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) = 2 xx 10^(-3) "mol" dm^(-3) s^(-1)`
`[A_(0)] = 0.05 ` mol `dm^(-3)`
`"rate"_(2) = 1 . 6 xx 10^(-2)` mol `dm^(-3) s^(-1)`
`(A_(0)) = 0.1` mol `dm^(-3)`
`therefore 2 xx 10^(-3) = k [0.05]^(n) " ".... (i)`
`1.6 xx 10^(-2) = k[0.1]^(n) " ".... (ii)`
Divide (i) by (ii)
`implies (2 xx 10^-3)/(1.6 xx 10^(-2)) = ([0.05]^(n))/([0.1]^(n)) = ([0.05]^(n))/(2^(n)[0.05]^(n))`
`implies (1)/(2^(n)) = (1)/(8) implies n = 3, " " therefore` Order of reaction = 3 .