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.

Explanation

If n is the order of reaction
Rate `=k[A_(0)]^(n)=kC^(n)`
For two different initial concentrations, we have
`(r_(1))/(r_(2))=((C_(1))/(C_(2)))^(n):.log""(r_(1))/(r_(2))=nlog""(C_(1))/(C_(2))`
or `" "n=(log(r_(1)//r_(2)))/(log(C_(1)//C_(2)))=(log(2.4xx10^(-4)//0.6xx10^(-4)))/(log(2xx10^(-3)//1xx10^(-3)))=(log4)/(log2)=2`
Hence, order of reaction = 2
Rate `=k[A_(0)]^(2)`
`k=("Rate")/([A_(0)]^(2))=(2.4xx10^(-4)"mol L"^(-1)s^(-1))/((2xx10^(-3)"mol L"^(-1))^(2))=0.6xx10^(2)"L mil"^(-1)s^(-1)`