The half-life periof of a substance is `50 min` at a certain initial concentration. When the concentration is reduced to one-half of its initial concentration, the half-life periof is found to be `25 min`. Calculate the order of reaction.

Explanation

Suppose the initial concentration in the first case is a `mol L^(-1)`. Then `[A_(0)]_(1) = a, (t_(1//2))_(1) = 50 min`
`[A_(0)]_(2) = (a)/(2), (t_(1//2))_(2) = 25 min`
We know that for a reaction of `nth` order
`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)`
Substituting the values, we get
`(50)/(25) = ((a//2)/(a))^(n-1)` or `(2)/(1) = ((1)/(2))^(n-1) = ((2)/(1))^(n-1)`
or `1 - n = 1` or `n = 0`
Hence, the reaction is of zero order.