Find Time Interval for First Order Reaction Concentration Decay

For a first-order reaction $\text{A} \rightarrow \text{B}$, the integrated rate law is given by: $k = \frac{1}{t} \ln \left(\frac{[\text{A}]_0}{[\text{A}]_t}\right)$

From the given data table:

• At initial time $t = 0\text{ min}$, the initial concentration is $[\text{A}]_0 = 0.6500\text{ M}$.
• At time $t = 20\text{ min}$, the concentration is $[\text{A}]_{20} = 0.00065\text{ M}$.
• At time $t = x\text{ min}$, the concentration is $[\text{A}]_x = 0.0650\text{ M}$.

First, we determine the rate constant $k$ using the data at $t = 20\text{ min}$: $k = \frac{1}{20} \ln \left(\frac{0.6500}{0.00065}\right)$ $k = \frac{1}{20} \ln(1000) = \frac{3 \ln(10)}{20}\text{ min}^{-1}$

Next, we write the equation for $k$ using the data at $t = x\text{ min}$: $k = \frac{1}{x} \ln \left(\frac{0.6500}{0.0650}\right)$ $k = \frac{1}{x} \ln(10)\text{ min}^{-1}$

Equating the two expressions for $k$: $\frac{1}{x} \ln(10) = \frac{3}{20} \ln(10)$

Solving for $x$: $\frac{1}{x} = \frac{3}{20} \implies x = \frac{20}{3}\text{ min} \approx 6.67\text{ min}$

Rounding off to the nearest integer gives: $x = 7$