For the reaction A+B+C D+E, the initial reaction rate wasmeasured for various in
ID: 685782 • Letter: F
Question
For the reaction A+B+C D+E, the initial reaction rate wasmeasured for various initial concentrations of reactants. Thefollowing data were collected: Trial [A](M) [B]
(M) [C]
(M) Initial Rate
(M/C) 1 0.10 0.10 0.10 3x10^-5 2 0.10 0.10 0.30 9*10^-5 3 0.20 0.10 0.10 1.2*10^-4 4 0.20 0.20 0.10 1.2*10^-4 What is the value of the rate constant k forthis reaction? What is the value of the rate constant k forthis reaction? Trial [A]
(M) [B]
(M) [C]
(M) Initial Rate
(M/C) 1 0.10 0.10 0.10 3x10^-5 2 0.10 0.10 0.30 9*10^-5 3 0.20 0.10 0.10 1.2*10^-4 4 0.20 0.20 0.10 1.2*10^-4
Explanation / Answer
For A+B+C .........>D+E rate =k*[A]x[B]y[C]z (general expression) for trial 1; rate1 =k*[0.10]x[0.10]y[0.10]z =3*10^-5 for trial 2 ; rate2 =k*[0.10]x[0.10]y[0.30]z =9*10^-5 for trail 3 ; rate3 =k*[0.20]x[0.10]y[0.10]z =1.2*10^-4 for trail 4 ; rate4 =k*[0.20]x[0.20]y[0.10]z =1.2*10^-4 strategy to find x,y, z: trail2 / trail1 :{k*[0.10]x[0.10]y[0.30]z }/{k*[0.10]x[0.10]y[0.10]z } = ( 9*10^-5) / ( 3*10^-5 ) (0.30/0.10)z = 3 zlog( 0.30/0.10) = log3 z = log3 / log(0.30/0.10) = 0.477 / 0.477 = 1 trail3 /trail1: k*[0.20]x[0.10]y[0.10]z /k*[0.10]x[0.10]y[0.10]z= 1.2*10^-4 / 3*10^-5 (0.20/ 0.10)x = 4 x = log4 / log(0.20 /0.10) = 0.6020 / 0.3010 = 2 trail4 /trail3: k*[0.20]x[0.20]y[0.10]z/ k*[0.20]x[0.10]y[0.10]z = 1.2*10^-4 /1.2*10^-4 (0.20/ 0.10)y = 1 y = log1 / log(0.20 /0.10) =0 / 0.3010 = 0 so actual rate =k*[A]2[B]0[C]1 = k*[A]2[C]1 now we can calculate the rate as fallows from trail 1 : 3*10^-5 M/C = k(0.10) 2(0.10) k = (3*10^-5 M/C ) / (0.10) 2(0.10) =3*10^-2Related Questions
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