A first-order reaction has a half-life of 29.2 s. how long does it take for the concentration of the reactant in the reaction to fall to one-sixteenth of its initial value?

Answer is: it takes 116,8 seconds to fall to one-sixteenth of its initial value

The half-life for the chemical reaction is 29,2 s and is independent of initial concentration.

c ₀ - initial concentration the reactant.

c - concentration of the reactant remaining at time.

t = 29,2 s.

First calculate the rate constant k:

k = 0,693 : t = 0,693 : 29,2 s = 0,0237 1/s.

ln (c/c ₀) = - k·t₁.

ln (1/16 : 1) = - 0,0237 1/s · t₁.

t₁ = 116,8 s.