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)=(1)/(k.a)` for `2^(nd)` order,
where a: initial conc. of reactant.