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CH. 16 KINETICS Study of rxn rates, D es in concen over time

CH. 16 KINETICS Study of rxn rates, D es in concen over time. Look at: How fast is a rxn? [react] & [pdts] at completion Rxn spontaneous (release E) or nonspont (absorb E). Examine: 1. Factors of influence 2. Express rates, determined experimentally

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CH. 16 KINETICS Study of rxn rates, D es in concen over time

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  1. CH. 16 KINETICSStudy of rxn rates, Des in concen over time Look at: How fast is a rxn? [react] & [pdts] at completion Rxn spontaneous (release E) or nonspont (absorb E) Examine: 1. Factors of influence 2. Express rates, determined experimentally 3. Effects of [ ] & T 4. Develop mechanisms 5. Catalysts

  2. Equations Rate = k[A]m[B]n gen. rate law (no pdts) 1st Order 2nd Order 0 Order

  3. FACTORS OF INFLUENCE RXN RATE How fast is a rxn? characteristic of each reactant H2(g) + F2(g) <----> 2 HF(g) FAST 3 H2(g) + N2(g) <----> 2 NH3(g) SLOW (rev. favored) Factors Can Control: [reacts] physical state of reacts Temp Catalyst 1. [REACTS] higher [ ] > is more particles > is more collisions > more rxn occurs

  4. 2. STATE surface area (SA) same phase: thermal motion creates contact diff phase: contact at surface contact, more finely divided solid (liquid) increases SA/V more SA > more collisions > faster rxn occurs 3. Temperature higher T > incr speed > more collisions > faster rxn higher T > incr KE > incr E of collisions SUMMARY

  5. RATE EXPRESSION rxn rate is: Des in [react] & [pdts] per unit time decr reacts while incr pdts 1st determine initial [ ] 2nd determine [ ] later at time, t [A]1> [A]2 rate always “+”; moles/L-s (mol-L-1-s-1) (M/s) (Ms-1) PRODUCT A -------> B as [B]2> [B]1

  6. RATES: Average, Instantaneous, Initial rate for specific rxn will vary over time of reacting coeff = 1 A + X <---> B + Y notice: [A] & [X] decr at same rate if measure D[ ] of 1 react, can follow rate ave rate: rate of rxn over a period of time [ ]0 ----> [ ]20 t0 ----> t20 [ ]4 ----> [ ]8 t4 ----> t8

  7. [ ]12 ----> [ ]15 t12 ----> t15 rate1 > rate2 ave rate interval decr from rate1 to rate2 as expected: [reacts] decr, fewer collisions, rate decr Instantaneous Rate: rate at a particular instant in time; tangent line Initial Rate: instantaneous rate at moment when reacts start rxn; no pdts present fwd & rev rxns need to be considered, as fwd slows, rev quickens, becomes mess to calculate difference in ratefwd <---> raterev

  8. Plot [A] vs t initial rate ave rate [A] . t4 . t8 ave rate .t14 instantaneous rate to t20 time

  9. express rate: [react] & [pdt] A + X <---> B + Y for every 1 mol A disappears, 1 mol Y appears rate same ratio for reacts decr and pdts incr SUMMARY ave rate: D[ ] over D time rate slow as rxn progresses, reacts decr, fewer collisions instantaneous: line tangent to curve at a specific time initial: instantaneous rate at t=0, only reacts no pdts

  10. Varied Coeff 1 A + 1 X <---> 2 B [A] decr same rate as [X] & 2 [B]; rate [A] & [X] = 0.5 rate [B] OR depends on what is reference subst

  11. Rate related to [react] or [pdt] aA + xX <----> bB + yY EX. __ NO(g) + __ O2(g) <----> __ N2O3(g) 1) Express rate in terms change in [NO], [O2], [N2O3] 2) [O2] decreases at 0.50 mol/L-s, find rate [NO] 3) [NO] decr at 1.60*10-4 mol/L-s, find rate [N2O3]

  12. EX. 4 NO(g) + 1 O2(g) <----> 2 N2O3(g)

  13. RATELAW Look at: rxn rate depends only on [react] & T @ fixed T, (no pdts) FORM: Rate = k[A]m[B]n k: proportionality constant, rate constant specific for a given rxn, at a specific T not D as rxn progresses Des w/ T aA + bB ----> cC + dD Exponents: m & n reaction order relates how rate affected by [ ]

  14. aA + bB ----> cC + dD Rate = k[A]m[B]n Rxn coeff (a, b) not related to rxn orders (m, n) RATE, RXN ORDERS, RATE CONSTANT Found experimentally; Not related to balanced eqn coeff Find parts by: 1) use [ ] to find initial rate 2) use initial rates to find rxn orders 3) use values to find rate const, k

  15. Use initial rate data to find rxn orders, rate const RXN ORDER individual order & overall order exponent on [ ] in rate law Order in A refers to exponent on [A] in rate law Overall order refers to S exponent [reacts] in law Single Reactant 1st Order Overall, if rate is dir. proportional to [A] Rate = k[A]1 2nd Order Overall, if rate dir. proportional to sq [A] Rate = k[A]2 0 Order Overall, if rate not dependant to [A] Rate = k[A]0 = k(1) = k

  16. NO + O3 -------> NO2 + O2 rate law determinted exprtmtly Rate = k[NO][O3] 1st Order, respect to NO; [NO]1 1st Order, respect to O3; [O3]1 Overall Order, sum of individual orders 1 + 1 =2 2nd Order Overall 2 NO + 2 H2 ----> N2 + 2 H2O Rate law = k[NO]2[H2] Find --- Order respect to NO: Order in H2: Order Overall: 2nd 1st 3rd2 + 1

  17. (CH3)3CBr + H2O -----> (CH3)3COH + H+ + Br- Order in (CH3)3CBr ---> Order Overall ----> 1st Order 1st Order Rate law = k[(CH3)3CBr] Notice H2O not shown in rate law????? Mean 0 order in H2O ([H2O]0); rate not dependent on [H2O] Also expressed as: rate = [(CH3)3CBr ][H2O]0 Recall, not related to balanced rxn

  18. Reaction Order can be +, 0 usually (or -, fraction) CHCl3 + Cl2 ----> CCl4 + HCl States: rate depends on sq.root of [Cl2], if incr [Cl2] by 4, rate incr by 2 Neg. exponent states rate decr as its [ ] incr for rate law that includes pdt 2 O3<---> 3 O2 Rate = k[O3]2[O2]-1 rate is 1/2 If dbl [O2],

  19. Reaction Order What is the order of each reactant & overall order for each a) 2 H2 + O2 ----> 2 H2O rate =k[H2][O2] b) 2 P + 3 Cl2 ----> 2 PCl3 rate = k[Cl2]2 c) HCl + NH3 ----> NH4Cl rate= k[HCl][NH3]2 1st order in [H2] & [O2] 2nd ovrall 2nd [Cl2] & zero in [P] 2nd overall 1st [HCl] 2nd [NH3] 3rd overall

  20. Rate Constants units depend on overall order units on rate constate: M1-order * time-1 What are the units on the rate constants for the following rate laws, rate is M/s a) rate = k[A]3 b) rate = k c) rate = [R][S] order = 3 order = 0 order = 2 label: M-2/s label: M/s label: M-1/s What is the order of rxn for the following rate constants? a) k = 7.5 s-3 b) k = 66.2 M-2/min c) k = 2.52*102 M-1/s 1 - order, 1 - (-1), so order = 2 1 - order, 1 - (-2), so order = 3 M not appear, exponent = 0, so 0 pwr is 1, so order = 1

  21. Describe the difference between rate of a reaction & rate constant. Rate rxn: D in [ ] over a specific time period Rate Const: is proportional constant in the rxn that measures reactivity of reactants What is the order of the following reaction? 4 Au + O2 + 2 H2O + 8 CN- <----> 4 Au(CN)2 + 4 OH- fwd & rev rxns need to be considered, as fwd slows, rev quickens, becomes mess to calculate difference in ratefwd<---> raterev If a reaction order =3, write 3 different rate laws to describe the reaction. Rate = k[A]3 = k[A]2[B] = k[A][B]2 = k[A][B][C]

  22. Rate Law from Experiments w/ Varying Concentrations of Reactants Order of rxn determinedby comparing rate when [react] ;  1 react at a time to determine rate of react on rate; hint: use higher [ ] on top Determine the rate law & rate constant for: A + B ---> C 1st: determine order for each reactant by comparing 2 trials, [ ] only 1 reactant changes

  23. Use #1 & #2 General Form: rate = k[A]m[B]n #2 (8.1*10-3) = k[0.3]m [0.1]n #1 (9.0*10-4) = k[0.1]m [0.1]n Find m: rate2/rate1k & [B] cancel (8.1*10-3)/(9.0*10-4) = [0.3]m/[0.1]m 9 = [0.3/0.1]m 9 = 3m m = 2 rate = k[A]2[B]n

  24. Determine order of B, use 2 trials where [B] only changes, #1 & #3 Use #1 & #3 General Form: rate = k[A]2[B]n #3 (1.8*10-3) = k[0.1]2 [0.2]n #1 (9.0*10-4) = k[0.1]2 [0.1]n Find n: rate3/rate1k & [A] cancel (1.8*10-3)/(9.0*10-4) = [0.2]n/[0.1]n 2 = [0.2/0.1]n 2 = 2n n = 1 rate = k[A]2[B]1

  25. Determine Rate Constant; use values from any trial & solve for k #1: 9.0*10-4 M/s = k[0.1]2[0.1] 9.0*10-4 M/s = k(1*10-3 M3) 0.9 M-2 /s = k #2: 8.1*10-3 M/s = k[0.3]2[0.1] 0.9 M-2 /s = k #3: 1.8*10-3 M/s = k[0.1]2[0.2] 0.9 M-2 /s = k Any trial can be used to determine rate constant; all give same answer

  26. Trial [AB2] [C] rate CB (M/min) 1 1.0 1.0 1.4*10-7 2 0.5 1.0 3.5*10-8 3 1.0 2.0 5.5*10-7 Determine the rate law & rate constant for: 2 AB2 + C ---> CB + BA Gen form: RATE = k [AB2]m [C]n (1.4*10-7)/(3.5*10-8) =[1.0]m/[0.5]m 4 = 2m 2 =m Compare: #1 & #2 1.4*10-7 = k[1.0]m [1.0]n 3.5*10-8 = k[0.5]m[1.0]n

  27. Compare: #3 & #1 5.5*10-7 = k[1.0]m [2.0]n 1.4*10-7 = k[1.0]m [1.0]n (5.5*10-7)/(1.4*10-7) =[2.0]n/[1.0]n 3.93round to4 = 2n 2 =n RATE = k [AB2]2 [C]2 Rate constant, k use any trial, #1 1.4*10-7 = k [1.0]2 [1.0]2 1.4*10-7 M-3/min = k

  28. INTEGRATED RATE LAWS Using initial rates - disadvantage ^ need multiple experiments ^ works well, slow rxns ^ too fast, large uncertainty Overcome these: use data on [ ] vs t Compare [ ] vs t to prediction made by integrated rate laws Assume: 1. Rxn has certain order 2. Make plot of data 3. Plot is linear if assumptions of rate law correct

  29. Order Rate Law Integrated straight line 0 rate = k [A]=-kt+[A]o [A] vs t 1 = k[A] ln[A]=-kt+ln[A]o ln[A] vs t 2 = k[A]2 1/[A]=2kt+1/[A]o 1/[A] vs t Guess Correctly: Guess Wrong: graph linear graph curve Look at 0, 1st,2nd order as most common Rate laws of form: Rate = [A]n Each order has unique “input” & “output” for straight line plot

  30. H2O2<----> H2O + O2 [H2O2] t(min) 0 order, [A] 1st order, ln[A] 2nd order,1/[A] 0.300 0 0.221 1 0.107 5 0.065 10 0.047 15 0.037 20 0.300 -1.21 3.33 0.221 -1.51 4.52 0.107 -2.23 9.35 0.065 -2.73 15.38 0.047 -3.06 21.28 0.037 -3.30 27.03

  31. Do: graph 0 order; if linear, conclude 0 order but not, check 1st order (ln [A]); not 1st order, check 2nd order (1/[A]); 2nd is linear, rate = k[A]2 slope of line twice rate constant What about determining; rate = k[A][B] ?????

  32. Trial [NH4] [NO2] rate 1 0.24 0.10 7.2*10-6 2 0.12 0.10 3.6*10-6 3 0.12 0.15 5.4*10-6 NH4+1(aq) + NO2-1(aq) <---> N2(aq) + 2 H2O(l) Rate = k[NH4]m[NO2]n Determine m, hold [NO2] constant rate1 =k[NH4]m rate2 = k[NH4]m log r1 = log k + m log[NH4] log r2 = log k + m log[NH4] log r2 = log k + m log[NH4] - log r1 = log k + m log[NH4] log r2 - log r1 = m log[NH4] - m log[NH4] log r2 - log r1 = m (log[NH4] - log[NH4]) log r2 -log r1. log[NH4] - log[NH4] =m

  33. REVIEW Factors that influence Expressions: aA+bB--->cC+dD rate law: rate = k[ ]m[ ]n rate order: m, n overall: m+n rate const: k 0: mol/L-s 1st: 1/s 2nd: L/mol-s Initial rates & integrated law: plot graph; t vs [A], ln[A], 1/[A] find k Half-life time required for [react] to reach 0.5[initial] 0: [A]o/2k 1st: ln 2/k, indep. of [ ] 2nd: 1/k[A]

  34. TEMP EFFECTS k: rate const A: frequency factor e: base e, 2.72 Ea: activation E R: gas const, 8.31 J/mol T: temp, K A: pz, (orientation probability)*(collision frequency) Ea: min E required for react Graph;ln k vs 1/T is linear T effects rate, es k incr T ---> larger k value ---> incr rate

  35. At 2 diff Temp (k1, T1) & (k2, T2) 2 NOCl ---> 2 NO + Cl2 Ea = 1.00*102 kJ/mol k1 = 0.286 L/mol-s @ 500. K 1.00*105 J/mol Given: Ea, k1, T2, T1 Find: k2 @ 490. K

  36. ln k2 - (-1.25) = -0.491 ln k2 = -1.75 k2 = 0.175 L/mol-s Know k = 0.175 L/mol-s find Ea

  37. Ea = (-0.491)*(-8.31)*(2.45*104) = 1.00*105 J/mol

  38. Graphing Method The constant, k, changes with changes in temp. As temp incr, the rate of rxn also incr. Use the data to find the following: a) calculate Ea using all data points b) calculate Ea from k values at 283oC & 508oC c) calcualte k value at 293oC from value in part b at 283oC a) Ea determined from slope of linear plot: ln k vs 1/TK b) Ea from: ln(k2/k1) = c) Ea as in part b, same form:

  39. Slope = -Ea/R ln k 1/T ln k vs 1/T linear relationship!!!! b, y-inter ln [A]

  40. Part a) calculate slope: slope = m = y/x = (-6.83 - -10.41)/(.00143 - .00159) m = -2.24*104 K m = -Ea/R Ea = -mR = -(-2.24*104)*(8.31) = 1.86*105 J (or 186 kJ)

  41. ln k2 - ln 3.52*10-7 = 0.716 ln k2 = 0.716 - 14.86 = 7.20*10-7 L/mol-s

  42. TRANSITION STATE -- Activated Complex unstable, sys highest in E if PE < Ea no rxn takes place E Diagram ----- Rxn Diagram -Hrxn, EXO +Hrxn, ENDO Ea,fwd> Ea,rev Ea,fwd< Ea,rev

  43. Hrxn Ea,fwd Ea,rev reactants pdts Transition state, intermediates form E ---------------> RXN progess ------------->

  44. H = 83 kJ Ea= 210 kJ E ---------------> RXN progess -------------> NO + Cl2 -----> NOCl + Cl 1. Draw & label rxn diagram a) 1-step, H = 83 kJ Ea,fwd = 210 kJ c) catalyst added = 210 - 83 = 127 kJ b) calculate Ea,rev Intermediates N+2; O-2; Cl-1; Cl N-O; Cl- Cl-O pdts react

  45. E ---------------> RXN progess -------------> 2-STEP REACTION 2. Draw & label 2-step rxn diagram pdts react

  46. Catalyst Rate of rxn must be increased to be useful; new pdt formed less $ as high Temp is $$$$$; some subst heat sensitive, decompose Catalyst: 1) increase rate w/o being consumed in rxn 2) enzymes speed bio processes; 3) lowers Ea, incr k, incr rate Homogeneous: catalyst same phase as react & pdt Heterogeneous: catalyst diff phase

  47. RXN MECHANISMS Rxn not a single step, but sequence of single rxn steps that sum overall rxn rate-determining step - RDS - the steps that determines the overall rate of rxn - usually 1st step of multistep rxn - if not, base certain assumptions, base elementary steps on exprmntly found rate law

  48. Find the molecularity & rate law eqn for each elementary rxn: MOLECULARITYRATE LAW a) 2 Cl -------> Cl2 b) O3 ------> O2 + O c) HI + O -----> I2 + H2O “Molecularity” Unimolecular: A ----> pdt Bimolecular: 2A ----> pdt A + B ----> pdt Termolecular: 2A + B ----> pdt A + B + C ----> pdt Bimolec Unimolec Termolec 2

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