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.