Step 1: Understanding the Concept:
For a first-order reaction, the rate of reaction is proportional to the concentration of the reactant. The integrated rate equation allows us to calculate the time required for a specific fraction of the reactant to remain. A unique property of first-order kinetics is that the time required for a certain fractional change is independent of the initial concentration. Here, we need to compare the times taken to reach two different levels of remaining concentration.
Step 2: Key Formula or Approach:
The integrated rate expression for a first-order reaction is:
\[ t = \frac{2.303}{k} \log_{10} \left( \frac{[A]_0}{[A]_t} \right) \]
Where:
- $[A]_0$ is the initial concentration.
- $[A]_t$ is the concentration at time $t$.
- $k$ is the rate constant.
Step 3: Detailed Explanation:
Let's find the expression for $t_{1/8}$ (time when $[A]_t = \frac{1}{8} [A]_0$):
\[ t_{1/8} = \frac{2.303}{k} \log_{10} \left( \frac{[A]_0}{\frac{1}{8}[A]_0} \right) = \frac{2.303}{k} \log_{10}(8) \]
Since $8 = 2^3$, we have $\log_{10}(8) = 3 \log_{10}(2)$.
\[ t_{1/8} = \frac{2.303}{k} \times 3 \times 0.3 = \frac{2.303}{k} \times 0.9 \]
Now, let's find the expression for $t_{1/10}$ (time when $[A]_t = \frac{1}{10} [A]_0$):
\[ t_{1/10} = \frac{2.303}{k} \log_{10} \left( \frac{[A]_0}{\frac{1}{10}[A]_0} \right) = \frac{2.303}{k} \log_{10}(10) \]
Since $\log_{10}(10) = 1$, we have:
\[ t_{1/10} = \frac{2.303}{k} \times 1 \]
Now, find the ratio:
\[ \frac{t_{1/8}}{t_{1/10}} = \frac{\frac{2.303}{k} \times 0.9}{\frac{2.303}{k} \times 1} = 0.9 \]
Step 4: Final Answer:
The ratio of the times is 0.9. The correct option is (A).