Chapter 14 Homework Answers

Transcription

1 4. Suden responses will vary. (a) combusion of gasoline (b) cooking an egg in boiling waer (c) curing of cemen Chaper 4 Homework Answers 4. A collision beween only wo molecules is much more probable han he simulaneous collision of hree molecules. We herefore conclude ha a reacion involving a wo body collision is faser (i.e. will occur more frequenly) han one requiring a hree body collision. 4.3 The insananeous rae of reacion is he rae of he reacion a a paricular momen. The average rae of reacion is he average rae for he reacion over he ime of he whole reacion. This includes he very rapid raes when he concenraion of reacans is high and he very slow raes when he concenraion of reacans is low. 4.4 A angen line o he curve of concenraion as a funcion of ime a ime zero is drawn. The slope of his line is deermined which is he iniial insananeous rae of reacion. 4.5 A homogeneous reacion is one in which all reacans and producs are in he same phase. An example would be: H (g) + O (g) H O(g) A heerogeneous reacion is one in which all reacans and producs are no in he same phase. An example would be: 5O (g) + C 8 H 8 (l) 6CO (g) + 8H O(g) 4.6 Chemical reacions ha are carried ou in soluion ake place smoohly because he reacans can mingle effecively a he molecular level. 4.7 Heerogeneous reacions are mos affeced by he exen of surface conac beween he reacan phases. 4.8 In a heerogeneous reacion, he smaller he paricle size, he faser he rae. This is because decreasing he paricle size increases he surface area of he maerial, hereby increasing conac wih anoher reacan phase. 4.9 This illusraes he effec of concenraion. 4. Reacion rae generally increases wih increasing emperaure. 4. In cool weaher, he raes of meabolic reacions of cold blooded insecs decrease, because of he effec of emperaure on rae. 4. The low emperaure causes he rae of meabolism o be very low. 4.3 Reacion rae has he unis mol L s, or molar per second (M s ). Reacion raes are repored as posiive values regardless of wheher he species increases or decreases during he recacion. 4.4 The unis are, in each case, whaever is required o give he unis of rae (mol L s ) o he overall rae law: (a) s (b) L mol s (c) L mol s

2 4.5 A zero order reacion has no dependence on concenraion while he firs order reacion is linearly dependen on concenraion and he second-order reacion is inversely dependen on concenraion. 4.6 No, he coefficiens of a balanced chemical reacion do no predic wih any cerainy he exponens in a rae law. These exponens mus be deermined hrough experimen. 4.7 This is he firs case in Table 4.4 of he ex, ha is a zero-order reacion. 4.8 This is he fourh case Table 4.4 of he ex, ha is a firs-order reacion. 4.9 This is he sevenh case in Table 4.4, and he rae increases by a facor of = This is he elevenh case in Table 4.4, and he order of he reacion wih respec o he reacan is Since he subsrae concenraion does no influence rae, is concenraion does no appear in he rae law, and he order of he reacion wih respec o subsrae is zero. 4. The half life of a firs order reacion is unaffeced by he iniial concenraion. 4.3 The half life of a second order reacion is inversely proporional o he iniial concenraion, as expressed in equaion The half-life of a zeroh-order reacion is direcly proporional o iniial concenraion. 4.5 Firs order [ A] ln k [ ] A = [ A ] = [ A ] Second order = k A A [ ] [ ] / = k A [ ] / = k [ A] [ A] / ln [ A] [ A] = k/ / ln.693 = = k k = k = = / [ A] [ A] [ A] [ A] [ A] Zeroh order A k A [ A] = k + [ A] [ ] = / + [ ] k = [ A ] [ A ] / / = [ A] k 4.6 For he zero order reacion, subsiue half he value of [A] for he value of [A] in he equaion [A] [A] = k [A] /[A] = k / /[A] = k / [A] = k 4.7 To answer his quesion refer o he inegraed equaions for each order. Graph (b) represens a kineics plo for a firs order reacion. Graph (c) represens a kineics plo for a second order reacion. Graph (a) represens a kineics plo for a zeroh order reacion.

3 4.8 According o collision heory, he rae is proporional o he number of collisions per second among he reacans. 4.9 The effeciveness of collisions is influenced by he orienaion of he reacans and by he acivaion energy. 4.3 This happens because a larger fracion of he reacan molecules possess he minimum energy necessary o surpass E a. 4.3 Poenial Energy E a (forward) H reacion E a (reverse) Producs Reacans Reacion Coordinae 4.3 Transiion Sae Poenial Energy Reacans H Producs Reacion Coordinae 4.33 Breaking a srong bond requires a large inpu of energy, hence a large E a An elemenary process is an acual collision even ha occurs during he reacion. I is one of he key evens ha moves he reacion along in he sepwise process ha leads o he overall reacion ha is observed. I is hus one sep in a poenially muli sep mechanism The rae deermining sep in a mechanism is he slowes sep The rae law for a reacion is based on he rae deermining sep.

4 4.37 Adding all of he seps gives: NO + H N + H O The inermediaes are N O and N O For such a mechanism, he rae law should be: rae = k[no ][CO] Since his is no he same as he observed rae law, his is no a reasonable mechanism o propose Add all of he seps ogeher: CO + O + NO CO + NO 4.4 On adding ogeher all of he seps in each separae mechanism, we ge, in each case: OCl + I Cl + OI 4.4 The prediced rae law is based on he rae deermining sep: rae = k[no ] 4.4 A caalys changes he mechanism of a reacion, and provides a reacion pah having a smaller acivaion energy A homogeneous caalys is one ha is presen in he same phase as he reacans. I is used in one sep of a cycle, bu regeneraed in a subsequen sep, so ha in a ne sense, i is no consumed Adsorpion is a clinging o a surface. Absorpion involves a peneraion below he surface, as in he acion of a sponge. Heerogeneous caalysis involves adsorpion The caalyic converer promoes oxidaion of unburned hydrocarbons, as well as he decomposiion of nirogen oxide polluans. Lead poisons he caalys and renders i ineffecive.

5 4.46 Since hey are in a -o- mol raio, he rae of formaion of SO is equal and opposie o he rae of consumpion of SO Cl. This is equal o he slope of he curve a any poin on he graph (see below). A min, we obain a value of abou 4 M/min. A 6 minues, his has decreased o abou 7 5 M/min This is deermined by he coefficiens of he balanced chemical equaion. For every mole of N ha reacs, 3 mol of H will reac. Thus he rae of disappearance of hydrogen is hree imes he rae of disappearance of nirogen. Similarly, he rae of disappearance of N is half he rae of appearance of NH 3, or NH 3 appears wice as fas as N disappears From he coefficiens in he balanced equaion we see ha, for every mole of B ha reacs, mol of A are consumed, and hree mol of C are produced. This means ha A will be consumed wice as fas as B, and C will be produced hree imes faser han B is consumed. rae of disappearance of A = (.3) =.6 mol L s rae of appearance of C = 3(.3) =.9 mol L s 4.5 We rewrie he balanced chemical equaion o make he problem easier o answer: N O 5 NO + /O. Thus, he raes of formaion of NO and O will be, respecively, wice and one half he rae of disappearance of N O 5. rae of formaion of NO = (.5 6 ) = 5. 6 mol L s rae of formaion of O = /(.5 6 ) =.3 6 mol L s 4.55 rae = (.3 L mol s )(. 7 mol/l)(. 7 mol/l) rae =.3 3 mol L s 4.6 The reacion is firs order in OCl, because an increase in concenraion by a facor of 4.75, while holding he concenraion of I consan (compare he firs and second experimens of he able), has caused an increase in rae by a facor of 4.75 = The order of reacion wih respec o I is also, as is demonsraed by a comparison of he firs and hird experimens. rae = k[ocl ][I ] Using he las daa se:.5 5 mol L s = k[.6 3 mol/l][9.6 3 mol/l] k = L mol s

6 4.65 In he firs, fourh, and fifh experimens, he concenraion of OH has been made o increase while he [(CH 3 ) 3 CBr] is lef unchanged. In each of hese experimens, here is no change in rae. This means ha he rae is independen of [OH ], and he order of reacion wih respec o OH is zero. The concenraion of (CH 3 ) 3 CBr increases by a facor of.4 from he firs o he second experimen, and by a facor of 3.84 from he firs o he hird experimen, as he OH concenraion is held consan. There is a corresponding.4 fold (i.e..4 ) increase in rae from he firs o he second experimen, and here is a 3.8 fold (i.e. 3.8 ) increase in rae from he firs o he hird experimen. In boh cases, we conclude ha he order wih respec o (CH 3 ) 3 CBr is one. rae = k[(ch 3 ) 3 CBr] Using he hird se of daa gives: mol L s = k[7.3 mol/l] k = 5. 3 s 4.67 Since i is he plo of /conc. ha gives a sraigh line, he order of he reacion wih respec o CH 3 CHO is wo. The rae consan is given by he slope direcly: k =.77 M s 4.73 ln [ A] [ ] k A = 6. x g L ln = k x 8 s 6.3 x g L 3 k = 7. x s 4.75 = k [ A] [ A] 4.79 Since / of he Sr 9 decays every half life, i will ake 5 half lives, or 5 8 yrs = 4 yrs, for he Sr 9 o decay o /3 of is presen amoun We can use he half-life o deermine he value of k x 3 s =.693/k k =.4 x -4 s - ln. M 4 k A = ln =.4 x x. M 4 =.3 x s [ A] [ ]

7 4.86 Use equaion 4.8 and 4.9. r ln = k r. 4 3 ln = (. y )( 9. y ) r. 4 3 = exp (. y )( 9. y ) r. 3 = r = The graph is prepared exacly as in example 4. of he ex. The slope is found using linear regression, o be: K. Thus K = E a /R E a = ( K)(8.34 J K mol ) = J/mol = 79 kj/mol Using he equaion, we proceed as follows: k Ea ln = k R T T 3.94 L mol s Ea 4.88 L mol s 8.34 J mol K 673 K 593 K ln = 4. K.97 = E a 8.34 J mol K E a = J/mol = 79.3 kj/mol 4.9 Using he equaion we have: k Ea ln = k R T T 3. L mol s Ea ln = L mol s 8.34 J mol K 43 K 373 K 4. K.37 = E a 8.34 J mol K E a = J/mol = 99 kj/mol Equaion saes k = A exp Ea RT

8 k A = E exp a RT = L mol s exp J mol ( 8.34 )( 373 K) J mol K 9 = 6.6 L mol s J / mol k = 6.6 L mol s exp 8.34 J mol K x 473 K k = 7.9 x - L mol - s An inermediae is produced in one sep and used up in anoher. The inermediae in his reacion is N O. The overall reacion is he sum of he wo seps NO(g) + O (g) NO (g) The rae law is deermined from he slow sep. Rae = k[n O ][O ] An inermediae canno be par of he rae law expression. We can use he firs sep, which is an equilibrium sep o solve for he concenraion of N O. Rae = k f [NO] = kr[no ] Solve for [N O ] o ge k f [NO ] = [NO] k r Rae = [NO] [O ] k where all of he consans have been combined ino a new consan

9 4.98 From he angen lines we can obain he insananeous rae a s and 9 s. For he decomposiion of cyclobuane: A s he slope is 9.6 x -7 M s - which is he rae of decomposiion. A 9 s he slope is 3.8 x -7 M s - which is he rae of decomposiion For he formaion of ehylene: A s he slope is.9 x -6 M s - which is he rae of formaion of. A 9 s he slope is 7.5 x -7 M s (a) The reacion rae is firs order so as he concenraion of reacans riples he rae will riple. (b) The reacion rae is second order so as he concenraion of reacans riples he rae will increase by a facor of nine. (c) The reacion rae is zeroh order so he rae is independen of concenraion. There is no change in he rae. (d) The reacion rae is second order so as he concenraion of reacans riple he rae increases by a facor of nine (e) The reacion rae is hird order so as he concenraion of reacans riple he rae increases by a facor of weny seven.

10 4.5 (a) rae = k [A] (b) rae = k [A ] (c) rae = k [A ] [E] (d) A + E B + C (e) The raes for he forward and reverse direcions of sep one are se equal o each oher in order o arrive a an expression for he inermediae [A ] in erms of he reacan [A]: k [A] = k [A ] k [A ] = [A] k This is subsiued ino he rae law for quesion (c) above, giving a rae expression ha is wrien k using only observable reacans: rae = k [A] [E] k 4. To solve his problem, plo he daa provided as /T vs / where T is he absolue emperaure and / is proporional o he rae consan. (min) T (K) /T ln(/) The slope of he graph is equal o E a /R, herefore: 7,74 = E a /R 7,74R = E a 7,74 K(8.34 J/mol K) = E a E a = 64,5 J/mol E a = 64 kj/mol From he sraigh line equaion, we can deermine he ime needed o develop he film a 5 C is 4 min.

11 4.7 We can deermine he emperaure ha he reacion mixure would achieve if all of he hea is supplied a once by using he enhalpy of reacion and he hea capaciy of he sysem. q = C cal T 4 kj mol - x J kj - x, mol = 7.5 x 6 J o C - T T = 53.3 o C Thus, he final emperaure of he mixure would be 78 o C or 35 K k Ea ln = k R T T J 5 kj mol x k kj ln = 5 x s 8.34 J K mol 35 K 98 K k ln = x s k = x -4 s - If he emperaure of he reacion mixure was 8 o C and all of he hea released a once he final emperaure would be: 8 o C + 53 o C = 35 o C so he reacion would indeed reach 99 o C.