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 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 [tex]\times [/tex]initial rate of reaction
Therefore, rate of reaction after substituting values
Rate of reaction =[tex]2\times 0.0090[/tex]=0.018 M/s
Rate of reaction when concentration of A is doubled=0.018 M/s.
Based on the rate of the reaction, and the concentration of the reactants, if A was doubled, the new rate would be 0.018 M/s.
The concentration of the reactant, A, determines the rate of the reaction.
The doubling of A will therefore double the rate of reaction to:
= 0.0090 x 2
= 0.018 M/s
In conclusion, the rate of reaction would be 0.018 M/s.
Find out more on rates of reaction at https://brainly.com/question/24795637.
