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

B

Explanation

Let the rate equation for the reaction be
rate `=k[A]^(n)`
where `k=` rate constant, n=order of reaction
[A]= concenration of reactant
given`"rate"_(1)=2xx10^(-3)"mol dm^(-3)s^(-1)`
`[A_(0)'=0.05"mol"dm^(-3)`
`"rate"_(II)=1.6xx10^(-2)"mol"dm^(-3)s^(-1)`
`[A_(0)]=0.1"mol"dm^(-3)`
`:.2xx10^(-3)=k[0.1]^(n)`...............i
`1.6xx10^(-2)=k[0.1]^(n)`.............ii
Divide i by ii
`implies(2xx10^(-3))/(1.6xx10^(-2))=([0.05]^(n))/([0.1]^(n))=([0.05]^(n))/(2^(n)[0.05]^(n))`
`implies1/(2^(n))=1/8`
`impliesn=3, :.` Order of reaction =3