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

`(t_(1//2))_(1) =40 =(0.693)/(k_(1))rArr k_(1)=(0.693)/(40)`
`(t_(1//2))_(o) =20 =(C_(o))/(2k_(o)) rArr k_(o) =(C_(o))/(40) rArr (k_(1))/(k_(o))=(0.693)/(C_(o))=(0.693)/(1.386)=0.5" mol"^(-1)" dm"^(-3)`