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Equations

Equations. Rate = k [A] m [B] n gen. rate law (no pdts). 1st Order 2nd Order 0 Order. FACTORS OF INFLUENCE RXN RATE. How fast is a rxn? characteristic of each reactant.

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Equations

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

  2. 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: physical state [reacts] of reacts Temp Catalyst incr [ ] incr collisions incr rate w/ incr T; KE (collisions) incr rxn w/o being involved; extra E source g-l-s; collisions; SA

  3. 1. [REACTS] higher [ ] > is more particles > is more collisions > more rxn occurs 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 Lets state: rates depend on collisions (E & orientation)

  4. 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)

  6. rxn: A ------> B as [B]2? [B]1 >

  7. CP 1 #1 Calculate the average rate of disappearance of A over time interval from 20s to 4s based on the figures below: t =0 s t = 20 s t = 40s 1 mol A 0.54 mol A 0.30 mol A 0 mol B 0.46 mol B 0.70 mol B Show complete steps: ??? M/s rxn rate = -[A]/t = - (0.30 - 0.54)M/(40 - 20)s = 0.24/20 = 0.012 M/sA #2 What is the rate of appearance of B from time interval of 0 to 40 s ?? ??? M/s rxn rate = +[B]/t = (0.70 - 0.00)M/(40 - 0)s = 0.70/40 = 0.0175 M/s

  8. RATES: Change, 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

  9. [ ]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

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

  11. C4H9Cl + H2O -----> C4H9OH + HCl Notice trend in Rate value as time increases

  12. Notice as time increases, slope of line flattens, smaller value SOoooooooo

  13. CP 2 Using graph 14.4 (pg. 577) #3) What is the instantaneous rate at t = 0. ?? M/s #4) Calculate the instantaneous rate of the disappearance of reactant at t = 300 s. ?? M/s #5) What is the rxn rate for the time interval of 250 s to 500 s. ?? M/s #3) At t = 0, instantaneous rate is line tangent (0, 0.100) & (210, 0.060) Rate = -Y/X = - (0.060 - 0.100)/(210 - 0) = 1.9*10-4 M/s #4) (100, 0.078) (500, 0.030) Rate =- (0.030 - 0.078)/(500 - 100) = 1.2*10-4 M/s #5) (250 , 0.060) (500, 0.0368) Rate =- (0.0368 - 0.060)/(500 - 250) = 9.3*10-5 M/s

  14. 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

  15. SUMMARY ave rate: D[ ] over D time rate slows as rxn progresses, reacts decr, fewer collisions instantaneous: line tangent to curve at a specific time initial: instantaneous rate @ t = 0, only reacts no pdts

  16. RXN RATES & EQUATIONS 1:1 C4H9Cl + H2O -----> C4H9OH + HCl Rate: - [C4H9Cl]/t = [C4H9OH]/t 2:1 2 HI -----> H2 + I2 - 0.5[HI]/t = [H2]/t - 0.5[HI]/t = [I2]/t [A] decr same rate as [X] & 2 [B]; rate [A] & [X] = 0.5 rate [B] OR Varied Coeff 1 A + 1 X <---> 2 B depends on what is reference subst

  17. GENERAL FORM Rate related to [react] or [pdt] aA + xX <----> bB + yY

  18. CP 3 EX. __ N2O5(g) ----> __ NO2(g) + __ O2(g) #6) How is the rate at which [N2O5] related to: a) the rate of [oxygen] b) the rate of [NO2] #7) At a particular time, the instantaneous rate of N2O5 decreases at 4.2*10-7 mol/L-s, find rate appearance of NO2 #8) What would be the rate of disappearance of N2O5 if the rate of NO2 is 2.1*10-7 M/s?

  19. EX. ___ N2O5(g) ----> ___ NO2(g) + ___ O2(g)

  20. CP 4 EX. __ NO(g) + __ O2(g) <----> __ N2O3(g) #9) Express rate in terms change in [NO], [O2], [N2O3] #10) [O2] decreases at 0.50 mol/L-s, find rate [NO] #11) [NO] decr at 1.60*10-4 mol/L-s, find rate [N2O3]

  21. EX. ___ NO(g) + ___ O2(g) <----> ___ N2O3(g)

  22. RATE LAW 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 k = rate/[reacts] aA + bB ----> cC + dD Exponents: m & n reaction order relates how rate affected by [ ]

  23. 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

  24. Practice Problem 2 NO + 2 H2 ----> N2 + 2 H2O Exp # [react 1] [react 2] Rate 1 0.10 0.10 1.23*10-3 2 0.10 0.20 2.46*10-3 3 0.40 0.10 4.92*10-3 Determine the rate constant, k, assume: rate =k [NO][H2] Use any experiment data values, will obtain same constant value Exp 1: k = 1.23*10-3/(0.10 *0.10) = Exp 2: k = 2.46*10-3/(0.10 *0.20) = Exp 3: k = 4.92*10-3/(0.40 *0.10) = 0.123 M-1/s

  25. 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

  26. 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

  27. (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

  28. 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

  29. 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],

  30. CP 5 Reaction Order What is the order of each reactant & overall order for each #12) 2 H2 + O2 ----> 2 H2O rate = k[H2][O2] #13) 2 P + 3 Cl2 ----> 2 PCl3 rate =k[Cl2]2 #14) HCl + NH3 ----> NH4Cl rate= k[HCl][NH3]2 order in ovrall order in overall order in overall

  31. 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; M-2/s order = 0; M/s order = 2; 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 - (-2), so order = 3 M not appear, exponent = 0, so 0 pwr is 1, so order = 1 1 - order, 1 - (-1), so order = 2

  32. Practice Problems 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 CP 6 #15) 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]

  33. 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 Trial [A] M [B] M rate C (M/s) -4 1 0.1 0.1 9.0*10 -3 2 0.3 0.1 8.1*10 -3 3 0.1 0.2 1.8*10 1st: determine order for each reactant by comparing 2 trials, [ ] only 1 reactant changes

  34. 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

  35. 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

  36. 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

  37. 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

  38. 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

  39. 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

  40. Guess Correctly: Guess Wrong: graph linear graph curve Look at 0, 1st,2nd order as most common Rate laws of form: Rate = [A]n

  41. 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 Each order has unique “input” & “output” for straight line plot Decomp of : HI -- 0.5 H2 + 0.5 I2 @ 443oC; k = 30.0 L/molmin rate eqn of –([HI]/t) = k[HI]2 What is the time required for [HI] to decrease from 0.010 M to 0.0050 M? Notice: 2nd Order Rxn = use integrated 2nd Order Rate Law solve for “t”: t = [1/[A] - 1/[A]o]/k (1/[0.0050] - 1/[0.010])/30.0 (200 M – 100 M)/(30.0 L/molmin) t = 3.33 min

  42. Determine half-life of HI at this temp? How much is left after 3 half-lives? t0.5 = 1/(k[A]o) = 1/(30.0[0.010]) = 1/(0.30) = 3.33 min amt left = [A]o/2n n = # half-lives 23 = 8 amt = [0.010]/8= 1.25*10-3 mol

  43. Now, if this were a 1st order rxn, –([HI]/t) = k[HI], what time is required? use integrated 1st Order Rate Law solve for “t”: ln[A] = ln[A]O - kt t = (ln[A]O - ln[A])/k = ln([0.010]/[0.0050])/30.0 = ln(2 M)/(30.0 L/molmin) t = 0.023 min t0.5 = ln 2/k = 0.693/30.0 = 0.0231 min amt left after 3 half-lives is the same, since time does not affect the outcome amt = [0.010]/8= 1.25*10-3 mol

  44. 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 [A] 0.35 0.3 0.25 0.2 [A] 0.15 0.1 0.05 0 0 5 10 15 20 25

  45. ln[A] 0 0 5 10 15 20 25 -0.5 -1 -1.5 ln[A] -2 -2.5 -3 -3.5 1/[A] 30 25 20 15 1/[A] 10 5 0 0 5 10 15 20 25

  46. 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] ?????

  47. 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]

  48. 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

  49. 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] amt remaining = (initial amt)(1/2n) n = # half-lives

  50. PRACTICE PROBLEMS 1. How many grams are left after 3 half-lives of a 10.0 g solution that is 1st order? 2. C4H9Cl react w/ water is 1st order. Estimate the half-life and use this to calculate k.

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