CHAPTER 6 KINETICS: RATES AND MECHANISMS OF CHEMICAL REACTIONS FOLLOW UP PROBLEMS 6.A Plan: Balance he equaion. The rae in ers of he change in concenraion wih ie for each subsance is [ A] expressed as, where a is he coefficien of he reacan or produc A. The rae of he reacans is given a a negaive sign. a) The balanced equaion is 4NO(g) + O (g) N O 3. Choose O as he reference because is coefficien is. Four olecules of NO (nirogen onoxide) are consued for every one O olecule, so he rae of O disappearance is /4 he rae of NO decrease. By siilar reasoning, he rae of O disappearance is / he rae of N O 3 (dinirogen rioxide) increase. [ NO] [ O ] NO 3 Rae + 4 b) Plan: Because NO is decreasing; is rae of concenraion change is negaive. Subsiue he negaive value ino he expression and solve for [O ]/. Rae ( 4 [ O ].60 x 0 ol/l s) 4.00x0 5 ol/l s 4 6.B Plan: Exaine he equaion ha expresses he rae in ers of he change in concenraion wih ie for each subsance. The nuber in he denoinaor of each fracion is he coefficien for he corresponding subsance in he balanced equaion. The ers ha are negaive in he rae equaion represen reacans in he balanced equaion, while he ers ha are posiive represen producs in he balanced equaion. For exaple, he following [ A] er,, describes a produc (he er is posiive) ha has a coefficien of a in he balanced equaion. In a par b), use he rae equaion o copare he rae of appearance of H O wih he rae of disappearance of O. a) The balanced equaion is 4NH 3 + 5O 4NO + 6H O. b) Plan: Because H O is increasing; is rae of concenraion change is posiive. Subsiue is rae ino he expression and solve for [O ]/. Rae 6 (.5x0 ol/l s) [O ] 5 5 6 (.5x0 ol/l s) [O ].0x0 ol/l s The negaive value indicaes ha [O ] is decreasing as he reacion progresses. The rae of reacion is always expressed as a posiive nuber, so [O ] is decreasing a a rae of.0x0 ol/l s. 6.A Plan: The reacion orders of he reacans are he exponens in he rae law. Add he individual reacion orders o obain he overall reacion order. Use he rae law o deerine how he changes lised in he proble will affec he rae. a) The exponen of [I ] is, so he reacion is firs order wih respec o I. Siilarly, he reacion is firs order wih respec o BrO 3, and second order wih respec o H +. The overall reacion order is ( + + ) 4, or fourh order overall. 6-
b) Rae k[i ][BrO 3 ][H + ]. If [BrO 3 ] and [I ] are ripled and [H + ] is doubled, rae k[3 x I ][3 x BrO 3 ][ x H + ], hen rae increases o 3 x 3 x or 36 ies is original value. The rae increases by a facor of 36. 6.B Plan: The reacion orders of he reacans are he exponens in he rae law. Add he individual reacion orders o obain he overall reacion order. Use he rae law o deerine how he changes lised in he proble will affec he rae. a) The exponen of [ClO ] is, so he reacion is second order wih respec o ClO. Siilarly, he reacion is firs order wih respec o OH. The overall reacion order is ( + ) 3, or hird order overall. b) Rae k[clo ] [OH ]. If [ClO ] is halved and [OH ] is doubled, rae k[/ x ClO ] [ x OH ], hen rae increases o (/) x or / is original value. The rae decreases by a facor of /. 6.3A Plan: Assue ha he rae law akes he general for rae k[h ] [I ] n. To find how he rae varies wih respec o [H ], find wo experiens in which [H ] changes bu [I ] reains consan. Take he raio of rae laws for hose wo experiens o find. To find how he rae varies wih respec o [I ], find wo experiens in which [I ] changes bu [H ] reains consan. Take he raio of rae laws for hose wo experiens o find n. Add and n o obain he overall reacion order. Use he rae law o solve for he value of k. For he reacion order wih respec o [H ], copare Experiens and 3: rae 3 H 3 [ ] rae [ H ] 9.3 x 0.9 x 0 3 3 3 [ 0.0550 ] [ 0.03] 4.8947 (4.867566) Therefore, If he reacion order was ore coplex, an alernae ehod of solving for is: log (4.8947) log (4.867566); log (4.8947)/log (4.867566) For he reacion order wih respec o [I ], copare Experiens and 4: rae 4 [ ] n rae I 4 n I.9 x 0. x 0 [ ] n 4 n [ 0.0056 ] [ 0.0033].777 (.69697) n Therefore, n The rae law is rae k[h ][I ] and is second order overall. Calculaion of k: k Rae/([H ][I ]) k (.9x0 3 ol/l s)/[(0.03 ol/l)(0.00 ol/l)].5 x 0 8 L/ol s k (.x0 ol/l s)/[(0.00 ol/l)(0.0033 ol/l)].5 x 0 8 L/ol s k 3 (9.3x0 3 ol/l s)/[(0.0550 ol/l)(0.00 ol/l)].5 x 0 8 L/ol s k 4 (.9x0 ol/l s)/[(0.00 ol/l)(0.0056 ol/l)].5 x 0 8 L/ol s Average k.5x0 8 L/ol s 6.3B Plan: Assue ha he rae law akes he general for rae k[h SeO 3 ] [I ] n [H + ] p. To find how he rae varies wih respec o [H SeO 3 ], find wo experiens in which [H SeO 3 ] changes bu [I ] and [H + ] reain consan. Take he raio of rae laws for hose wo experiens o find. To find how he rae varies wih respec o [I ], find wo experiens in which [I ] changes bu [H SeO 3 ] and [H + ] reain consan. Take he raio of rae laws for hose wo experiens o find n. To find how he rae varies wih respec o [H + ], find wo experiens in which 6-
[H + ] changes bu [H SeO 3 ] and [I ] reain consan. Take he raio of rae laws for hose wo experiens o find p. Add, n, and p o obain he overall reacion order. Use he rae law o solve for he value of k. For he reacion order wih respec o [H SeO 3 ], copare Experiens and 3: rae 3 rae [H SeO 3 ] 3 [H SeO 3 ] 3.94x0 6 ol/l s 9.85x0 7 [.0x0 ol/l]3 ol/l s [.5x0 3 ol/l] 4 (4) Therefore, For he reacion order wih respec o [I ], copare Experiens and : rae rae [I n ] [I n ] 7.88x0 6 ol/l s 9.85x0 7 [3.0x0 n ol/l] ol/l s [.5x0 n ol/l] 8 () n Therefore, n 3 For he reacion order wih respec o [H + ], copare Experiens and 4: rae 4 rae [H + p ] 4 [H + p ] 3.5x 0 5 ol/l s 7.88x0 6 [3.0x0 p ol/l]4 ol/l s [.5x0 p ol/l] 4 () p Therefore, p The rae law is rae k[h SeO 3 ][I ] 3 [H + ] and is sixh order overall. b) Calculaion of k: k Rae/([H SeO 3 ][I ] 3 [H + ] ) k (9.85x0 7 ol/l s)/[(.5x0 3 ol/l)(.5x0 ol/l) 3 (.5x0 ol/l) ] 5.x0 5 L 5 /ol 5 s 6.4A Plan: The reacion is second order in X and zero order in Y. For par a), copare he wo aouns of reacan X. For par b), copare he wo rae values. a) Since he rae law is rae k[x], he reacion is zero order in Y. In Experien, he aoun of reacan X has no changed fro Experien and he aoun of Y has doubled. The rae is no affeced by he doubling of Y since Y is zero order. Since he aoun of X is he sae, he rae has no changed. The iniial rae of Experien is also 0.5x0 5 ol/l s. b) The rae of Experien 3 is four ies he rae in Experien. Since he reacion is second order in X, he concenraion of X us have doubled o cause a four-fold increase in rae. There should be 6 black spheres and 3 green spheres in Experien 3. 5.0 x 0 [ ] x 5 0.5 x 0 3 [ ] 4 [ ] x 9 x 6 black spheres 6-3
6.4B 6.4B Plan: Exaine how a change in he concenraion of he differen reacans affecs he rae in order o deerine he rae law. Then use he rae law o deerine he nuber of paricles in he scene for Experien 4. a) Coparing Experiens and, we can see ha he nuber of blue A spheres does no change while he nuber of yellow B spheres changes fro 4 o. Alhough he nuber of yellow spheres changes, he rae does no change. Therefore, B has no effec on he rae, which suggess ha he reacion is zero order wih respec o B. Now ha we know ha he yellow B spheres do no affec he rae, we can look a he influence of he blue A spheres. Coparing Experiens and 3, we can see ha he nuber of blue spheres changes fro 4 o (experien 3 concenraion of A is half of ha of experien ) while he rae changes fro.6x0 3 ol/l s o 8.0x0 4 ol/l s (he rae of experien 3 is half of he rae of experien ). The fac ha he concenraion of blue spheres changes in he sae way he rae changes suggess ha he reacion is firs order wih respec o he blue A spheres. Therefore, he rae law is: rae k[a]. b) The rae of Experien 4 is wice he rae in Experien. Since he reacion is firs order in A, he concenraion of A us have doubled o cause a wo-fold increase in rae. There should be x 4 paricles 8 paricles of A in he scene for Experien 4. 6.5A Plan: The rae expression indicaes ha he reacion order is wo (exponen of [HI] ), so use he inegraed second-order law. Subsiue he given concenraions and he rae consan ino he expression and solve for ie. k A A [ ] [ ] 0 [ 0.00900 ol/ L] [ 0.000 ol/ L] 0 0.00900 ol/ L 0.000 ol/ L.4 x0 L/ol s 4.696x0 s 4.6x0 s (.4x0 L/ol s)() 6.5B Plan: The proble saes ha he decoposiion of hydrogen peroxide is a firs-order, reacion, so use he firsorder inegraed rae law. Subsiue he given concenraions and he ie ino he expression and solve for he rae consan. a) ln [A] 0 k [A] ln.8 M (k)(0.0 in) 0.85M k 0.4094 0.04094 0.04 in 0.0 in b) If you sar wih an iniial concenraion of hydrogen peroxide of.0 M ([A] 0.0 M) and 5% of he saple decoposes, 75% of he saple reains ([A] 0.75 M). ln.00 M 0.75M (0.04 in )() 0.877 7.07 7.0 in 0.04 in 6-4
6.6A Plan: The iniial scene conains paricles of Subsance X. In he second scene, which occurs afer.5 inues have elapsed, half of he paricles of Subsance X (6 paricles) reain. Therefore,.5 in is he half-life. The half-life is used o find he nuber of paricles presen a 5.0 in and 0.0 in. To find he olariy of X, oles of X is divided by he given volue. a) Since.5 inues is he half-life, 5.0 inues represens wo half-lives: paricles of X Afer 5.0 inues, 3 paricles of X reain; 9 paricles of X have reaced o produce 9 paricles of Y. Draw a scene in which here are 3 black X paricles and 9 red Y paricles..5 in 6 paricles of X.5 in 3 paricles of X b) 0.0 inues represens 4 half-lives:.5 in.5 in.5 in paricles of X 6 paricles of X 3 paricles of X.5 in.5 paricles of X 0.75 paricle of X 0.0 ol Moles of X afer 0.0 in ( 0.75 paricle) 0.5 ol paricle Molariy ol X 0.5 ol volue 0.50 L 0.30 M 6.6B Plan: The iniial scene conains 6 paricles of Subsance A. In he second scene, which occurs afer 4 inues have elapsed, half of he paricles of Subsance A (8 paricles) reain. Therefore, 4 in is he half-life. The half-life is used o find he aoun of ie ha has passed when only one paricle of Subsance A reains and o find he nuber of paricles of A presen a 7 inues. To find he olariy of A, oles of A is divided by he given volue. a) The half-life is 4 inues. We can use ha inforaion o deerine he aoun of ie ha has passed when one paricle of Subsance A reains: 4 in 4 in 6 A paricles 8 A paricles 4 A paricles Afer 4 half-lives, only one paricle of Subsance A reains. 4 half-lives 4 x (4 in/half-life) 96 in b) 4 in A paricles 7 in Nuber of half-lives a 7 in 4 in/half-life 3 half-lives According o he schee above in par a), here are A paricles lef afer hree half-lives. 0.0 ol A Aoun (ol) of A afer 7 inues ( paricles A) 0.0 ol A paricle A 0.0 ol A M 0.80 M 0.5 L 6.7A Plan: Rearrange he firs-order half-life equaion o solve for k. ln ln k 0.0599985 0.059 h 3. h / 4 in A paricle 6.7B Plan: Rearrange he firs-order half-life equaion o solve for half-life. Deerine he nuber of half-lives ha pass in he 40 day period described in he proble and use his nuber o deerine he aoun of pesicide reaining. 6-5
a) / ln k ln 9 x 0 7.706 8 days day 40 days b) Nuber of half-lives a 40 days 8 days/half-life 5 half-lives Afer each half-life, ½ of he saple reains. Afer five half-lives, ½ x ½ x ½ x ½ x ½ (½) 5 of he saple reains. 3 6.8A Plan: The acivaion energy, rae consan a T, and a second eperaure, T, are given. Subsiue hese values ino he Arrhenius equaion and solve for k, he rae consan a T. k ln Ea k R T T k 0.86 L/ol s T 500. K E a.00x0 kj/ol k? L/ol s T 490. K k 3.00 x0 kj/ ol 0 J ln 0.86 L/ol s 8.34 J/ ol K 500.K 490.K kj k ln 0.86 L/ol s 0.49093 k 0.6057 0.86 L/ol s k (0.6057)(0.86 L/ol s) 0.75048 0.75 L/ol s 6.8B Plan: The acivaion energy and rae consan a T are given. We are asked o find he eperaure a which he rae will be wice as fas (i.e., he eperaure a which k x k ). Subsiue he given values ino he Arrhenius equaion and solve for T. k ln k Ea R T T k 7.0x0 3 L/ol s T 500. K E a.4x0 5 J/ol k (7.0x0 3 ) L/ol s T unknown ln 7.0x0 3 L/ol s.4x0 5 J/ol 7.0x0 3 L/ol s 8.34 J/ol K 5.0550x0 5 K 500. K T T. 9494489x0 3 K T 53 K 500. K T 6.9A Plan: Begin by using Saple Proble 6.9 as a guide for labeling he diagra. The reacion energy diagra indicaes ha O(g) + H O(g) OH(g) is an endoheric process, because he energy of he produc is higher han he energy of he reacans. The highes poin on he curve indicaes he 6-6
ransiion sae. In he ransiion sae, an oxygen ao fors a bond wih one of he hydrogen aos on he H O olecule (hashed line) and he O-H bond (dashed line) in H O weakens. E a(fwd) is he su of H rxn and E a(rev). Transiion sae: O H O H Energy E a(fwd) +78 kj E a(rev) +6 kj H rxn +7 kj Reacion progress 6.9B Plan: E a(fwd) is he su of H rxn and E a(rev), so H rxn can be calculaed by subracing E a(fwd) E a(rev). Use Saple Proble 6.9 as a guide for drawing he diagra. H rxn E a(fwd) E a(rev) H rxn 7 kj 7 kj 65 kj The negaive sign of he enhalpy indicaes ha he reacion is exoheric. This eans ha he energy of he producs is lower han he energy of he reacans. The highes poin on he curve indicaes he ransiion sae. In he ransiion sae, a bond is foring beween he chlorine and he hydrogen (hashed line) while he bond beween he hydrogen and he broine weakens (anoher hashed line): Transiion sae: Cl H Br 6.0A Plan: The overall reacion can be obained by adding he hree seps ogeher. The oleculariy of each sep is he oal nuber of reacan paricles; he oleculariies are used as he orders in he rae law for each sep. a) () H O (aq) OH(aq) () H O (aq) + OH(aq) H O(l) + HO (aq) (3) HO (aq) + OH(aq) H O(l) + O (g) Toal: H O (aq) + OH(aq) + HO (aq) OH(aq) + H O(l) + HO (aq) + O (g) (overall) H O (aq) H O(l) + O (g) b) () Uniolecular 6-7
() Biolecular (3) Biolecular c) () Rae k [H O ] () Rae k [H O ][OH] (3) Rae 3 k 3 [HO ][OH] 6.0B Plan: The overall reacion can be obained by adding he hree seps ogeher. The oleculariy of each sep is he oal nuber of reacan paricles; he oleculariies are used as he orders in he rae law for each sep. a) () NO(g) N O (g) () N O (g) + H (g) N O(g) + H O(g) (3) N O(g) +H (g) N (g) + H O(g) Toal: NO(g) + N O (g) + H (g) + N O(g) N O (g) + N O(g) + H O(g) + N (g) (overall) NO(g) + H (g) H O(g) + N (g) b) () Biolecular () Biolecular (3) Biolecular c) () Rae k [NO] () Rae k [N O ][H ] (3) Rae 3 k 3 [N O][H ] 6.A Plan: The overall reacion can be obained by adding he hree seps ogeher. An inerediae is a subsance ha is fored in one sep and consued in a subsequen sep. The overall rae law for he echanis is deerined fro he slowes sep (he rae-deerining sep) and can be copared o he experienal rae law. a) () H O (aq) OH(aq) () H O (aq) + OH(aq) H O(l) + HO (aq) (3) HO (aq) + OH(aq) H O(l) + O (g) Toal: H O (aq) + OH(aq) + HO (aq) OH(aq) + H O(l) + HO (aq) + O (g) (overall) H O (aq) H O(l) + O (g) OH(aq) and HO (aq) are inerediaes in he given echanis. OH(aq) are produced in he firs sep and consued in he second and hird seps; HO (aq) is produced in he second sep and consued in he hird sep. Noice ha he inerediaes were no included in he overall reacion. b) The observed rae law is: rae k[h O ]. In order for he echanis o be consisen wih he rae law, he firs sep us be he slow sep. The rae law for sep one is he sae as he observed rae law. 6.B Plan: The overall reacion can be obained by adding he hree seps ogeher. An inerediae is a subsance ha is fored in one sep and consued in a subsequen sep. The overall rae law for he echanis is deerined fro he slowes sep (he rae-deerining sep) and can be copared o he experienal rae law. a) () NO(g) N O (g) () N O (g) + H (g) N O(g) + H O(g) (3) N O(g) +H (g) N (g) + H O(g) Toal: NO(g) + N O (g) + H (g) + N O(g) N O (g) + N O(g) + H O(g) + N (g) (overall) NO(g) + H (g) H O(g) + N (g) N O (g) and N O(g) are inerediaes in he given echanis. N O (g) is produced in he firs sep and consued in he second sep; N O(g) is produced in he second sep and consued in he hird sep. Noice ha he inerediaes were no included in he overall reacion. b) The observed rae law is: rae k[no] [H ], and he second sep is he slow, or rae-deerining, sep. The rae law for sep wo is: rae k [N O ][H ]. This rae law is NOT he sae as he observed rae law. However, since N O is an inerediae, i us be replaced by using he firs sep. For an equilibriu, rae forward rxn rae reverse rxn. For sep hen, k [NO] k [N O ]. Rearranging o solve for [N O ] gives [N O ] (k /k )[NO]. Subsiuing his value for [N O ] ino he rae law for he second sep gives he overall rae law as rae (k k /k )[NO] [H ] or rae k[no] [H ], which is consisen wih he observed rae law. 6-8
CHEMICAL CONNECTIONS BOXED READING PROBLEMS B6. Plan: Add he wo equaions, canceling subsances ha appear on boh sides of he arrow. The rae law for each sep follows he fora of rae k[reacans]. An iniial reacan ha appears as a produc in a subsequen sep is a caalys; a produc ha appears as a reacan in a subsequen sep is an inerediae (produced in one sep and consued in a subsequen sep). a) Add he wo equaions: () X(g) + O 3 (g) XO(g) + O (g) () XO(g) + O(g) X(g) + O (g) Overall O 3 (g) + O(g) O (g) Rae laws: Sep Rae k [X][O 3 ] Sep Rae k [XO][O] b) X acs as a caalys; i is a reacan in sep and a produc in sep. XO acs as an inerediae; i was produced in sep and consued in sep. B6. Plan: Replace X in he echanis in B6. wih NO, he caalys. To find he rae of ozone depleion a he given concenraions, use sep ) since i is he rae-deerining (slow) sep. a) () NO(g) + O 3 (g) NO (g) + O (g) () NO (g) + O(g) NO(g) + O (g) Overall O 3 (g) + O(g) O (g) b) Rae k [NO][O 3 ] (6x0 5 c 3 /olecule s)[.0x0 9 olecule/c 3 ][5x0 olecule/c 3 ] 3x0 7 olecule/s B6.3 Plan: The p facor is he orienaion probabiliy facor and is relaed o he srucural coplexiy of he colliding paricles. The ore coplex he paricles, he saller he probabiliy ha collisions will occur wih he correc orienaion and he saller he p facor. a) Sep wih Cl will have he higher value for he p facor. Since Cl is a single ao, no aer ` how i collides wih he ozone olecule, he wo paricles should reac, if he collision has enough energy. NO is a olecule. If he O 3 olecule collides wih he N ao in he NO olecule, reacion can occur as a bond can for beween N and O; if he O 3 olecule collides wih he O ao in he NO olecule, reacion will no occur as he bond beween N and O canno for. The probabiliy of a successful collision is saller wih NO. b) The ransiion sae would have weak bonds beween he chlorine ao and an oxygen ao in ozone, and beween ha oxygen ao and a second oxygen ao in ozone. Cl O O O END OF CHAPTER PROBLEMS 6. Changes in concenraions of reacans (or producs) as funcions of ie are easured o deerine he reacion rae. 6. Rae is proporional o concenraion. An increase in pressure will increase he nuber of gas olecules per uni volue. In oher words, he gas concenraion increases due o increased pressure, so he reacion rae increases. Increased pressure also causes ore collisions beween gas olecules. 6.3 The addiion of ore waer will dilue he concenraions of all solues dissolved in he reacion vessel. If any of hese solues are reacans, he rae of he reacion will decrease. 6-9
6.4 An increase in solid surface area would allow ore gaseous coponens o reac per uni ie and hus would increase he reacion rae. 6.5 An increase in eperaure affecs he rae of a reacion by increasing he nuber of collisions, bu ore iporanly, he energy of collisions increases. As he energy of collisions increases, ore collisions resul in reacion (i.e., reacans producs), so he rae of reacion increases. 6.6 The second experien proceeds a he higher rae. I in he gaseous sae would experience ore collisions wih gaseous H. 6.7 The reacion rae is he change in he concenraion of reacans or producs per uni ie. Reacion raes change wih ie because reacan concenraions decrease, while produc concenraions increase wih ie. 6.8 a) For os reacions, he rae of he reacion changes as a reacion progresses. The insananeous rae is he rae a one poin, or insan, during he reacion. The average rae is he average of he insananeous raes over a period of ie. On a graph of reacan concenraion vs. ie of reacion, he insananeous rae is he slope of he angen o he curve a any one poin. The average rae is he slope of he line connecing wo poins on he curve. The closer ogeher he wo poins (shorer he ie inerval), he ore closely he average rae agrees wih he insananeous rae. b) The iniial rae is he insananeous rae a he poin on he graph where ie 0, ha is when reacans are ixed. 6.9 The calculaion of he overall rae is he difference beween he forward and reverse raes. This coplicaion ay be avoided by easuring he iniial rae, where produc concenraions are negligible, so he reverse rae is negligible. Addiionally, he calculaions are siplified as he reacan concenraions can easily be deerined fro he volues and concenraions of he soluions ixed. 6.0 A ie 0, no produc has fored, so he B(g) curve us sar a he origin. Reacan concenraion (A(g)) decreases wih ie; produc concenraion (B(g)) increases wih ie. Many correc graphs can be drawn. Two exaples are shown below. The graph on he lef shows a reacion ha proceeds nearly o copleion, i.e., [producs] >> [reacans] a he end of he reacion. The graph on he righ shows a reacion ha does no proceed o copleion, i.e., [reacans] > [producs] a reacion end. B(g) Concenraion A(g) Concenraion A(g) B(g) Tie Tie 6. a) Calculae he slope of he line connecing (0, [C] o ) and ( f, [C] f ) (final ie and concenraion). The negaive of his slope is he average rae. b) Calculae he negaive of he slope of he line angen o he curve a x. c) Calculae he negaive of he slope of he line angen o he curve a 0. d) If you ploed [D] vs. ie, you would no need o ake he negaive of he slopes in a)-c) since [D] would increase over ie. 6. Plan: The average rae is he oal change in concenraion divided by he oal change in ie. a) The average rae fro 0 o 0.0 s is proporional o he slope of he line connecing hese wo poins: [AX ] ( 0.0088 ol/l 0.0500 ol/l) Rae 0.0003 0.000 ol/l s 0.0 s 0 s ( ) 6-0
The negaive of he slope is used because rae is defined as he change in produc concenraion wih ie. If a reacan is used, he rae is he negaive of he change in reacan concenraion. The / facor is included o accoun for he soichioeric coefficien of for AX in he reacion. b) [AX ] vs ie 0.06 0.05 0.04 [AX ] 0.03 0.0 0.0 0 0 5 0 5 0 5 ie, s The slope of he angen o he curve (dashed line) a 0 is approxiaely 0.004 ol/l s. This iniial rae is greaer han he average rae as calculaed in par a). The iniial rae is greaer han he average rae because rae decreases as reacan concenraion decreases. 6.3 Plan: The average rae is he oal change in concenraion divided by he oal change in ie. [AX ] ( 0.0088 ol/l 0.049 ol/l) a) Rae 6.70833x0 ( 0.0 s 8.0 s) 4 6.7x0 4 ol/l s b) The rae a exacly 5.0 s will be higher han he rae in par a). 0.06 0.05 0.04 [AX ] 0.03 0.0 0.0 0 0 5 0 5 0 5 ie, s The slope of he angen o he curve a 5.0 s (he rae a 5.0 s) is approxiaely.8x0 3 ol/l s. 6-
6.4 Plan: Use Equaion 6. o describe he rae of his reacion in ers of reacan disappearance and produc appearance. A negaive sign is used for he rae in ers of reacan A since A is reacing and [A] is decreasing over ie. Posiive signs are used for he rae in ers of producs B and C since B and C are being fored and [B] and [C] increase over ie. Reacan A decreases wice as fas as produc C increases because wo olecules of A disappear for every olecule of C ha appears. Expressing he rae in ers of each coponen: [A] [B] [C] Rae Calculaing he rae of change of [A]: [A] [C] ol A/L s ( ol C/L s) ol C/L s 4 ol/l s The negaive value indicaes ha [A] is decreasing as he reacion progresses. The rae of reacion is always expressed as a posiive nuber, so [A] is decreasing a a rae of 4 ol/l s. 6.5 Plan: Use Equaion 6. o describe he rae of his reacion in ers of reacan disappearance and produc appearance. A negaive sign is used for he rae in ers of reacan D since D is reacing and [D] is decreasing over ie. Posiive signs are used for he rae in ers of producs E and F since E and F are being fored and [E] and [F] increase over ie. For every 3/ ole of produc E ha is fored, 5/ ole of F is produced. Expressing he rae in ers of each coponen: [D] [E] [F] Rae 3 5 Calculaing he rae of change of [F]: 5/ ol F/L s ( 0.5 ol E/L s) 3/ ol E/L s 0.46667 0.4 ol/l s 6.6 Plan: Use Equaion 6. o describe he rae of his reacion in ers of reacan disappearance and produc appearance. A negaive sign is used for he rae in ers of reacans A and B since A and B are reacing and [A] and [B] are decreasing over ie. A posiive sign is used for he rae in ers of produc C since C is being fored and [C] increases over ie. The / facor is included for reacan B o accoun for he soichioeric coefficien of for B in he reacion. Reacan A decreases half as fas as reacan B decreases because one olecule of A disappears for every wo olecules of B ha disappear. Expressing he rae in ers of each coponen: [A] [B] [C] Rae Calculaing he rae of change of [A]: ol A/L s ( 0.5 ol B/L s) ol B/L s 0.5 ol/l s 0. ol/l s The negaive value indicaes ha [A] is decreasing as he reacion progresses. The rae of reacion is always expressed as a posiive nuber, so [A] is decreasing a a rae of 0. ol/l s. 6.7 Plan: Use Equaion 6. o describe he rae of his reacion in ers of reacan disappearance and produc appearance. A negaive sign is used for he rae in ers of reacans D, E, and F since hese subsances are reacing and [D], [E], and [F] are decreasing over ie. Posiive signs are used for he rae in ers of producs G and H since hese subsances are being fored and [G] and [H] increase over ie. Produc H increases half as fas as reacan D decreases because one olecule of H is fored for every wo olecules of D ha disappear. 6-
Expressing he rae in ers of each coponen: [D] [E] [F] Rae [G] 3 Calculaing he rae of change of [H]: ol H/L s ( 0. ol D/L s) ol D/L s 0.05 ol/l s [H] 6.8 Plan: A er wih a negaive sign is a reacan; a er wih a posiive sign is a produc. The inverse of he fracion becoes he coefficien of he olecule. N O 5 is he reacan; NO and O are producs. N O 5 (g) 4NO (g) + O (g) 6.9 Plan: A er wih a negaive sign is a reacan; a er wih a posiive sign is a produc. The inverse of he fracion becoes he coefficien of he olecule. CH 4 and O are he reacans; H O and CO are producs. CH 4 + O H O + CO 6.0 Plan: The average rae is he oal change in concenraion divided by he oal change in ie. The iniial rae is he slope of he angen o he curve a 0.0 s and he rae a 7.00 s is he slope of he angen o he curve a 7.00 s [NOBr] 0.0033 0.000 ol/l a) Rae 6.7x0 4 ol/l s 0.00 0.00 s 0.0055 0.007 ol/l b) Rae 8.0x0 4 ol/l s 4.00.00 s c) Iniial Rae y/ x [(0.0040 0.000) ol/l]/[4.00 0.00) s].5x0 3 ol/l s d) Rae a 7.00 s [(0.0030 0.0050) ol/l]/[.00 4.00) s].857x0 4.9x0 4 ol/l s e) Average beween 3 s and 5 s is: 6-3
Rae [(0.0050 0.0063) ol/l]/[5.00 3.00) s] 6.5x0 4 ol/l s Rae a 4 s 6.7x0 4 ol/l s, hus he raes are equal a abou 4 seconds. 6. Plan: Use Equaion 6. o describe he rae of his reacion in ers of reacan disappearance and produc appearance. A negaive sign is used for he rae in ers of reacans N and H since hese subsances are reacing and [N ] and [H ] are decreasing over ie. A posiive sign is used for he rae in ers of he produc NH 3 since i is being fored and [NH 3 ] increases over ie. [N ] [H ] [NH 3] Rae 3 6. Plan: Use Equaion 6. o describe he rae of his reacion in ers of reacan disappearance and produc appearance. A negaive sign is used for he rae in ers of he reacan O since i is reacing and [O ] is decreasing over ie. A posiive sign is used for he rae in ers of he produc O 3 since i is being fored and [O 3 ] increases over ie. O 3 increases /3 as fas as O decreases because wo olecules of O 3 are fored for every hree olecules of O ha disappear. [O ] [O 3] a) Rae 3 b) Use he ole raio in he balanced equaion: 5.7x0 ol O /L s ol O 3 /L s.45x0 5 ol/l s 3 ol O /L s 6.3 a) k is he rae consan, he proporionaliy consan in he rae law. k represens he fracion of successful collisions which includes he fracion of collisions wih sufficien energy and he fracion of collisions wih correc orienaion. k is a consan ha varies wih eperaure. b) represens he order of he reacion wih respec o [A] and n represens he order of he reacion wih respec o [B]. The order is he exponen in he relaionship beween rae and reacan concenraion and defines how reacan concenraion influences rae. The order of a reacan does no necessarily equal is soichioeric coefficien in he balanced equaion. If a reacion is an eleenary reacion, eaning he reacion occurs in only one sep, hen he orders and soichioeric coefficiens are equal. However, if a reacion occurs in a series of eleenary reacions, called a echanis, hen he rae law is based on he slowes eleenary reacion in he echanis. The orders of he reacans will equal he soichioeric coefficiens of he reacans in he slowes eleenary reacion bu ay no equal he soichioeric coefficiens in he overall reacion. c) For he rae law rae k[a] [B] subsiue in he unis: Rae (ol/l in) k[a] [B] rae ol/l in ol/l in k 3 [A] [B] ol ol ol L L 3 L 3 ol L k 3 L in ol k L /ol in 6.4 a) Plo eiher [A ] or [B ] vs. ie and deerine he negaive of he slope of he line angen o he curve a 0. b) A series of experiens a consan eperaure bu wih differen iniial concenraions are run o deerine differen iniial raes. By coparing resuls in which only he iniial concenraion of a single reacan is changed, he order of he reacion wih respec o ha reacan can be deerined. c) When he order of each reacan is known, any one experienal se of daa (reacan concenraion and reacion rae) can be used o deerine he reacion rae consan a ha eperaure. 6-4
6.5 a) The rae doubles. If rae k[a] and [A] is doubled, hen he rae law becoes rae k[ x A]. The rae increases by or. b) The rae decreases by a facor of four. If rae k[b] and [B] is halved, hen he rae law becoes rae k[/ x B]. The rae decreases o (/) or /4 of is original value. c) The rae increases by a facor of nine. If rae k[c] and [C] is ripled, hen he rae law becoes rae k[3 x C]. The rae increases o 3 or 9 ies is original value. 6.6 Plan: The order for each reacan is he exponen on he reacan concenraion in he rae law. The individual orders are added o find he overall reacion order. The orders wih respec o [BrO 3 ] and o [Br ] are boh since boh have an exponen of. The order wih respec o [H + ] is (is exponen in he rae law is ). The overall reacion order is + + 4. firs order wih respec o BrO 3, firs order wih respec o Br, second order wih respec o H +, fourh order overall 6.7 Plan: The order for each reacan is he exponen on he reacan concenraion in he rae law. The individual orders are added o find he overall reacion order. The rae law ay be rewrien as rae k[o 3 ] [O ]. The order wih respec o [O 3 ] is since i has an exponen of. The order wih respec o [O ] is since i has an exponen of. The overall reacion order is + ( ). second order wih respec o O 3, ( ) order wih respec o O, firs order overall 6.8 a) The rae is firs order wih respec o [BrO 3 ]. If [BrO 3 ] is doubled, rae k[ x BrO 3 ], hen rae increases o or ies is original value. The rae doubles. b) The rae is firs order wih respec o [Br ]. If [Br ] is halved, rae k[/ x Br ], hen rae decreases by a facor of (/) or / ies is original value. The rae is halved. c) The rae is second order wih respec o [H + ]. If [H + ] is quadrupled, rae k[4 x H + ], hen rae increases o 4 or 6 ies is original value. 6.9 a) The rae is second order wih respec o [O 3 ]. If [O 3 ] is doubled, rae k[ x O 3 ], hen rae increases o or 4 ies is original value. The rae increases by a facor of 4. b) [O ] has an order of. If [O ] is doubled, rae k[ x O ], hen rae decreases o or / ies is original value. The rae decreases by a facor of. c) [O ] has an order of. If [O ] is halved, rae k[/ x O ], hen rae decreases by a facor of (/) or ies is original value. The rae increases by a facor of. 6.30 Plan: The order for each reacan is he exponen on he reacan concenraion in he rae law. The individual orders are added o find he overall reacion order. The order wih respec o [NO ] is, and he order wih respec o [Cl ] is. The overall order is: + 3 for he overall order. 6.3 Plan: The order for each reacan is he exponen on he reacan concenraion in he rae law. The individual orders are added o find he overall reacion order. The rae law ay be rewrien as rae k[hno ] 4 [NO]. The order wih respec o [HNO ] is 4, and he order wih respec o [NO] is. The overall order is: 4 + ( ) for he overall order. 6.3 a) The rae is second order wih respec o [NO ]. If [NO ] is ripled, rae k[3 x NO ], hen rae increases o 3 or 9 ies is original value. The rae increases by a facor of 9. b) The rae is second order wih respec o [NO ] and firs order wih respec o [Cl ]. If [NO ] and [Cl ] are doubled, rae k[ x NO ] [ x Cl ], hen he rae increases by a facor of x 8. c) The rae is firs order wih respec o [Cl ]. If Cl is halved, rae k[/ x Cl ], hen rae decreases o / ies is original value. The rae is halved. 6-5
6.33 a) The rae is fourh order wih respec o [HNO ]. If [HNO ] is doubled, rae k[ x HNO ] 4, hen rae increases o 4 or 6 ies is original value. The rae increases by a facor of 6. b) [NO] has an order of. If [NO] is doubled, rae k[ x NO], hen rae increases o or /() /4 ies is original value. The rae decreases by a facor of 4. c) The rae is fourh order wih respec o [HNO ]. If [HNO ] is halved, rae k[/ x HNO ] 4, hen rae decreases o (/) 4 or /6 ies is original value. The rae decreases by a facor of 6. 6.34 Plan: The rae law is rae [A] [B] n where and n are he orders of he reacans. To find he order of each reacan, ake he raio of he rae laws for wo experiens in which only he reacan in quesion changes. Once he rae law is known, any experien can be used o find he rae consan k. a) To find he order for reacan A, firs idenify he reacion experiens in which [A] changes bu [B] is consan. Use experiens and (or 3 and 4 would work) o find he order wih respec o [A]. Se up a raio of he rae laws for experiens and and fill in he values given for raes and concenraions and solve for, he order wih respec o [A]. rae exp [A] exp rae exp [A] exp 45.0 ol/l in 0.300 ol/l 5.00 ol/l in 0.00 ol/l 9.00 (3.00) log (9.00) log (3.00) Using experiens 3 and 4 also gives second order wih respec o [A]. To find he order for reacan B, firs idenify he reacion experiens in which [B] changes bu [A] is consan. Use experiens and 3 (or and 4 would work) o find he order wih respec o [B]. Se up a raio of he rae laws for experiens and 3 and fill in he values given for raes and concenraions and solve for n, he order wih respec o [B]. n rae exp 3 [B] exp 3 rae exp [B] exp n 0.0 ol/l in 0.00 ol/l 5.00 ol/l in 0.00 ol/l.00 (.00) n log (.00) n log (.00) n The reacion is firs order wih respec o [B]. b) The rae law, wihou a value for k, is rae k[a] [B]. c) Using experien o calculae k (he daa fro any of he experiens can be used): Rae k[a] [B] rae k [A] [B] 5.00 ol/l in [0.00 ol/l] [0.00 ol/l] 5.00x03 L /ol in 6.35 Plan: The rae law is rae k [A] [B] n [C] p where, n, and p are he orders of he reacans. To find he order of each reacan, ake he raio of he rae laws for wo experiens in which only he reacan in quesion changes. Once he rae law is known, any experien can be used o find he rae consan k. a) To find he order for reacan A, firs idenify he reacion experiens in which [A] changes bu [B] and [C] are consan. Use experiens and o find he order wih respec o [A]. Se up a raio of he rae laws for experiens and and fill in he values given for raes and concenraions and solve for, he order wih respec o [A]. 6-6
rae exp [A] exp rae exp [A] exp.5x0 ol/l in 0.000 ol/l 3 6.5x0 ol/l in 0.0500 ol/l.00 (.00) log (.00) log (.00) The order is firs order wih respec o A. To find he order for reacan B, firs idenify he reacion experiens in which [B] changes bu [A] and [C] are consan. Use experiens and 3 o find he order wih respec o [B]. Se up a raio of he rae laws for experiens and 3 and fill in he values given for raes and concenraions and solve for n, he order wih respec o [B]. n rae exp 3 [B] exp 3 rae exp [B] exp n 5.00 x 0 ol/l in 0.000 ol/l.5 x 0 ol/l in 0.0500 ol/l 4.00 (.00) n log (4.00) n log (.00) n The reacion is second order wih respec o B. To find he order for reacan C, firs idenify he reacion experiens in which [C] changes bu [A] and [B] are consan. Use experiens and 4 o find he order wih respec o [C]. Se up a raio of he rae laws for experiens and 4 and fill in he values given for raes and concenraions and solve for p, he order wih respec o [C]. p rae exp 4 [C] exp 4 rae exp [C] exp 3 p 6.5x0 ol/l in 0.000 ol/l 3 6.5x0 ol/l in 0.000 ol/l.00 (.00) p log (.00) p log (.00) p 0 The reacion is zero order wih respec o C. b) Rae k[a] [B] [C] 0 Rae k[a][b] c) Using he daa fro experien o find k: Rae k[a][b] 3 rae 6.5 x 0 ol/l in k 50.0 L /ol s [A][B] [0.0500 ol/l][0.0500 ol/l] 6.36 Plan: Wrie he appropriae rae law and ener he unis for rae and concenraions o find he unis of k. The unis of k are dependen on he reacion orders and he uni of ie. a) A firs-order rae law follows he general expression, rae k[a]. The reacion rae is expressed as a change in concenraion per uni ie wih unis of ol/l ie. Since [A] has unis of ol/l, k has unis of ie : Rae k[a] ol ol k L ie k ol L ie ol L L ol x L ie L ol ie ie 6-7
b) A second-order rae law follows he general expression, rae k[a]. The reacion rae is expressed as a change in concenraion per uni ie wih unis of ol/l ie. Since [A] has unis of ol /L, k has unis of L/ol ie: Rae k[a] ol ol k L ie L ol k L ie ol L L x ol L ie ol ol ie L c) A hird-order rae law follows he general expression, rae k[a] 3. The reacion rae is expressed as a change in concenraion per uni ie wih unis of ol/l ie. Since [A] has unis of ol 3 /L 3, k has unis of L /ol ie: Rae k[a] 3 3 ol ol k L ie L ol 3 k L ie ol L L x 3 3 ol L ie ol ol ie 3 L d) A 5/-order rae law follows he general expression, rae k[a] 5/. The reacion rae is expressed as a change in concenraion per uni ie wih unis of ol/l ie. Since [A] has unis of ol 5/ /L 5/, k has unis of L 3/ /ol 3/ ie: 5/ ol ol k L ie L ol 5/ 3/ k L ie ol L L x 5/ 5/ 3/ ol L ie ol ol ie 5/ L 6.37 Plan: Wrie he appropriae rae law and ener he unis for rae and he rae consan o find he unis of concenraion. The unis of concenraion will give he reacion order. a) Rae k[a] ol ol ol L s L s L ol ol L s L ol L s ol L us be 0. The reacion is zero order. b) Rae k[a] ol ol L yr yr L 6-8
ol L ol L ol L yr yr ol L ol yr x L yr us be. The reacion is firs order. c) Rae k[a] / ol ol ol / L s L s L ol / ol L s ol L s / L x / ol L s ol / L s ol L ol L / / us be /. The reacion is / order. d) Rae k[a] 5/ ol ol ol 5/ L in L in L ol 5/ ol L in ol L in 5/ L x 5/ ol L in ol 5/ L in 7/ ol ol 7/ L us be 7/. The reacion is 7/ order. L 6.38 Plan: The rae law is rae k [CO] [Cl ] n where and n are he orders of he reacans. To find he order of each reacan, ake he raio of he rae laws for wo experiens in which only he reacan in quesion changes. Once he rae law is known, he daa in each experien can be used o find he rae consan k. a) To find he order for CO, firs idenify he reacion experiens in which [CO] changes bu [Cl ] is consan. Use experiens and o find he order wih respec o [CO]. Se up a raio of he rae laws for experiens and and fill in he values given for raes and concenraions and solve for, he order wih respec o [CO]. rae exp [CO] exp rae exp [CO] exp 9.9x0 ol/l in.00 ol/l 30.33x0 ol/l in 0.00 ol/l 9.699 (0.0) log (9.699) log (0.0) 0.9867 The reacion is firs order wih respec o [CO]. To find he order for Cl, firs idenify he reacion experiens in which [Cl ] changes bu [CO] is consan. Use experiens and 3 o find he order wih respec o [Cl ]. Se up a raio of he rae laws for experiens and 3 and fill in he values given for raes and concenraions and solve for n, he order wih respec o [Cl ]. 6-9
n rae exp 3 [Cl ] exp 3 rae exp [Cl ] exp 9 n.30x0 ol/l in.00 ol/l 30.33x0 ol/l in 0.00 ol/l 9.774 (0.0) n log (9.774) n log (0.0) n 0.990 The reacion is firs order wih respec o [Cl ]. Rae k[co][cl ] b) k rae/[co][cl ] Exp : k (.9x0 9 ol/l s)/[.00 ol/l][0.00 ol/l].9x0 8 L/ol s Exp : k (.33x0 30 ol/l s)/[0.00 ol/l][0.00 ol/l].33x0 8 L/ol s Exp 3: k 3 (.30x0 9 ol/l s)/[0.00 ol/l][.00 ol/l].30x0 8 L/ol s Exp 4: k 4 (.3x0 3 ol/l s)/[0.00 ol/l][0.000 ol/l].3x0 8 L/ol s k avg (.9x0 8 +.33x0 8 +.30x0 8 +.3x0 8 ) L/ol s/4.3x0 8 L/ol s 6.39 The inegraed rae law can be used o plo a graph. If he plo of [reacan] vs. ie is linear, he order is zero. If he plo of ln[reacan] vs. ie is linear, he order is firs. If he plo of inverse concenraion (/[reacan]) vs. ie is linear, he order is second. a) The reacion is firs order since ln[reacan] vs. ie is linear. b) The reacion is second order since /[reacan] vs. ie is linear. c) The reacion is zero order since [reacan] vs. ie is linear. 6.40 The half-life ( / ) of a reacion is he ie required o reach half he iniial reacan concenraion. For a firs-order process, no olecular collisions are necessary, and he rae depends onlyon he fracion of he olecules having sufficien energy o iniiae he reacion. 6.4 Plan: The rae expression indicaes ha his reacion is second order overall (he order of [AB] is ), so use he second-order inegraed rae law o find ie. We know k (0. L/ol s), [AB] 0 (.50 M), and [AB] (/3[AB] 0 /3(.50 M) 0.500 M), so we can solve for. k AB AB [ ] [ ] 0 [ AB] [ AB] 0 k 0.500 M.50 M 0. L/ol s 6.6667 7 s 6.4 Plan: The rae expression indicaes ha his reacion is second order overall (he order of [AB] is ), so use he second-order inegraed rae law. We know k (0. L/ol s), [AB] 0 (.50 M), and (0.0 s), so we can solve for [AB]. k AB AB [ ] [ ] 0 [ ] k + AB [ AB ]0 6-0
(0. L/ol s) (0.0 s) + [ AB] [ AB].66667 M [AB] 0.375 0.4 M.50 M 6.43 Plan: This is a firs-order reacion so use he firs-order inegraed rae law. In par a), we know (0.5 in). Le [A] 0 M and hen [A] 50% of M 0.5 M. Solve for k. In par b), use he value of k o find he ie necessary for 75.0% of he copound o reac. If 75.0% of he copound has reaced, 00 75 5% reains a ie. Le [A] 0 M and hen [A] 5% of M 0.5 M. a) ln [A] ln [A] 0 k ln [0.5] ln [] k(0.5 in) 0.69347 0 k(0.5 in) 0.69347 k(0.5 in) k 0.0660 in Alernaively, 50.0% decoposiion eans ha one half-life has passed. Thus, he firs-order half-life equaion ay be used: ln ln ln / k 0.06604 0.0660 in k / 0.5 in b) ln [A] ln [A] 0 k ln[a] ln[a] 0 k ln[0.5] ln[] 0.0660 in.0045.0 in If you recognize ha 75.0% decoposiion eans ha wo half-lives have passed, hen (0.5 in).0 in. 6.44 Plan: This is a firs-order reacion so use he firs-order inegraed rae law (he unis of k, yr, indicaes firs order). In par a), he firs-order half-life equaion ay be used o solve for half-life since k is known. In par b), use he value of k o find he ie necessary for he reacan concenraion o drop o.5% of he iniial concenraion. Le [A] 0.00 M and hen [A].5% of M 0.5 M. ln ln a) / 577.6 5.8x0 yr k 0.00 yr b) ln [A] ln [A] 0 k [A] ln 0 k [A] [A] 0.00 M [A] 0.5 M k 0.00 yr ln.00 M 0.5 M (0.00 yr ) 73.86795.7x0 3 yr If he suden recognizes ha.5% reaining corresponds o hree half-lives; hen siply uliply he answer in par a) by hree. 6.45 Plan: In a firs-order reacion, ln [NH 3 ] vs. ie is a sraigh line wih slope equal o k. The half-life can be deerined using he firs-order half-life equaion. 6-
a) A new daa able is consruced: (Noe ha addiional significan figures are reained in he calculaions.) x-axis (ie, s) [NH 3 ] y-axis (ln [NH 3 ]) 0 4.000 M.3869.000 3.986 M.3879.000 3.974 M.37977 ln[nh 3 ].387.386.385.384.383.38.38.38.379 0 0.5.5 Tie, s k slope rise/run (y y )/(x x ) k (.37977.3869)/(.000 0) (0.0065)/() 3.60x0 3 s 3x0 3 s (Noe ha he saring ie is no exac, and hence, liis he significan figures.) ln ln b) / 3 k 3.60x0 s.6 x0 s 6.46 The cenral idea of collision heory is ha reacans us collide wih each oher in order o reac. If reacans us collide o reac, he rae depends on he produc of he reacan concenraions. 6.47 No, collision frequency is no he only facor affecing reacion rae. The collision frequency is a coun of he oal nuber of collisions beween reacan olecules. Only a sall nuber of hese collisions lead o a reacion. Oher facors ha influence he fracion of collisions ha lead o reacion are he energy and orienaion of he collision. A collision us occur wih a iniu energy (acivaion energy) o be successful. In a collision, he orienaion, ha is, which ends of he reacan olecules collide, us bring he reacing aos in he olecules ogeher in order for he collision o lead o a reacion. 6.48 A any paricular eperaure, olecules have a disribuion of kineic energies, as will heir collisions have a range of energies. As eperaure increases, he fracion of hese collisions which exceed he hreshold energy, increases; hus, he reacion rae increases. / 6.49 k Ae E a RT The Arrhenius equaion indicaes a negaive exponenial relaionship beween eperaures and he rae consan, k. In oher words, he rae consan increases exponenially wih eperaure. / 6.50 The Arrhenius equaion, k A e E a RT, can be used direcly o solve for acivaion energy a a specified eperaure if he rae consan, k, and he frequency facor, A, are known. However, he frequency facor is usually no known. To find E a wihou knowing A, rearrange he Arrhenius equaion o pu i in he for of a linear plo: ln k ln A E a /RT where he y value is ln k and he x value is /T. Measure he rae consan a a series of eperaures and plo ln k vs. /T. The slope equals E a /R. 6-
6.5 a) The value of k increases exponenially wih eperaure. b) A plo of ln k vs. /T is a sraigh line whose slope is E a /R. a) b) The acivaion energy is deerined fro he slope of he line in he ln k vs. /T graph. The slope equals E a /R. 6.5 a) As eperaure increases, he fracion of collisions which exceed he acivaion energy increases; hus, he reacion rae increases. b) A decrease in acivaion energy lowers he energy hreshold wih which collisions us ake place o be effecive. A a given eperaure, ore collisions occur wih he lower energy so rae increases. 6.53 No. For 4x0 5 oles of EF o for, every collision us resul in a reacion and no EF olecule can decopose back o AB and CD. Neiher condiion is likely. All collisions will no resul in produc as soe collisions will occur wih an energy ha is lower han he acivaion energy. In principle, all reacions are reversible, so soe EF olecules decopose. Even if all AB and CD olecules did cobine, he reverse decoposiion rae would resul in an aoun of EF ha is less han 4x0 5 oles. 6.54 Collision frequency is proporional o he velociy of he reacan olecules. A he sae eperaure, boh reacion ixures have he sae average kineic energy, bu no he sae velociy. Kineic energy equals / v, where is ass and v velociy. The ehylaine (N(CH 3 ) 3 ) olecule has a greaer ass han he aonia olecule, so ehylaine olecules will collide less ofen han aonia olecules, because of heir slower velociies. Collision energy hus is less for he N(CH 3 ) 3 (g) + HCl(g) reacion han for he NH 3 (g) + HCl(g) reacion. Therefore, he rae of he reacion beween aonia and hydrogen chloride is greaer han he rae of he reacion beween ehylaine and hydrogen chloride. 6-3
The fracion of successful collisions also differs beween he wo reacions. In boh reacions he hydrogen fro HCl is bonding o he nirogen in NH 3 or N(CH 3 ) 3. The difference beween he reacions is in how easily he H can collide wih he N, he correc orienaion for a successful reacion. The groups (H) bonded o nirogen in aonia are less bulky han he groups bonded o nirogen in riehylaine (CH 3 ). So, collisions wih correc orienaion beween HCl and NH 3 occur ore frequenly han beween HCl and N(CH 3 ) 3 and he reacion NH 3 (g) + HCl(g) NH 4 Cl(s) occurs a a higher rae han N(CH 3 ) 3 (g) + HCl(g) (CH 3 ) 3 NHCl(s). Therefore, he rae of he reacion beween aonia and hydrogen chloride is greaer han he rae of he reacion beween ehylaine and hydrogen chloride. 6.55 Each A paricle can collide wih hree B paricles, so (4 x 3) unique collisions are possible. 6.56 Plan: Use Avogadro s nuber o conver oles of paricles o nuber of paricles. The nuber of unique collisions is he produc of he nuber of A paricles and he nuber of B paricles. 3 6.0x0 A paricles Nuber of paricles of A (.0 ol A) 6.08x0 ol A 3 paricles of A 3 6.0x0 B paricles Nuber of paricles of B (. ol B).79997x0 ol B 4 paricles of B Nuber of collisions (6.08 x 0 3 paricles of A)(.79997 x 0 4 paricles of B) 7.76495x0 47 7.76x0 47 unique collisions 6.57 / Plan: The fracion of collisions wih a specified energy is equal o he e E a er in he Arrhenius equaion. / f e E a T 5 C + 73 98 K E a 00. kj/ol R 8.34 J/ol K 8.34x0 3 kj/ol K a 00. kj/ ol 8.34x0 3 kj/ ol K 98 K 40.36096 E RT ( )( ) / Fracion e E a RT e 40.36096.9577689x0 8.96x0 8 6.58 / Plan: The fracion of collisions wih a specified energy is equal o he e E a er in he Arrhenius equaion. / f e E a T 50. C + 73 33 K E a 00. kj/ol R 8.34 J/ol K 8.34x0 3 kj/ol K a 00. kj/ ol 8.34x0 3 kj/ ol K 33 K 37.38095 E RT ( )( ) / Fracion e E a RT e 37.38095 6.753x0 7 The fracion increased by (6.753x0 7 )/(.9577689x0 8 ).73775.7 6.59 Plan: You are given one rae consan k a one eperaure T and he acivaion energy E a. Subsiue hese values ino he Arrhenius equaion and solve for k a he second eperaure. k 4.7x0 3 s T 5 C + 73 98 K k? T 75 C + 73 348 K E a 33.6 kj/ol 33,600 J/ol ln k Ea k R T T k 33,600 J/ol ln 3 4.7x0 s 8.34 J/ol K 98 K 348 K 6-4
k ln.94855 (unrounded) Raise each side o ex 3 4.7x0 s k 3 4.7x0 s 7.08577 k (4.7x0 3 s )(7.08577) 0.039858 0.033 s 6.60 Plan: You are given he rae consans, k and k, a wo eperaures, T and T. Subsiue hese values ino he Arrhenius equaion and solve for E a. k 4.50x0 5 L/ol s T 95 C + 73 468 K k 3.0x0 3 L/ol s T 58 C + 73 53 K E a? ln k Ea k R T T k 3 J 3.0x0 L/ ol s R ln k 8.34 ln 5 ol K 4.50x0 L/ ol s E a T T 468 K 53 K E a.3984658x0 5 J/ol.40x0 5 J/ol 6.6 Plan: The reacion is exoheric (ΔH is negaive), so he energy of he producs us be lower han ha of he reacans. Use he relaionship H rxn E a(fwd) E a(rev) o solve for E a(rev). To draw he ransiion sae, noe ha he bond beween B and C will be breaking while a bond beween C and D will be foring. a) Energy E a (fwd) 5 kj/ol E a (rev) ABC + D H rxn 55 kj/ol AB + CD Reacion coordinae b) H rxn E a(fwd) E a(rev) E a(rev) E a(fwd) H rxn 5 kj/ol ( 55 kj/ol).70x0 kj/ol c) bond foring B A C D bond weakening 6-5
6.6 Plan: The forward acivaion energy E a(fwd) is larger han he reverse acivaion energy E a(rev) which indicaes ha he energy of he producs us be higher han ha of he reacans. Use he relaionship H rxn E a(fwd) E a(rev) o solve for H rxn. To draw he ransiion sae, noe ha he bonds in he A and B olecules will be breaking while bonds beween A and B will be foring. a) b) H rxn E a(fwd) E a(rev) 5 kj/ol 85 kj/ol 40 kj/ol c) A..... B A..... B 6.63 Plan: You are given he rae consans, k and k, a wo eperaures, T and T. Subsiue hese values ino he Arrhenius equaion and solve for E a. k 0.76/s T 77 C + 73 000. K k 0.87/s T 757 C + 73 030. K E a? ln k Ea k R T T k J 0.87 / s R ln k 8.34 ln ol K 0.76 / s E a T T 000. K 030. K E a 3.8585x0 4 J/ol 3.9x0 4 J/ol 6.64 Plan: The reacion is endoheric (ΔH is posiive), so he energy of he producs us be higher han ha of he reacans. Use he relaionship H rxn E a(fwd) E a(rev) o solve for E a(rev). To draw he ransiion sae, noe ha he bond in Cl will be breaking while he bond beween N and Cl will be foring. 6-6