The rate of a reaction starting with initial concentration of `2 xx 10^(-3)` and `1 xx 10^(-3) M` are equal to `2.40 xx 10^(-40)` and `0.60 xx 10^(-4) M s^(-1)`, respectively. Calculate the order or reaction w.r.t. reactant and also the rate constant.

Explanation

`(r_(0))_(1) = k[A_(0)]_(1)^(a)` (`a` is order of reaction)
`(r_(0))_(2) = k[A_(0)]_(2)^(a)`
`((r_(0))_(1))/((r_(0))_(2)) = {([A_(0)]_(1))/([A_(0)]_(2))}^(a)`
or `a = (log(r_(0))_(1) - log(r_(0))_(2))/(log[A_(0)]_(1)-log[A_(0)]_(2))`
`= (log[2.40 xx 10^(-4)]-log [0.60 xx 10^(-4)])/(log[2 xx 10^(-3)]-log[1xx10^(-3)])`
`= (-3.62 + 4.22)/(-2.70 + 3) = 2`
`r = k[A]^(2)`
Also, `k = (r )/([A]^(2)) = (2.40 xx 10^(-4))/([2 xx 10^(-3)]^(2)) = 60 mol^(-1) L s^(-1)`