Step 1: Understanding the Concept:
For a first-order reaction, the time required for the concentration to reduce to half its initial value is constant, known as the half-life ($t_{1/2}$).
The concentration after $n$ half-lives is given by the formula: $[A]_t = \frac{[A]_0}{2^n}$.
Once the number of half-lives and the total time are known, we can calculate the half-life and subsequently the rate constant ($k$).
Step 2: Key Formula or Approach:
1. Find $n$ (number of half-lives) using $[A]_t = [A]_0 / 2^n$.
2. $t_{1/2} = \text{Total time} / n$.
3. $k = \frac{\ln(2)}{t_{1/2}} = \frac{0.693}{t_{1/2}}$.
Step 3: Detailed Explanation:
- Initial concentration $[A]_0 = 400$ mol/L.
- Final concentration $[A]_t = 50$ mol/L.
- Total time $T = 7.5 \times 10^3$ s.
First, find how many times the concentration was halved:
$400 \rightarrow 200$ (1st half-life)
$200 \rightarrow 100$ (2nd half-life)
$100 \rightarrow 50$ (3rd half-life)
Alternatively: $400/50 = 8 = 2^3$. Thus, $n = 3$.
Now, find the duration of one half-life:
$3 \times t_{1/2} = 7.5 \times 10^3$ s
$t_{1/2} = 2.5 \times 10^3$ s.
Now, calculate the rate constant $k$:
$k = 0.693 / (2.5 \times 10^3)$
$k = 0.2772 \times 10^{-3}$
$k = 2.772 \times 10^{-4}$ $s^{-1}$.
This value corresponds to approximately $2.77 \times 10^{-4}$ $s^{-1}$.
Step 4: Final Answer:
The rate constant $k$ is $2.77 \times 10^{-4}$ $s^{-1}$. The correct option is (D).