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Review

Review. Differential Rate Laws. rate (M s -1 ) =. k. [A] a. [B] b. A + 3 B  2 C. rate =. - [ A ] t. = - [B] t. 1/3. Assume that A is easily detected. initial rate =. 1x10 -3 M s -1. a ) 1x10 -3 Ms -1 b ) 3x10 -3 Ms -1 c ) 0.33x10 -3 Ms -1.

charity-hebert
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Review

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  1. Review Differential Rate Laws ... rate (M s-1) = k [A]a [B]b A + 3B 2C rate = -[A] t = - [B] t 1/3 Assume that A is easily detected initial rate = 1x10-3 M s-1 a) 1x10-3Ms-1 b) 3x10-3Ms-1 c) 0.33x10-3Ms-1 -[B] t what is

  2.            Integrated rate laws differential rate laws are differential equations t = t zero order reaction rate = k = -d[A]/dt t = 0 t = t first order reaction rate = k[A] = -d[A]/dt t = 0 t = t rate = k[A]2 = -d[A]/dt second order reaction t = 0 differential rate eqn integrated rate eqn

  3.    Integrated rate laws differential rate laws are differential equations zero order reaction rate = k = -d[A]/dt -kt [A] = + [A]0 y = mx + b slope = -k intercept = [A]0 differential rate eqn integrated rate eqn

  4. t (min) [ethanol] M • 0 0.065 • 15 0.058 • 0.051 • 45 0.044 • 60 0.037 CH3CH2OH + NAD+ CH3CHO + NADH + H+ zero order in ethanol [CH2CH2OH] = -kt + [CH3CH2OH]0 = 4.7 x 10-4 k = - .037 - .065 rate = k 60 - 0 rate = 4.7 x 10-4 M min-1 t1/2 = [CH3CH2OH]0 = 70 min 2 k

  5.    Integrated rate laws differential rate laws are differential equations first order reaction rate = k[A] = -d[A]/dt -kt ln [A] = + ln [A]0 y = mx + b slope = -k intercept = ln [A]0 differential rate eqn integrated rate eqn

  6. t (min) [cis-platin] M 0 0.0060 200 0.0044 400 0.0033 600 0.0024 800 0.0018 cis-[Pt(NH3)2Cl2] + H2O  [Pt(NH3)2Cl(H2O)]Cl- first order in cis-platin ln [cis-platin] = -kt +ln[cis-platin]0 = 1.5x10-3 k = - (-6.32)-(-5.12) rate = k [cis-platin] 800 - 0 rate = 1.5 x 10-3 min-1 ln 2 t1/2 = = 462 min k

  7. t (h) [cis-platin] M 0 0.0060 200 0.0044 400 0.0033 600 0.0024 800 0.0018 cis-[Pt(NH3)2Cl2] + H2O  [Pt(NH3)2Cl(H2O)]Cl- first order in cis-platin ln [cis-platin] = -kt +ln[cis-platin]0 = 1.5x10-3 k = - (-6.32)-(-5.12) rate = k [cis-platin] 800 - 0 rate = 1.5 x 10-3 min-1 radioactive decay 1st order ln 2 t1/2 = = 462 min k 14C dating t1/2 = 5730 years

  8.    Integrated rate laws differential rate laws are differential equations second order reaction rate = k[A]2 = -d[A]/dt kt 1/[A] = + 1/[A]0 y = mx + b slope = k intercept = 1/[A]0 differential rate eqn integrated rate eqn

  9. t (s) [X] (M) • 0.392 • 4 0.336 • 0.294 • 0.261 • 10 0.235 1/.235 – 1/.393 = k = .213 10 - 2 1/.235 = (.213)(10) + 1/[X]0 1 [X]0 = 0.471 t1/2 = k[X]0 second order in X 1/[X] = kt + 1[X]0 rate = [X]2 0.213 M-1s-1

  10. Integrated rate laws secondorder reactions rate = k [A]2 1 = kt + 1 [A]t [A]0 many second order reactions rate = k [A] [B] A and B consumed stoichiometrically A + B  C [A]0 = [B]0 if not, no analytical solution

  11. Pseudo order reactions high order reactions difficult to analyze put in large excess of all but one reagent [A]0 (M) [B]0 (M) rate =k [A]a [B]b 1.0 1.0 x 10-3 [B]  constant -0.5 x 10-3 mol -0.5x 10-3 mol rate = k’ [A]a 0.5 x 10-3 0.999 k’ k = k[A]a[B]b = k’[A] [B]b

  12. X + 2Y  2Z • t (s) [X] (M) [Y] (M) • 0 0.470 2.0 • 0.448 • 0.427 • 0.409 • 0.392 • 10 0.376 second order in X • t (s) [X] (M) [Y] (M) • 0 0.470 4.0 • 0.427 • 0.391 • 0.361 • 0.335 • 10 0.313

  13. X + 2Y  2Z rate = k [X]2[Y] rate = k’ [X]2 k = k’ k = .028 M-2 [Y] s 2.70 - 2.13 k’ = = 0.057 10 - 0 second order in X [Y] = 2.0 M first order in Y .106 y =1 = (4.0)y .057 (2.0)y 4.26 - 2.13 k’’ = = .106 [Y] = 4.0 M 10 - 0

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