For a chemical reaction `Ararr B`,the rate of the reaction is `2.0xx10^(-3) sec^(-1)`, when the initial concentration is `0.05 mol dm^(-3)`. The rate of the same reaction is `1.6xx10^(-2) mol dm^(-3) sec^(-1)`. When the initial concentration is `0.1` mol `dm^(3)`, find the order of reaction.

Answer

3

Explanation

Let the rate equation for the reaction be rate `=K[A]^(n)`
`r_(1)=2.0xx10^(-3) mol dm^(3) mol dm^(-3) sec^(-1)`,
`[A_(0)]=0.05 mol dm^(-3)`
`:. 2.0xx10^(-3)=K[0.05]^(n) …(1)`
and `r_(2)=1.6xx10^(-2) mol dm^(-3) sec^(-1)`,
`[A_(0)]=0.1 mol dm^(-3)`
`:. 1.6xx10^(-2)=K[0.1]^(n)`
eq. (1) divide by. `(2)`,
`(2.0xx10^(-3))/(1.6xx10^(-2))=([0.05]^(n))/([0.1]^(n))implies 1/8=(1/2)^(n)`
`implies n=3`
`:.` Order of reaction `=3`