A graph between `log t_((1)/(2))` and log a (abscissa), a being the initial concentration of A in the reaction For reaction `Ato`Product, 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

C

Explanation

`t_(1//2)=Ka^(1-n)`,n being order K=Rate constant
log `t_(1//2)=log K+(1-n) log a`
1-n=-1
n=2