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?

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

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 initial  rate of reaction

Therefore, rate of reaction after substituting values

Rate of reaction ==0.018 M/s

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