Chaper 3 Homework Answers 3.. The answer is c, doubling he [C] o while keeping he [A] o and [B] o consan. 3.2. 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 = 0.693 0.35 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. 3.3. 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. 3.4. 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.
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.] = 7 5.57 x 0 M/s 3 2.72 x 0 M = 2.047 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 = 0.0050 M, [Cyb.] = 0.009 M, and = 455 s. [0.009 M] k(455 s) = ln [0.0050 M] = 0.235 Solving gives 5.088 x 0 4 /s. Now le [Cyb.] = he concenraion afer 750 s and [Cyb.] o = 0.0050 M, and calculae [Cyb.]. [Cyb.] ln = (5.088 x 0 4 /s)(750 s) = 0.386 [0.0050 M] Taking he anilog of boh sides yields [Cyb.] = 0.68275 [0.0050 M] Solving for [Cyb.] gives [Cyb.] = 0.68275 x [0.0050 M] =.024 x 0 3 =.02 x 0 3 M 3.63. 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) = 0.3928 = 0.393 L/(mol s)
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 + 0.50 M 0.50 M 0.25 M 2 4.3 x 0 s = 5.8 x 0 4 = 5.8 x 0 4 M/s 3.65. 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] 0.00 0.0000 00.0 3.00 0.00740 35. 4.00 0.00683 46.4 5.00 0.00634 57.7 0.00 0.00463 25.9 20.00 0.00304 328.9 30.00 0.00224 446.4 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 = [446.4 00.0]/ M.54 = [30.00 0.00]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). 3.66. For ploing ln k versus /T, he daa below are used: k ln k /T (K) 2.69 x 0 3 5.98.402 x 0 3 6.2 x 0 3 5.08.364 x 0 3.40 x 0 2 4.268.328 x 0 3 3.93 x 0 2 3.236.294 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 0.402 x 0 ]/K = 2.483 x 04 K Because he slope = E a /R, you can solve for E a using R = 8.3 J/K mol: Ea = 2.483 x 0 4 K 8.3 J/ (K mol) E a = 2.483 x 0 4 K x 8.3 J/K mol = 2.063 x 0 5 J/mol (2. x 0 2 kj/mol)
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: (0.076 0.03) 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 3.68..0 4.50 x 0 5 = 4.5 x 0 5 M/s 2.0 4.33 x 0 5 = 4.3 x 0 5 M/s 3.0 4.00 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) + 0.50 mol/l k x (57 s) = 0.30 mol/l 0.50 mol/l =.333 L/mol.333 L/mol 57 s = 0.0233 = 0.023 L/(mol s) 3.69. 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. 3.29. 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.
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) = 0.8880 [H O ] 2 2 o [H2O 2] [H O ] 2 2 o = e 0.8880 = 0.447 Fracion of H 2 O 2 decomposed =.00 0.447 = 0.58852 Since here is.00 mol of H 2 O 2 presen a he sar, he moles of H 2 O 2 decomposed are 0.58852 mol. Thus, Moles of O 2 formed = ( O 2 /2 H 2 O 2 ) x 0.58852 mol H 2 O 2 = 0.29426 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 (0.29426 mol)(0.08206 L am/(k mol))(298 K) = (740/ 760) am = 7.39 = 7.4 L