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"mol dm"^(-3)`
C. `1.5 "mol dm"^(-3)`
D. `2 "mol"^(-1) dm^(3)`

Answer

A

Explanation

`t_(1//2)=(0.693)/(k_(1))" " t_(1//2)=(a_(0))/(2k_(0))`
`40=(0.693)/(k_(1))," " 20=(1.386)/(2k_(0))=(0.693)/(k_(0))`
`(20)/(40)=(0.693//k_(0))/(0.693//k_(1))=(k_(1))/(k_(0))`
`(k_(1))/(k_(0))=0.5 (sec^(-1))/("mol dm"^(-3)sec^(-1))=0.5 "mol"^(-1) "dm"^(3)`