For a chemical raction `AtoB` the rate of the reaction is `2xx10^(-3)` mol `dm^(-3)s^(-1)`, when the initial concentration is 0.05 mol `dm^(-3)`. The rate of the same reaction is `1.6xx10^(-2)` mol `dm^(-3)s^(-1)` when the initial concentration is 0.1 mol `dm^(-3)`. The order of the reaction is

Options

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

Answer

C

Explanation

(B) Let Rate =k `[A]^(n)`
`2xx10^(-3) =k (0.05)^(n)`
`1.6xx10^(2) = k (0.05)^(n) `
`1.6xx10^(2) = k (0.1)^(n)`
Dividing (i) by (ii)
`=((0.5)^(n))/(2^(n)(0.05)^(n))`
` (1)/(2^(n))=(1)/(8)`
`:. n=3`