CH14- KINETICS

# x y this is the rate law not expression a

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Unformatted text preview: ate = k[F2 ][ClO2 ] we can use the data from any of the three trials to find the value of the rate constant, k if we take the first trial: Keep the units!! you can pick any trial and solve for k rate = k[F2 ][ClO2 ] rate 0.0012 M/s k= = = 1.2 M!1s!1 [F2 ][ClO2 ] (0.10 M)(0.01 M) CHM1311 Chemical Kinetics 24 Example: Deriving Rate Laws The initial rates of the reaction A + 2 B C are followed at 25C. Using the data below, derive the rate law for the reaction and the value of the rate constant. Trial initial [A] initial [B] initial rate (M/s) 1 0.100 0.100 5.50 x 10-6 from trial 1-->2, a is doubled, but rate is quadrupled, rate prop [A]^2 -5 2 0.200 0.100 2.20 x 10 3 0.400 0.100 8.80 x 10-5 trail 1-->4, [B] is tripled and so is the rate, rate prop [B]^1 -5 overall 3rd order 4 0.100 0.300 1.65 x 10 5 0.100 0.600 3.30 x 10-5 CHM1311 Chemical Kinetics 25 Example: Deriving Rate Laws The initial rates of the reaction A + 2 B C are followed at 25oC. Using the data below, derive the rate law for the reaction and the value of the rate constant. The rate law is therefore: rate = k[A]2[B] Note that the same law can be derived using other pairs of data. Next, we find the value of k using the data from the first trial: v (5.50 !10"6 M/s) k= 2 = [A] [B] (0.100 M)2 (0.100 M) k = 5.50 !10"3 M"2s"1 CHM1311 Chemical Kinetics 26 Concentrations as a function of time the rate laws allow us to determine the rate of the reaction at any point in the reaction to determine an equation that will link the concentration of a reactant at any point in time ([A]t) to the initial concentration ([A]o), k and the time (t), we must integrate the rate law now, let's examine in detail: zero order reactions first order reactions second order reactions CHM1311 Chemical Kinetics 27 Zero Order Reactions the sum of exponents is zero ( x + y + ... = 0) the graph of [A] as a function of time is a straight line A products rate = k [A]0 rate = k value of k = slope units of k = mol L1 s1 CHM1311 Chemical Kinetics 28 Zero Order Reactions Integrating the rate law from time = 0 to time = t : d[A] rate = =k dt [A] = [A]0 at t = 0 [A] = [A]t at t = t ! " d[A] = [A]0 [A]t t " k dt 0 [A]t + [A]0 = kt [A]t = [A]0 kt CHM1311 Chemical Kinetics 29 Zero Order Half-Life the half-life of a reaction, t1/2, is the time needed for the concentration of a reactant to decrease by half half life is dependant on concentration for a zero order reaction: [A]o [A]t t= k at t 1 [A]t = 1 [A]o 2 [A]o 2k [A] = [A]0 at t = 0 [A] = [A]t at t = t 2 therefore t1 = 2 for zero order reactions, the half-life is dependent on the concentration CHM1311 Chemical Kinetics 30 First Order Reactions the sum of exponents is one ( x + y + ... = 1) the graph of [A] as a function of time is a monoexponential curve A products rate = k [A]1 [A] o units of k = s1 Temps Time [A] CHM1311 Chemical Kinetics 31 First Order Reactions Integrating the rate law from time = 0 to time = t : [A] d[A] rate = = k[A] dt o [A] = [A]0 at t = 0...
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## This note was uploaded on 01/16/2013 for the course CHM 1311 taught by Professor Mayer during the Fall '08 term at University of Ottawa.

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