Review Differential Rate Laws rate M s1 k
- Slides: 13
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 = -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 = 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+ 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 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-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-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 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. 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 + 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 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] (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 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
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