The half-life periofs of a reaction at initial concentration `0.1 mol L^(-1)` and `0.5 mol L^(-1)` are `200s` and `40 s`, respectively. The order of the reaction is

Options

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

Answer

C

Explanation

(c ) `[A]_(0) = 0.1 M rarr t_(1//2) = 200s` …(i)
`[A]_(0) = 0.5 M rarr t_(1//2) = 40 s` …(ii)
Form cases (i) and (ii), we can conclude
`t_(1//2) prop [A]_(0)^(-1)`
Therefore, it is a second order reaction.
Alternatively,
`((t_(2))/(t_(1))) = ((a_(1))/(a_(2)))^(n-1) rArr ((40)/(200)) = ((0.1)/(0.5))^(n-1)`
`rArr ((1)/(5))^(1) = ((1)/(5))^(n-1)`
`:. n-1 = 1`
`n = 2`