The half-life of a reaction is halved as the initial concentration of the reaction is doubled. The order of the reaction is

Options

A. `0.5`
B. 1
C. 2
D. 0

Answer

C

Explanation

`t_(1//2) prop (1)/([A_(0)]^(n-1)), ((t_(1//2))1)/((t_(1//2))2) = ([A_(0)]_(2)^(n-1))/([A_(0)]_(1)^(n-1)) = {([A_(0)]_(2))/([A_(0)]_(1))}^(n-1)`
`(t)/(t//2) = (2a)/(a)^(n-1) or 2 = (2)^(n-1)`
or ` n -1 = 1 or n =2 `