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

Answer

A

Explanation

For first order reaction
`k_(1)=(0.693)/(t_(1//2))=(0.693)/(40s)`
For a zero order reaction: `k_(0)=([R]_(0))/(2t_(1//2))=(1.386)/(2xx20_(s))`
Therefore
`(k_(1))/(k_(0))=(0.693)/(40s)xx(40s)/(1.386)`
`=(0.693(s^(-1))/(1.386(moldm^(-3)s^(-1)))`
`=0.5mol^(-1)dm^(3)`