A graph between log `t_(1//2)` and log a (abscissa), a being the initial concentration of A in the reaction A `rarr` product, is depicted in the figure. The rate law is

Options

A. `-(d[A])/(dt)=k`
B. `-(d[A])/(dt)=k[A]`
C. `-(d[A])/(dt)=k[A]^(2)`
D. `-(d[A])/(dt)=k[A]^(3)`

Answer

d

Explanation

`t_(1//2)=ka^(1-n)`, n being order, k=rate constant log `t_(1//2)`=log k+(1-n)log a
Slope, (1-n) = -2, n = 3