Under the same reaction condition, initial concentration of 1.386 mol `dm^(-3)` of a substance becomes half in 40s and 20s through first order and zero order kinetics respectively. Find out the `k_(1)/k_(0)` ratio for first order `(k_(1))` and zero order `(k_(0))` of the reaction.

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

a) For first order reaction,
`k_(1) = 0.693/t_(1//2) = 0.693/(40)`
For zero order reaction
`k_(0) = ([A]_(0))/(2t_(1//2)) = (1.386 dm^(-3))/(2 xx20)`
`k_(1)/k_(0) = 0.693/(40) xx 40/(1.386 "mol"dm^(-3))`
`0.5 mol^(-1)dm^(3)`