CHAPTER 13 CHEMICAL KINETICS

Transcription

1 CHAPTER 3 CHEMICAL KINETICS 3.5 In general for a reacion aa + bb cc + dd rae Δ[A] Δ[B] Δ[C] Δ[D] a Δ b Δ c Δ d Δ (a) rae rae Δ[H ] Δ[I ] Δ[HI] Δ Δ Δ + Δ 3 Δ Δ Δ[Br ] [BrO ] [H ] [Br ] 5 Δ Δ 6 Δ 3 Δ Noe ha because he reacion is carried ou in he aqueous phase, we do no monior he concenraion of waer. 3.6 (a) rae rae Δ[H ] Δ[O ] Δ[HO] Δ Δ Δ Δ[NH 3] Δ[O ] Δ[NO] Δ[HO] 4 Δ 5 Δ 4 Δ 6 Δ 3.7 Rae Δ[NO] Δ Δ[NO] Δ.66 M/s Δ[NO] Δ[NO ] Δ Δ (a) Δ[NO ] Δ Δ[NO] Δ.66 M /s Δ[O ] Δ Δ[O ].66 M/s.33 M /s Δ 3.8 Sraegy: The rae is defined as he change in concenraion of a reacan or produc wih ime. Each change in concenraion erm is divided by he corresponding soichiomeric coefficien. Terms involving reacans are preceded by a minus sign. Δ[N ] Δ[H ] Δ[NH 3] rae Δ 3 Δ Δ Soluion: (a) If hydrogen is reacing a he rae of.74 M/s, he rae a which ammonia is being formed is

2 34 CHAPTER 3: CHEMICAL KINETICS or Δ[NH 3] Δ[H ] Δ 3 Δ Δ[NH 3] Δ[H ] Δ 3 Δ Δ [NH 3] (.74 /s) Δ M 3.49 M /s The rae a which nirogen is reacing mus be: Δ[N ] Δ[H ] (.74 M/s) Δ 3 Δ 3.5 M /s Will he rae a which ammonia forms always be wice he rae of reacion of nirogen, or is his rue only a he insan described in his problem? 3.5 rae [NH 4 + ][NO ] (3. 4 /M s)(.6 M)(.8 M) 6. 6 M/s 3.6 Assume he rae law has he form: rae [F ] x [ClO ] y To deermine he order of he reacion wih respec o F, find wo experimens in which he [ClO ] is held consan. Compare he daa from experimens and 3. When he concenraion of F is doubled, he reacion rae doubles. Thus, he reacion is firs-order in F. To deermine he order wih respec o ClO, compare experimens and. When he ClO concenraion is quadrupled, he reacion rae quadruples. Thus, he reacion is firs-order in ClO. The rae law is: rae [F ][ClO ] The value of can be found using he daa from any of he experimens. If we ae he numbers from he second experimen we have: 3 rae 4.8 M/s. M s [F ][ClO ] (. M)(.4 M) Verify ha he same value of can be obained from he oher ses of daa. Since we now now he rae law and he value of he rae consan, we can calculae he rae a any concenraion of reacans. rae [F ][ClO ] (. M s )(. M)(. M).4 4 M/s 3.7 By comparing he firs and second ses of daa, we see ha changing [B] does no affec he rae of he reacion. Therefore, he reacion is zero order in B. By comparing he firs and hird ses of daa, we see ha doubling [A] doubles he rae of he reacion. This shows ha he reacion is firs order in A. rae [A]

3 CHAPTER 3: CHEMICAL KINETICS 343 From he firs se of daa: 3. M/s (.5 M).3 s Wha would be he value of if you had used he second or hird se of daa? Should be consan? 3.8 Sraegy: We are given a se of concenraions and rae daa and ased o deermine he order of he reacion and he iniial rae for specific concenraions of X and Y. To deermine he order of he reacion, we need o find he rae law for he reacion. We assume ha he rae law aes he form rae [X] x [Y] y How do we use he daa o deermine x and y? Once he orders of he reacans are nown, we can calculae for any se of rae and concenraions. Finally, he rae law enables us o calculae he rae a any concenraions of X and Y. Soluion: (a) Experimens and 5 show ha when we double he concenraion of X a consan concenraion of Y, he rae quadruples. Taing he raio of he raes from hese wo experimens x y rae5.59 M/s (.4) (.3) 4 rae.7 M/s x y (.) (.3) Therefore, x (.4) x (.) x 4 or, x. Tha is, he reacion is second order in X. Experimens and 4 indicae ha doubling [Y] a consan [X] doubles he rae. Here we wrie he raio as x y rae4.54 M/s (.) (.6) rae.7 M/s x y (.) (.3) Therefore, y (.6) y (.3) y or, y. Tha is, he reacion is firs order in Y. Hence, he rae law is given by: rae [X] [Y] The order of he reacion is ( + ) 3. The reacion is 3rd-order. The rae consan can be calculaed using he daa from any one of he experimens. Rearranging he rae law and using he firs se of daa, we find: rae.53 M/s.6 M s [X] [Y] (. M) (.5 M) Nex, using he nown rae consan and subsiuing he concenraions of X and Y ino he rae law, we can calculae he iniial rae of disappearance of X. rae (.6 M s )(.3 M) (.4 M).38 M/s

4 344 CHAPTER 3: CHEMICAL KINETICS 3.9 (a) second order, zero order, (c).5 order, (d) hird order 3. (a) For a reacion firs-order in A, Rae [A].6 M/s (.35 M).46 s For a reacion second-order in A, Rae [A].6 M/s (.35 M).3 /M s 3. The graph below is a plo of ln P vs. ime. Since he plo is linear, he reacion is s order ln P vs. ime ln P y -.9E-4x E ime (s) Slope.9 4 s 3. Le P be he pressure of ClCO CCl 3 a, and le x be he decrease in pressure afer ime. Noe ha from he coefficiens in he balanced equaion ha he loss of amosphere of ClCO CCl 3 resuls in he formaion of wo amospheres of COCl. We wrie: ClCO CCl 3 COCl Time [ClCO CCl 3 ] [COCl ] P P x x

5 CHAPTER 3: CHEMICAL KINETICS 345 Thus he change (increase) in pressure (ΔP) is x x x. We have: (s) P (mmhg) ΔP x P ClCOCCl3 ln P ClCOCCl3 PClCO CCl If he reacion is firs order, hen a plo of ln P ClCOCCl vs. would be linear. If he reacion is second order, 3 a plo of / P vs. would be linear. The wo plos are shown below. ClCOCCl ln P vs. ime. lnp y -.8E-3x +.74E ime (s).5. /P vs. ime.5 /P ime (s) From he graphs we see ha he reacion mus be firs-order. For a firs-order reacion, he slope is equal o. The equaion of he line is given on he graph. The rae consan is:.8 3 s.

6 346 CHAPTER 3: CHEMICAL KINETICS 3.7 We now ha half of he subsance decomposes in a ime equal o he half-life, /. This leaves half of he compound. Half of wha is lef decomposes in a ime equal o anoher half-life, so ha only one quarer of he original compound remains. We see ha 75% of he original compound has decomposed afer wo half lives. Thus wo half-lives equal one hour, or he half-life of he decay is 3 min. / % saring compound / 5% saring compound 5% saring compound Using firs order ineics, we can solve for using Equaion (3.3) of he ex, wih [A] and [A] 5, [A] ln [A] 5 ln (6 min) ln(.5) 6 min.3 min Then, subsiuing ino Equaion (3.6) of he ex, you arrive a he same answer for / min 3 min 3.8 (a) Sraegy: To calculae he rae consan,, from he half-life of a firs-order reacion, we use Equaion (3.6) of he ex. Soluion: For a firs-order reacion, we only need he half-life o calculae he rae consan. From Equaion (3.6) s.98 s Sraegy: The relaionship beween he concenraion of a reacan a differen imes in a firs-order reacion is given by Equaions (3.3) and (3.4) of he ex. We are ased o deermine he ime required for 95% of he phosphine o decompose. If we iniially have % of he compound and 95% has reaced, hen wha is lef mus be (% 95%), or 5%. Thus, he raio of he percenages will be equal o he raio of he acual concenraions; ha is, [A] /[A] 5%/%, or.5/.. Soluion: The ime required for 95% of he phosphine o decompose can be found using Equaion (3.3) of he ex. [A] ln [A] (.5) ln (.98 s ) (.) ln(.5).98 s 5 s

7 CHAPTER 3: CHEMICAL KINETICS (a) Since he reacion is nown o be second-order, he relaionship beween reacan concenraion and ime is given by Equaion (3.7) of he ex. The problem supplies he rae consan and he iniial (ime ) concenraion of NOBr. The concenraion afer s can be found easily. + [NOBr] [NOBr] (.8 / M s)(s) + [NOBr].86 M [NOBr] 9 M [NOBr].34 M If he reacion were firs order wih he same and iniial concenraion, could you calculae he concenraion afer s? If he reacion were firs order and you were given he /, could you calculae he concenraion afer s? The half-life for a second-order reacion is dependen on he iniial concenraion. The half-lives can be calculaed using Equaion (3.8) of he ex. [A] (.8 / s)(.7 ) M M 7 s For an iniial concenraion of.54 M, you should find 3 s. Noe ha he half-life of a secondorder reacion is inversely proporional o he iniial reacan concenraion [A] [A] s

8 348 CHAPTER 3: CHEMICAL KINETICS 3.37 Graphing Equaion (3.3) of he ex requires ploing ln versus /T. The graph is shown below ln vs /T ln y -.4E+4x + 3.6E (K ) T The slope of he line is.4 4 K, which is E a /R. The acivaion energy is: E a slope R (.4 4 K) (8.34 J/K mol) E a.3 5 J/mol 3 J/mol Do you need o now he order of he reacion o find he acivaion energy? Is i possible o have a negaive acivaion energy? Wha would a poenial energy versus reacion coordinae diagram loo lie in such a case? 3.38 Sraegy: A modified form of he Arrhenius equaion relaes wo rae consans a wo differen emperaures [see Equaion (3.4) of he ex]. Mae sure he unis of R and E a are consisen. Since he rae of he reacion a 5 C is.5 3 imes faser han he rae a 5 C, he raio of he rae consans,, is also.5 3 :, because rae and rae consan are direcly proporional. Soluion: The daa are: T 5 C 53 K, T 5 C 43 K, and /.5 3. Subsiuing ino Equaion (3.4) of he ex, Ea T T ln R TT 3 Ea 53K 43K ln(.5 ) 8.34 J/mol K (53 K)(43 K) Ea J 8.34 K mol K E a.35 5 J/mol 35 J/mol

9 CHAPTER 3: CHEMICAL KINETICS The appropriae value of R is 8.34 J/K mol, no.8 L am/mol K. You mus also use he acivaion energy value of 63 J/mol (why?). Once he emperaure has been convered o Kelvin, he rae consan is: 63 J/mol Ea/ RT (8.34 J/mol K)(348 K) Ae (8.7 s ) e (8.7 s )(3.5 ) 3. 3 s Can you ell from he unis of wha he order of he reacion is? 3.4 Use a modified form of he Arrhenius equaion o calculae he emperaure a which he rae consan is s. We carry an exra significan figure hroughou his calculaion o minimize rounding errors. Ea ln R T T s.4 J/mol ln s 8.34 J/mol K T 63 K 4 ln(.57) (.5 K) T 63 K K T 9.43T.5 4 K T 644 K 37 C 3.4 Le be he rae consan a 95 K and he rae consan a 35 K. We wrie: Ea T ln T R TT Ea 95 K 35 K J/K mol (95 K)(35 K) E a J/mol 5.8 J/mol 3.4 Since he raio of raes is equal o he raio of rae consans, we can wrie: rae ln ln rae ln. Ea (3 K 78 K) ln J/K mol (3 K)(78 K) E a 5. 4 J/mol 5. J/mol

10 35 CHAPTER 3: CHEMICAL KINETICS 3.5 (a) The order of he reacion is simply he sum of he exponens in he rae law (Secion 3. of he ex). The order of his reacion is. The rae law reveals he ideniy of he subsances paricipaing in he slow or rae-deermining sep of a reacion mechanism. This rae law implies ha he slow sep involves he reacion of a molecule of NO wih a molecule of Cl. If his is he case, hen he firs reacion shown mus be he rae-deermining (slow) sep, and he second reacion mus be much faser. 3.5 (a) Sraegy: We are given informaion as o how he concenraions of X, Y, and Z affec he rae of he reacion and are ased o deermine he rae law. We assume ha he rae law aes he form rae [X ] x [Y] y [Z] z How do we use he informaion o deermine x, y, and z? Soluion: Since he reacion rae doubles when he X concenraion is doubled, he reacion is firsorder in X. The reacion rae riples when he concenraion of Y is ripled, so he reacion is also firsorder in Y. The concenraion of Z has no effec on he rae, so he reacion is zero-order in Z. The rae law is: rae [X ][Y] (c) If a change in he concenraion of Z has no effec on he rae, he concenraion of Z is no a erm in he rae law. This implies ha Z does no paricipae in he rae-deermining sep of he reacion mechanism. Sraegy: The rae law, deermined in par (a), shows ha he slow sep involves reacion of a molecule of X wih a molecule of Y. Since Z is no presen in he rae law, i does no ae par in he slow sep and mus appear in a fas sep a a laer ime. (If he fas sep involving Z happened before he rae-deermining sep, he rae law would involve Z in a more complex way.) Soluion: A mechanism ha is consisen wih he rae law could be: X + Y XY + X (slow) X + Z XZ (fas) The rae law only ells us abou he slow sep. Oher mechanisms wih differen subsequen fas seps are possible. Try o inven one. Chec: The rae law wrien from he rae-deermining sep in he proposed mechanism maches he rae law deermined in par (a). Also, he wo elemenary seps add o he overall balanced equaion given in he problem The firs sep involves forward and reverse reacions ha are much faser han he second sep. The raes of he reacion in he firs sep are given by: forward rae [O 3 ] reverse rae [O][O ] I is assumed ha hese wo processes rapidly reach a sae of dynamic equilibrium in which he raes of he forward and reverse reacions are equal: [O 3 ] [O][O ]

11 CHAPTER 3: CHEMICAL KINETICS 35 If we solve his equaliy for [O] we have: [O] [O 3] [O ] The equaion for he rae of he second sep is: rae [O][O 3 ] If we subsiue he expression for [O] derived from he firs sep, we have he experimenally verified rae law. [O 3] [O 3] overall rae [O ] [O ] The above rae law predics ha higher concenraions of O will decrease he rae. This is because of he reverse reacion in he firs sep of he mechanism. Noice ha if more O molecules are presen, hey will serve o scavenge free O aoms and hus slow he disappearance of O The experimenally deermined rae law is firs order in H and second order in NO. In Mechanism I he slow sep is bimolecular and he rae law would be: Mechanism I can be discarded. rae [H ][NO] The rae-deermining sep in Mechanism II involves he simulaneous collision of wo NO molecules wih one H molecule. The rae law would be: rae [H ][NO] Mechanism II is a possibiliy. In Mechanism III we assume he forward and reverse reacions in he firs fas sep are in dynamic equilibrium, so heir raes are equal: f [NO] r [N O ] The slow sep is bimolecular and involves collision of a hydrogen molecule wih a molecule of N O. The rae would be: rae [H ][N O ] If we solve he dynamic equilibrium equaion of he firs sep for [N O ] and subsiue ino he above equaion, we have he rae law: f rae [H ][NO] [H ][NO] r Mechanism III is also a possibiliy. Can you sugges an experimen ha migh help o decide beween he wo mechanisms? 3.6 Higher emperaures may disrup he inricae hree dimensional srucure of he enzyme, hereby reducing or oally desroying is caalyic aciviy.

12 35 CHAPTER 3: CHEMICAL KINETICS 3.6 The rae-deermining sep involves he breadown of ES o E and P. The rae law for his sep is: rae [ES] In he firs elemenary sep, he inermediae ES is in equilibrium wih E and S. The equilibrium relaionship is: [ES] [E][S] or [ES] [E][S] Subsiue [ES] ino he rae law expression. rae [ES] [E][S] 3.63 In each case he gas pressure will eiher increase or decrease. The pressure can be relaed o he progress of he reacion hrough he balanced equaion. In (d), an elecrical conducance measuremen could also be used Temperaure, energy of acivaion, concenraion of reacans, and a caalys Sricly, he emperaure mus be given whenever he rae or rae consan of a reacion is quoed Firs, calculae he radius of he. cm 3 sphere. V 4 3 πr cm πr 3 r.34 cm The surface area of he sphere is: area 4πr 4π(.34 cm).6 cm Nex, calculae he radius of he.5 cm 3 sphere. V 4 πr cm 4 πr r.668 cm The surface area of one sphere is: area 4πr 4π(.668 cm) 5.6 cm The oal area of 8 spheres 5.6 cm cm

13 CHAPTER 3: CHEMICAL KINETICS 353 Obviously, he surface area of he eigh spheres (44.9 cm ) is greaer han ha of one larger sphere (.6 cm ). A greaer surface area promoes he caalyzed reacion more effecively. I can be dangerous o wor in grain elevaors, because he large surface area of he grain dus can resul in a violen explosion X P vs. ime P (mmhg) 5 P 7 mmhg 3 s P 35 mmhg 83 s 5 X ime (s) / (83 3)s 7 s 3.68 The overall rae law is of he general form: rae [H ] x [NO] y (a) Comparing Experimen # and Experimen #, we see ha he concenraion of NO is consan and he concenraion of H has decreased by one-half. The iniial rae has also decreased by one-half. Therefore, he iniial rae is direcly proporional o he concenraion of H ; x. Comparing Experimen # and Experimen #3, we see ha he concenraion of H is consan and he concenraion of NO has decreased by one-half. The iniial rae has decreased by one-fourh. Therefore, he iniial rae is proporional o he squared concenraion of NO; y. The overall rae law is: rae [H ][NO], and he order of he reacion is + 3. Using Experimen # o calculae he rae consan, rae [H ][NO] rae [H ][NO] 6.4 M/s.38 / M s (. M)(.5 M)

14 354 CHAPTER 3: CHEMICAL KINETICS (c) Consuling he rae law, we assume ha he slow sep in he reacion mechanism will probably involve one H molecule and wo NO molecules. Addiionally he hin ells us ha O aoms are an inermediae. H + NO N + H O + O O + H H O H + NO N + H O slow sep fas sep 3.69 Since he mehanol conains no oxygen 8, he oxygen aom mus come from he phosphae group and no he waer. The mechanism mus involve a bond breaing process lie: O CH 3 O P O H H O 3.7 If waer is also he solven in his reacion, i is presen in vas excess over he oher reacans and producs. Throughou he course of he reacion, he concenraion of he waer will no change by a measurable amoun. As a resul, he reacion rae will no appear o depend on he concenraion of waer. 3.7 Mos ransiion meals have several sable oxidaion saes. This allows he meal aoms o ac as eiher a source or a recepor of elecrons in a broad range of reacions. 3.7 Since he reacion is firs order in boh A and B, hen we can wrie he rae law expression: rae [A][B] Subsiuing in he values for he rae, [A], and [B]: 4. 4 M/s (.6 )(.4 3 ).7 /M s Knowing ha he overall reacion was second order, could you have prediced he unis for? 3.73 (a) To deermine he rae law, we mus deermine he exponens in he equaion rae [CH 3 COCH 3 ] x [Br ] y [H + ] z To deermine he order of he reacion wih respec o CH 3 COCH 3, find wo experimens in which he [Br ] and [H + ] are held consan. Compare he daa from experimens () and (5). When he concenraion of CH 3 COCH 3 is increased by a facor of.33, he reacion rae increases by a facor of.33. Thus, he reacion is firs-order in CH 3 COCH 3. To deermine he order wih respec o Br, compare experimens () and (). When he Br concenraion is doubled, he reacion rae does no change. Thus, he reacion is zero-order in Br. To deermine he order wih respec o H +, compare experimens () and (3). When he H + concenraion is doubled, he reacion rae doubles. Thus, he reacion is firs-order in H +.

15 CHAPTER 3: CHEMICAL KINETICS 355 The rae law is: rae [CH 3 COCH 3 ][H + ] Rearrange he rae law from par (a), solving for. rae + [CH3COCH 3][H ] Subsiue he daa from any one of he experimens o calculae. Using he daa from Experimen (), M/s (.3 M)(.5 M) / M s (c) Le be he rae consan for he slow sep: + OH rae [CH 3 C CH 3 ][H O] () Le and be he rae consans for he forward and reverse seps in he fas equilibrium. + OH [CH 3 COCH 3 ][H 3 O + ] [CH 3 C CH 3 ][H O] () Therefore, Equaion () becomes + rae [CH3COCH 3][H3O ] which is he same as (a), where /. ] 3.74 Recall ha he pressure of a gas is direcly proporional o he number of moles of gas. This comes from he ideal gas equaion. P nrt V The balanced equaion is: N O(g) N (g) + O (g) From he soichiomery of he balanced equaion, for every one mole of N O ha decomposes, one mole of N and.5 moles of O will be formed. Le s assume ha we had moles of N O a. Afer one halflife here will be one mole of N O remaining and one mole of N and.5 moles of O will be formed. The oal number of moles of gas afer one half-life will be: nt nn O + nn + n O mol + mol +.5 mol.5 mol A, here were mol of gas. Now, a, here are.5 mol of gas. Since he pressure of a gas is direcly proporional o he number of moles of gas, we can wrie:. am.5 mol gas a.63 am afer one half -life mol gas ( )

16 356 CHAPTER 3: CHEMICAL KINETICS 3.75 Fe 3+ undergoes a redox cycle: Fe 3+ Fe + Fe 3+ Fe 3+ oxidizes I : Fe 3+ + I Fe + + I Fe + reduces S O 8 : + 3+ Fe + S O 8 Fe + SO4 I + S O 8 I + SO 4 The uncaalyzed reacion is slow because boh I and S O 8 are negaively charged which maes heir muual approach unfavorable The rae expression for a hird order reacion is: The unis for he rae law are: Δ[A] rae Δ M M s 3 M s 3 [A] 3.77 For a rae law, zero order means ha he exponen is zero. In oher words, he reacion rae is jus equal o a consan; i doesn' change as ime passes. (a) The rae law would be: rae [A] The inegraed zero-order rae law is: [A] + [A]. Therefore, a plo of [A] versus ime should be a sraigh line wih a slope equal o.

17 CHAPTER 3: CHEMICAL KINETICS 357 [A] [A] [A] A, [A]. Subsiuing ino he above equaion: [A] [A] [A] [A] (c) When [A], [A] [A] Subsiuing for, [A] [A] This indicaes ha he inegraed rae law is no longer valid afer wo half-lives Boh compounds, A and B, decompose by firs-order ineics. Therefore, we can wrie a firs-order rae equaion for A and also one for B. [A] ln [A] A [B] ln [B] B [A] [A] [B] A e B e [B] A e [B] [B] B e [A] [A] We can calculae each of he rae consans, A and B, from heir respecive half-lives. A min.39 min B min.385 min The iniial concenraion of A and B are equal. [A] [B]. Therefore, from he firs-order rae equaions, we can wrie: A A [A] [A] ( B A) ( ) 4 [B] e [B] e B e e B e e.46 4 e

18 358 CHAPTER 3: CHEMICAL KINETICS ln min 3.79 There are hree gases presen and we can measure only he oal pressure of he gases. To measure he parial pressure of azomehane a a paricular ime, we mus wihdraw a sample of he mixure, analyze and deermine he mole fracions. Then, P azomehane P T Χ azomehane This is a raher edious process if many measuremens are required. A mass specromeer will help (see Secion 3.4 of he ex). 3.8 (a) Changing he concenraion of a reacan has no effec on. 3.8 (c) (d) If a reacion is run in a solven oher han in he gas phase, hen he reacion mechanism will probably change and will hus change. Doubling he pressure simply changes he concenraion. No effec on, as in (a). The rae consan changes wih emperaure. (e) A caalys changes he reacion mechanism and herefore changes. 3.8 Mahemaically, he amoun lef afer en half lives is: (a) A caalys wors by changing he reacion mechanism, hus lowering he acivaion energy. A caalys changes he reacion mechanism. (c) A caalys does no change he enhalpy of reacion. (d) A caalys increases he forward rae of reacion. (e) A caalys increases he reverse rae of reacion.

19 CHAPTER 3: CHEMICAL KINETICS The ne ionic equaion is: Zn(s) + H + (aq) Zn + (aq) + H (g) (a) (c) (d) Changing from he same mass of granulaed zinc o powdered zinc increases he rae because he surface area of he zinc (and hus is concenraion) has increased. Decreasing he mass of zinc (in he same granulaed form) will decrease he rae because he oal surface area of zinc has decreased. The concenraion of proons has decreased in changing from he srong acid (hydrochloric) o he wea acid (aceic); he rae will decrease. An increase in emperaure will increase he rae consan ; herefore, he rae of reacion increases A very high [H ], [H ] >> rae [NO] [H ] [H ] [NO] A very low [H ], [H ] << [NO] [H ] rae [NO] [H ] The resul from Problem 3.68 agrees wih he rae law deermined for low [H ] If he reacion is 35.5% complee, he amoun of A remaining is 64.5%. The raio of [A] /[A] is 64.5%/% or.645/.. Using he firs-order inegraed rae law, Equaion (3.3) of he ex, we have [A] ln [A].645 ln (4.9 min)..439 (4.9 min).896 min

20 36 CHAPTER 3: CHEMICAL KINETICS 3.87 Firs we plo he daa for he reacion: N O 5 4NO + O.5 Iniial Rae vs. Conc. Iniial Rae x, (M/s) [Dinirogen Penoxide] (M) The daa in linear, wha means ha he iniial rae is direcly proporional o he concenraion of N O 5. Thus, he rae law is: Rae [N O 5 ] The rae consan can be deermined from he slope of he graph daa.. 5 s Δ(Iniial Rae) Δ[NO 5] or by using any se of Noe ha he rae law is no Rae [N O 5 ], as we migh expec from he balanced equaion. In general, he order of a reacion mus be deermined by experimen; i canno be deduced from he coefficiens in he balanced equaion The firs-order rae equaion can be arranged o ae he form of a sraigh line. ln[a] + ln[a] If a reacion obeys firs-order ineics, a plo of ln[a] vs. will be a sraigh line wih a slope of. The slope of a plo of ln[n O 5 ] vs. is min. Thus, min The equaion for he half-life of a firs-order reacion is: min 3. min

21 CHAPTER 3: CHEMICAL KINETICS The red bromine vapor absorbs phoons of blue ligh and dissociaes o form bromine aoms. Br Br The bromine aoms collide wih mehane molecules and absrac hydrogen aoms. Br + CH 4 HBr + CH 3 The mehyl radical hen reacs wih Br, giving he observed produc and regeneraing a bromine aom o sar he process over again: CH 3 + Br CH 3 Br + Br Br + CH 4 HBr + CH 3 and so on (a) In he wo-sep mechanism he rae-deermining sep is he collision of a hydrogen molecule wih wo iodine aoms. If visible ligh increases he concenraion of iodine aoms, hen he rae mus increase. If he rue rae-deermining sep were he collision of a hydrogen molecule wih an iodine molecule (he one-sep mechanism), hen he visible ligh would have no effec (i migh even slow he reacion by depleing he number of available iodine molecules). To spli hydrogen molecules ino aoms, one needs ulraviole ligh of much higher energy. 3.9 For a firs order reacion: decay rae a ln decay rae a.86 4 ln (. yr ) yr 3.9 (a) We can wrie he rae law for an elemenary sep direcly from he soichiomery of he balanced reacion. In his rae-deermining elemenary sep hree molecules mus collide simulaneously (one X and wo Y's). This maes he reacion ermolecular, and consequenly he rae law mus be hird order: firs order in X and second order in Y. The rae law is: rae [X][Y] The value of he rae consan can be found by solving algebraically for. 3 rae 3.8 M/s [X][Y] (.6 M)(.88 M).9 M s Could you wrie he rae law if he reacion shown were he overall balanced equaion and no an elemenary sep? 3.93 (a) O + O 3 O Cl is a caalys; ClO is an inermediae. (c) The C F bond is sronger han he C Cl bond. (d) Ehane will remove he Cl aoms: Cl + C H 6 HCl + C H 5

22 36 CHAPTER 3: CHEMICAL KINETICS (e) The overall reacion is: O + O 3 O. o o o o rxn f f f 3 Δ H ΔH (O ) [ Δ H (O) + ΔH (O )] Δ Hrxn o () [()(49.4 J/mol) + ()(4. J/mol)] Δ Hrxn o 39.6 J/mol The reacion is exohermic /[ClO] vs. ime 8 /[ClO] 4 y.36e+7x +.7E+5.E+.E-3.E-3 3.E-3 4.E-3 ime (s)

23 CHAPTER 3: CHEMICAL KINETICS 363 Reacion is second-order because a plo of /[ClO] vs. ime is a sraigh line. The slope of he line equals he rae consan,. Slope.4 7 /M s 3.95 We can calculae he raio of / a 4 C using he Arrhenius equaion. / Ea RT ( Ea E a )/ RT ΔEa/ RT e e Ea / RT Ae Ae 8. e ΔEa (8.34 J/K mol)(33 K) ln(8.) ΔEa (8.34 J/K mol)(33 K) ΔE a J/mol Having calculaed ΔE a, we can subsiue bac ino he equaion o calculae he raio / a 3 C (573 K). e J/mol (8.34 J/K mol)(573 K) During he firs five minues or so he engine is relaively cold, so he exhaus gases will no fully reac wih he componens of he caalyic converer. Remember, for almos all reacions, he rae of reacion increases wih emperaure The acual appearance depends on he relaive magniudes of he rae consans for he wo seps (a) E a has a large value. E a. Orienaion facor is no imporan.

24 364 CHAPTER 3: CHEMICAL KINETICS 3.99 A plausible wo-sep mechanism is: NO + NO NO 3 + NO NO 3 + CO NO + CO (slow) (fas) 3. Firs, solve for he rae consan,, from he half-life of he decay yr yr 5.44 yr Now, we can calculae he ime for he pluonium o decay from 5. g o. g using he equaion for a firs-order reacion relaing concenraion and ime. [A] ln [A]. 6 ln (.84 yr ) 5..6 (.84 6 yr ) yr 3. A high pressure of PH 3, all he sies on W are occupied, so he rae is independen of [PH 3 ]. 3. (a) Caalys: Mn + ; inermediae: Mn 3+ Firs sep is rae-deermining. (c) Wihou he caalys, he reacion would be a ermolecular one involving 3 caions! (Tl + and wo Ce 4+ ). The reacion would be slow. The caalys is a homogeneous caalys because i has he same phase (aqueous) as he reacans.

25 CHAPTER 3: CHEMICAL KINETICS (a) Since a plo of ln (sucrose) vs. ime is linear, he reacion is s order ln [sucrose] vs. ime ln [sucrose] y -3.68E-3x E ime (min) Slope min min [A] ln [A].5 3 ln (3.68 ) 84 min (c) [H O] is roughly unchanged. This is a pseudo-firs-order reacion. 3.4 Iniially, he number of moles of gas in erms of he volume is: n PV (.35 am) V V RT L am.8 ( )K mol K We can calculae he concenraion of dimehyl eher from he following equaion. [(CH 3) O] ln [(CH 3 ) O] [(CH 3) O] [(CH 3) O] e Since, he volume is held consan, i will cancel ou of he equaion. The concenraion of dimehyl eher afer 8. minues (48 s) is: [(CH ) O] e V [(CH 3 ) O] M s V s 4 ( )

26 366 CHAPTER 3: CHEMICAL KINETICS Afer 8. min, he concenraion of (CH 3 ) O has decreased by ( )M or M. Since hree moles of produc form for each mole of dimehyl eher ha reacs, he concenraions of he producs are (3)(8.4 4 M).5 3 M. The pressure of he sysem afer 8. minues is: nrt n P RT MRT V V P [(5.6 3 ) + (.5 3 )]M (.8 L am/mol K)(73 K) P.45 am 3.5 This is a uni conversion problem. Recall ha cm 3 L cm L 6. molecules 7.9 molecule s cm mol L/mol s /M s Δ[B] 3.6 (a) [A] [B] Δ Δ[B] If, Δ Then, from par (a) of his problem: [A] [B] [B] [A] 3.7 (a) Drining oo much alcohol oo fas means all he alcohol dehydrogenase (ADH) acive sies are ied up and he excess alcohol will damage he cenral nervous sysem. Boh ehanol and mehanol will compee for he same sie a ADH. An excess of ehanol will replace mehanol a he acive sie, leading o mehanol s discharge from he body. 3.8 (a) The firs-order rae consan can be deermined from he half-life yr.47 yr See Problem 3.8. Mahemaically, he amoun lef afer en half lives is: 9.8 4

27 CHAPTER 3: CHEMICAL KINETICS 367 (c) If 99.% has disappeared, hen.% remains. The raio of [A] /[A] is.%/% or./.. Subsiue ino he firs-order inegraed rae law, Equaion (3.3) of he ex, o deermine he ime. [A] ln [A]. ln (.47 yr ). 4.6 (.47 yr ) 86 yr 3.9 () Assuming he reacions have roughly he same frequency facors, he one wih he larges acivaion energy will be he slowes, and he one wih he smalles acivaion energy will be he fases. The reacions raned from slowes o fases are: () Reacion (a): ΔH 4 J/mol Reacion : ΔH J/mol Reacion (c): ΔH J/mol < (c) < (a) (a) and (c) are exohermic, and is endohermic. 3. (a) There are hree elemenary seps: A B, B C, and C D. There are wo inermediaes: B and C. (c) (d) The hird sep, C D, is rae deermining because i has he larges acivaion energy. The overall reacion is exohermic. 3. The fire should no be doused wih waer, because ianium acs as a caalys o decompose seam as follows: H O(g) H (g) + O (g) H gas is flammable and forms an explosive mixure wih O. 3. Le ca unca Then, Ea(ca) Ea(unca) Ae RT Ae RT Since he frequency facor is he same, we can wrie: Ea(ca) Ea(unca) e RT e RT

28 368 CHAPTER 3: CHEMICAL KINETICS Taing he naural log (ln) of boh sides of he equaion gives: or, Ea(ca) Ea(unca) RT RT Ea(ca) Ea(unca) T T Subsiuing in he given values: 7.J/mol 93 K T.8 3 K 4J/mol T This emperaure is much oo high o be pracical. 3.3 Firs, le's calculae he number of radium nuclei in. g. 3 mol Ra 6. Ra nuclei. g.7 Ra nuclei 6.3 g Ra mol Ra We can now calculae he rae consan,, from he aciviy and he number of nuclei, and hen we can calculae he half-life from he rae conan. aciviy N aciviy 3.7 nuclear disinegraions/s N.7 nuclei The half-life is:.4 /s /s 5. s Nex, le's conver 5 years o seconds. Then we can calculae he number of nuclei remaining afer 5 years. 365 days 4 h 36 s 5 yr.58 s yr day h Use he firs-order inegraed rae law o deermine he number of nuclei remaining afer 5 years. N ln N N ln (.4 /s)(.58 s).7 N.7 e. N. Ra nuclei

29 CHAPTER 3: CHEMICAL KINETICS 369 Finally, from he number of nuclei remaining afer 5 years and he rae consan, we can calculae he aciviy. aciviy N aciviy (.4 /s)(. nuclei) 3. nuclear disinegraions/s 3.4 (a) The rae law for he reacion is: rae [Hb][O ] We are given he rae consan and he concenraion of Hb and O, so we can subsiue in hese quaniies o solve for rae. rae (. 6 /M s)(8. 6 M)(.5 6 M) rae.5 5 M/s (c) If HbO is being formed a he rae of.5 5 M/s, hen O is being consumed a he same rae,.5 5 M/s. Noe he : mole raio beween O and HbO. The rae of formaion of HbO increases, bu he concenraion of Hb remains he same. Assuming ha emperaure is consan, we can use he same rae consan as in par (a). We subsiue rae, [Hb], and he rae consan ino he rae law o solve for O concenraion. rae [Hb][O ].4 4 M/s (. 6 /M s)(8. 6 M)[O ] [O ] M 3.5 Iniially, he rae increases wih increasing pressure (concenraion) of NH 3. The sraigh-line relaionship in he firs half of he plo shows ha he rae of reacion is direcly proporional o he concenraion of ammonia. Rae [NH 3 ]. The more ammonia ha is adsorbed on he ungsen surface, he faser he reacion. A a cerain pressure (concenraion), he rae is no longer dependen on he concenraion of ammonia (horizonal porion of plo). The reacion is now zero-order in NH 3 concenraion. A a cerain concenraion of NH 3, all he reacive sies on he meal surface are occupied by NH 3 molecules, and he rae becomes consan. Increasing he concenraion furher has no effec on he rae. 3.6 n [A] C n [A], where C is a proporionaliy consan. Subsiuing in for zero, firs, and second-order reacions gives: n C [A] [A] C n C C [A]

30 37 CHAPTER 3: CHEMICAL KINETICS n C [A] Compare hese resuls wih hose in Table 3.3 of he ex. Wha is C in each case? 3.7 (a) The relaionship beween half-life and rae consan is given in Equaion (3.6) of he ex min.35 min Following he same procedure as in par (a), we find he rae consan a 7 C o be.58 3 min. We now have wo values of rae consans ( and ) a wo emperaures (T and T ). This informaion allows us o calculae he acivaion energy, E a, using Equaion (3.4) of he ex. Ea T T ln R TT.35 min Ea 373 K 343 K ln 3.58 min (8.34 J/mol K) (373 K)(343 K) E a. 5 J/mol J/mol (c) Since all he above seps are elemenary seps, we can deduce he rae law simply from he equaions represening he seps. The rae laws are: Iniiaion: rae i [R ] Propagaion: rae p [M][M ] Terminaion: rae [M'][M"] The reacan molecules are he ehylene monomers, and he produc is polyehylene. Recalling ha inermediaes are species ha are formed in an early elemenary sep and consumed in a laer sep, we see ha hey are he radicals M', M", and so on. (The R species also qualifies as an inermediae.) (d) The growh of long polymers would be favored by a high rae of propagaions and a low rae of erminaion. Since he rae law of propagaion depends on he concenraion of monomer, an increase in he concenraion of ehylene would increase he propagaion (growh) rae. From he rae law for erminaion we see ha a low concenraion of he radical fragmen M' or M" would lead o a slower rae of erminaion. This can be accomplished by using a low concenraion of he iniiaor, R. 3.8 (a) The unis of he rae consan show he reacion o be second-order, meaning he rae law is mos liely: Rae [H ][I ] We can use he ideal gas equaion o solve for he concenraions of H and I. We can hen solve for he iniial rae in erms of H and I and hen conver o he iniial rae of formaion of HI. We carry an exra significan figure hroughou his calculaion o minimize rounding errors.

31 CHAPTER 3: CHEMICAL KINETICS 37 n PV RT n M V P RT Since he oal pressure is 658 mmhg and here are equimolar amouns of H and I in he vessel, he parial pressure of each gas is 89 mmhg. am 89 mmhg 76 mmhg [H ] [I ].974 M Lam.8 (4 + 73) K mol K Le s conver he unis of he rae consan o /M min, and hen we can subsiue ino he rae law o solve for rae. 6 s.4.45 M s min M min We now ha, Rae [H ][I ] 4 Rae.45 (.974 M)(.974 M) M/min M min or Δ[HI] Rae Δ Δ[HI] 4 Rae ()(5.658 M/min) Δ 3.3 M/min We can use he second-order inegraed rae law o calculae he concenraion of H afer. minues. We can hen subsiue his concenraion bac ino he rae law o solve for rae. + [H ] [H ].45 (. min) + [H ] M min.974 M [H ].534 M We can now subsiue his concenraion bac ino he rae law o solve for rae. The concenraion of I afer. minues will also equal.534 M. We now ha, Rae [H ][I ] 4 Rae.45 (.534 M)(.534 M) 3.47 M/min M min Rae Δ[HI] Δ

32 37 CHAPTER 3: CHEMICAL KINETICS or Δ[HI] Δ 4 Rae ()(3.47 M/min) M/min The concenraion of HI afer. minues is: [HI] ([H ] [H ] ) [HI] (.974 M.534 M) M 3.9 Firs, we wrie an overall balanced equaion. P P * P * ½P P ½P The average molar mass is given by: mol g ([P] M + [P ] M) M L mol () mol ([P] + [P ] ) L where M is he molar mass of P and [P] is he concenraion of P a a laer ime in he reacion. Noe ha in he numeraor [P ] is muliplied by because he molar mass of P is double ha of P. Also noe ha he unis wor ou o give unis of molar mass, g/mol. Based on he soichiomery of he reacion, he concenraion of [P ] is: [P ] [P] [P] Subsiuing bac ino Equaion () gives: [P] [P] [P] + M M ([P] M + [P] M [P] M) M[P] [P] [P] [P] + [P] [P] + [P] + [P] [P] M () In he proposed mechanism, he denauraion sep is rae-deermining. Thus, Rae [P] Because we are looing a change in concenraion over ime, we need he firs-order inegraed rae law, Equaion (3.3) of he ex. [P] ln [P] [P] [P] e [P] [P] e

33 CHAPTER 3: CHEMICAL KINETICS 373 Subsiuing ino Equaion () gives: or M[P] M M [P] + [P] e + e M M e M ln M M M The rae consan,, can be deermined by ploing ln M M M a slope of. versus. The plo will give a sraigh line wih 3. The half-life is relaed o he iniial concenraion of A by n [A] According o he daa given, he half-life doubled when [A] was halved. This is only possible if he half-life is inversely proporional o [A]. Subsiuing n ino he above equaion gives: [A] Looing a his equaion, i is clear ha if [A] is halved, he half-life would double. The reacion is secondorder. We use Equaion (3.8) of he ex o calculae he rae consan. [A] [A] (. M)(. min).4 / M min 3. (a) The half-life of a reacion and he iniial concenraion are relaed by C n [A] where C is a consan. Taing he common logarihm of boh sides of he equaion, log log C ( n )log[a] Because pressure is proporional o concenraion a consan emperaure, he above equaion can also be wrien as ' log ( n ) log P + log C

34 374 CHAPTER 3: CHEMICAL KINETICS A plo of log vs. logp gives a slope of (n ). The daa used for he plo are: logp log y.x log / log P There are clearly wo ypes of behavior exhibied in he graph. A pressures above 5 mmhg, he graph appears o be a sraigh line. Fiing hese hree poins resuls in a bes fi line wih an equaion of y.x +.5. The slope of he line is.; herefore,. (n ), or n, and he reacion is zero-order. Alhough he daa are limied, i is clear ha here is a change in slope below 5 mmhg, indicaing a change in reacion order. I does appear ha he limiing slope as pressure approaches zero is iself zero. Thus, (n ), or n, and he limiing behavior is ha of a firs-order reacion. (c) As discovered in par (a), he reacion is firs-order a low pressures and zero-order a pressures above 5 mmhg. The mechanism is acually he same a all pressures considered. A low pressures, he fracion of he ungsen surface covered is proporional o he pressure of NH 3, so he rae of decomposiion will have a firs-order dependence on ammonia pressure. A increased pressures, all he caalyic sies are occupied by NH 3 molecules, and he rae becomes independen of he ammonia pressure and hence zero-order in NH 3.

35 CHAPTER 3: CHEMICAL KINETICS From Equaion (3.4) of he ex, Ea ln R T T 5.4 J/mol ln 8.34 J/mol K 66 K 6 K ln e.6 The rae consan a 66 K is.6 imes greaer han ha a 6 K. This is a 6% increase in he rae consan for a % increase in emperaure! The resul shows he profound effec of an exponenial dependence. In general, he larger he E a, he greaer he influence of T on. 3.3 λ (he absorbance of A) decreases wih ime. This would happen for all he mechanisms shown. Noe ha λ (he absorbance of B) increases wih ime and hen decreases. Therefore, B canno be a produc as shown in mechanisms (a) or. If B were a produc is absorbance would increase wih ime and level off, bu i would no decrease. Since he concenraion of B increases and hen afer some ime begins o decrease, i mus mean ha i is produced and hen i reacs o produce produc as in mechanisms (c) and (d). In mechanism (c), wo producs are C and D, so we would expec o see an increase in absorbance for wo species. Since we see an increase in absorbance for only one species, hen he mechanism ha is consisen wih he daa mus be (d). λ 3 is he absorbance of C.