At 298K, the kinetics of a chemical reacation were studied andthe following init

Question

At 298K, the kinetics of a chemical reacation were studied andthe following inital rate data were obtained. A+ B -> C Experiment#               [A](M)              [B](M)                   InitialRate (M/s) 1                                  0.020                0.010                      3.2x 10-4 2                                  0.030                 0.010                      7.2x 10-4 3                                   0.020                 0.020                      6.4x 10-4 What is the rate law for this reaction and what is the valueof the rate constant at 298K? Thanks! At 298K, the kinetics of a chemical reacation were studied andthe following inital rate data were obtained. A+ B -> C Experiment#               [A](M)              [B](M)                   InitialRate (M/s) 1                                  0.020                0.010                      3.2x 10-4 2                                  0.030                 0.010                      7.2x 10-4 3                                   0.020                 0.020                      6.4x 10-4 What is the rate law for this reaction and what is the valueof the rate constant at 298K? Thanks! What is the rate law for this reaction and what is the valueof the rate constant at 298K? Thanks!

Explanation / Answer

We Know that :         The given Reactionis :           A+ B-> C           Rate= K [ A ]a [ B ]b           3.2 x10-4 = K [ 0.020 ]a [ 0.010]b        7.2 x10-4   = K [ 0.030]a [ 0.010 ]b                7.2 x10-4   = K [ 0.030]a [ 0.010 ]b           Dividethe experimental values 1 by 2 we get :           32 / 72 =   ( 2 / 3 )a               a = 2            3.2x 10-4 = K [ 0.020 ]a [ 0.010]b                        6.4x 10-4 = K [ 0.020 ]a [ 0.020]b               Dividethe experimental values 1 by 3 we get :                32 / 64 = ( 1 / 2 )b                    b = 1          Rate law forthe above equation is :           Rate = K [ A ] 2 [ B ]             K = 3.2 x 10-4 / [  0.020 ]2 [ 0.010 ]                   = 80 M-2s-1                    b = 1          Rate law forthe above equation is :           Rate = K [ A ] 2 [ B ]             K = 3.2 x 10-4 / [  0.020 ]2 [ 0.010 ]                   = 80 M-2s-1

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