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 1st order reaction,
`t_(1//2) = (0*693)/(k) or k_(1) = (0.0693)/(t_(1//2)) = (0*693)/(40s)`
For zero order reaction,
`t_(1//2) = ([A]_(0))/(2k) or k_(0) = ([A]_(0))/(2t_(1//2)) = (1*386 mol dm^(-3))/(2xx 20s)`
`therefore (k_(1))/(k_(0)) = (0*693)/(40s) xx (2xx 20s)/(1*386 mol dm^(-3)) = 0*5 mol^(-1) dm^(3)`