Question

Consider the following reaction at 309 K. 1 A + 1 B → C + D where rate = rate=k[A]2[B]. An experiment was performed for a certain number of seconds where [A]o = 1.07 M and [B]o = 0.000167 M. A plot of ln[B] vs time had a slope of -9.63. What will the rate of this reaction be if a new experiment is preformed when [A] = [B] = 0.212 M?

Answer 1

Given rate= K[A]² [B]

Compared to the initial concentration of A, the initial concentration of B is very very small. This is an example of pseudo order reaction and rate can be approximated as

-d[B]/dt= K’ [B] (1)

, where K’= K[A]²,

When equation is integrated ( 1^st order equation)

lnB= ln[B]0-K’t

where K= rate constant and [B] = concentration of B at any time t and [B]0= initial concentration of B

so the plot of lnB vs t gives a straight line whose slope =- K’

given K’=9.63/(sec

since K’= K[A]2

concentration of A does not change significantly through out the course of reaction

K’ = K[A]o²

9.63= K[ 1.07]² , K= 8.41/M2.sec

Rate = 8.41[A]² [B]

[A] =0.212= [B]

Rate= 8.41*0.212*0.212*0.212= 0.080M/sec