Question

Consider the following reaction at 325 K.

2 A + 2 B → C + D

where rate = rate=k[A][B]^2. An experiment was performed for a certain number of seconds where [A]o = 0.000533 M and [B]o = 1.55 M. A plot of ln[A] vs time had a slope of -6.55. What will the rate of this reaction be if a new experiment is preformed when [A] = [B] = 0.185 M?

Answer 1

The rate of reaction is given by

-rA= K[A] [B]²

since concentration of A <<<B^, even all the A is consumed, it is as good as no B is consumed. As a result

[B]= [B]0

-rA= K’[A], where K’= K[B]²

-rA= -dCA/dt= K’[A]

When integrated noting that t=0, CA= CAO and at t=t, CA=CA

lnCA= lnCAO-K’t

The plot of ln[A] vs time gives slope of -K’

-K’ is given as -6.55

K’=6.55/sec

But K’= K[B]², 6.55= K[1.55]²

K= 2.73/M2.sec

Hence the rate of reaction –rA =2.73[A] [B]²

when [A] =[B] =0.185

-rA= 2.73*0.185*(0.185)²= 0.0173 M/sec