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 the first order reaction, rate constant
`k_(1) = (0.693)/(t_(1//2)) = (0.693)/(40)`
For zero order reaction, rate constant
`k_(0) = (A_(0))/(2 xx t_(1//2)) = (1.386)/(2 xx 20)`
Hence,
`(k_(1))/(k_(0)) = (0.693)/(1.386) = 0.5`