In: Chemistry
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?
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 ( 1st 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