CHAPTER 4 CHEMICAL KINETICS PRACTICE EXAMPLES A B (E) The rae f cnsumpin fr a reacan is expressed as he negaive f he change in mlariy divided by he ime inerval. The rae f reacin is expressed as he rae f cnsumpin f a reacan r prducin f a prduc divided by is sichimeric cefficien. A.387 M.369 M min 5 rae f cnsumpin f A = = = 8.93 M s 8.5 min 6 s 5 8.93 M s 5 rae f reacin = rae f cnsumpin f A = 4.46 M s (E) We use he rae f reacin f A deermine he rae f frmain f B, ning frm he balanced equain ha 3 mles f B frm (+3 mles B) when mles f A reac ( mles A). (Recall ha M means mles per lier. ).55M A.5684M A 3mles B 4 rae f B frmain=.6 M s 6s.5min mles A min A (M) The 4-s angen line inersecs he -s verical line a.75 M and reaches M a 35 s. The slpe f ha angen line is hus M.75 M 4 slpe = = 3.3 M s = insananeus rae f reacin 35 s s 4 The insananeus rae f reacin = 3.3 M s. (b) A 4 s, HO =.39 M. A 45 s, HO =.39 M + rae 4 A 45 s, H O =.39 M + 3.3 ml H O L s 5s =.39 M.7 M =.37 M B (M) Wih nly he daa f Table 4. we can use nly he reacin rae during he firs 4 s, 4 HO / = 5. M s, and he iniial cncenrain, HO =.3M. We calculae he change in HO and add i HO deermine HO. 4 H O = rae f reacin f H O = 5. M s s =.5 M H O = H O + H O =.3 M +.5 M =.7 M This value differs frm he value f.5 M deermined in ex Example 4-b because 4 he ex used he iniial rae f reacin 7. M s, which is a bi faser han he average rae ver he firs 4 secnds. 6
Chaper 4: Chemical Kineics 3A (M) We wrie he equain fr each rae, divide hem in each her, and slve fr n. n 5 = NO 5 = 5.45 M s = 3.5 M n 5 = NO 5 =.35 M s =.78 M 5 n n n 5.45 M s NO 5 3.5 M 3.5 5 n n.35 M s NO 5.78 M.78 R R R R = = 4.4 = = = = 4.4 n n We ep an exra significan figure (4) emphasize ha he value f n =. Thus, he reacin is firs-rder in NO 5. rae = HgCl C O 3B (E) Fr he reacin, we nw ha Exp. find he rae. 4. Here we will cmpare Exp. 4 HgCl CO 4 4.5 M.45 M rae4 = = =.4 = 5 HgCl.5 M.5 M CO 4 5 7 rae 4 =.4.8 M min = 3.9 M min. rae rae.8 M min The desired rae is 4A (E) We place he iniial cncenrains and he iniial raes in he rae law and slve fr. rae = A B = 4.78 M s =. M.87 M 4.78 M s = =4.4 M s. M.87 M n 4B 5A (E) We nw ha rae = HgCl CO 4 and =7.6 3 M min. Thus, inserin f he saring cncenrains and he value in he rae law yields: 3 7 Rae = 7.6 M min.5 M.5 M =.4 M min (E) Here we subsiue direcly in he inegraed rae law equain. 3 ln A = + ln A = 3. s 35 s + ln.8 =.98 +.3 =.48.48 A = e =. M b g 5B (M) This ime we subsiue he prvided values in ex Equain 4.3. HO.49 M.443 ln = = 6 s = ln =.443 = = 7.38 s H O.3 M 6 s 4 Nw we chse H O =.49 M, H O =.6, = 8 s 6 s = s HO.6 M.88 ln = = s = ln =.88 = = 7.3 s 4 HO.49 M s These w values agree wihin he limis f he experimenal errr and hus, he reacin is firsrder in [H O ]. 6
Chaper 4: Chemical Kineics 6A 6B 7A (M) We can use he inegraed rae equain find he rai f he final and iniial cncenrains. This rai equals he fracin f he iniial cncenrain ha remains a ime. A 3 ln = =.95 s 5 s =.443 A A A =.443 e =.64; 64.% f A remains. (M) Afer w-hirds f he sample has decmpsed, ne-hird f he sample remains. HO = HO 3, and we have Thus HO HO 3 ln = = ln = ln / 3 =.99 = 7.3 4 s HO HO.99 min 7.3 s 6 s 3 = =.5 s = 5. min 4 (M) A he end f ne half-life he pressure f DTBP will have been halved, 4 mmhg. A he end f anher half-life, a 6 min, he pressure f DTBP will have halved again, mmhg. Thus, he pressure f DTBP a 5 min will be inermediae beween he pressure a 8. min (4 mmhg) and ha a 6 min ( mmhg). T bain an exac answer, firs we deermine he value f he rae cnsan frm he half-life..693.693 = = =.866 min / 8. min PDTBP ln = =.866 min 5 min =.8 P DTBP PDTBP.8 = e =.34 PDTBP P P =.34 =.348 mmhg = 7 mmhg DTBP DTBP 7B (M) We use parial pressures in place f cncenrains in he inegraed firs-rder rae equain. Nice firs ha mre han 3 half-lives have elapsed, and hus he ehylene xide b g 3 pressure has declined a ms.5 = 9 f is iniial value. P 36 s P ln = =.5 s 3. h =. = e =.4 P h P P 3 4 3. =.4 P =.4 78 mmhg =.9 mmhg 7 3 63
Chaper 4: Chemical Kineics (b) P ehylene xide iniially 78 mmhg.9 7 mmhg (~ ). Essenially all f he ehylene xide is cnvered CH 4 and CO. Since pressure is prprinal mles, he final pressure will be wice he iniial pressure ( mle gas mles gas; 78 mmhg 564 mmhg). The final pressure will be.56 3 mmhg. 8A (D) We firs begin by ling fr a cnsan rae, indicaive f a zer-rder reacin. If he rae is cnsan, he cncenrain will decrease by he same quaniy during he same ime perid. If we chse a 5-s ime perid, we ne ha he cncenrain decreases.88 M.74 M =.4 M during he firs 5 s,.74 M.6 M =. M during he secnd 5 s, during he hird 5 s, and.6 M.5 M =. M.5 M.44 M =.8 M during he furh 5-s perid. This is hardly a cnsan rae and we hus cnclude ha he reacin is n zer-rder. We nex l fr a cnsan half-life, indicaive f a firs-rder reacin. The iniial cncenrain f.88 M decreases ne half f ha value,.44 M, during he firs s, indicaing a -s half-life. The cncenrain halves again. M in he secnd s, anher -s half-life. Finally, we ne ha he cncenrain halves als frm.6 M a 5 s.3 M a 5 s, ye anher -s half-life. The rae is esablished as firs-rder. The rae cnsan is =.693 =.693 3 =6.93 s. / s Tha he reacin is firs-rder is made apparen by he fac ha he ln[b] vs ime pl is a sraigh line wih slpe = - ( = 6.85 3 s )..85.75 Pl f [B] versus Time Pl f ln([b]) versus Time -. -.4 4 7 6 Pl f /[B] versus Time.65 -.6 5 [B] (M).55.45.35 ln([b]) -.8 - -. -.4 /[B] (M - ) 4 3.5 -.6.5 3 Time(s) -.8 - Time(s) 3 Time (s) 64
Chaper 4: Chemical Kineics 8B (D) We pl he daa in hree ways deermine he rder. () A pl f [A] vs. ime is linear if he reacin is zer-rder. () A pl f ln [A] vs. ime will be linear if he reacin is firs-rder. (3) A pl f /[A] vs. ime will be linear if he reacin is secnd-rder. I is bvius frm he pls belw ha he reacin is zer-rder. The negaive f he slpe f he line equals 3 =.83 M.5 M8. min = 9.8 M/min ( = 9.3 3 M/min using a graphical apprach)..5.3 Pl f [A] versus Time -.5 Pl f ln([a]) versus Time Pl f /[A] versus Time [A] (M)..9.7.5.3. ln([a]) -.7 -.9 -. /[A] (M - ) 9 8 7 6.9.7 y = -.93x +.494 Time (min) -.3 -.5 Time (min) 5 4 Time (min) 9A 9B (M) Firs we cmpue he value f he rae cnsan a 75. C wih he Arrhenius equain. We nw ha he acivain energy is E a =.6 5 J/ml, and ha =3.46 5 s a 98 K. The emperaure f 75. C = 348. K. ln = ln 3.46 s F HG I KJ 5 Ea.6 J/ml = = R T T 8.345 J ml K 98. K 348. K 5 = 3.46 s e = 3.46 s 4.6 =.6 s 5 +6.4 5.693.693 = = = 43 s a 75 C.6 s / F HG I K J =6.4 (M) We use he inegraed rae equain deermine he rae cnsan, realizing ha ne-hird remains when w-hirds have decmpsed. NO NO 3 ln = ln = ln = =.5 h =.99 NO NO 3 5 5 5 5.99 h = =.3 s.5 h 36 s 4 65
Chaper 4: Chemical Kineics Nw use he Arrhenius equain deermine he emperaure a which he rae cnsan is.4 4 s. F HG 4 5.4 s Ea.6 J/ml ln = ln =.77= 5 = 3.46 s R T T 8.345 J ml K 98 K T.77 8.345 K = 3 = 3. K T = 3 5 K T 98 K.6 A (M) The w seps f he mechanism mus add, in a Hess's law fashin, prduce he verall reacin. Overall reacin: CO + NO CO + NO r CO + NO CO + NO I KJ 3 3 Secnd sep: NO + CO NO + CO r + NO + CO NO + CO Firs sep: NO NO + NO F HG 3 If he firs sep is he slw sep, hen i will be he rae-deermining sep, and he rae f ha sep will be he rae f he reacin, namely, rae f reacin = NO. B (M) () The seps f he mechanism mus add, in a Hess's law fashin, prduce he verall reacin. This is dne belw. The w inermediaes, NOFb ggand F(g), are each prduced in ne sep and cnsumed in he nex ne. Slw: NOFbgg NO Fbgg + Fbgg Fas: Fbgg + NO bgg NOFbgg Ne: NO g + F g NO F g Fas: NO g + F g NO F g b g b g b g () The prpsed mechanism mus agree wih he rae law. We expec he rae-deermining sep deermine he reacin rae: Rae = 3NOF. T eliminae NO F, we recgnize ha he firs elemenary reacin is very fas and will have he same rae frward as reverse: R f = NOF = NOF = R r. We slve fr he cncenrain f inermediae:nof = NOF /. We nw subsiue his expressin fr NO F in he rae equain: Rae = 3/ NOF. Thus he prediced rae law agrees wih he experimenal rae law. I KJ 66
Chaper 4: Chemical Kineics INTEGRATIVE EXAMPLE A. (M) The ime required fr he fixed (c) prcess f suring is hree imes as lng a 3 C refrigerar emperaure (76 K) as a C rm emperaure (93 K). c/ 64 h E E a a Ea 4 ln ln ln. (. ) c/ 3 64 h R 93 K 76 K T T R R. R.8.345 J ml K 4 Ea 4.4 J/ml 44 J/ml 4 4. K. K (b) Use he E a deermined in par calculae he suring ime a 4 C = 33 K. 4 E a 4.4 J/ml ln.5 ln R T T 8.345 J ml K 33 K 93 K 64 h.5 e.37.3764 h. h 64 h B. (M) The species A * is a reacive inermediae. Le s deal wih his species by using a seady sae apprximain. d[a * ]/d = = [A] - [A * ][A] [A * ]. Slve fr [A * ]. - [A * ][A] + [A * ] = [A] [A] [A*] = The rae f reacin is: Rae = [A * [A] ] = - [A] -[A] A lw pressures ([A] ~ and hence >> - [A]), he denminar becmes ~ and he rae law is [A] Rae = = [A] Secnd-rder wih respec [A] A high pressures ([A] is large and - [A] >> K ), he denminar becmes ~ - [A] and he rae law is [A] [A] Rae = = Firs-rder wih respec [A] [A] - - EXERCISES Raes f Reacins. (M) A + B C + 3D [A] = 6. 4 M s Rae = [A] = /(6. 4 M s ) = 3. 4 M s (b) Rae f disappearance f B = [A] = /(6. 4 M s ) = 3. 4 M s (c) Rae f appearance f D = 3 [A] = 3(6. 4 M s ) = 9.3 4 M s 67
Chaper 4: Chemical Kineics. (M) In each case, we draw he angen line he pled curve. HO.7 M.6 M The slpe f he line is = = 9. M s 4 s 6 s HO 4 reacin rae = = 9. M s (b) 4 Read he value where he hriznal line [H O ] =.5 M inersecs he curve, 5 s r 36 minues 3. (E) 4. (M) (b) A.474 M.485 M 3 Rae = = =. M s 8.4 s 7.5 s A.498 M.565 M Rae = = =.67 Mmin. min. min A.433 M.498 M Rae = = =.65 Mmin. min. min The raes are n equal because, in all excep zer-rder reacins, he rae depends n he cncenrain f reacan. And, f curse, as he reacin prceeds, reacan is cnsumed and is cncenrain decreases, s he rae f he reacin decreases. 5. (M) A = A + A =.588 M.3 M =.575 M i (b) A =.565 M.588 M =.3 M.3 M = A = =. min A. M/min ime = + = 4.4 +. min = 5.4 min 6. (M) Iniial cncenrains are HgCl =.5 M and CO 4 =.3 M. The iniial rae f he reacin is 7. 5 M min. Recall ha he reacin is: HgCl (aq) + CO 4 (aq) Cl (aq) + CO (g) + HgCl (aq). The rae f reacin equals he rae f disappearance f C O 4. Then, afer hur, assuming ha he rae is he same as he iniial rae, 5 ml CO4 ml HgCl 6 min HgCl =.5 M 7. h.96 M L s ml CO4 h F I HG K J 5 ml 6min (b) CO 4 =.3 M 7. h Lmin h =.96 M 68
Chaper 4: Chemical Kineics 7. (M) (b) (c) 8. (M) (b) (c) A C Rae = = =.76 M s 5 C =.76 5 M s = 3.5 5 M/s A C = =.76 M s Assume his rae is cnsan. 5 6 s A =.358 M +.76 M s.min =.357 M min A =.76 5 M s A.35 M.358 M = = = 4.5 s 5 5.76 M/s.76 M/s 4 n O 5.7 ml HO ml O 4 =. L sln =.9 ml O /s L sln s ml H O n O 4 ml O 6 s =.9 =.7 ml O / min s min V O ml O =.7 min,44 ml O a STP 3.8 = ml O ml O a STP min 9. (M) Nice ha, fr every mmhg drp in he pressure f A(g), here will be a crrespnding mmhg rise in he pressure f B(g) plus a mmhg rise in he pressure f C(g). We se up he calculain wih hree lines f infrmain belw he balanced equain: () he iniial cndiins, () he changes ha ccur, which are relaed each her by reacin sichimery, and (3) he final cndiins, which simply are iniial cndiins + changes. A(g) B(g) + C(g) Iniial. mmhg. mmhg. mmhg Changes. mmhg +. mmhg +. mmhg Final. mmhg. MmHg. mmhg Tal final pressure =. mmhg +. mmhg +. mmhg = 3. mmhg (b) A(g) B(g) + C(g) Iniial. mmhg. mmhg. mmhg Changes. mmhg +4. mmhg +. mmhg Final 8 mmhg 4. mmhg. mmhg Tal pressure = 8. mmhg + 4. mmhg +. mmhg = 4. mmhg 69
Chaper 4: Chemical Kineics. (M) (b) (c) We will use he ideal gas law deermine N O 5 pressure mln O 5 L am. g.86 73 + 65 K nrt 8.g ml K 76 mmhg P{N O } = = = 3 mmhg 5 V 5 L am Afer.38 min, ne half-life passes. The iniial pressure f NO 5 decreases by half 6.5 mmhg. Frm he balanced chemical equain, he reacin f ml NO5bgg prduces 4 ml NObgg and ml Obgg. Tha is, he cnsumpin f ml f reacan gas prduces 5 ml f prduc gas. When measured a he same emperaure and cnfined he same vlume, pressures will behave as amuns: he reacin f mmhg f reacan prduces 5 mmhg f prduc. 5 mmhg(prduc) P = 3 mmhg N O ( iniially) 6.5 mmhg N O ( reacan) + 6.5 mmhg(reacan) al 5 5 mmhg(reacan) (3 6.5 6) mmhg 3 mmhg Mehd f Iniial Raes. (M) Frm Exp. Exp. 3, [A] is dubled, while [B] remains fixed. This causes he rae 4 6.75 M s increases by a facr f =.. 4 3.35 M s Thus, he reacin is firs-rder wih respec A. Frm Exp. Exp., [B] dubles, while [A] remains fixed. This causes he rae 3.35 M s increases by a facr f =4.34. 4 3.35 M s Thus, he reacin is secnd-rder wih respec B. (b) Overall reacin rder rder wih respec A rder wih respec B = = 3. The reacin is hird-rder verall. 4 (c) Rae = 3.35 M s = b.85 M gb.33 Mg 4 3.35 M s = =. M s.85 M.33 M b gb g. (M) Frm Exp. and Exp. we see ha [B] remains fixed while [A] riples. As a resul, he iniial rae increases frm 4. 3 M/min.3 M/min, ha is, he iniial reacin rae riples. Therefre, he reacin is firs-rder in [A]. Beween Exp. and Exp. 3, we see ha [A] dubles, which wuld duble he rae, and [B] dubles. As a cnsequence, he iniial rae ges frm.3 M/min 5. M/min, ha is, he rae quadruples. Since an addiinal dubling f he rae is due he change in [B], he reacin is firs-rder in [B]. Nw we deermine he value f he rae cnsan. 6
Chaper 4: Chemical Kineics Rae 5. M/ min Rae = A B = = =5.8 A B 3. M 3. M The rae law is Rae = 5.8 3 L ml min A B c h. L ml min 3 3. (M) Frm Experimen, [NO] remains cnsan while [Cl ] is dubled. A he same ime he iniial rae f reacin is fund duble. Thus, he reacin is firs-rder wih respec [Cl ], since dividing reacin by reacin gives = x when x =. Frm Experimen 3, [Cl ] remains cnsan, while [NO] is dubled, resuling in a quadrupling f he iniial rae f reacin. Thus, he reacin mus be secnd-rder in [NO], since dividing reacin 3 by reacin gives 4 = x when x =. Overall he reacin is hird-rder: Rae = [NO] [Cl ]. The rae cnsan may be calculaed frm any ne f he experimens. Using daa frm Exp., = 5 Rae.7 M s = [NO] [Cl ] (.5 M) (.55 M) = 5.7 M s 4. (M) Frm Exp. Exp., [B] remains cnsan a.4 M and [C] remains cnsan a. M, bu [A] is halvedb.5g. A he same ime he rae is halvedb.5g. Thus, he x reacin is firs-rder wih respec A, since.5 =.5 when x =. Frm Exp. Exp. 3, [A] remains cnsan a.7 M and [C] remains cnsan a. M, bu [B] is halvedb.5g, frm.4 M.7 M. A he same ime, he rae is y quarered.5 when y =. b g. Thus, he reacin is secnd-rder wih respec B, since.5 =.5 Frm Exp. Exp. 4, [A] remains cnsan a.4 M and [B] remains cnsan a.5, frm. M.5 M. A he same ime, he rae is.4 M, bu [C] is halved increased by a facr f.. Rae 4 = 6 Rae 3 = 6 Rae = 4 Rae = 4 Rae = Rae. 4 z Thus, he rder f he reacin wih respec C is, since.5 =. when z =. (b).4 M.4 M. M rae 5 =.7 M.7 M.5 M = +.4 M.4 M. M = rae = rae = 4rae This is based n rae =.4 M.4 M. M 6
Chaper 4: Chemical Kineics Firs-Order Reacins 5. (E) TRUE The rae f he reacin des decrease as mre and mre f B and C are frmed, bu n because mre and mre f B and C are frmed. Raher, he rae decreases because he cncenrain f A mus decrease frm mre and mre f B and C. (b) FALSE The ime required fr ne half f subsance A reac he half-life is independen f he quaniy f A presen. 6. (E) FALSE Fr firs-rder reacins, a pl f ln [A] r lg [A] vs. ime yields a sraigh line. A graph f [A] vs. ime yields a curved line. (b) TRUE The rae f frmain f C is relaed he rae f disappearance f A by he sichimery f he reacin. 7. (M) Since he half-life is 8 s, afer 9 s five half-lives have elapsed, and he riginal quaniy f A has been cu in half five imes. final quaniy f A = b.5g 5 iniial quaniy f A =.35 iniial quaniy f A Abu 3.3% f he riginal quaniy f A remains unreaced afer 9 s. r Mre generally, we wuld calculae he value f he rae cnsan,, using ln.693 = = =.385 s Nw ln(% unreaced) = - = -.385 s - (9s) = - / 8 s 3.465 (% unreaced) =.33 % = 3.3% f he riginal quaniy. (b) Rae = A =.385 s.5 M =.93 M / s 8. (M) (b) The reacin is firs-rder, hus A. M.8 ln = = ln = 54 min( ) = =.385 min A.8 M 54 min We may nw deermine he ime required achieve a cncenrain f.5 M A.5 M 3.47 ln = = ln =.385 min ( ) = = 9. min A.8 M.385 min Since we nw he rae cnsan fr his reacin (see abve), 4 Rae = A =.385 min.5 M = 9.6 M/min 6
Chaper 4: Chemical Kineics 9. (M) (b) The mass f A has decreased ne furh f is riginal value, frm.6 g.4 g. Since 4 =, we see ha w half-lives have elapsed. Thus, / = 38 min, r / =9 min..693 A =.693/ / = =.36 min ln = =.36 min 6 min =. 9 min A A = e. =. r A = A e =.6 g A. =.8 g A A. (M) A.63 M.56 ln = = ln =.56 = =.6 min A.86 M 6. min.693.693 = = = 43.3 min.6 min (b) / (c) We need slve he inegraed rae equain find he elapsed ime. A.35 M ln = = ln =.45 =.6 min A.86 M A = (d) ln A = becmes A A e which in urn becmes.45 = = 77.8min.6 min 6 min h A = A e =.86 M exp.6 min.5 h =.86.97 =.74 M. (M) We deermine he value f he firs-rder rae cnsan and frm ha we can calculae he half-life. If he reacan is 99% decmpsed in 37 min, hen nly % (.) f he iniial cncenrain remains. A. 4.6 ln = = ln = 4.6 = 37min = =.336 min A. 37 min =.693.693 =.336 min / =.6 min 63
Chaper 4: Chemical Kineics. (E) If 99% f he radiaciviy f 3 P is ls, % (.) f ha radiaciviy remains. Firs we cmpue he value f he rae cnsan frm he half-life. =.693 =.693 =.485 d / 4.3 d Then we use he inegraed rae equain deermine he elapsed ime. ln A ln A = ln days A =. = A.485 d. =95 3. (D) (b) 4. (M) 35 [A] ln = ln(.35) = = (4.8 3 min ) = 8 min. [A] Ne: We did n need nw he iniial cncenrain f aceaceic acid answer he quesin. Le s assume ha he reacin aes place in a.l cnainer. ml aceaceic acid. g aceaceic acid =.9795 ml aceaceic acid..9 g aceaceic acid Afer 575 min. (~ 4 half lives, hence, we expec ~ 6.5% remains as a rugh apprximain), use inegraed frm f he rae law find [A] = 575 min. [A] ln = = (4.8 3 min )(575 min) =.766 [A] [A] = e.766 [A] =.693 (~ 6.3% remains) =.63 [A] = 6. 3 [A].9795 mles mles. [A] reaced = [A] [A] = (.98 6. 3 ) mles =.9 mles aceaceic acid. The sichimery is such ha fr every mle f aceaceic acid cnsumed, ne mle f CO frms. Hence, we need deermine he vlume f.98 mles CO a 4.5 C (97.65 K) and 748 rr (.984 am) by using he Ideal Gas law. Lam.98 ml.86 97.65 K nrt Kml V = = =.3 L CO P.984 am A.5 g 4 ln = = ln = 3.47 = 6. s A 8. g 3.47 3 = = 5.6 s 93 min 4 6. s We subsiued masses fr cncenrains, because he same subsance (wih he same mlar mass) is presen iniially a ime, and because i is a clsed sysem. 64
Chaper 4: Chemical Kineics (b) amun O = 77.5 g N O 5 ml N O5 ml O 8. g N O ml N O 5 5 =.359 ml O Lam.359 ml O.86 (4573) K nrt ml K V 9.56 L O P am 745 mmhg 76 mmhg 5. (D) (b) (c) 6. (D) If he reacin is firs-rder, we will bain he same value f he rae cnsan frm several ses f daa. A.497 M.88 3 ln = = ln = s =.88, = =.88 s A.6 M s A.344 M.556 ln = = ln = 3 s =.556, = =.85 s A.6 M 3 s 3 A.85 M.744 ln = = ln = 4 s =.744, = =.86 s A.6 M 4 s 3 A.98 M.9 ln = = ln = 6 s =.9, = =.85 s A.6 M 6 s 3 A.94 M.854 ln = = ln = s =.854, = =.85 s A.6 M s 3 The virual cnsancy f he rae cnsan hrughu he ime f he reacin cnfirms ha he reacin is firs-rder. Fr his par, we assume ha he rae cnsan equals he average f he values bained in par..88 +.85+.86 +.85 3 3 = s =.86 s 4 We use he inegraed firs-rder rae equain: 3 A = A exp =.6 M exp.86 s 75 s 75.4 A =.6 M e =.48 M 75 If he reacin is firs-rder, we will bain he same value f he rae cnsan frm several ses f daa. P 64 mmhg ln = = ln = 39 s =.67, =.67 P 3 mmhg 39 =4.8 4 s s P 4 mmhg ln = = ln = 777 s =.33, =.33 P 3 mmhg 777 =4.6 4 s s P 87 mmhg ln = = ln = 95 s =.5, =.5 P 3 mmhg 95 =4.8 4 s s 65
Chaper 4: Chemical Kineics (b) (c) P 78.5 mmhg.38 ln = = ln = 355 s =.38, = = 4.37 s P 3 mmhg 355 s The virual cnsancy f he rae cnsan cnfirms ha he reacin is firs-rder. Fr his par we assume he rae cnsan is he average f he values in par : 4.3 4 s. 4 A 39 s, he pressure f dimehyl eher has drpped 64 mmhg. Thus, an amun f dimehyl eher equivalen a pressure f b3 mmhg 64 mmhg = g 48 mmhg has decmpsed. Fr each mmhg pressure f dimehyl eher ha decmpses, 3 mmhg f pressure frm he prducs is prduced. Thus, he increase in he pressure f he prducs is 3 48=44 mmhg. The al pressure a his pin is 64 mmhg +44 mmhg = 48 mmhg. Belw, his calculain is dne in a mre sysemaic fashin: bch 3g Obgg CH g 4b g + Hbgg + CObgg Iniial 3 mmhg mmhg mmhg mmhg Changes 48 mmhg + 48 mmhg + 48 mmhg + 48 mmhg Final 64 mmhg 48 mmhg 48 mmhg 48 mmhg Pal = PDME + Pmehane + Phydrgen + PCO 64 mmhg 48 mmhg 48 mmhg 48 mmhg 48 mmhg (d) This quesin is slved in he same manner as par (c). The resuls are summarized belw. bch 3g Obg g CH g H g CO g 4bg bg + bg Iniial 3 mmhg mmhg mmhg mmhg Changes 3 mmhg 3 mmhg 3 mmhg + 3 mmhg Final mmhg 3 mmhg 3 mmhg 3 mmhg P = P + P + P + P al DME mehane hydrgen CO mmhg 3 mmhg 3 mmhg 3 mmhg 936 mmhg P 4 (e) We firs deermine P DME a s. ln = = 4.3 s s =.43 P P.43 = e =.65 P = 3 mmhg.65 = 3 mmhg P Then we use he same apprach as was used fr pars (c) and (d) CH Og CH g + H g + COg 3 4 Iniial 3 mmhg mmhg mmhg mmhg Changes 9 mmhg 9 mmhg 9 mmhg 9 mmhg Final 3 mmhg 9 mmhg 9 mmhg 9 mmhg P = P + P + P + P al DME mehane hydrgen CO = 3 mmhg +9 mmhg +9 mmhg +9 mmhg = 53. mmhg 66
Chaper 4: Chemical Kineics Reacins f Varius Orders 7. (M) Se II is daa frm a zer-rder reacin. We nw his because he rae f se II is cnsan..5 M/5 s =. M s. Zer-rder reacins have cnsan raes f reacin. (b) A firs-rder reacin has a cnsan half-life. In se I, he firs half-life is slighly less han 75 sec, since he cncenrain decreases by slighly mre han half (frm. M.47 M) in 75 s. Again, frm 75 s 5 s he cncenrain decreases frm.47 M. M, again by slighly mre han half, in a ime f 75 s. Finally, w half-lives shuld see he cncenrain decrease ne-furh f is iniial value. This, in fac, is wha we see. Frm s 5 s, 5 s f elapsed ime, he cncenrain decreases frm.37 M.8 M, i.e., slighly less han ne-furh f is iniial value. Nice ha we cann mae he same saemen f cnsancy f half-life fr se III. The firs half-life is s, bu i aes mre han 5 s (frm s 5 s) fr [A] again decrease by half. (c) Fr a secnd-rder reacin,/ A / A =. Fr he iniial s in se III, we have =. L ml = s, =. L ml s.5 M. M Fr he iniial s, we have =. L ml = s, =. L ml s.33 M. M Since we bain he same value f he rae cnsan using he equain fr secnd-rder ineics, se III mus be secnd-rder. 8. (E) Fr a zer-rder reacin (se II), he slpe equals he rae cnsan: = A / =. M/ s =. M/s 9. (M) Se I is he daa fr a firs-rder reacin; we can analyze hse iems f daa deermine he half-life. In he firs 75 s, he cncenrain decreases by a bi mre han half. This implies a halflife slighly less han 75 s, perhaps 7 s. This is cnsisen wih he her ime perids ned in he answer Review Quesin 8 (b) and als he fac ha in he 5-s inerval frm 5 s s, he cncenrain decreases frm.6 M.4 M, which is a bi mre han a facr-f-fur decrease. The facr-f-fur decrease, ne-furh f he iniial value, is wha we wuld expec fr w successive half-lives. We can deermine he half-life mre accuraely, by baining a value f frm he relain ln A / A = fllwed by / =.693/ Fr insance, ln(.78/.) = - (5 s); = 9.94-3 s -. Thus, / =.693/9.94-3 s - = 7 s. 3. (E) We can deermine an apprximae iniial rae by using daa frm he firs 5 s. A.8 M. M Rae = = =.8 M s 5 s s 67
Chaper 4: Chemical Kineics 3. (M) The apprximae rae a 75 s can be aen as he rae ver he ime perid frm 5 s s. A. M.5 M RaeII = = =. M s s 5 s (b) A.37 M.6 M RaeI = = =.48 M s s 5 s A.5 M.67 M (c) RaeIII = = =.34 M s s 5 s Alernaively we can use [A] a 75 s (he values given in he able) in he m relainship Rae = A, where m =,, r. Rae =. M s.5 ml/l =. M s II (b) Since 7s,.693 / 7s.99s I / s Rae.99 (.47 ml/l).47 M s Rae =. L ml s.57 ml/l =.3 M s (c) III 3. (M) We can cmbine he apprximae raes frm Exercise 3, wih he fac ha s have elapsed, and he cncenrain a s. A II =. M There is n reacan lef afer s. (b) A = A s rae =.37 M s.47 M s =.3 M I b g c h b g c h (c) A = A s rae =.5 M s.3 M s =.47 M III 33. (E) Subsiue he given values in he rae equain bain he rae f reacin. 34. (M) c hb g b g 4 Rae = A B =.3 M min.6 M 3.83 M =.39 M / min (b) A firs-rder reacin has a cnsan half-life. Thus, half f he iniial cncenrain remains afer 3. minues, and a he end f anher half-life 6. minues al half f he cncenrain presen a 3. minues will have reaced: he cncenrain has decreased ne-quarer f is iniial value. Or, we culd say ha he reacin is 75% cmplee afer w half-lives 6. minues. A zer-rder reacin prceeds a a cnsan rae. Thus, if he reacin is 5% cmplee in 3. minues, in wice he ime 6. minues he reacin will be % cmplee. (And in ne-fifh he ime 6. minues he reacin will be % cmplee. Alernaively, we can say ha he rae f reacin is %/6. min.) Therefre, he ime required fr he 6. min reacin be 75% cmplee = 75% = 45 min. % 68
Chaper 4: Chemical Kineics 35. (M) Fr reacin: HI(g) / H (g) / I (g) (7 K) Time [HI] (M) ln[hi] /[HI](M ) (s)...9.5..8..35 3.74.3.35 4.68.386.47 /[HI] (M - ).6.4. Pl f /[HI] vs ime y =.8x +.997 Frm daa abve, a pl f /[HI] vs. yields a sraigh line. The reacin is secnd-rder in HI a 7 K. Rae = [HI]. The slpe f he line = =.8 M s. 5 5 Time(s) 36. (D) We can graph /[ArSO H] vs. ime and bain a sraigh line. We can als graph [ArSO H] vs. ime and ln([arso H]) vs. ime demnsrae ha hey d n yield a sraigh line. Only he pl f /[ArSO H] versus ime is shwn. /[ArSOH] (M - ) /[ArSOH] (M - ) 5 3 Pl f /[ArSO H] versus Time Pl f /[ArSO H] versus Time y =.37x + 9.464 5 5 5 3 Time (min) (b) The lineariy f he line indicaes ha he reacin is secnd-rder. We slve he rearranged inegraed secnd-rder rae law fr he rae cnsan, using he lnges ime inerval. = A A A A F HG = =.37 L ml min 3 min.96 M. M (c) We use he same equain as in par (b), bu slved fr, raher han. = = = 73. min A A.37 L ml min.5 M. M I KJ = 69
Chaper 4: Chemical Kineics (d) We use he same equain as in par (b), bu slve fr, raher han. = = = 9 min A A.37 L ml min.5 M. M (e) We use he same equain as in par (b), bu slve fr, raher han. = = = 36 min A A.37 L ml min.35 M. M 37. (M) [A] (ml/l) Pl [A] vs, ln[a] vs, and /[A] vs and see which yields a sraigh line..8.7.6.5.4.3.. y = -.5x +.75 R =. Time(s) 5 ln[a] -.5 - -.5 - -.5-3 -3.5 5 y = -.9x -.9 R =.99 Time(s) 5 y =.8x -.965 R =.795 /[A] 5 Time(s) 5 Clearly we can see ha he reacin is zer-rder in reacan A wih a rae cnsan f 5. -3. (b) The half-life f his reacin is he ime needed fr ne half f he iniial [A] reac..358 M Thus, A =.75 M =.358M and / = 5. 3 =7 s. M/s 63
Chaper 4: Chemical Kineics 38. (D) We can eiher graph / C4H 6 vs. ime and bain a sraigh line, r we can deermine he secnd-rder rae cnsan frm several daa pins. Then, if indeed is a cnsan, he reacin is demnsraed be secnd-rder. We shall use he secnd echnique in his case. Firs we d a bi f algebra. = F I A A A A = (b) F I = =.843 L ml min.8 min HG.44 M.69 M K J F = =.875 L ml min 4.55 min.4 M.69 M = 4.5 min M M HG I HG K J F I =.89 L ml min HG.3.69 K J F I =.87 L ml min HG.845.69 K J = 68.5 min M M The fac ha each calculain generaes similar values fr he rae cnsan indicaes ha he reacin is secnd-rder. The rae cnsan is he average f he values bained in par..843+.875 +.89 +.87 = L ml min =.87 L ml min 4 KJ (c) We use he same equain as in par, bu slve fr, raher han. = = =. min A A.87 L ml min.43 M.69 M (d) We use he same equain as in par, bu slve fr, raher han. = = =.6 min A A.87 L ml min.5 M.69 M 39. (E) A.49 M.5 M iniial rae = = = +. M / min. min. min A.935 M 3.4 M iniial rae = = = +.89 M / min. min. min (b) When he iniial cncenrain is dubledb.g, frm.5 M 3.4 M, he iniial rae x quadruples 4. when x =). b g. Thus, he reacin is secnd-rder in A (since. = 4. 63
Chaper 4: Chemical Kineics 4. (M) Le us assess he pssibiliies. If he reacin is zer-rder, is rae will be cnsan. During.6 M.8 M /8 min =.3 M/min. Then, during he firs he firs 8 min, he rae is 4 min, he rae is.35 M.8 M /4 min =.9 M/min. Thus, he reacin is n zer-rder. If he reacin is firs-rder, i will have a cnsan half-life ha is cnsisen wih is rae cnsan. The half-life can be assessed frm he fac ha 4 min elapse while he cncenrain drps frm.8 M. M, ha is, ne-furh f is iniial value. Thus, 4 min equals w half-lives and / = min. This gives =.693/ / =.693/ min =.35 min. Als.35 = ln A M = ln =.87 = 4 min =.87 =.34 min A.8 M 4 min The cnsancy f he value f indicaes ha he reacin is firs-rder. (b) The value f he rae cnsan is =.34 min. (c) Reacin rae = (rae f frmain f B) = A Firs we need [A] a =3. min A A ln = =.34 min 3. min =. = e =.36 A. A A =.36.8 M =.9 M rae f frmain f B =.34 min.9 M =. M min 4. (M) The half-life f he reacin depends n he cncenrain f A and, hus, his reacin cann be firs-rder. Fr a secnd-rder reacin, he half-life varies inversely wih he reacin rae: / =/ c A h r =/ c A h. Le us aemp verify he secnd-rder naure f his / reacin by seeing if he rae cnsan is fixed. = =. L ml min. M5 min = =. L ml min. M 5 min = =. L ml min.5 M min The cnsancy f he rae cnsan demnsraes ha his reacin indeed is secnd-rder. The rae equain is Rae = A and =. L ml min. 63
Chaper 4: Chemical Kineics 4. (M) The half-life depends n he iniial NH 3 and, hus, he reacin cann be firs-rder. Le us aemp verify secnd-rder ineics. = fr a secnd-rder reacin = = 4 M min NH.3 M 7.6 min 3 / = = 8 M min = = 865 M min.5 M 3.7 min.68 M.7min The reacin is n secnd-rder. Bu, if he reacin is zer-rder, is rae will be cnsan. A /.3 M 4 Rae = = =. M/min / 7.6 min.5 M 4 Rae = =. M/min 3.7 min.68 M 4 Rae =. M / min Zer-rder reacin.7 min (b) The cnsancy f he rae indicaes ha he decmpsiin f ammnia under hese cndiins is zer-rder, and he rae cnsan is =. 4 M/min. [A] 43. (M) Zer-rder: / = Secnd-rder: / = [A] A zer-rder reacin has a half life ha varies prprinally [A], herefre, increasing [A] increases he half-life fr he reacin. A secnd-rder reacin's half-life varies inversely prprinal [A], ha is, as [A] increases, he half-life decreases. The reasn fr he difference is ha a zer-rder reacin has a cnsan rae f reacin (independen f [A] ). The larger he value f [A], he lnger i will ae reac. In a secnd-rder reacin, he rae f reacin increases as he square f he [A], hence, fr high [A], he rae f reacin is large and fr very lw [A], he rae f reacin is very slw. If we cnsider a bimlecular elemenary reacin, we can easily see ha a reacin will n ae place unless w mlecules f reacans cllide. This is mre liely when he [A] is large han when i is small. 44. (M) [A] =.693 [A] Hence, =.693 r [A] =.39 M [A] (b), Hence, [A] (c) [A].693, Hence,.693 = [A] = r [A] =. M [A] =.44 M [A] r [A] =.44 M 633
Chaper 4: Chemical Kineics Cllisin Thery; Acivain Energy 45. (M) (b) (c) 46. (M) (b) 47. (M) The rae f a reacin depends n a leas w facrs her han he frequency f cllisins. The firs f hese is wheher each cllisin pssesses sufficien energy ge ver he energy barrier prducs. This depends n he acivain energy f he reacin; he higher i is, he smaller will be he fracin f successful cllisins. The secnd facr is wheher he mlecules in a given cllisin are prperly riened fr a successful reacin. The mre cmplex he mlecules are, r he mre freedm f min he mlecules have, he smaller will be he fracin f cllisins ha are crrecly riened. Alhugh he cllisin frequency increases relaively slwly wih emperaure, he fracin f hse cllisins ha have sufficien energy vercme he acivain energy increases much mre rapidly. Therefre, he rae f reacin will increase dramaically wih emperaure. The addiin f a caalys has he ne effec f decreasing he acivain energy f he verall reacin, by enabling an alernaive mechanism. The lwer acivain energy f he alernaive mechanism, (cmpared he uncaalyzed mechanism), means ha a larger fracin f mlecules have sufficien energy reac. Thus he rae increases, even hugh he emperaure des n. The acivain energy fr he reacin f hydrgen wih xygen is quie high, high, in fac, be supplied by he energy rdinarily available in a mixure f he w gases a ambien emperaures. Hwever, he spar supplies a suiably cncenraed frm f energy iniiae he reacin f a leas a few mlecules. Since he reacin is highly exhermic, he reacin f hese firs few mlecules supplies sufficien energy fr ye her mlecules reac and he reacin prceeds cmplein r he eliminain f he limiing reacan. A larger spar simply means ha a larger number f mlecules reac iniially. Bu he evenual curse f he reacin remains he same, wih he iniial reacin prducing enugh energy iniiae sill mre mlecules, and s n. The prducs are J/ml clser in energy he energy acivaed cmplex han are he reacans. Thus, he acivain energy fr he reverse reacin is 84 J / ml J / ml = 63 J / ml. 634
Chaper 4: Chemical Kineics (b) The reacin prfile fr he reacin in Figure 4- is seched belw. A.... B Transin Sae Penial Energy (J) E a (reverse) = 63 J E a (frward) = 84 J H = + J Prducs C + D Reacans A + B Prgress f Reacin 48. (M) In an endhermic reacin (righ), E a mus be larger han he H fr he reacin. Fr an exhermic reacin (lef), he magniude f E a may be eiher larger r smaller han ha f H. In her wrds, a small acivain energy can be assciaed wih a large decrease in he enhalpy, r a large E a can be cnneced a small decrease in enhalpy. reacans prducs ΔH< prducs reacans ΔH> 49. (E) (b) (c) (d) There are w inermediaes (B and C). There are hree ransiin saes (peas/maxima) in he energy diagram. The fases sep has he smalles E a, hence, sep 3 is he fases sep in he reacin wih sep a clse secnd. Reacan A (sep ) has he highes E a, and herefre he slwes smalles cnsan (e) Endhermic; energy is needed g frm A B. (f) Exhermic; energy is released mving frm A D. 635
Chaper 4: Chemical Kineics 5. (E) (b) (c) (d) There are w inermediaes (B and C). There are hree ransiin saes (peas/maxima) in he energy diagram. The fases sep has he smalles E a, hence, sep is he fases sep in he reacin. Reacan A (sep ) has he highes E a, and herefre he slwes smalles cnsan (e) Endhermic; energy is needed g frm A B. (f) Endhermic, energy is needed g frm A D. Effec f Temperaure n Raes f Reacin 5. (M) 4 E F a I 5.4 L ml s Ea ln = = F I ln = RHG T TKJ.8 L ml s R HG 683 K 599 K K J 3.95 R= E.5 4 a R Ea = 3.95 4 5 =.93 K 8.345 J ml K =.6 J / ml = 6 J / ml 4.5 5. (M) 3 5 E 5. L ml s.6 J/ml ln = a = ln = R T T.8 L ml s 8.345 J ml K 683 K T 4.7 5.7 =.9 = = 8.96 4 683 K T 683 K T.9 = 8.96 5 +.46 3 =.55 3 T = 645 K T 53. (D) Firs we need cmpue values f ln and /T. Then we pl he graph f ln versus /T. T,C C C C 3 C T, K 73 K 83 K 93 K 33 K / T, K.366.353.34.33, s 5.6 6 3. 5.6 4 7.6 4 ln.9.35 8.74 7.8 636
Chaper 4: Chemical Kineics Pl f ln versus /T.35 /T (K - ).345.365-7. -8. ln -9. -. -. y = -35x + 37.4 -. (b) The slpe = Ea/ R. J 4 J J Ea = R slpe = 8.345.35 K = ml K J ml (c) We apply he Arrhenius equain, wih =5.6 6 s a C (73 K), =? a 4C (33 K), and E a = 3 3 J/ml. 3 E a J/ml ln = = = 6.36 6 5.6 s R T T 8.345 J ml K 73 K 33 K 6.36 6 3 e = 548 = = 5485.6 s = 3.7 s 6 5.6 s 54. (D) Here we pl ln vs./t. The slpe f he sraigh line equals E / a R. Firs we abulae he daa pl. (he pl is shwn belw). T, C 5.83 3. 59.75 9.6 T, K 88.98 35.7 33.9 363.76 /T, K.3464.3769.339.749, M s 5.3 5 3.68 4 6.7 3.9 ln 9.898 7.97 5.4.9 - -4.693.693 3.7 s / = = =.3 s 3 Pl f ln versus /T.7.9 /T (K - ).3.33.35 ln -6-8 - y = -89x + 7.77-637
Chaper 4: Chemical Kineics 4 The slpe f his graph =.9 K = Ea / R 4 J 4 J J J Ea =.89 K8.345 = 9.54 = 9.5 ml K ml J ml (b) We calculae he acivain energy wih he Arrhenius equain using he w exreme daa pins..9 E a Ea ln = ln = +7.77 = = 5 5.3 R T T R 88.98 K 363.76 K 4 Ea 7.7698.345 J ml K 4 = 7.38 K Ea = = 9.8 J/ml 4 R 7.38 K Ea =9J/ml. The w E a values are in quie gd agreemen, wihin experimenal limis. (c) We apply he Arrhenius equain, wih E a = 9.8 4 J/ml, =5.3 5 M s a 5.83 C (88.98 K), and =? a. C (373. K). 3 E a 9.8 J/ml ln = = 5 5.3 M s R T T 8.345 J ml K 88.98 K 373. K 8.58 3 ln = 8.58 e = 5.5 = 5 5 5.3 M s 5.3 M s 3 5 = 5.5 5.3 M s =.54 M s 55. (M) The half-life f a firs-rder reacin is inversely prprinal is rae cnsan: =.693/ /. Thus we can apply a mdified versin f he Arrhenius equain find E a. (b) / E a 46. min E a ln = ln = = ln = / R T T.6 min R 98 K + 73 K Ea 4.888.345 J ml K J.88 = 6.89 Ea = = 34.8 J/ml 4 R 6.89 K J 3. min 34.8 J/ml F I = 46. 8.345 98 =.53 = 4.9 3 ln min J ml K HG K J F I HG 98 K J T T.53 4 3 = = 3.65 =.99 T = 334 K = 6 C 3 T 98 4.9 T 56. (M) The half-life f a firs-rder reacin is inversely prprinal is rae cnsan: =.693/ /. Thus, we can apply a mdified versin f he Arrhenius equain. / E a.5 h E a R T T.5 h R 93 K 4 + 73 K Ea 4.78.345 J ml K J.7 =.8, Ea = = 3 J/ml 4 R.8 K J ln = ln = = ln = / 638
b Chaper 4: Chemical Kineics (b) The relainship is = A exp Ea / RT 3 3 3 J ml 3 4.9 5 =.5 s exp =.5 s e 3.5 s 8.345 J ml K 73 + 3 K 57. (M) I is he change in he value f he rae cnsan ha causes he reacin g faser. Le be he rae cnsan a rm emperaure, C (93 K). Then, en degrees higher (3 C r 33 K), he rae cnsan =. E a Ea 4 E ln = ln =.693 = = =.3 K a R T T R 93 33 K R.693 8.345 J ml K 4 E a = =5. J / ml =5J / ml 4.3 K (b) Since he acivain energy fr he depiced reacin (i.e., N O + NO N + NO ) is 9 J/ml, we wuld n expec his reacin fllw he rule f humb. 58. (M) Under a pressure f. am, he biling pin f waer is apprximaely C r 394 K. Under a pressure f am, he biling pin f waer is C r 373 K. We assume an acivain energy f 5. 4 J / ml and cmpue he rai f he w raes. ln Rae 4 E F I a 5. J/ml = = F I =.88 Rae RHG T TKJ 8.345 J ml K HG 373 394 K K J.88 Rae = e Rae =.4 Rae. Cing will ccur.4 imes faser in he pressure cer. Caalysis 59. (E) (b) g Alhugh a caalys is recvered unchanged frm he reacin mixure, i des ae par in he reacin. Sme caalyss acually slw dwn he rae f a reacin. Usually, hwever, hese negaive caalyss are called inhibirs. The funcin f a caalys is change he mechanism f a reacin. The new mechanism is ne ha has a differen (lwer) acivain energy (and frequenly a differen A value), han he riginal reacin. 6. (M) If he reacin is firs-rder, is half-life is min, fr in his ime perid [S] decreases frm. M.5 M, ha is, by ne half. This gives a rae cnsan f =.693/ / =.693/ min =.693 min. The rae cnsan als can be deermined frm any w f he her ses f daa. A. M.357 = ln = ln =.357 = 6 min = =.595 min A.7 M 6 min This is n a very gd agreemen beween he w values, s he reacin is prbably n firsrder in [A]. Le's ry zer-rder, where he rae shuld be cnsan. 639
Chaper 4: Chemical Kineics.9 M. M.5 M. M Rae = =.5 M/min Rae = =.5 M/min min min. M.9 M.5 M.9 M Rae = =.5 M/min Rae = =.5 M/min 6 min min min min Thus, his reacin is zer-rder wih respec [S]. 6. (E) Bh plainum and an enzyme have a meal cener ha acs as he acive sie. Generally speaing, plainum is n disslved in he reacin sluin (heergeneus), whereas enzymes are generally sluble in he reacin media (hmgeneus). The ms impran difference, hwever, is ne f specificiy. Plainum is raher nnspecific, caalyzing many differen reacins. An enzyme, hwever, is quie specific, usually caalyzing nly ne reacin raher han all reacins f a given class. 6. (E) In bh he enzyme and he meal surface cases, he reacin ccurs in a specialized lcain: eiher wihin he enzyme pce r n he surface f he caalys. A high cncenrains f reacan, he limiing facr in deermining he rae is n he cncenrain f reacan presen bu hw rapidly acive sies becme available fr reacin ccur. Thus, he rae f he reacin depends n eiher he quaniy f enzyme presen r he surface area f he caalys, raher han n hw much reacan is presen (i.e., he reacin is zer-rder). A lw cncenrains r gas pressures he reacin rae depends n hw rapidly mlecules can reach he available acive sies. Thus, he rae depends n cncenrain r pressure f reacan and is firs-rder. 63. (E) Fr he sraigh-line graph f Rae versus [Enzyme], an excess f subsrae mus be presen. 64. (E) Fr human enzymes, we wuld expec he maximum in he curve appear arund 37C, i.e., nrmal bdy emperaure (r pssibly a slighly elevaed emperaures aid in he cnrl f diseases (37-4 C). A lwer emperaures, he reacin rae f enzyme-acivaed reacins decreases wih decreasing emperaure, fllwing he Arrhenius equain. Hwever, a higher emperaures, hese emperaure sensiive bichemical prcesses becme inhibied, prbably by emperaure-induced srucural mdificains he enzyme r he subsrae, which preven frmain f he enzyme-subsrae cmplex. Reacin Mechanisms 65. (E) The mleculariy f an elemenary prcess is he number f reacan mlecules in he prcess. This mleculariy is equal he rder f he verall reacin nly if he elemenary prcess in quesin is he slwes and, hus, he rae-deermining sep f he verall reacin. In addiin, he elemenary prcess in quesin shuld be he nly elemenary sep ha influences he rae f he reacin. 66. (E) If he ype f mlecule ha is expressed in he rae law as being firs-rder cllides wih her mlecules ha are presen in much larger cncenrains, he reacin will seem depend nly n he amun f hse ypes f mlecules presen in smaller cncenrain, since he larger cncenrain will be essenially unchanged during he curse f he reacin. Such a siuain is quie cmmn, and has been given he name pseud firs-rder. I is als pssible have mlecules which, d n paricipae direcly in he reacin including prduc mlecules 64
Chaper 4: Chemical Kineics srie he reacan mlecules and impar hem sufficien energy reac. Finally, cllisins f he reacan mlecules wih he cnainer walls may als impar adequae energy fr reacin ccur. 67. (M) The hree elemenary seps mus sum give he verall reacin. Tha is, he verall reacin is he sum f sep sep sep 3. Hence, sep = verall reacin sep sep 3. Ne ha all species in he equains belw are gases. Overall: NO + H N + HO NO + H N + HO NO N O N O NO Firs: Third NO+H N +HO r N +HO NO+H The resul is he secnd sep, which is slw: H + NO HO + NO The rae f his rae-deermining sep is: Rae = H NO Since N O des n appear in he verall reacin, we need replace is cncenrain wih he cncenrains f species ha d appear in he verall reacin. T d his, recall ha he firs sep is rapid, wih he frward reacin ccurring a he same rae as he reverse reacin. NO = frward rae = reverse rae = NO. This expressin is slved fr NO, which hen is subsiued in he rae equain fr he verall reacin. NO NO = Rae = H NO The reacin is firs-rder in H and secnd-rder in [NO]. This resul cnfrms he experimenally deermined reacin rder. 68. (M) Prpsed mechanism: - I(g) I(g) Observed rae law: I(g) + H (g) HI(g) Rae = [I ][H ] I (g) + H (g) HI(g) The firs sep is a fas equilibrium reacin and sep is slw. Thus, he prediced rae law is Rae = [I] [H ]. In he firs sep, se he rae in he frward direcin fr he equilibrium equal he rae in he reverse direcin. Then slve fr [I]. Rae frward = Rae reverse Use: Rae frward = [I ] and Rae reverse = - [I] Frm his we see: [I ] = - [I]. Rearranging (slving fr [I] ) [I] [I ] = Subsiue in Rae = [I] [I ] [H ] = [H ] = bs [I ][H ] - - Since he prediced rae law is he same as he experimenal rae law, his mechanism is plausible. 64
Chaper 4: Chemical Kineics 69. (M) Prpsed mechanism: - Cl (g) Cl(g) Observed rae law: Cl(g) + NO(g) NOCl(g) Rae = [Cl ][NO] Cl (g) + NO(g) NOCl(g) The firs sep is a fas equilibrium reacin and sep is slw. Thus, he prediced rae law is Rae = [Cl] [NO] In he firs sep, se he rae in he frward direcin fr he equilibrium equal he rae in he reverse direcin. Then express [Cl] in erms f, - and [Cl ]. This mechanism is alms cerainly n crrec because i invlves a era mlecular secnd sep. Rae frward = Rae reverse Use: Rae frward = [Cl ] and Rae reverse = - [Cl] Frm his we see: [Cl ] = - [Cl]. Rearranging (slving fr [Cl] ) [Cl] [Cl ] = Subsiue in Rae = [Cl] [NO] [Cl ] = [NO] = bs [Cl ][NO ] - - There is anher plausible mechanism. Cl (g) + NO(g) NOCl(g) + Cl(g) - Cl(g) + NO(g) NOCl(g) Rae frward = Rae reverse - Cl (g) + NO(g) NOCl(g) Use: Rae frward = [Cl ][NO] and Rae reverse = - [Cl][NOCl] Frm his we see: [Cl ][NO] = - [Cl][NOCl]. Rearranging (slving fr [Cl]) [Cl ][NO] [Cl ][NO] [Cl] = Subsiue in Rae = [Cl][NO] = -[NOCl] -[NOCl] If [NOCl], he prduc is assumed be cnsan (~ M using mehd f iniial raes), hen cnsan = bs Hence, he prediced rae law is bs [Cl ][NO] which agrees wih -[NOCl] he experimenal rae law. Since he prediced rae law agrees wih he experimenal rae law, bh his and he previus mechanism are plausible, hwever, he firs is dismissed as i has a eramlecular elemenary reacin (exremely unliely have fur mlecules simulaneusly cllide). 7. (M) A pssible mechanism is: Sep : O3 O + Ofas Sep : O+ O 3 O slw 3 The verall rae is ha f he slw sep: Rae = 3 O O 3. Bu O is a reacin inermediae, whse cncenrain is difficul deermine. An expressin fr [O] can be fund by assuming ha he frward and reverse fas seps prceed wih equal speed. O3 Rae = Rae O 3= OO O = O3 O 3 3 Rae = O 3 O 3 = O O Then subsiue his expressin in he rae law fr he reacin. This rae equain has he same frm as he experimenally deermined rae law and hus he prpsed mechanism is plausible. 64
Chaper 4: Chemical Kineics 7. (M) S S S :S S:S S:S SS S :S S :S d ds:s S S S :S S:S SS 7. (M) ds:s S S S :S d CH CO (aq) + OH CH C O CH (aq) + H O (l) 3 3 3 3 CH C O CH (aq) + CH CO (aq) Pr d We ne ha CH 3 C(O)CH is an inermediae species. Using he seady sae apprximain, while is cncenrain is n nwn during he reacin, he rae f change f is cncenrain is zer, excep fr he very beginning and wards he end f he reacin. Therefre, d CH3COCH CH3 CO OH CH3COCH HO d CH3COCH CH3 CO Rearranging he abve expressin slve fr CH 3 C(O)CH gives he fllwing expressin CH3 COOH CH3COCH HO CH3 CO The rae f frmain f prduc, herefre, is: dprd CH3COCH CH3 CO d CH3 COOH CH3 CO H O CH 3 CO CH3 CO OH HO CH3 CO 643
Chaper 4: Chemical Kineics INTEGRATIVE AND ADVANCED EXERCISES 73. (M) The daa fr he reacin saring wih. M being firs-rder r secnd-rder as well as ha fr he firs-rder reacin using. M is shwn belw Time (min) [A] =. M (secnd rder) [A] =. M (firs rder) [A] =. M (secnd rder) [A] =. M (firs rder).... 5.63.55.9..46.3.59.6 5.36.65.435.33 5.5.5.86. Clearly we can see ha when [A] =. M, he firs-rder reacin cncenrains will always be lwer han ha fr he secnd-rder case (assumes magniude f he rae cnsan is he same). If, n he her hand, he cncenrain is abve. M, he secnd-rder reacin decreases faser han he firs-rder reacin (remember ha he half-life shrens fr a secnd-rder reacin as he cncenrain increases, whereas fr a firs-rder reacin, he half-life is cnsan). Frm he daa, i appears ha he crssver ccurs in he case where [A] =. M a jus ver minues. Secnd-rder a minues:. ( min) [A] M min [A] =.549 M Firs-rder a minues:. ln[a] ln() ( min) min [A] =.534 M A quic chec a.5 minues reveals,.(.5 min) Secnd-rder a.5 minues: [A] M min [A] =.568 M Firs-rder a.5 minues:.(.5 min) ln[a] ln() M min [A] =.567 M Hence, a apprximaely.5 minues, hese w pls will share a cmmn pin (pin a which he cncenrain versus ime curves verlap). 74. (M) The cncenrain vs. ime graph is n linear. Thus, he reacin is bviusly n zer-rder (he rae is n cnsan wih ime). A quic l a varius half lives fr his reacin shws he ~.37 min (. M.5 M), ~.3 min (.8 M.4 M), and ~.38 min(.4 M. M). Since he half-life is cnsan, he reacin is prbably firs-rder. (.37.3.38).693.693 (b) average /.36 min.94min 3 /.36 min r perhaps beer expressed as =.9 min due imprecisin. 644
Chaper 4: Chemical Kineics (c) When 3.5 min,[a].35 M. Then, rae = [A] =.94 min.35 M.3 M/min. (d) [A].48 M.339 M Slpe Rae.637 M / min Rae.64 M/min. 6. min3. min (e) Rae = [A] =.94 min. M =.9 4 M/min. 75. (M) The reacin being invesigaed is: - + + MnO 4 (aq) +5HO (aq) +6H (aq) Mn (aq) +8HO(l) +5O (g) We use he sichimeric cefficiens in his balanced reacin deermine [H O ]. -. mml MnO4 5 mml HO 37. ml iran - ml iran mml MnO4 [HO ].86 M 5. ml 76. (D) We assume in each case ha 5. ml f reacing sluin is iraed. - -.3 mml HO mml MnO4 ml iran vlume MnO 5. ml ml 5 mml H O. mml MnO -..3 M HO 46.4 ml. M MnO 4 iran a s A s V iran =.. M H O = 4. 38.5 ml A 4 s V iran =..7 M H O = 34.433 ml 7.5 A 6 s V iran =..49 M H O = 9.8 ml A s V iran =..98 M H O = 9. 6.5 6 ml Tangen line A 8 s V iran =..6 M H O =. 4 ml 5.5 A 3 s V iran =..5 M H O = 5. ml 4-4 7 4 8 The graph f vlume f iran vs. elapsed ime is given abve. This graph is f apprximaely he same shape as Figure 4-, in which [H O ] is pled agains ime. In rder deermine he rae, he angen line a 4 s has been drawn n he graph. The inerceps f he angen line are a 34 ml f iran and 8 s. Frm his infrmain we deermine he rae f he reacin. - 34 ml L. ml MnO4 5 ml HO - 8 s ml L iran ml MnO4 4 Rae 6. M/s.5 L sample This is he same as he value f 6. 4 bained in Figure 4- fr 4 s. The discrepancy is due, n dub, he carse naure f ur pl. 645
Chaper 4: Chemical Kineics 77. (M) Firs we cmpue he change in [H O ]. This is hen used deermine he amun, and ulimaely he vlume, f xygen evlved frm he given quaniy f sluin. Assume he O (g) is dry. [HO ] 3 6 s [H O ].7 M/s. min. M min. ml HO ml O amun O.75 L sln.89 ml O L ml HO Lam.89 ml O.86 (734) K nrt Vlume O ml K. L O P am 757 mmhg 76 mmhg 78. (M) We nw ha rae has he unis f M/s, and als ha cncenrain has he unis f M. The generalized rae equain is Rae A. In erms f unis, his becmes M/s M/s = {unis f } M. Therefre {Unis f } M s M 79. (M) Cmparing he hird and he firs lines f daa, [I - ] and [OH - ]say fixed, while [OCl ] dubles. Als he rae fr he hird ineics run is ne half f he rae fund fr he firs run. Thus, he reacin is firs-rder in [OCl - ]. Cmparing he furh and fifh lines, [OCl ] and [I - ] say fixed, while [OH - ] is halved. Als, he fifh run has a reacin rae ha is wice ha f he furh run. Thus, he reacin is minus firs-rder in [OH - ]. Cmparing he hird and secnd lines f daa, [OCl - ] and [OH - ] say fixed, while he [I - ] dubles. Als, he secnd run has a reacin rae ha is duble ha fund fr he hird run. Thus, he reacin is firsrder in [I - ]. (b) The reacin is firs-rder in [OCl - ] and [I - ] and minus firs-rder in [OH - ]. Thus, he verall rder = + =. The reacin is firs-rder verall. (c) [OCl ][I ] Rae [OH ] 4 Rae [OH ] 4.8 M/s. M using daa frm firs run: [OCl ][I ].4 M. M 6. s 8. (M) We firs deermine he number f mles f N O prduced. The parial pressure f N O(g) in he we N O is 756 mmhg.8 mmhg = 743 mmhg. am 743 mmhg.5 L PV 76 mmhg amun N O.7 ml N O RT.86 L am ml K (73 5) K Nw we deermine he change in [NH NO ]. 646
Chaper 4: Chemical Kineics ml NHNO.7 ml NO ml NO [NH NO ].5 M.65 L sln.693 [NHNO ] final.5 M.5 M.93 M.563 min 3 min [A].93 M ln ln min elapsed ime [A].563 min.5 M 8. (D) We need deermine he parial pressure f ehylene xide a each ime in rder deermine he rder f he reacin. Firs, we need he iniial pressure f ehylene xide. The pressure a infinie ime is he pressure ha resuls when all f he ehylene xide has decmpsed. Because w mles f prduc gas are prduced fr every mle f reacan gas, his infinie pressure is wice he iniial pressure f ehylene xide. P iniial = 49.88 mmhg = 4.94 mmhg. Nw, a each ime we have he fllwing. (CH ) O(g) CH 4(g) CO(g) Iniial: 4.94 mmhg Changes: x mmhg +x mmhg +x mmhg Final: 4.94 + x mmhg Thus, x = P 4.94 and P EO = 4.94 x = 4.94 (P 4.94) = 49.88 P Hence, we have, he fllwing values fr he parial pressure f ehylene xide., min 4 6 P EO, mmhg 4.94.74 98. 77.3 6.73 37.54. Fr he reacin be zer-rder, is rae will be cnsan. (.74 4.94) mmhg The rae in he firs min is: Rae.4 mmhg/min min (77.3 4.94) mmhg The rae in he firs 4 min is: Rae.9 mmhg/min 4 min We cnclude frm he nn-cnsan rae ha he reacin is n zer-rder. Fr he reacin be firs-rder, is half-life mus be cnsan. Frm 4 min min a perid f 6 min he parial pressure f ehylene xide is apprximaely halved, giving an apprximae half-life f 6 min. And, in he firs 6 min, he parial pressure f ehylene xide is apprximaely halved. Thus, he reacin appears be firs-rder. T verify his enaive cnclusin, we use he inegraed firsrder rae equain calculae sme values f he rae cnsan. P.74 mmhg ln ln. min P min 4.94 mmhg 37.54 mmhg 6.73 mmhg ln. min ln. min min 4.94 mmhg 6 min 4.94 mmhg The cnsancy f he firs-rder rae cnsan suggess ha he reacin is firs-rder. P 8. (M) Fr his firs-rder reacin P ln Elapsed ime is cmpued as: ln P P We firs deermine he pressure f DTBP when he al pressure equals mmhg. 647