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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.

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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. [A] [A] Concentration (M) Concentration (M) t (ms) t (ms) Exp. 1 [B]i initial rate [A]i (M) (M) (M s-1) 1.0 1.0 1.0 x 10-3 Exp. 2 [A]i [B]I initial rate (M)(M)(M s-1) 2.0 1.0 2.0 x 10-3 Exp. 3 [A]i [B]I initial rate (M)(M)(M s-1) 1.0 x 10-3 1.0 2.0

  3. Exp. 1 k rate = [A]a [B]b [A]i [B]I initial rate (M)(M)(M s-1) 1.0 1.0 a = 0 a = 1 a = 2 2 x 10-3 = 1 x 10-3 [2.0]a [1.0]a rate 2 = rate 1 1.0 x 10-3 Exp. 2 1 x 10-3 = 1 x 10-3 rate 3 = rate 1 [2.0]b [1.0]b b = 0 b = 1 b = 2 [A]i [B]I initial rate (M)(M)(M s-1) 2.0 1.0 rate = k [A] 2.0 x 10-3 1storder reaction Exp. 3 [A]i [B]I initial rate (M)(M)(M s-1) 1.0 2.0 1x10-3(M s-1) = k [1.0 M] k = 1 x 10-3 s-1 1.0 x 10-3

  4. Concentration (M) t (ms) Exp [A]i [B]i initial rate (M) (M) (M s-1) 1 1.0 1.0 1.0 x 10-3 2 2.0 1.0 2.0 x 10-3 3 1.0 2.0 2.0 x 10-3 4 2.02.0 4.0 x 10-3 2.0 x 10-3 = 1.0 x 10-3 rate 2 = rate 1 [2.0]a [1.0]a 1st order in [A] a = 1 2.0 x 10-3 = 1.0 x 10-3 rate 3 = rate 1 [2.0]b [1.0]b 1st order in [B] b = 1 rate = k [A] [B] 2nd order reaction

  5. ln         Integrated rate laws differential rate laws are differential equations t = t rate = k[A] = -d[A]/dt t = 0 rate = k[A]2 = -d[A]/dt differential rate eqn integrated rate eqn rate = k[A] [A]t = - kt [A]0

  6. ln [A]t = - kt [A]0 Integrated rate laws ln[A]t = ln[A]t - ln[A]0 -kt + ln[A]0 = - kt y = mx + b y = ln[A]t x = t m = -k b = ln[A]0 ln[A]t plot v.s. t linear

  7. [A] Concentration (M) t (ms) ln [A] ln [A] rate = k[A] ln[A]t = - kt + ln[A]0 y = ln[A]t x = t m = -k b = ln[A]0

  8. ln [A]t = - kt [A]0 ln 1st order reactions ln[A]t = - kt + ln[A]0  2N2O5(g) 4NO2(g) + O2(g) rate = (M s-1) k [N2O5 ] (M) k = 5.1 x 10-4 s-1 What is [N2O5] after 3.2 min if [N2O5] = 0.25M 0 [N2O5] = - (5.1x10-4s-1) (192 s) 3.2 [0.25] [N2O5]3.2 = 0.23 M

  9. ln 1st order reactions ln [A]t = - kt [A]0 ln[A]t = - kt + ln[A]0 How long will it take for [N2O5] to go from 0.25 M to 0.125 M ? [0.125] = t = 23 min - (5.1x10-4s-1) t [0.25] The half-life (t1/2) is the time required for [reactant] to decrease to 1/2 [reactant]i t1/2= ln 2 k

  10. 1st order reactions t1/2=ln 2 k 700 ms = ln 2 k k 1 s-1 Radioactive decay 1st order 14C dating t1/2 = 5730 years

  11. 1 1 [A]0 [A]t Integrated rate laws differential rate eqn integrated rate eqn ________________________ ________________________ rate = k[A] ln [A]t = - kt [A]0 1st order = kt + rate = k[A]2 2nd order y = mx + b y = 1 [A]t x = t m = k b = 1 [A]0 A plot of v.s. is linear 1/[A]t t

  12. [A] rate = k[A]2 t1/2 = 1 k[A]0 1 = kt + 1 [A]t [A]0 1/[A]

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

  14. Pseudo 1st order reactions high order reactions difficult to analyze put in large excess of all but one reagent [A]0 [B]0 rate =k [A]a [B]b 1.0 x 10-3 M 1.0 M [B]  constant -0.5 x 10-3 mol -0.5x 10-3 mol rate = k’ [A]a 0.5 x 10-3 M 0.999 M k = k’ k[A]a[B]b = k’[A] [B]b

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