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. (a) `0.5 mol^(-1) dm^(3)`
B. (b) `1.0 mol dm^(-3)`
C. (c ) `1.5 mol dm^(-3)`
D. (d) `2.0 mol^(-1) dm^(3)`

Answer

a

Explanation

For first order reaction,
`K=2.303/t_(1//2)"log"a/(0.5a)`
`=2.303/t_(1//2)log_(10)2=0.693/t_(1//2)`
`:. t_(1//2)=0.693/K`,
`:. K_(1)=0.693/t_(1//2)=0.693/40`
for zero order reaction,
`:. K_(0)=A_(0)/(2xxt_(1//2))=1.386/(2xx20)`
Now, `K_(1)/K_(0)=0.693/40xx40/1.386=0.5 mol^(-1) dm^(3)`