The half-lives for two samples are 0.1 and 0.8 s whose concentrations are 400 and 50 mol `L^(-1)` respectively. The order of the reaction is

Options

A. 0
B. 1
C. 2
D. 3

Answer

C

Explanation

The ratio of two half-lives and their concentrations are related as `((t_(1//2))_(1))/((t_(1//2))_(2))=[(a_(2))/(a_(1))]^(n-1)` [Here, n = order]
Therefore, `(0.1)/(0.8)=[(50]/(400)]^(n-1)`
On taking log of both sides
`log .(0.1)/(0.8)=(n-1)log. (50)/(400)`
`1 = n-1implies n=2`