/5/4 Chaper 4 Chemical Kineics Chemical Kineics Raes of Reacions Chemical Kineics is he sudy of he rae of reacion. How fas does i ake place? Very Fas Reacions Very Slow Reacions Acid/Base Combusion Rusing Explosive Decomposiion Decomposiion of H O Radioacive Decay
# of paricles /5/4 Collision Theory Reacion raes are no consan. They are acually dependan on he condiions under which he reacion akes place. How can we make a reacion go faser? How do we increase he rae of a reacion? Rule #: For a reacion o ake place, reacans mus collide!! Bu even if hey do collide, i does no guaranee a reacion!! Energy These wo reacans canno reac because hey are no in conac wih one anoher. They may evenually run ino each oher, bu hey need ime o diffuse Number of paricles ha acually reac Collision Theory Rule #: When molecules collide, hey mus be in an opimal orienaion. RXN! (maybe) Unfruiful Collisions (no reacions) Correc orienaion for a reacion Rule #3: When molecules collide, hey mus have a cerain minimum amoun of energy in order for he reacion o occur. Alhough his collision seems o be in he correc orienaion, he reacans do no have enough energy RXN! No enough energy (no reacions) Enough energy for a reacion
Energy Energy Energy /5/4 Collision Theory and Acivaion Energy (E a ) *Acivaed Complex/Transiion Sae The acivaion energy (E a ) for a reacion is he energy barrier ha mus be overcome for he reacion o ake place. Similar o a roller coaser Reacans E a ΔH Reacion Progress Producs uli-sep reacion * * E a E a A muli-sep reacion has wo E a and wo ransiion saes ΔH Exohermic vs. Endohermic Reacions E a ΔH E a ΔH Reacion Progress Producs have lower energy han reacans Energy released during reacion -ΔH EXOTHERIC Reacion Progress Producs have higher energy han reacans Energy absorbed during reacion +ΔH ENDOTHERIC 3
/5/4 Concenraion and Rae Lower concenraion of reacans Higher concenraion of reacans Higher concenraion of reacans allow more opporuniies for collisions. Surface Area and Rae Can only collide wih paricles on he surface of he solid If he paricles are made finer, here are many more opporuniies for collisions Higher surface area of reacans increase he rae of a reacion 4
# of paricles /5/4 Temperaure and Rae Why does emperaure affec rae? ore collisions: Since paricles move faser a higher emperaures here is more opporuniy o inerac/collide. ore energy: ore paricles will have enough energy o overcome E a Disribuion of energies of paricles a differen emperaures T E a Area = number of paricles ha reac a lower emp Area = number of paricles ha reac a higher emp T Noice ha increasing he emperaure does no change he E a Lower Energy Higher Energy ore reacan paricles reach E a = faser reacion Caalyss and rae Reacans *Acivaed Complex/Transiion Sae E a E a ΔH Reacion Progress Producs Since he reacion akes a differen roue, a differen ransiion sae and differen mechanism will be involved Noice ha he ΔH of he caalyzed and unanalyzed reacions are he same Caalyss allow a differen pah for he reacion o ale place. A pah wih a lower E a. Caalyss are regeneraed in a reacion. Tha is, hey are no consumed 5
# of paricles /5/4 Caalyss Caalyss allow more paricles o reac per uni of ime Caalyzed E a (new rxn roue) E a Disribuion of paricles ha reach E a in a caalyzed vs. non-caalyzed reacion Caalysis lowers E a allowing more paricles o reac (less energy is needed) Energy ore reacan paricles reach E a = faser reacion Area = number of paricles ha reac a lower emp Area = number of paricles ha reac a higher emp Biological Caalyss / Caalysis any reacions in he human body are caalyzed by proeins called enzymes. Enzymes caalyze reacions by binding o reacan paricles called subsraes. Allows unfavorable reacions o ake place under biological condiions. Kineics of enzymes are much more complicaed!! 6
Amoun of reacan /5/4 Chemical Kineics Wha does a graphical represenaion of differen reacion rae look like? Le s say we have wo differen decomposiion reacions; one faser han he oher. Reacion A Reacion B Slower Reacion Faser Reacion ime While fas and slow are good qualiaive words when comparing raes, how do we quaniaively deermine his? [reacan( s)] [produc(s)] Raeof reacion os ofen, especially for aqueous soluions, we define amoun as concenraion ( or mol/l) denoed by brackes, [ ]. Time can be in any unis. Very ofen, we use seconds (s). How do we deermine rae of reacion? We can measure he concenraions of reacans or producs various ways. In some soluions, we can use visible specroscopy o deermine concenraion of reacans or producs, usually expressed in or mol/l. Gases can be expressed in parial pressures (am). [reacan( s)] [produc(s)] Raeof reacion Δ = final - iniial A B. = s.5 = 5s.5 = 5s (.5.) Rae (5s- s).5 5s. s (.5.5) Rae (5s- 5s).5 5s. s 7
Concenraion () /5/4 Chemical Kineics Raes of Reacions A B [reacan( s)] [produc(s)] Raeof reacion Δ = final - iniial Time (min) [Reacan] () 5..5.5..5.5.63 (.5 5.) Avg Rae (.5min-.min).5.5min 5. (.6.63).47 Avg Rae.47 (.5min-.5min).min min min..3.5.6 3..8 Avg Rae (.8.6 ) (3. min -.5 min).8.5 min. min We can calculae an average rae of he reacion beween any wo daa poins during he reacion Overall Avg Rae (5..8 ) ( min -3. min) 4.9 3. min.6 min Chemical Kineics Raes of Reacions 5. /min 6. Δ Amoun of reacan 5. Δ Time 4. 3...6 /min...5.5.5 3 Time (min).73 /min Wha abou an insananeous rae? Le s say a.5 minues? 8
[Phenyl Aceae] () /5/4 Chemical Kineics Raes of Reacions [Phenyl Aceae] Time (s) ()..55 5..4 3..3 45..3 6..7 75.. 9..85 Wha is he overall average reacion rae 5.7x -3 /s Wha is he average rae from =.s o =5.s 8.67x -3 /s.6.5.4.3.. 4 6 8 Time (s) Should he average from =5.s o 3.s be higher or lower han =.s o =5.s??? The rae from =5.s o 3.s should be slower Rae of Disappearance of Reacans and Appearance of Producs aa + bb c C Δ Raeof reacion a Δ Δ[B] b Δ Δ[C] c Δ Generic rae of reacion Δ Raeof reacion a Δ Δ[B] b Δ Δ[C] c Δ Rae of change in A Rae of change in B Rae of change in C Noice ha he reacans have a - and he producs do no. If he coefficiens are differen, he raes will be differen 9
/5/4 Rae of disappearance of reacans and appearance of producs based on coefficiens A + B C Δ Raeof reacion Δ Δ[B] Δ Δ[C] Δ If C is appearing a 5.x -5 /s a wha rae is A disappearing? Δ Δ[C] Δ Δ Δ Δ 5 5.x /s Δ 5.x 5 /s Δ If A is disappearing a -. /min a wha rae is B disappearing? Δ Δ[B] Δ Δ Δ[B] Δ[B] Δ[B]. /s./s. /s Δ Δ Δ If B is disappearing a -.44 /min a wha is he (generic) reacion rae? Δ[B] Raeof reacion Raeof reacion -.44/min Δ Raeof reacion. /min Rae of Disappearance of Reacans and Appearance of Producs In he reacion N O 5 (g) 4NO (g) + O (g) he appearance of NO is 3.x - /s. Wha is he rae of disappearance of N O 5?.56 x - /s In he synhesis of Ammonia, N (g) + 3H (g) NH 3 (g), if he rae of disappearance of H is 4.5x -4 /s, wha is he rae of appearance of ammonia, NH 3? 3.x -4 /s
/5/4 Chemical Kineics and Concenraion Rae Law Each reacion can be expressed by wha is called a Rae Law The rae law describes how each reacan s concenraion has an effec on he reacion rae General Rae Law Rae = k x [B] y = concenraion () of reacan A [B] = concenraion () of reacan B k = rae consan (depends on oher unis) Rae = usually in /s or mol/l s x and y = order of each reacan Only reacans ha affec he rae of he reacion are included in he rae law. In oher words, no every reacan ha appears in he balanced equaion will be in he rae law! Deermining Rae Law Deermining he Order of each reacan While i seems like i should work, we should never deermine he Rae Law based on he balanced equaion!! This is because reacions can have muliple inermediae seps no accouned for in he balanced equaion. We can deermine he Rae Law using experimenaion daa of iniial raes. A + B Z Trial Iniial Iniial [B] Iniial Rae..3. /s..6. /s 3.3.3 9. /s Rae = k x [B] y Firs, find wo rials where one reacan has he same concenraion. This why we can cancel he effec of ha reacan has on he rae (For example, rials &). Plug everyhing you know ino he rae equaion. Trial Trial. /s..6 k. /s..3 x y k x y y y = Ignore he reacan wih he same concenraion and ignore k, for now. Solve for he exponen
/5/4 Deermining Rae Law Deermining he Order of Each Reacan So Far Rae = k x [B] We normally do no show he as an exponen Now, we go hrough he same process for A: Trial Iniial Iniial [B] Iniial Rae..3. /s..6. /s 3.3.3 9. /s Trial 3 9. /s.3.3 k. /s..3 x Trial 9 k 3 x y x y 9 3 x = Therefore, he rae law can be wrien as: Rae = k [B] The reacion is second-order wih respec o A and firs-order wih respec o B + = 3 This reacion is 3 rd order overall Deermining Rae Law Shor Cu Trial Iniial Iniial [B] Iniial Rae..3. /s..6. /s 3.3.3 9. /s Trial Iniial Iniial [B] Iniial Rae..3. /s..6. /s 3.3.3 9. /s.6.3.. s s.3. 9.. s s concenraion doubles rae doubles concenraion riples rae nonuples (9x) y x 3 9 s order nd order Therefore, he rae law can be wrien as: Rae = k [B] The reacion is second-order wih respec o A and firs-order wih respec o B + = 3 This reacion is 3 rd order overall
/5/4 Deermining Rae Law Deermining he Rae Law Consan (k) Since k does no change during he reacion, we can use any rial o calculae he value of k Trial Iniial.. Iniial [B].3.6 Iniial Rae. /s. /s 3.3.3 9. /s Cancel unis (./s) Trial Rae = k [B]. /s = k [.] [.3] k (.) (.3) - - k 333 or s s Trial Rae = k [B]. /s = k [.] [.6] (./s) k (.) (.6) - - k 333 or s s Rae Law Rae 333 s [B] A = nd -order B = s -order Overall reacion = 3 rd order k = 333 / s Deermining Rae Law Pracice Use he following daa o deermine he rae law and he value of k for he reacion beween NO and O. Iniial [NO] Iniial [O ] Iniial Rae Trial () () (/s)...8...57 3..4.4 4.4..7 5...4 Trial Iniial () Iniial [B] () Iniial [C] () Rae (/s)....8x -6.3.. 5.4x -6 3.3 3.. 4.86x -3 4.. 3..8x -6 3
/5/4 Deermining Rae Law Tricks Up y Sleeve Trial Iniial () Iniial [B] () Rae (/s)...67... 3.3. 3.67 Trial Iniial () Iniial [B] () Rae (/s)...67... 3.3. 3.67....67 s s.3. 3.67.67 s s concenraion changes.x. y. s order rae changes.x concenraion changes.3x.3 x. WTF?! rae changes.x Deermining Rae Law Tricks Up y Sleeve # ln(a x ) = x ln(a).3 x. Take naural log of boh sides ln Rearrange using ln(a x ) = x ln(a) x.3 ln..3 ln. x ln.3 ln. x ln x.6.778 solve for x x =.96 3 4
/5/4 Deermining Rae Law Group Pracice Trial Iniial P A (am) Iniial P B (am) Rae (am/min) 5.97.5.33.5.5 7.8x -4 3.5 7.6.38x - Iniial Iniial [B] Rae Trial () () (/s)...5x -4...x -4 3.3.3.x -3 *This one may iniially look impossible, bu i s no. Hin: Think abou wha you already know. Inegraed Rae Laws - Rae and Time Wih he help of calculus, we can use anoher form of he rae law ha allows us o calculae how concenraion of a reacan changes wih ime. I is called he inegraed rae law. Remember s -order rae law Rae = k ln k -or- ln k ln nd -order rae law Rae = k k Wha abou a reacion where he concenraion of he reacans ha do no affec he rae. Does i exis? Yup. h -order rae law Rae = k k 5
/5/4 Inegraed Rae Laws s order A cerain decomposiion reacion is s order overall. Wha is he rae consan if he concenraion of A goes from an iniial concenraion of 5.55 o. in 3. minues? ln k k k. ln k(3.min) 5.55 -.69 k(3.min) k =.54min - How long would i ake for he concenraion o reach 5% of is original concenraion? 5% ln (.536min ) % -.693 (.536min ) =.3min Using % direcly in he equaion will only work wih s order since k doesn include a concenraion uni. For oher orders ( nd and h ), use acual concenraions in he equaions. Inegraed Rae Laws s order ore naural log ricks Wha would be he concenraion afer 3 minues? x - ln (.536 min )(3min) 5.55 How do we ge rid of naural log (ln)? ln k k k e ln(x) = x x ln - 5.55 -(.536min )(3min) e e x 5.55-4 9.4x x =.5 x - ln (.536 min )(3min) 5.55 x ln lnx -lny y x ln y CAUTION! DOES NOT lnx lny - lnx-ln5.55 (.536min )(3min) lnx 5.54 x =.5 6
/5/4 Inegraed Rae Laws nd order A second-order decomposiion has a rae consan of.5 - min -. Assuming you sar wih 3., how long will i ake for only 5.% of he reacan o remain. ln k k k Unlike s order, you canno simply use % s. You mus use specific concenraions or amouns for your reacan. 5.% 3..6.% So.. - (.5 min ).6 3. - 5.94 (.5 min ) min. If he same reacion ook 65s o reach., wha was he original concenraion? Wach for ime unis. UST be he same in k and for - (.5 min )(.4min) min 65s.4 min 6s 3.98.5 Half-lives The ime i akes for a reacan o reach half of is iniial amoun 6 8 4 Original concenraion % 5 % 5 %.5 % 6.5 % / / / / half-life half-lives 3 half-lives 4 half-lives / General Equaion for Half-life x Where x = number of half-lives 7
(mol/l) (mol/l) (mol/l) /5/4 Half lives 7 6 5 7 6 s Order nd Order 5 4 3 s half-life 4 3 s half-life nd half-life nd half-life 3 rd half-life 4 h half-life 3 rd half-life 4 h half-life.5.5.5 3 Time (s).5.5.5 3 Time (s) Half lives are equal Half-lives increase over ime ln() k / / k 7 6 5 Inegraed Rae Laws Half lives h Order 4 s half-life 3 nd half-life 3 rd half-life 4 h half-life.5.5.5 3 Time (s) Half lives decrease / k 8
/5/4 Half-life equaions General Equaion for Half-life x Where x = number of half-lives Half-lives are relaed o he rae consan, k, by he following equaions s Order nd Order h Order ln k k k / / / ln() k k k Half-Life Examples A cerain reacion has a rae consan of.5 hr -. Wha is he half life of his reacion in hours? inues?.555hr / / / ln() k k k How many half lives will i ake for he same reacion above o go from 6. o.? 8. half-lives If a nd order reacion has a half life of 3. days, how many hours will i ake for.% of he reacan o disappear if he concenraion sars a.? 9
ln / /5/4 Graphing Inegraed Rae Laws Equaion of a sraigh line slope y = m x + b y-inercep ln s -order rae law ln k ln ln vs. ime / nd -order rae law k / vs. ime h -order rae law k vs. ime Deermining order by graphing To deermine he order of he reacan, you mus graph he daa hree differen ways: Time vs. Conc.; Time vs. ln(conc.); and Time vs. /Conc. Time (s) ().5.333333 3.5 4. 5.66667 6.4857 7.5 8. Time (s) ln -.6935 -.986 3 -.3869 4 -.6944 5 -.7976 6 -.9459 7 -.7944 8 -.97 Time (s) / / 3 3 4 4 5 5 6 6 7 7 8 8 9 Time vs. Conc. Time vs. ln(conc.) Time vs. /(Conc.)
/5/4 ulisep Reacions Reacion echanisms NO + CO NO + CO NO + CO NO + CO? This reacion happens in muliple seps. Bu wha are hey? And how can we figure i ou? Firs we need o learn abou hese seps. ulisep Reacions Reacion echanisms Reacions ha have muliple seps are made of individual seps called elemenary seps. Elemenary seps are single (concered) seps. Sep : NO NO + NO 3 Sep : NO 3 + CO NO + CO NO + CO NO + CO Inermediaes are chemical species ha are creaed during he reacion, bu no a par of he overall reacion equaion. Inermediaes are creaed in an elemenary sep and consumed in anoher Elemenary seps mus add up o he overall reacion equaion. Sep : NO N O 4 Sep : N O 4 + CO NO + CO These are no plausible seps because he seps add up o a differen overall reacion CO CO
/5/4 oleculariy and Rae Laws of Elemenary seps Rae laws for elemenary seps can be deermined direcly from he balanced equaion The moleculariy of an elemenary sep is based on he rae law for ha sep oleculariy Equaion Rae Law Unimolecular Bimolecular Termolecular A Produc(s) A + A Produc(s) A + B Produc(s) A + A +A Produc(s) A + B + B Produc(s) A + B + C Produc(s) Rae = k Rae = k Rae = k[b] Rae = k 3 Rae = k[b] Rae = k[b][c] Since elemenary seps are single seps, rae laws for each sep can be deermined from he balanced chemical equaion Sep : NO NO + NO 3 Sep : NO 3 + CO NO + CO Elemenary Sep Rae = k [NO ] Elemenary Sep Rae = k [NO 3 ][CO] Boh of hese elemenary seps are bimolecular Overall reacion raes of muli-sep reacions Two-sep reacion Elemenary Seps 6, kisses per hour 5, kisses per hour Slow Sep Fas Sep How many kisses can be processed per hour overall??? The overall reacion can go no faser han he slow sep. This sep is called he rae-deermining or rae-limiing sep. Jus remember ha reacion raes are no saic; reacion raes are variable based on concenraion.
/5/4 uliple-sep Reacions Elemenary Seps Rae-limiing Sep s A X Slow-sep s B + X C Overall Chemical Formula A + A X B + X C A + B C Rae < Rae The elemenary-sep formulas mus add up o he overall chemical formula Rae laws for elemenary seps can be deermined direcly from he balanced equaion A X + C B + X C Sep Rae Law Rae = k Sep Rae Law Rae = k [B][X] If he slow, rae-deermining, sep is firs, he overall rae law is he same as he slow sep Sep Rae Law Rae = k Overall Rae Law Rae = k Noice ha he concenraion of B does no affec he rae of reacion = Overall reacion raes of muli-sep reacions Two-sep reacion Elemenary Seps 5, kisses per hour Fas RXN, kisses per hour Slow RXN How many kisses can be processed per hour overall??? 5, kisses per hour Alhough i seems like he overall rae should = he slow sep, in realiy, reacion raes are variable based on concenraions of reacans. 3
/5/4 uliple-sep Reacions Elemenary Seps Rae-limiing Sep nd Slow-sep nd 3 rd A X B + X C Rae > Rae Overall Chemical Formula A + A X B + X C A + B C The elemenary-sep formulas will add up o he overall chemical formula A X + C B + X C However, unlike he rae-limiing sep being firs, If he slow, raedeermining, Sep Rae Law Sep Rae Law sep is second i s much more complicaed Rae = k Rae = k [B][X] uliple-sep Reacions Elemenary Seps Rae-limiing Sep nd If he slow sep is second, he boleneck creaes an equilibrium in he firs sep. Slow-sep nd 3 rd A X B + X C Rae > Rae A X Sep Rae Law Rae = k = k - [X] Sep Rae Law Rae = k [B][X] Equilibrium Some producs in Sep rever back ino reacans before coninuing o sep Inermediaes canno be in he overall rae law. The overall rae law canno be based solely on he Sep rae law. We mus combine boh rae laws. Sep Rae Law Rae = k = k - [X] k [X] k k Rae k [B] k Overall Rae Law Rae k [B] Solve for he inermediae Subsiue Combine k s 4
/5/4 uliple-sep Reacions Elemenary Seps Rae-limiing Sep Noice ha he same reacion wih he same elemenary seps can give us differen overall rae laws because of he locaion of he slow sep. A + B C Slow-sep s Slow sep nd 3 rd A X B + X C A X B + X C Rae < Rae Rae > Rae Overall Rae Law Rae = k Overall Rae Law Rae k [B] How do we know which one is correc for his reacion? Experimenally Deermined Rae Law uliple-sep Reacions Elemenary Seps NO + CO NO + CO has an experimenally deermined rae law of Rae = k[no ] NO + CO NO + CO? Which of he following is mos likely o represen he mechanism for he reacion? NO + CO NO + CO NO NO + NO 3 (slow) NO 3 + CO NO + CO (fas) NO NO + NO 3 (fas) NO 3 + CO NO + CO (slow) 5
# of paricles /5/4 Caalyss # echanisms Caalyzed E a (new rxn roue) E a H O H O + O H O + I - IO - +H O H O + IO - I - + H O + O Energy Area = number of paricles ha reac a lower emp Area = number of paricles ha reac a higher emp ore reacan paricles reach E a = faser reacion Noice ha he caalys, I -, is regeneraed in he end. There is no ne change in caalys in he reacion. The inermediae, IO -, is compleely used in he reacion (as i should be). Arrhenius Equaion Svane Arrhenius deermined ha acivaion energy (E a ) of a reacion is relaed o he reacion consan. Tha is a any emperaure he lower he acivaion energy, he faser a reacion will proceed. k Ae -E RT a lnk lna- E a R T E a lnk - lna R T y = m x + b 6
/5/4 Arrhenius Equaion We can also do calculaions wihou graphing, bu combining wo equaions ino one. If we know boh k and k, along wih T and T we calculae he energy of acivaion Ea lnk lna- RT Ea lnk lna- RT k ln k Ea R J R 8.34 molk T T In a cerain equaion, he rae consan a 7K was measured as.57 - s - and ha a 895 K was measured as 556 - s -. Find he acivaion energy of he reacion. 556 ln.57 s s Ea 8.34 J mol K 7K 895K J mol K Ea -4 5.38 3.9x K.45x 8.34 5 J/mol 7