For a first-order reaction, the rate of the reaction is proportional to the concentration of the reactant:
Rate = k[A]
Where [A] is the concentration of the reactant and k is the rate constant. The integrated rate law for a first-order reaction is:
ln([A]t/[A]0) = -kt
Where [A]t is the concentration of the reactant at time t and [A]0 is the initial concentration.
To find the first and second half-lives, we need to use the fact that the reaction is 75% complete after 395 s. This means that [A]t/[A]0 = 0.25, and we can substitute this value into the integrated rate law:
ln(0.25) = -k(395)
Solving for k, we get:
k = ln(0.25) / (-395) ≈ 0.00226 s^-1
The first half-life is the time it takes for the concentration of the reactant to decrease to half of its initial value. We can use the integrated rate law to solve for the first half-life:
ln(0.5) = -k(t1/2)
Solving for t1/2, we get:
t1/2 = ln(2) / k ≈ 307 s
The second half-life is the time it takes for the concentration of the reactant to decrease to one-fourth of its initial value. We can use the same equation and substitute [A]t/[A]0 = 0.25 again:
ln(0.25) = -k(t2/2)
Solving for t2/2, we get:
t2/2 = ln(4) / k ≈ 1229 s
Therefore, the first half-life for this reaction is approximately 307 s, and the second half-life is approximately 1229 s.