The half-life of a first order reaction is determined to be 92.0 years. How long will it take for the concentration of the reactant to reach 2% of its initial value?

(1/2)^n = .02

n ln .5 = ln .02
n = -3.91/-.693
n = 5.644 half lives
times 92 = 519 years

92.0 years X (multiply) with 5.64386

To determine the time it takes for the concentration of the reactant to reach a certain value in a first-order reaction, we can use the following formula:

t = (ln(C₀/C))/(k)

Where:
t = time
C₀ = initial concentration of the reactant
C = final concentration of the reactant
k = rate constant

In this case, we are given the half-life, which is the time it takes for the concentration of the reactant to decrease by half. Since a first-order reaction has a constant rate constant, we can use this information to calculate the rate constant (k).

Given:
Half-life (t₁/₂) = 92.0 years

The half-life of a first-order reaction can be related to the rate constant (k) using the formula:

t₁/₂ = (0.693)/k

Substituting the given half-life value:

92.0 years = (0.693)/k

Now, let's solve for k:

k = (0.693)/92.0

k ≈ 0.00753 year⁻¹

Next, we need to find the time it takes for the concentration of the reactant to reach 2% (0.02) of its initial value. Using the formula mentioned earlier:

t = (ln(C₀/C))/(k)

Substituting the known values:

t = (ln(1.00/0.02))/(0.00753)

Now, calculate t:

t ≈ (ln(50))/(0.00753)

Using a calculator or software, we find that:

t ≈ 277.1 years

Therefore, it will take approximately 277.1 years for the concentration of the reactant to reach 2% of its initial value.