Final answer:
The time for 80% completion of the reaction is 20.09 minutes. If the initial concentration of the reactant is doubled, the time for 80% completion is approximately 10.34 minutes.
Step-by-step explanation:
The half-life of a first-order reaction is a constant that is related to the rate constant for the reaction: t₁/2 = 0.693/k. In this case, the half-life of the reaction is given as 15 minutes. To calculate the time for 80% completion of the reaction, we can use the formula:
time = t₁/2 * ln(1 / (1 - x))
where x is the fraction remaining after time t.
For 80% completion, x = 0.2.
Plugging in the values, we get:
time = 15 min * ln(1 / (1 - 0.2)) = 15 min * ln(5)
Using a calculator, the time is approximately 20.09 minutes for 80% completion of the reaction.
If the initial concentration of the reactant is doubled, the rate constant remains the same, but the reaction proceeds faster. To calculate the time taken for 80% completion in this case, we can use the same formula:
time = t₁/2 * ln(1 / (1 - x))
However, the initial concentration used will be twice the original concentration. Plugging in the values, we get:
time = 15 min * ln(2) ≈ 10.34 minutes.