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 of reaction w.r.t. reactant and also the rate constant.

Answer

2

Explanation

`A to B`
`{:("initial concentration", "Rate of reaction"),(2 xx 10^(-3) M, 2.40 xx 10^(-4) Ms^(-1)),(1 xx 10^(-3) M, 0.60 xx 10^(-4) Ms^(-1)):}`
rate of reaction
`r=k[A]^(x)`
where x = order of reaction
hence
`2.40 xx 10^(-4) = k[2 xx 10^(-3)]^(x)`......(i)
`0.60 xx 10^(-4) = k[1 xx 10^(-3)]^(x)`.............(ii)
On dividing eqn (i) from eqn (ii) we get
`4=(2)^(x)`
`therefore x=2`
i.e. order of reaction =2