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

Relation Between half-life time and initial concentration is
`t_(1//2)prop(1)/((a_(0))^(n-1)` Then,
`(t_(1//2))_(1)prop(1)/((a_(0))_(1)^(-1))" "...(i)`
`(t_(1//2))/(2)prop(1)/((2a_(0))^(n-1)" "...(ii)` Divide by Eq(ii)
`(t_(1//2)xx2)/t_(1//2)=((2)^(n-1)(a_(0))^(n-1))/((a_(0))^(n-1))`
`2=(2)^(n-1)rArr -1=1`
`n=2`