Question

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 1

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.
c₀ - 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/c₀) = -k·t₁.
ln(1/16 ÷ 1) = -0,0237 1/s · t₁.

t₁ = 116,8 s.