Question

For the multi‑step reaction A+B⟶C+D the rate‑limiting step is unimolecular, with A as the sole reactant. If [A] and [B] are both 0.110 M, then the rate of reaction is 0.0090 M/s. What is the rate of the reaction if [A] is doubled?

Answer 1

Answer:

Rate of reaction when concentration of A is doubled=0.018 M/s.

Step-by-step explanation:

We are given that for the multi-step reaction

$A +B\rightarrow C+D$

It is given that the rate-limiting step is unimolecular  with A as the sole reactant

It means rate of reaction depend upon the concentration of reactant A only

Rate of reaction =k[A]

[A]=0.110 M

[B]=0.110 M

Rate of reaction =0.0090 M/s

According to rate law

Rate of reaction=k[A]

If concentration of A is doubled then the rate of reaction is given by

Rate of reaction =k[2A]= 2 k[A]=2 $\times$initial  rate of reaction

Therefore, rate of reaction after substituting values

Rate of reaction =$2\times 0.0090$=0.018 M/s

Rate of reaction when concentration of A is doubled=0.018 M/s.