Question

A first-order reaction has a half-life of 20.8 s . How long does it take for the concentration of the reactant in the reaction to fall to one-fourth of its initial value?

Answer 1

ln A = -kt + ln A_o

k = rate constant                   t = time

Ao = Initial concentration                A = concentration at time t

We can calculate the rate constant using the formula

Rate constant = 0.693 / half life = 0.693 / 20.8 = 0.0333 s-1

let us take initial concentration = 1 M

concentration at last = 0.25 M

ln 0.25 = - 0.0333 x t + ln 1

t = 41.63 seconds

Hence it will take 41.63 seconds the concentration of the reactant in the reaction to fall to one-fourth of its initial value