Question

After 75.0 min, 19.0% of a compound has decomposed. What is the half-life of this reaction assuming first-order kinetics?

Answer 1

Let us consider,

Initial amount = 100

remaining amount = 100 19.0 = 81.0

Time = 75.0 min

First order integrated rate constant equation is,

k = (1/t) * ln[A]0/[A]t

k = (1/75.0) * ln(100 / 81.0)

k = 0.00281 min-1

THerefore, Half life time of first order reaction is,

t1/2 = 0.693 / k

t1/2 = 0.693 / 0.00281

t1/2 = 246.6 min