Chapter 13 Homework Answers

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1 Chaper 3 Homework Answers 3.. The answer is c, doubling he [C] o while keeping he [A] o and [B] o consan a. Since he graph is no linear, here is no way o deermine he reacion order by inspecion. A secondorder reacion should exhibi linear behavior if a plo is made of /[A] vs ime. b. The daa seem o beer suppor a firsorder reacion. As he ime changes by 2 seconds, he concenraion of A decreases by 0 2. This is he behavior expeced for an exponenial rae law like a firsorder reacion rae law, [A] = [A] 0 e k. c. The rae consan is ln 2 /2 = s =.980 =.98 /s d. A possible mechanism wih inermediaes X and Y is A X + Y (slow sep) X + Y B + C (fas sep) e. The behavior of he rae consan k as a funcion of emperaure is described by he Arrhenius equaion. k E ln 2 a = k R T T2 As he emperaure increases he rae consan also increases. More molecules collide wih sufficien energy o overcome he poenial energy barrier so he rae of reacion increases a. E mus be a produc since is concenraion increases wih ime. If E were a reacan, you would expec he concenraion o decrease over ime. b. The average rae is faser beween poins A and B because he slope of he curve is seeper in his region. Remember, he seeper he curve, he greaer he rae of change For he reacion 3I + H 3 AsO 4 + 2H + I 3 + H 3 AsO 3 + H 2 O, he rae of decomposiion of I and he rae of formaion of I 3 are, respecively, Rae = [I ]/ Rae = [I 3 ]/ To relae he wo raes, divide each rae by he coefficien of he corresponding subsance in he chemical equaion and equae hem. /3 [I ] = [I 3 ] 3.5. If he rae law is Rae = k[no] 2 [Cl 2 ], he order wih respec o NO is 2 (second order), and he order wih respec o Cl 2 is (firsorder). The overall order is 2 + = 3, hirdorder.

2 3.6. The reacion rae doubles when he concenraion of ehylene oxide is doubled, so he reacion is firs order in ehylene oxide. The rae equaion should have he form Rae = k[c 2 H 4 O] Subsiuing values for he rae and concenraion yields a value for k: rae [E. Ox.] = x 0 M/s x 0 M = x 0 4 = 2.05 x 0 4 /s 3.7. By comparing Experimens 2 and 3, you see ha doubling [I ] doubles he rae, so he reacion is firs order in I. From Experimens and 3, you see ha doubling [ClO ] also doubles he rae, so he reacion is firs order in ClO. From Experimens 3 and 4, you see ha doubling [OH ] halves he rae; ha is, 2 m = /2. Hence m =, and he rae is inversely proporional o he firs power of OH. The rae law is Rae = k[i ][ClO ]/[OH ] Subsiuing values for he rae and concenraions yields a value for k: rae [OH ] [ClO ][I ] = 2 6. x 0 M/s [0.00 M ] [0.00 M] [0.00 M] = 6.0 = 6. /s 3.8. Firs, find he rae consan, k, by subsiuing experimenal values ino he firsorder rae equaion. Le [Cyb.] o = M, [Cyb.] = M, and = 455 s. [0.009 M] k(455 s) = ln [ M] = Solving gives x 0 4 /s. Now le [Cyb.] = he concenraion afer 750 s and [Cyb.] o = M, and calculae [Cyb.]. [Cyb.] ln = (5.088 x 0 4 /s)(750 s) = [ M] Taking he anilog of boh sides yields [Cyb.] = [ M] Solving for [Cyb.] gives [Cyb.] = x [ M] =.024 x 0 3 =.02 x 0 3 M The rae consan for a secondorder reacion is relaed o he halflife by [A] /2 o Using a halflife of 425 s and an iniial A concenraion of 5.99 x 0 3 mol/l, he rae consan is 3 (425 s)(5.99 x 0 mol/l) = = L/(mol s)

3 3.64. The rae law for a zeroorder reacion is [A] = k + [A] o. Using a ime of 4.3 x 0 2 s o go from an iniial concenraion of 0.50 M o 0.25 M, he rae consan is 0.25 M = k x 4.3 x 0 2 s M 0.50 M 0.25 M x 0 s = 5.8 x 0 4 = 5.8 x 0 4 M/s For he firsorder plo, follow Figure 3.9, and plo ln [MA], he mehyl aceae concenraion, versus ime in minues. This plo does no yield a sraigh line, so he reacion is no firs order. For he secondorder plo, follow Figure 3.0, and plo /[MA] versus ime in minues. The daa used for ploing are, min [MA], M /[MA] The plo requires a graph wih oo many lines o be reproduced here, bu i yields a sraigh line, demonsraing ha he reacion is second order in [MA]. The slope of he line may be calculaed from he difference beween he las poin and he firs poin: Slope = [ ]/ M.54 = [ ]min M min = 0.92 s M In his case, he slope equals he rae consan, so.5/(m m), or 0.92/(M s) For ploing ln k versus /T, he daa below are used: k ln k /T (K) 2.69 x x x x x x x x 0 3 The plo yields an approximaely sraigh line. The slope of he line may be calculaed from he difference beween he las poin and he firs poin: (3.236) (5.98) Slope = 3 3 [.294 x x 0 ]/K = x 04 K Because he slope = E a /R, you can solve for E a using R = 8.3 J/K mol: Ea = x 0 4 K 8.3 J/ (K mol) E a = x 0 4 K x 8.3 J/K mol = x 0 5 J/mol (2. x 0 2 kj/mol)

4 3.67. All raes of reacion are calculaed by dividing he decrease in concenraion by he difference in imes; only he seup for he firs rae afer.0 min is given below: ( ) M Rae (.0 min) = (.0 0) min x min 60 s = 4.50 x 0 5 = 4.5 x 0 5 M/s A summary of he imes and raes is given in he able. Time, min Rae x 0 5 = 4.5 x 0 5 M/s x 0 5 = 4.3 x 0 5 M/s x 0 5 = 4.0 x 0 5 M/s a. A plo of /[A] versus ime is a sraigh line for a secondorder reacion. b. The secondorder inegraed rae law is [A] = k + [A] o Afer 57 s, he concenraion of A dropped 40% of is iniial value, so [A] = 0.60[A] o, or (0.60)(0.50 M) = 0.30 M. Using hese values gives 0.30 mol/l = k x (57 s) mol/l k x (57 s) = 0.30 mol/l 0.50 mol/l =.333 L/mol.333 L/mol 57 s = = L/(mol s) Facors ha affec he raes of reacions: i) Concenraions of he reacans ii) Temperaure iii) Caalyss A higher concenraions, more molecules can undergo effecive collisions and give more produc per uni of ime. The emperaure affecs he number of molecules ha have enough energy o reac. The higher he emperaure, he larger he fracion of molecules ha have enough energy o reac. Caalyss provide anoher pahway for reacion ha has a lower acivaion energy, so more molecules have he minimum energy o reac. The value of he rae consan for a paricular reacion depends on he emperaure and he acivaion energy. The concenraion of caalys and of he solven, if he reacion occurs in soluion, can affec E a a. The overall reacion is 2H 2 O 2 2H 2 O + O 2 b. The caalys is I, and he inermediae is IO. c. No, he rae law can no be specified unil he raedeermining sep is esablished.

5 3.30. The balanced equaion is 2H 2 O 2 2H 2 O + O 2. Noe ha he moles of O 2 formed will be equal o onehalf ha of he moles of H 2 O 2 decomposed. Now, use he inegraed form of he firsorder rae law o calculae he fracion of he.00 mol H 2 O 2 decomposing in 20.0 min, or 200 s. [H2O 2] ln = k = (7.40 x 0 4 /s)(200 s) = [H O ] 2 2 o [H2O 2] [H O ] 2 2 o = e = Fracion of H 2 O 2 decomposed = = Since here is.00 mol of H 2 O 2 presen a he sar, he moles of H 2 O 2 decomposed are mol. Thus, Moles of O 2 formed = ( O 2 /2 H 2 O 2 ) x mol H 2 O 2 = mol O 2 Now, use he ideal gas law o calculae he volume of his number of moles of O 2 gas a 25 C and 740 mmhg. V = nrt P ( mol)( L am/(k mol))(298 K) = (740/ 760) am = 7.39 = 7.4 L

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