1.

For a given reaction, `Ararr"Product, rate is" 1xx10^(-4)"M s^(-1)` when [A] = `0.01` M and rate is `1.41xx10^(-4)"M s"^(-1)` when[A] = `0.02` M. Hence, rate law is :A. `-(d[A])/(dt)=k[A]^(2)`B. `-(d[A])/(dt)=k[A]`C. `-(d[A])/(dt)=(k)/(4)[A]`D. `-(d[A])/(dt)=k[A]^(1//2)`

Answer» Correct Answer - D


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