The half-life of a 1st order reaction is 1 hr. What is the fraction of the reactant remaining after 3 hours?
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For a first-order reaction, the fraction remaining after time \( t \) is given by the equation \( \frac{[A]_t}{[A]_0} = e^{-kt} \), where \( k \) is the rate constant and \( t \) is the time. The half-life for a first-order reaction is constant and is independent of the initial concentration.
In a first-order reaction, the rate of the reaction is directly proportional to the concentration of the reactant. The half-life (\( t_{1/2} \)) for a first-order reaction is constant and does not depend on the initial concentration. The formula for the fraction of reactant remaining at any time \( t \) is:
\[
\frac{[A]_t}{[A]_0} = e^{-kt}
\]
where:
- \( [A]_t \) is the concentration of the reactant at time \( t \),
- \( [A]_0 \) is the initial concentration,
- \( k \) is the rate constant, and
- \( t \) is the time elapsed.
For a first-order reaction, the relationship between the half-life and the rate constant is given by:
\[
t_{1/2} = \frac{0.693}{k}.
\]
Given that the half-life \( t_{1/2} = 1 \, \text{hr} \), we can calculate the rate constant \( k \):
\[
k = \frac{0.693}{t_{1/2}} = \frac{0.693}{1} = 0.693 \, \text{hr}^{-1}.
\]
Now, to find the fraction remaining after 3 hours, we substitute into the formula for the fraction of reactant remaining:
\[
\frac{[A]_t}{[A]_0} = e^{-0.693 \times 3} = e^{-2.079} \approx \frac{1}{8}.
\]
Thus, the fraction of the reactant remaining after 3 hours is \( \frac{1}{8} \).
