For the reaction of I, II and III orders, `k_(1) = k_(2) = k_(3)` when concentrations are expressed in mole `L^(-1)`. What will be the relation in `k_(1), k_(2), k_(3)`, if the concentration are expressed in `mol mL^(-1)`?

Explanation

`k prop [a]^(1-n)`
For first order: `k prop [a]^(1-1) rArr k prop 1`
So `k` is constant and is independent of units of `"a"`.
`:. k_(1) = k'_(1)` …(i)
Fo second order: `k prop [a]^(1-2) prop [a]^(-1)`
`k prop [a]^(-1)` when units of a `mol L^(-1)`.
`k'_(2) prop [a]^(-1)` when units of a in `mol mL^(-1)`.
`rArr k'_(2) prop [a xx 10^(3)]^(-1)` when a in `mol ml^(-1)`.
`prop [a]^(-1)[10^(3)]^(-1)`
`k'_(2) prop k_(2) xx 10^(-3)`
`:. k_(2) prop (k'_(2))/(10^(-3)) prop k'_(2) xx 10^(3)`
`:. k_(2) prop k'_(2) xx 10^(3)`
For third order: `k prop [a]^(1-3) prop [a]^(-2)`
`k_(3) prop [a]^(-2)` when a is in `mol L^(-1)`
`k'_(3) prop [a]^(-2)` when a is in `mol ml^(-1)`
`rArr k'_(3) prop [a xx 10^(3)]^(-2)`
`k'_(3) prop [a]^(-2) [10^(-6)]`
`rArr k'_(3) prop [k_(3) xx 10^(-6)]`
`:. k_(3) prop (k'_(3))/(10^(-6))`
`k_(3) prop k'_(3) xx 10^(6)`
Since `k_(1) = k_(2) = k_(3)`
`k'_(1) = k'_(2) xx 10^(3) = k'_(3) xx 10^(6)`