A first-order reaction has a half-life of 2.10 hours.How much time is required for the reactant to be reduced to 33.0% of its initial concentration?

Question

Quizplus · Accepted Answer

The equation for a first-order reaction is: ln([A]/[A]0) = -kt
where [A]0 is the initial concentration, [A] is the concentration at time t, k is the rate constant, and t is time.

The half-life of a first-order reaction is given by: t1/2 = (ln2)/k

We are given that t1/2 = 2.10 hours. Therefore, we can calculate the rate constant k as:
k = (ln2)/t1/2 = ln2/2.10 = 0.33 h^-1

We need to find the time it takes for the reactant to be reduced to 33.0% of its initial concentration. This means [A]/[A]0 = 0.330. Substituting this value into the first-order reaction equation, we get:
ln(0.330) = -0.33t

Solving for t, we get:
t = [ln(0.330)]/(-0.33) = 3.36 hours

Therefore, the answer is choice C, 3.36 hours.

A First-Order Reaction Has a Half-Life of 2