This question asks for the rate constant ($k$) of a first-order reaction given the initial and final concentrations of the reactant and the time taken.

Reaction Rate Constant Calculation

For a first-order reaction, the relationship between concentration and time is given by the integrated rate law:

$ \ln \frac{[A]_0}{[A]_t} = kt $

Where:

• $[A]_0$ is the initial concentration of the reactant.
• $[A]_t$ is the concentration of the reactant at time $t$.
• $k$ is the rate constant.
• $t$ is the time elapsed.

Identify Given Values

From the question:

• Initial concentration, $[A]_0 = 400 \text{ moles lit}^{-1}$
• Final concentration, $[A]_t = 50 \text{ moles lit}^{-1}$
• Time, $t = 7.5 \times 10^3 \text{ s}$

Calculate Rate Constant ($k$)

1. Rearrange the integrated rate law to solve for $k$:

   $ k = \frac{1}{t} \ln \frac{[A]_0}{[A]_t} $

2. Substitute the given values into the equation:

   $ k = \frac{1}{7.5 \times 10^3 \text{ s}} \ln \frac{400 \text{ moles lit}^{-1}}{50 \text{ moles lit}^{-1}} $

3. Simplify the concentration ratio:

   $ \frac{400}{50} = 8 $

4. Calculate the natural logarithm of 8:

   $ \ln(8) \approx 2.0794 $

5. Compute the rate constant $k$:

   $ k = \frac{2.0794}{7.5 \times 10^3 \text{ s}} $

   $ k \approx \frac{2.0794}{7500} \text{ s}^{-1} $

   $ k \approx 0.00027725 \text{ s}^{-1} $

6. Express the result in scientific notation:

   $ k \approx 2.77 \times 10^{-4} \text{ s}^{-1} $

Final Answer

The calculated rate constant is approximately $2.77 \times 10^{-4} \text{ s}^{-1}$.