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Unformatted text preview: » u “We...” 2...,» 14.32 (a) Using Equation 14.13 for a ﬁrst order reaction: In[A]t = kt + ln[A]0
2.5 min = 150 s; [N205]0 = (0.0250 mol/2.0 L) = 0.0125 = 0.013 M
In[N205]150 = (6.82 x 10‘3 s“)(150 s) + In(0.0125)
ln[N205]15° = 1.0230 + (4.3820) = 5.4050 = 5.41
[N205]150 = 4.494 x 10“3 = 4.5 x 104 M; mol N205 = 4.494 x 10“3 M x 2.0 L = 9.0 x 10“3 mol
(b) [N205], = 0.010 moi/2.0 L = 00050 M; [r4205]o = 00125 M In (0.0050) = (6.82 x 104 s") (t) + In(0.0125) t: W = 134.35: 1.3 x 102sx 1mm =2.24=2.2 min
(6.82 x 10'35 “) 60 5 v. r (c) t“.2 = 0.693/k = 0.693/6.82 X 10'38'1 = 101.6 = 102 s or 1.69 min 14.38 (a) Make both ﬁrst and secondorder plots to see which Iihear. 2
time1mi“) [c12H22011] (M ) "‘[cquzou] 1I[c12H22011]
0 0.316 1.152 3.16
39 0.274 —1.295 3.65
80 0.238 1.435 4.20
140 0.190 1.661 5.26
210 0.146 1.924 6.85
«1.0 —  . 8 
2 1 2 \'\ 7 n
a; \\ E 7/ 5
§ 1.4 o 6 // g
N N
.— N‘5 / i
g —1 6 \ 2g /
E "4 /
—1.8 \ 333 _/'
2‘0 1 "‘ ‘ 2 . _ v.— r 0 50 100 150 200 250 0 :0 100 150 200 250
tiime (mm) time (min) The plot of In [C12H22011] in linear, so the reaction is ﬁrst order 10 C12H220“. (b) = slope =  [1.924  (—1.295)]I 171 min = 3.68 x 10“3 min‘1
(The sIope of the bestﬁt line is 3.67 x 10" min“.) 14.43 Analyze/Plan. Use the deﬁnitions of activation energy (Emax  Emma) and AE (Eprod  Ema) to
sketch the graphs and calculate Ea for the reverse reaction. Solve: (a) (b) Ea(reverse) = 18 kJ/mol Ea=154 kJ l AE = 136 kJ 14.46 E,i for the reverse reactions is: (a) 45  (25) = 70 kJ (b) 35  (~10) = 45 kJ (c) 55  10 = 45 kJ
Based on the magnitude of E3, the reverse of reactions (b) and (0) occur at the same rate,
which is faster than the reverse of reaction (a). 14.55 Analyze/Plan. Elementary processes occur as a single step, so theimolecﬁarity is
determined by the number of reactant molecules; the rate taw reﬂects reactant stoichiometry. Solve: (a) unimolecular, rate = kICIZ] (b) bimolecular, rate = k[OCl‘][H20]
(c) bimolecular, rate = k[NO][Cl2] 14.58 Two intermediates, three transition states. the C ) D step is fastest, the overall reaction is
endothermic. 14.59 (a) H2(9) + ICKQ) 9 H1(9) + HCKQ) H1(9)+ 101(9) 912(9)+ HCKQ) H2(9) + ZICKQ) '9 I2(9) + ZHCKQ) (b) lntennediates are produced and consumed during reaction. HI is the intermediate.
(c) Follow the logic in Sample Exercise 14.13. First step: rate = k[H2][ICl] Second step: rate = k[HI][ICl]
(d) The slow step determines the rate law for the overall reaction. If the ﬁrst step is slow, the observed rate law is: rate = k[H2][HCl]. 14.74 Let k and En equal the rate constant and activation energy for the uncatalyzed reaction. Let
kc and Eac equal the rate constant and activation energy of the catalyzed reaction. A is the
same for the uncataIyzed and catalyzed reactions. kc/ k = 1 x 105, T = 37°C = 310 K.
According to Equation [14.20], In k = EalRT + In A. Subtracting In k from In kc InA . E E
InkcInk= 3° +lnA _a
RT RT E E 
In (kc/k): am“; E,E,c=RT In (kc/k) 8.314J
Ea E =
ac KmoI X 310K X In (1 x 105) = 2.966 X1O4J = 29.7 kJ The enzyme must Iower the activation energy by 29.7 kJ in order to achieve a 1 x 105 fold
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This note was uploaded on 07/17/2008 for the course CHEM 122 taught by Professor Zellmer during the Winter '07 term at Ohio State.
 Winter '07
 Zellmer
 Chemistry

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