Step 1: Understanding the Concept:
For a first-order reaction, the time required for the concentration to drop by a specific ratio is constant regardless of the starting concentration. We can find the rate constant $k$ from the $t=20$ data point, and then use it to find the time $x$ corresponding to the intermediate drop. Alternatively, a simple logarithmic ratio scaling gives the answer directly. Step 2: Key Formula or Approach:
First-order kinetics equation: $k = \frac{1}{t} \ln \left(\frac{[A]_0}{[A]_t}\right)$
Rearranging for time: $t = \frac{1}{k} \ln \left(\frac{[A]_0}{[A]_t}\right)$ Step 3: Detailed Explanation:
Let's analyze the concentration drops:
Initial concentration $[A]_0 = 0.6500 \text{ M}$.
At $t = 20 \text{ min}$, $[A]_{20} = 0.00065 \text{ M}$.
The ratio of concentrations is: $\frac{[A]_0}{[A]_{20}} = \frac{0.6500}{0.00065} = 1000$.
So, $k = \frac{1}{20} \ln(1000) = \frac{1}{20} \ln(10^3) = \frac{3 \ln 10}{20} \text{ min}^{-1}$.
At $t = x \text{ min}$, $[A]_x = 0.0650 \text{ M}$.
The ratio of concentrations is: $\frac{[A]_0}{[A]_x} = \frac{0.6500}{0.0650} = 10$.
So, $x = \frac{1}{k} \ln(10)$.
Substitute the value of $k$ into the equation for $x$:
\[ x = \frac{1}{\frac{3 \ln 10}{20}} \times \ln 10 \]
\[ x = \frac{20}{3 \ln 10} \times \ln 10 \]
The $\ln 10$ terms cancel out perfectly:
\[ x = \frac{20}{3} = 6.666\dots \text{ min} \]
Rounding to the nearest integer, we get 7. Step 4: Final Answer:
The value of $x$ is 7.
