Under the same reaction conditions, the intial concentration of `1.386 mol dm^(-3)` of a substance becomes half in `40 s` and `20 s` theough first order and zero order kinetics, respectively.
The ratio `(k_(1)//k_(0))` of the rate constants for first order `(k_(1))` and zero order `(k_(0))` of the reaction is

Options

A. `0.5 mol^(-1)dm^(3)`
B. `1.0 mol dm^(-3)`
C. `1.5 mol dm^(-3)`
D. `2.0 mol^(-1) dm^(3)`

Answer

A

Explanation

For first order reaction, `t_(1//2) =(ln_(2))/(k_(1))=40s` …(i)
For zero order reaction, `t_(1//2) =([A_(0)])/(2k_(0)) =20s` ..(ii)
Divide eqn (i) by (ii), we get `(1)/(2) =([A]_(0))/(2k_(0)) xx (k_(1))/(ln_(2))`
`(k_(1))/(k_(0)) = (ln_(2))/([A]_(0)) = (0.693)/(1.386)=0.5`