Question

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.

Answer 1

For the first order kinetics,
`k_(1)= 0.693/t_(1//2) = 0.693/(40s)`
For zero order kinetics, `k_(0) = [A]_(0)/2t_(1//2)= (1.386 mol dm^(-3))/(2 xx 20s)`
`k_(1)/k_(0) = 0.693/(40s) xx (40s)/(1.386 mol dm^(-3))= 0.693/(1.386 mol dm^(-3)) = 0.5 mol ^(-1)dm^(3)`