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?
Accepted Solution
A:
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 [tex]A +B\rightarrow C+D [/tex]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 MRate of reaction =0.0090 M/sAccording to rate law Rate of reaction=k[A]If concentration of A is doubled then the rate of reaction is given byRate of reaction =k[2A]= 2 k[A]=2 [tex]\times [/tex]initial rate of reactionTherefore, rate of reaction after substituting values Rate of reaction =[tex]2\times 0.0090[/tex]=0.018 M/sRate of reaction when concentration of A is doubled=0.018 M/s.