Lect w2 152 - rate laws_alg
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  • Update for Tro.
  • Tier 1.5
  • Tier 1.5
  • Tier 1
  • Tier 1.5
  • Tier 2 introduction
  • Tier 2 introduction
  • Since we already went through an example – Let the students work this one. Tier 2
  • Tier 1.5
  • Tier 3
  • Tier 1
  • Tier 1, Tier 1.5
  • Point out that this only works (b) for 1 st order reactions where half-life is not dependent on initial concentration. Tier 1, Tier 1
  • Tier 2

Lect w2 152 - rate laws_alg Presentation Transcript

  • 1. General Chemistry II CHEM 152 Week 2
  • 2. Week 2 Reading Assignment Chapter 13 – Sections 13.2 (rate), 13.3 (rate law), 13.4 (integrated rate law)
  • 3. A more refined description The average reaction rate gives us information about the speed of the reaction over a certain period of time. But what if we are interested in a more detailed description. How do we calculate the instantaneous reaction rate at any particular time?
  • 4. START END Different Rates During a Reaction Overall average rate Initial Rate Instantaneous Rate (slope of the line At a point)
  • 5. . From your lab experience, how could we follow the course of this reaction? Measure rates? What’s changing? What could we measure? Measuring Rates Consider this reaction: Br 2 ( aq ) + HCOOH ( aq ) 2Br - ( aq ) + 2H + ( aq ) + CO 2 ( g ) time
  • 6.  [Br 2 ]   Absorption Br 2 ( aq ) + HCOOH ( aq ) 2Br - ( aq ) + 2H + ( aq ) + CO 2 ( g ) time 393 nm light Detector 393 nm Br 2 ( aq )
  • 7. Instantaneous rate = rate for specific tiny instance in time Br 2 ( aq ) + HCOOH ( aq ) 2Br - ( aq ) + 2H + ( aq ) + CO 2 ( g ) Average rate =   [Br 2 ]  t = - [Br 2 ] final – [Br 2 ] initial t final - t initial slope of tangent slope of tangent slope of tangent
  • 8. rate  [Br 2 ] rate = k [Br 2 ] + 0.0 = rate constant = 3.50 x 10 -3 s -1 Y=mX+b Constant = Slope = k How do the RATES change with [Br 2 ] ? Rate = k [Br 2 ] 1 this is called a 1 st order reaction Rate Law The rate law for a reaction tells us how rate varies with concentration of the reactants. Br 2 (aq) + HCOOH(aq)  HBr(aq) + CO 2 (g) R A T E k = rate [Br 2 ]
  • 9. SOLUTION: For each of the following reactions, determine the reaction order with respect to each reactant and the overall order from the given rate law. (a) The reaction is 2nd order in NO, 1st order in O 2 , and 3rd order overall. Your Turn (a) 2NO( g ) + O 2 ( g ) 2NO 2 ( g ); rate = k [NO] 2 [O 2 ]
    • CH 3 CHO( g ) CH 4 ( g ) + CO( g );
    • rate = k [CH 3 CHO] 3/2
    (c) H 2 O 2 ( aq ) + 3I - ( aq ) + 2H + ( aq ) I 3 - ( aq ) + 2H 2 O( l ); rate = k [H 2 O 2 ][I - ]
  • 10. SOLUTION: For each of the following reactions, determine the reaction order with respect to each reactant and the overall order from the given rate law. (a) The reaction is 2nd order in NO, 1st order in O 2 , and 3rd order overall. Your Turn (a) 2NO( g ) + O 2 ( g ) 2NO 2 ( g ); rate = k [NO] 2 [O 2 ]
    • CH 3 CHO( g ) CH 4 ( g ) + CO( g );
    • rate = k [CH 3 CHO] 3/2
    (c) H 2 O 2 ( aq ) + 3I - ( aq ) + 2H + ( aq ) I 3 - ( aq ) + 2H 2 O( l ); rate = k [H 2 O 2 ][I - ]
  • 11. What is the effect on the rate if the concentration of A is doubled? Consider the reaction: 2A + 2B + C -> 2D + E If the rate law for this reaction is Rate = k [A][B] 2
    • No change
    • x2
    • x4
    • x8
    • x16
  • 12. What is the effect on the rate if the concentration of A is doubled?
    • No change
    • x2
    • x4
    • x8
    • x16
    Rate = k [A][B] 2
  • 13. What is the effect on the rate if the concentrations of A, B & C are all doubled?
    • No change
    • x2
    • x4
    • x8
    • x16
    Rate = k [A][B] 2
  • 14. Measured Reaction Rates
    • The rate of reaction changes with time, as the concentrations of reactants and products change.
    • The rate constant , but not the rate, is independent of time.
    • The instantaneous rate is the rate at any given time and is equal to the slope of a line tangent to the plot of concentration vs. time.
  • 15. Determining Reaction Orders
    • The rate law for a given reaction must be determined experimentally.
    • The order must be found for each species involved.
    • The two major methods are
      • Initial Rate Method
      • Integrated Rate Law Method
  • 16. Rate Laws In general, for a A + b B  x X with a catalyst C Rate = k [A] m [B] n [C] p The exponents m, n, and p are the reaction order with respect each reactant Overall reaction order= m + n + p These numbers can be 0, 1, 2 or fractions and must be determined by experiment ! They are NOT related to the stoichiometric coefficients a,b,x
  • 17. Experiment Initial Reactant Concentrations (mol/L) Initial Rate (mol/L*s) 1 2 3 4 5 O 2 NO 1.10x10 -2 1.30x10 -2 3.21x10 -3 1.10x10 -2 3.90x10 -2 28.8x10 -3 2.20x10 -2 1.10x10 -2 3.30x10 -2 1.30x10 -2 2.60x10 -2 1.30x10 -2 6.40x10 -3 12.8x10 -3 9.60x10 -3 Your Turn
  • 18. Determining Reaction Orders Using Initial Rates Compare 2 experiments in which the concentration of one reactant varies and the concentration of the other reactant(s) remains constant. k [O 2 ] 2 m [NO] 2 n k [O 2 ] 1 m [NO] 1 n = = [O 2 ] 2 m [O 2 ] 1 m = 6.40x10 -3 mol/L*s 3.21x10 -3 mol/L*s = ; 2 = 2 m m = 1 Do a similar calculation for the other reactant(s). Run a series of experiments, each of which starts with a different set of reactant concentrations, and from each obtain an initial rate. O 2 ( g ) + 2NO( g ) 2NO 2 ( g ) rate = k [O 2 ] m [NO] n rate 2 rate 1 [O 2 ] 2 [O 2 ] 1 m 1.10x10 -2 mol/L 2.20x10 -2 mol/L m
  • 19. Experiment Initial Reactant Concentrations (mol/L) Initial Rate (mol/L*s) 1 2 3 4 5 O 2 NO 1.10x10 -2 1.30x10 -2 3.21x10 -3 1.10x10 -2 3.90x10 -2 28.8x10 -3 2.20x10 -2 1.10x10 -2 3.30x10 -2 1.30x10 -2 2.60x10 -2 1.30x10 -2 6.40x10 -3 12.8x10 -3 9.60x10 -3 Your Turn
  • 20. Expected Solution : Another Example Consider this reaction inside a car engine: How can we use this data to determine the rate law and the rate orders with respect to each reactant? rate = k [NO 2 ] m [CO] n Exp. Initial Rate (mol/L*s) Initial [NO 2 ] Initial [CO] 1 2 3 0.0050 0.080 0.0050 0.10 0.10 0.40 0.10 0.10 0.20
  • 21. 16 = 4 m and m = 2 The reaction is 2 nd order in NO 2 . First, choose two experiments in which [CO] remains constant and the [NO 2 ] varies . One Variable at a Time 0.080 0.0050 rate 2 rate 1 [NO 2 ] 2 [NO 2 ] 1 m = k [NO 2 ] m 2 [CO] n 2 k [NO 2 ] m 1 [CO] n 1 = 0.40 0.10 = m Exp. Initial Rate (mol/L*s) Initial [NO 2 ] Initial [CO] 1 2 3 0.0050 0.080 0.0050 0.10 0.10 0.40 0.10 0.10 0.20
  • 22. Now, choose two experiments in which [NO 2 ] remains constant and the [CO] varies. One Variable at a Time The reaction is zero order in CO . rate = k [NO 2 ] 2 [CO] 0 = k [NO 2 ] 2 Overall order Exp. Initial Rate (mol/L*s) Initial [NO 2 ] Initial [CO] 1 2 3 0.0050 0.080 0.0050 0.10 0.10 0.40 0.10 0.10 0.20 k [NO 2 ] m 3 [CO] n 3 k [NO 2 ] m 1 [CO] n 1 [CO] 3 [CO] 1 n = rate 3 rate 1 = 0.0050 0.0050 = 0.20 0.10 n ; 1 = 2 n and n = 0
  • 23. Your Turn rate = k [NO 2 ] 2 What is the value of the constant, k, in this reaction? Exp. Initial Rate (mol/L*s) Initial [NO 2 ] Initial [CO] 1 2 3 0.0050 0.080 0.0050 0.10 0.10 0.40 0.10 0.10 0.20
  • 24. What are the UNITS of k in this reaction?
    • s -1
    • L/(mol s)
    • mol/(L s)
    • L 2 /(mol 2 s)
  • 25. Overall Reaction Order Units of k (t in seconds) 0 mol/(L*s) (or mol L -1 s -1 )  M/s 2 L/(mol*s) (or L mol -1 s-1)  1/Ms 3 L 2 /(mol 2 *s) (or L 2 mol -2 s -1 )  1/M 2 s Units: Rate Constant k 1 1/s (or s -1 )
  • 26. Concentration-Time Relations Consider a FIRST ORDER REACTION: A  B The rate law is Rate= k[A] RATE= -d[A]/dt = k[A] How can we predict the concentration of reactants at any moment in time? How do you find [A] = f(t) ?
  • 27. First-Order Rate Law A  B k
  • 28. Integrating - (d[A]/d t) = k [A], we get [A] / [A] 0 =fraction remaining after time t has elapsed. Called the integrated first-order rate law Integrated Rate Laws
  • 29. Do you have to memorize all of the kinetics equations?
    • Yes
    • No
  • 30.
    • Plot the experimental ln[A] vs. time
    • If the graph is linear, the reaction is first-order.
    • The slope of this line = -k
    y = mx + b First-Order Rate Law ln [A] = - k t + ln [A] 0
  • 31. Examining a reaction
    • Consider the process in which methyl isonitrile is converted to acetonitrile.
    This reaction is suspected to be first order… Design an experiment to determine the order of the reaction. CH 3 NC CH 3 CN
  • 32. First-Order Processes
    • When ln P is plotted as a function of time, a straight line results.
      • The process is first-order.
      • k is the negative slope: 5.1  10 -5 s - 1 .
  • 33. First-Order Reactions The half-life , t ½ , is the time required for the concentration of a reactant to decrease to half of its initial concentration. t ½ = t when [A] = [A] 0 /2 = 1200 s = 20 minutes How do you know decomposition is first order? units of k (s -1 ) ln [A] 0 [A] 0 /2 k = t ½ ln2 k = 0.693 k = What is the half-life of N 2 O 5 if it decomposes with a rate constant of 5.7 x 10 -4 s -1 ? t ½ ln2 k = 0.693 5.7 x 10 -4 s -1 =
  • 34. A plot of [N 2 O 5 ] vs. time for three half-lives
  • 35. Second-Order Rate Law 2A  B k
  • 36.
    • Plot the experimental 1/[A] vs. time
    • If the graph is linear, the reaction is second-order.
    • The rate constant is k = slope
    Second-Order Rate Law y = mx + b
  • 37. Second-Order Half Life
  • 38. Zero-Order Reactions rate = k [A] 0 = k k = = M /s [A] is the concentration of A at any time t [A] 0 is the concentration of A at time t =0 t ½ = t when [A] = [A] 0 /2 [A] = [A] 0 - kt A product rate = - D[A] D t rate [A] 0 D[A] D t = k - t ½ = [A] 0 2 k
  • 39. Consider the following data for the reaction: N 2 O 5 -> 2NO 2 + ½ O 2 Graphical Analysis
  • 40. Summary ln[A] t = -kt + ln[A] 0 1/[A] t = kt + 1/[A] 0 [A] t = -kt + [A] 0
  • 41. At 1000 0 C, cyclobutane (C 4 H 8 ) decomposes in a first-order reaction, with the very high rate constant of 87 s -1 , to two molecules of ethylene (C 2 H 4 ). If the initial C 4 H 8 concentration is 2.00M, what is the concentration after 0.010 s? _____________ M
  • 42. At 1000 0 C, cyclobutane (C 4 H 8 ) decomposes in a first-order reaction, with the very high rate constant of 87 s -1 , to two molecules of ethylene (C 2 H 4 ). (b) What fraction of C 4 H 8 has decomposed in this time?
  • 43. SOLUTION: [C 4 H 8 ] = 0.84 mol/L = 0.58 ln 2.00 [C 4 H 8 ] = -(87s -1 ) (0.010s) ln [C 4 H 8 ] 0 [C 4 H 8 ] t = - k t (a) (b) [C 4 H 8 ] 0 - [C 4 H 8 ] t [C 4 H 8 ] 0 = 2.00M - 0.84M 2.00M
  • 44. PLAN: Cyclopropane is the smallest cyclic hydrocarbon. Because its 60 0 bond angles allow poor orbital overlap, its bonds are weak. As a result, it is thermally unstable and rearranges to propene at 1000 0 C via the following first-order reaction: The rate constant is 9.2s -1 , (a) What is the half-life of the reaction? (b) How long does it take for the concentration of cyclopropane to reach one-quarter of the initial value? One-quarter of the initial value means two half-lives have passed. Use t 1/2 = ln2/k to find the half-life
  • 45. SOLUTION: t 1/2 = 0.693/9.2s -1 = 0.075s (a) 2 t 1/2 = 2(0.075s) = 0.15 s (b)
  • 46. 0 th Order 1 st Order 2 nd Order Plot for straight line Slope, y-intercept Half-life Rate law rate = k rate = k [A] rate = k [A] 2 Units for k mol/(L*s) 1/s L/(mol*s) Int. rate law (straight-line form) [A] t = -k t + [A] 0 ln[A] t = - k t + ln[A] 0 1/[A] t = k t + 1/[A] 0 [A] t vs. t ln[A] t vs. t 1/[A] t vs t -k, [A] 0 -k, ln[A] 0 k, 1/[A] 0 [A] 0 /(2 k) (ln 2)/ k 1/( k [A] 0 ) Overview
  • 47. Summary Activity: N 2 O 5 (g)  NO 3 (g) + NO 2 (g) Consider the following graphs and reaction data to predict the concentration of N 2 O 5 at 275 sec. Time (s) [N 2 O 5 ] 0 1.000 25 0.822 50 0.677 75 0.557 100 0.458 125 0.377 150 0.310 175 0.255 200 0.210
  • 48. What is the concentration of N 2 O 5 after 275 sec? ______________ mol/L