Question

The following is a multi-step reaction. The rate-limiting step is unimolecular, with A as the sole reactant.

A+B=C+D

If A and B are both .125 M, then the rate of the reaction is .0090M/s.

a) What is the rate of the reaction if A is doubled?

b) Starting with the original concentrations, what is the rate of the reaction if B is halved?

c) Starting with the original concentrations, what is the rate of the reaction is A and B are both increased by a factor of four?

Answer 1

The reaction A+B----->C+D is unimolecular with respect to A suggesting thar the rate, r1 = K[ A]¹[ [B]⁰,

K is rate constant

when [A] =[B]= 0.125M, r1= 0.009 M/s

when [A] is doubled. the new rate= K[2A] [B]0 = 2* initial rate =2*r1=2*0.009=0.018 M/s

when [B] is halved, since the rate is indepedent of concentration of B, even concentration of B is halved ,there is no effect on rate of reaction, hence new rate= r1= 0.009 M/s

[A]=0.125*4= 0.5M =[B], new rate= K *0.5*1 =4*r1= 4*0.009= 0.036 M/s