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)`?

Options

A. `K_(1) = K_(2) = K_(3)`
B. `K_(1)=K_(2) xx 10^(-3) = K_(3) xx 10^(-6)`
C. `K_(1) = 2K_(2) = K_(3)`
D. `2K_(1) = 3K_(2) = 4K_(3)`

Answer

B

Explanation

rate `=K["Reactant"]^(m)`
Let Let concentration of reactant be a mol`"litre"^(1)` then for I order: `r_(1) = K_(1)[a]^(m)`.... (1) If concentration of reactant be a mol mL-1, then concentration in mol `"litre"^(-1) = a xx 10^(3)`
Thus, `r_(1) = K_(1)[a xx 10^(3)]^(1) , r_(1) = K_(1) xx 10^(3) [a]`
`r_(1)=K_(1)[a]`...........(2)
Similarly for II order , `r_(2) =K_(2) xx 10^(6)[a]^(2)`
`r_(2) =K_(2)^(.)[a]^(2)`.....(3)
Similarly for III order: `r_(3) = K_(3) xx 10^(9) [a]^(3)`
`r_(3) = K_(3)^(.)[a]^(3)`..........(4)
Given, `K_(1) = K_(2) = K_(3)`
`therefore K_(1)^(.)/10^(3) = K_(2)^(.)/10^(6) = K_(3)^(.)/10^(9)` or `K_(1)^(.) = K_(2)^(.) xx 10^(-3) = K_(2)^(.) xx 10^(-6)`