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?

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lublana lublana · Answer 1 · 2019-07-31T08:05:14+02:00

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.

Parrain Parrain · Answer 2 · 2022-04-05T16:03:45+02:00

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.

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