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.

Options

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

Answer

B

Explanation

Let the rate equation for the reaction be ,
Rate = `k[A]^(n)`
where , k=rate constant n,=order of reaction
[A] = concentration of reactant
given, `"rate"_(1)=2xx10^(-3) "mol dm"^(-3)s^(-1)`
`[A_(0)]=0.05"mol dm"^(-3)`
`"rate"_(2)=1.6xx10^(-2)"mol dm"^(-3)s^(-1)`
`(A_(0))=0.1 "mol dm"^(-3)`
:. `2xx10^(-3)=k(0.05)^(n)` .... (i)
`1.6xx10^(-2)=k(0.1)^(n)` ....(ii)
Divide (i) by (ii)
implies `(2xx10^(-3))/(1.6xx10^(-2))=((0.5)^(n))/((0.1)^(n))=((0.5)^(n))/(2^(n)(0.05)^(n))implies(1)/(2^(n))=(1)/(8)`
implies `n=3`, :. Order of reaction =3