For a chemical reaction, if the concentration of the reactant a doubled, its half-life is halved. Determine the order of the reaction.

Explanation

Half-life of 'n' th order reaction, `t_(1//2)prop(1)/([A]_(0)^(n-1))" " ...[1]`
where `[A]_(0)` = initial concentration of reactant and `"n"ne1`.
where the concentration of the reactant is `2[A]_(0)` , the half-life of the reaction becomes `(1)/(2)xxt_(1//2)`
`therefore(1)/(2)t_(1//2)prop(1)/([2A]_(0)^(n-1)) " or, "(1)/(2)t_(1//2)prop(1)/(2^(n-1)[A]_(0)^(n-1))" "...[2]`
Dividing equation [1] by [2] , we get `2=2^(n-1)`
or, n - 1 = 1 or, n = 2
Thus , the order of the reaction = 2.