ><tm=3 x233 min=69.9 min to: tlr‘l tm T CHAPTER 15 CHEMICAL KINETICS 491 ____________,____._____————— 24. 25. Or we can use the integrated rate law: In = _kt, In X ': M = _ x 102 mind) t, t:
[A10 2.00x10 M 111(0125)
—2.91 x104 min’1
= 70.0 min 3.. Since the [Qmom vs time plot was linear, then the reaction is zero order in CZHSOH. The
slope of the [GilLOB} vs time plot equals k. Therefore, the rate law, the integrated rate law
and the rate constant value are: Rate = k[C2HSOlI]° = k; [CZHSOH] = ~kt + [QEOI—IL; k = 4.00 X 10'5 mol L'1 s’1 b. The halflife expression for a zero order reaction is: t”z = [AL/2k. = [CZHSOHJG : 1.25 x 102 mom.
2k 2x4.00x10'5molL‘ls" Note: we could have used the integrated rate law to solve for tm where
[CZHSOH] = (1.25 x 104:?) molfL. =156s tin c. [CEHSOII} = kt + [cansomw 0 1nth = {4.00 x 105 mol L" 51) t + 1.25 x 102 molfL 2
t: 1.25x10 molfL =313s 4.00x10'5 mol L"1 s ‘1 The ﬁrst assumption to make is that the reaction is ﬁrst order. For a ﬁrst order reaction, a graph
of In [H202] vs time will yield a straight line. If this plot is not linear, then the reaction is not
ﬁrst order and we make another assumption. Time [11101] 111 [H202] one (5) (moUL) 0 1.00 0.000 120. 0.91 41.094 "°°
300. 0.78 41.25 '3'
600. 0.59 053 i“
1200. 0.37 0.99 5 4m
1800. 0.22 i.51
2400. 0.13 2.04
3000. 0.082 2.50
3600. 0.050 3.00 4°“ Note: We carried extra signiﬁcant ﬁgures in 0 50° “00 '500 2‘00 300° 3600 some of the ln values in order to reduce round "ms", off error. For the plots, we will do this most of
the time when the In function is involved. CHAPTER 15 CHENﬂCAL KINETICS 495 ________________—___.__————————~———— 29. 30. 31. From the data, the pressure of C2H50H decreases at a constant rate of 13 torr for every 100. s.
Since the rate of disappearance of CZHSOH is not dependent on concentration, the reaction is zero order in CzHSOH. __ 13torr latm
k  x
100. s 760torr The rate law and integrated rate law are: =13 X104atmfs Rate =k=1.7 x 10“ atmfs; PC HsOH =lct+ 250. torr[ 76m
2  orr 1 “I” ] = 4s + 0.329 atm At 900. s: PCZHSOH = 1.7 ><10“i annfs X 900. s + 0.329 atm = 0.176 mm = 0.18 am =130t0rr a. Since the 1! [A] vs. time plot is linear with a positive slope, the reaction is second order with
respect to A. The yintercept in the plot will equal 1l[A]o. Extending the plot, the y—intercept
will be about 10, so 1110 = 0.1 M= [A]o. b. The slope of the 121A] vs time plot will equal k. (60  20) Limo] slope = k = = 10 Umols
(5  1) S
i=kt+ 1 = ‘0L x95+ 1 =100, [A]=0.01M
[A] [A]o mol 5 0.1 M
c. For a secondorder reaction, the halflife does depend on concentration: tU2 = Mi] .
a
First halflife: tm=———1—ﬁ =1 5
10L x 0.1 moi
mol 5 L Second halflife ([A]° is now 0.05 M): tm = 13(10 x 0.05) = 2 5
Third halflife ([A]{) is now 0.025 M): t1}.2 = 1l(10 X 0.025) = 4 s
a. We check for ﬁrst order dependence by graphing 1n [concentration] vs. time for each set of data. The rate dependence on NO is determined from the ﬁrst set of data since the ozone
concentration is relatively large compared to the NO concentration, so it is effectively constant.
time (ms) [NO] (molecules/cm3) in [NO]
0 6.0 X 103 20.21
100. 5.0 X 108 20.03
500. 2.4 X 103 19.30
700. 1.7 X 103 18.95
1000. 9.9 X 101r 18.41 495 CHAPTER 15 CHEMICAL KINETICS ___________.____,__._._—..m—————— O 250 500 750 . 1000
time(ms) Since 111 [NO] vs. t is linear, the reaction is ﬁrst order with respect to NC). We follow the same procedure for ozone using the second set of data. The data and plot are: time (ms) [03] (moleculesfcm3) In [03]
0 1.0 X lCllo 23.03
50. 8.4 x 109 22.85
100. 7.0 X 109 22.67
200. 4.9 X 109 22.31
300. 3.4 x109 21.95
23
f—
to
O
""’ 22
E
21
U 100 200 300
time (m s) The plot of In [03} vs. t is linear. Hence, the reaction is ﬁrst order with respect to ozone.
b. Rate = k[NO}[03] is the overall rate law. c. For NO experiment, Rate = 1:“ [NO] and k” = {slope from graph of 111 [N0] vs. t). [8.41 — 20.21 W =13 s"
(1000. —O)><10"3 s k’ = slope =  CHAPTER 15 CHEMICAL KINETICS 497 M 32. For ozone experiment, Rate = k" [03] and k" = {slope from in [03] vs. t). (21.95 43.03)
(300. — 0))(10'3 s H... — slope =  =16 s" d. From NO experiment, Rate = k[NO][03] = k’ [NO] where k’ = k[03].
k’ = 1.8 s'1 = k(1.0 X 10” moleculesfcms), k= 1.8 X 10'14 cm3 molecules“1 5"
We can check this from the ozone data. Rate = k” [03] = k[NO][03] where k” = k[NO].
It” = 3.6 s“ = k(2.0 X 10“ moleculesfcnE), k = 1.8 x 10"4 cm3 molecules" 5"
Both values of k agree. This problem differs in two ways from the previous problems:
1. a product is measured instead of a reactant and
2. only the volume of a gas is given and not the concentration. We can ﬁnd the initial concentration of CﬁHsNzCl from the amount of N2 evolved aﬁer inﬁnite
time when all the CﬁlrlsNﬁl has decomposed (assuming the reaction goes to completion). 3
n: E}: : 1.00 atm x (58.3 x10 13:220 X103 molNz RT 0.08206 L atrn X 323 K
moi K Since each mole of CSHSNZCI that decomposes produces one mole of N2, then the initial
concentration (t = 0) of CﬁILNZCI was: —3
2.20x10 mol =OIOSSOM
40.0x 1031. We can similarly calculate the moles of N2 evolved at each point of the experiment, subtract that
from 2.20 x 10'3 mo] to get the moles of CEHSNZCI remaining, and then calculate [CSHSNQCI] at
each time. We would then use these results to make the appropriate graph to determine the order
of the reaction. Since the rate constant is related to the slope of the straight line, we would favor
this approach to get a value for the rate constant. There is a simpler way to check for the order of the reaction that saves doing a lot of math. The
quantity (Vm  Vt) where Vm = 58.3 mL N2 evolved and VI = mL of N2 evolved at time t will be
proportional to the moles of CGHSNZCI remaining; (Vm — V) will also be proportional to the
concentration of C6H5NZCI. Thus, we can get the same information by using (V00 — V!) as our
measure of [CsHsNzClL If the reaction is ﬁrst order, a graph of in (Va, — V!) vs. t would be
linear. The data for such a graph are: 493 CELAPTER 15 CHEMICAL KINETICS {(8) V. (mL) (V00  VI) 1110!; — V!)
0 0 58.3 4.066 6 19.3 39.0 3.664 9 26.0 ' 32.3 3.475 14 36.0 22.3 3.105 22 45.0 13.3 2.588 30. 50.4 1.9 2.07 We can see from the graph that this plot is linear, so the reaction is ﬁrst order. The differential
rate law is: —d[CSHSN2Cl]fdt = rate = kICGHSNZCI] and the integrated rate law is: ln[05}15N2Cl]
= kt + h1[CSI15N2Cl]a. From separate data, It was determined to be 6.9 X 10‘2 s". 33. For a ﬁrst order reaction, the integrated rate law is: ‘ In([A]f[A]D) = kt. Solving for k: In [ 0.250 01001. = k X 120. s, k=0.0116 s'1
1.00 moUL In M =0.0116s‘1xt, t=150.s
2.00molfL
34. In [A] =~kt; k= 113 = 06931 =4.5 x 102 (11
[A10 rm 14.30 If [A]o = 100.0, then after 95.0% completion, [A] = 5.0. In “549— =—4.ss x102 d" x t, t=62 days
1000 __ l
or k —
1 “AL: tIJ’IIAlo k = ————~—~—— = 0.12 L r1101" 5'1
143 5(0060 mOUL) 35. For a second order reaction: 1],; = T CHAPTER 15 CHEMICAL KINETICS 499 ____________,___.__.—.—————~—————————— 36. 3?. 38. 39. a. The integrated rate law for a second order reaction is: 111A] = kt + 1/ [161]” and the halflife
expression is: t1]2 = 1fk[A]o. We could use either to solve for tm. Using the integrated rate law:
_._—1———= “2.005 + m—1———, k= M = 0.555Lmol" s1
(090012) molfL 0.900 molfL 2.00 s
b. _Fl——=0.555an:.1‘15.l x t+ 1 ,t= 8'9L’m01 =165
0.100 molfL 0.900 molfL 0.555 L 11101 '1 s '1 a. Ifthe reaction is 3 8.5% complete, then 38.5% of the original concentration is consumed,
leaving 61.5%. [A] = 61.5% of [AL or [A] = 0.515 [A]; in 0615 [A10 [A] J =kt, in ] =k(480.s) EAL, ln(0.615) = —k(480. 3), 0.486 = k(480. s), k = 1.01 X 10‘3 s'1 0 b. tm = (In 2)/k 2 0.15931fl.01><10'3 s“ = 686 s c. 25% complete: [A] = 0.?5 [Ale ln(0.75) = —1.01 x 10'3 (t), t= 280 s
75% complete: [A] = 0.25 [A]; ln(0.25)= 1.01 x 10'3 (t), t= 1.4 x 103 5
Or, we know it takes 2 X In for reaction to be 75% complete. t = 2 X 686 s = 1370 s
95% complete: [A] = 0.05 [1010; 111(005) = ~1.01 x 10'3 (t), t = 3 x 103 s Successive halflives increase in time for a second order reaction. Therefore, assume reaction is
second order in A. t1”: _!_.—. k: ._.l_ .._.._._.......__l_._.—— = 1.0141110141126114 10A]; tm[A]D loommroM) L=kt+ 1 1'011 xao.0min+ l
[A] [A]o molmin 0.10M =90.M", [A]=1.1><10'2M b. 30.0 min = 2 halflives, so 25% of original A is remaining. [A] = 0.250110 M) = 0.025 M a. [A] = kt + [A]o; Ifk = 5.0 x 102 moi L1 s" and [A], = 1.00 x 103 M, then: [A =(5.0 X 10':a mol L’1 s“)t + 1.00 x 103 mom, A
b. [21° =—(s.0 >< 102)t,.1+ [A], since att=tm, [A] = [A],,12.
..3 A
0.50[A]., = (5.0 x 102) 1m, rm = w = 1.0 x 102 5; Or we can use t1.2 = L19.
5.0 x 10" 2 500 CHAPTER 15 CHEMICAL KINETICS ________________.__._._.——_———————————— c. [A] = ict + [A]o=(5.0 X 10~2 mol L" s")(5.0 x 10'3 s) + 1.00 X 10'3 moUL = 15 X 10“ mol/L
[A}Md=1.00 X103 moUL  7.5 X 10“ moHL = 2.5 X 10" moUL
[B1pmducecl =[A1maeted = 25 X 104 M 40. 100% —> 50% —* 25% *i' 12.5%; This process is 3 halflives = 304 h) = 42 hours. 41. 3. Since [A]o < < [B]o or [C]o, then the B and C concentrations remain constant at 1.00 M for
this experiment. So: rate = k[A]2[B][C] = k’[A]2 where k' = k[B] [C] For this pseudo second order reaction: J= 't+ 1 , _'.—L—=k'(3.00min)+ ._1__
[A] [A]o 3.26 X 105 M 1.00 x 104 M
k’ = 6890 Lrnol‘l min" = 115 L mol'I s‘1
r  1 1
.k' =k[B][C},k 1‘ — “3 “‘01 5 = 115 L3 mol“3 51 [BHC] (1.00 M) (1.00M) b. For this pseudo second order reaction: rate = k’[A]2, tm = k: 1 = __ﬁ——1———T = 8?.0 s
We 115 Lmol '15 “(1.00 x 10‘4 %
c. J" =k't+ 1 = 115 Lmol" s1 x 600. s + 1 = 7.90 x 10* Umol,
[A] [A30 1.00x 104 1‘59
[A] = #1—— =12? x 105 1’19
7.90 x 104 Limo] L From the stoichiometry in the balanced reaction, 1 mol of B reacts with every 3 mol of A. amountA reacted = 1.00 x 104M— 1.27 x 105 M: 8.? x105}; lmolB
3molA amount B reacted = 8.? X 10‘5 molfL x = 2.9 x 10‘5 M
[B] = 1.00M— 2.9 X104 M= 1.00M As we mentioned in part a, the concentration of B (and C) remain constant since the A
concentration is so small. ...
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 Fall '08
 CHIRIK,P
 Rate equation

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