122 HW_Ch14_15 key

122 HW_Ch14_15 key - » u “We...” 2...,» 14.32 (a)...

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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 second-order 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 \ 33-3 _/' -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 best-fit 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 Inkc-Ink= 3° +lnA- _a RT RT E E - In (kc/k): am“; E,-E,c=RT In (kc/k) 8.314J Ea- E = ac K-moI 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 increase in reaction rate. U>o>jfla7 :fi A737- : yzx/ofjg“ : a? Ely/04 )‘(k- LY; /0 5‘ I ‘ A J] 71> 23”] J}, at A/;J< (ff/rt: rz/cs 0/: / (ta/)J [#7 MM" .66 (flew/5K. x1... [g] 9%, = 6.: 52/9,, Mpg; 6. fl‘z ~ 7' y ‘ .. _ / Z fl/Uz. flax]. av I //’/LO ’gyz’bjflh [$71 [M37 07 C (“L Xflflpfl'K/ityci ( Ab “g/exéx/Uangflj /52> a) :2 [waif/g?) ifimdfi) +fir13> Crowd ) ‘ ,_ C 7 ~ / fl (1).. 6L I’Xfi (j/‘(Jl’ s-$=9g#l " :/’3;‘/0_Z : 76,7Z ,5 A rue/M m, 44%” ’ A? T '0 Z '2. 301) f 53 7 god: flag/fig) '2. 7/ a]— /5°> fl; 2(0. 3% r)(z>,/¢r)z(.’7$‘f>k z 0, 03%; “LA 03> Fa/Am) \/;:jlkol 76)! /S«,/‘52 , \5 - 45> J/JJA‘ €74“ >11 //;/h0/c fflg 9/0263) ¢ 5) fl/{UZ /(/”/0 6M {ya/AMK KM/u a/ta’réjrj A/v/0/( féyz M/‘//filml j tw//Q/é/"JC Aimee Ala)“: ‘// 4’ 65/<fl 36% /o/ wt flcouc #64,,” /p/, xvq J0 chu-i/ ‘flL @554 MOM [719/4 29 /j(5. 00m /VZL% I 29/00 14% . 9%. A 3M 74 M [W m; 4 0; ¢ ...
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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.

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122 HW_Ch14_15 key - » u “We...” 2...,» 14.32 (a)...

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