Final answer:
To calculate the rate constant for the first-order reaction, the integrated first-order rate law is used with the given completion percentage. Solving the equation, the rate constant is found to be 0.0025 s⁻¹, which is option B.
Step-by-step explanation:
The question is asking for the calculation of the rate constant for a first-order reaction given that it is 38.5% complete in 480 seconds. To calculate the rate constant (k) for a first-order reaction, we can use the formula derived from the integrated first-order rate law:
ln(rac{[A]_0}{[A]_t}) = kt
where [A]_0 is the initial concentration and [A]_t is the concentration at time t. Since the reaction is 38.5% complete, 61.5% (100% - 38.5%) of the reactant remains. Plugging this into the formula, we can solve for k. The complete set of steps is as follows:
- Calculate the fraction of reactant remaining: 100% - 38.5% = 61.5% or 0.615 in decimal form.
- Substitute the values into the integrated rate law: ln(rac{1}{0.615}) = kt.
- Consider that at t = 480 s and solve for k: ln(rac{1}{0.615}) / 480 s = k.
- Calculate k: k = ln(1 / 0.615) / 480 s = 0.0025 s⁻¹.
Therefore, the rate constant for the reaction is 0.0025 s⁻¹, which corresponds to option B.