Review Differential Rate Laws rate M s1 k

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Review Differential Rate Laws rate (M s-1) = k [A]a [B]b. . . A

Review Differential Rate Laws rate (M s-1) = k [A]a [B]b. . . A + 3 B 2 C rate = - [A] = - 1/3 [B] t t Assume that A is easily detected initial rate = 1 x 10 -3 M s-1 a) 1 x 10 -3 Ms-1 b) 3 x 10 what is - [B] -3 Ms-1 c) 0. 33 x 10 t

Integrated rate laws differential rate laws are differential equations t=t rate = k =

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

Integrated rate laws differential rate laws are differential equations zero order reaction rate =

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

CH 3 CH 2 OH + NAD+ CH 3 CHO + NADH + H+

CH 3 CH 2 OH + NAD+ CH 3 CHO + NADH + H+ t (min) [ethanol] M 0 0. 065 15 0. 058 30 0. 051 45 0. 044 60 0. 037 zero order in ethanol [CH 2 OH] = -kt + [CH 3 CH 2 OH]0 rate = k k = -. 037 -. 065 = 4. 7 x 10 -4 60 - 0 -1 rate = 4. 7 x 10 -4 M min t 1/2 = [CH 3 CH 2 OH]0 = 70 min 2 k

Integrated rate laws differential rate laws are differential equations rate = k[A] = -d[A]/dt

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

cis-[Pt(NH 3)2 Cl 2] + H 2 O [Pt(NH 3)2 Cl(H 2 O)]Clt (min)

cis-[Pt(NH 3)2 Cl 2] + H 2 O [Pt(NH 3)2 Cl(H 2 O)]Clt (min) [cis-platin] M 0 0. 0060 2000. 0044 400 0. 0033 600 0. 0024 800 0. 0018 first order in cis-platin ln [cis-platin] = -kt +ln [cis-platin]0 rate = k [cis-platin] k = - (-6. 32)-(-5. 12) = 1. 5 x 10 -3 800 - 0 rate = 1. 5 x 10 -3 min-1 t 1/2 = ln 2 = 462 min k

cis-[Pt(NH 3)2 Cl 2] + H 2 O [Pt(NH 3)2 Cl(H 2 O)]Clt (h)

cis-[Pt(NH 3)2 Cl 2] + H 2 O [Pt(NH 3)2 Cl(H 2 O)]Clt (h) [cis-platin] M 0 0. 0060 2000. 0044 400 0. 0033 600 0. 0024 800 0. 0018 first order in cis-platin ln [cis-platin] = -kt +ln [cis-platin]0 k = - (-6. 32)-(-5. 12) = 1. 5 x 10 -3 800 - 0 -3 -1 rate = 1. 5 x 10 min radioactive decay 1 st order t 1/2 = ln 2 = 462 min 14 C dating t k 1/2 = 5730 years rate = k [cis-platin]

Integrated rate laws differential rate laws are differential equations rate = k[A]2 = -d[A]/dt

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

t (s) 2 4 6 8 10 [X] (M) 0. 392 0. 336 0.

t (s) 2 4 6 8 10 [X] (M) 0. 392 0. 336 0. 294 0. 261 0. 235 1/. 235 – 1/. 393 =. 213 = k 10 - 2 1/. 235 = (. 213)(10) + 1/[X]0 = 0. 471 second order in X 1/[X] = kt + 1[X]0 rate = 0. 213 M-1 s-1[X]2 1 t 1/2 = k[X]0

Integrated rate laws second order reactions rate = k [A]2 1 = kt +

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

Pseudo order reactions high order reactions difficult to analyze put in large excess of

Pseudo order reactions high order reactions difficult to analyze put in large excess of all but one reagent [A]0 (M) rate = k [A]a [B]b [B] constant rate = k’ [A]a [B]0 (M) 1. 0 x 10 -3 -0. 5 x 10 -3 mol -0. 5 x 10 -3 mol 0. 5 x 10 -3 0. 999 k[A]a[B]b = k’[A] k = k’ [B]b

X + 2 Y 2 Z t (s) 0 2 4 6 8 10

X + 2 Y 2 Z t (s) 0 2 4 6 8 10 [X] (M) 0. 470 0. 448 0. 427 0. 409 0. 392 0. 376 [Y] (M) 2. 0 t (s) 0 2 4 6 8 10 [X] (M) 0. 470 0. 427 0. 391 0. 361 0. 335 0. 313 [Y] (M) 4. 0 second order in X

X + 2 Y 2 Z rate = k [X]2[Y] rate = k’ [X]2

X + 2 Y 2 Z rate = k [X]2[Y] rate = k’ [X]2 k = k’ k =. 028 M-2 [Y] s k’ = 2. 70 - 2. 13 = 0. 057 10 - 0 second order in X [Y] = 2. 0 M first order in Y. 106 = (4. 0)y . 057 (2. 0)y y =1 k’’ = 4. 26 - 2. 13 =. 106 10 - 0 [Y] = 4. 0 M