Ratio of First Order Decomposition Times

For a first-order decomposition reaction, the integrated rate law is given by: $k \cdot t = \ln \left( \frac{[A]_0}{[A]_t} \right) = 2.303 \log_{10} \left( \frac{[A]_0}{[A]_t} \right)$

where:

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

Step 1: Calculate $t_{1/8}$ When the concentration decomposes to $\frac{1}{8}$ of its initial concentration, $[A]_{t_{1/8}} = \frac{[A]_0}{8}$. $k \cdot t_{1/8} = \ln \left( \frac{[A]_0}{\frac{[A]_0}{8}} \right) = \ln(8) = 2.303 \log_{10}(8) = 2.303 \times 3 \log_{10}(2)$

Step 2: Calculate $t_{1/10}$ When the concentration decomposes to $\frac{1}{10}$ of its initial concentration, $[A]_{t_{1/10}} = \frac{[A]_0}{10}$. $k \cdot t_{1/10} = \ln \left( \frac{[A]_0}{\frac{[A]_0}{10}} \right) = \ln(10) = 2.303 \log_{10}(10) = 2.303 \times 1$

Step 3: Calculate the ratio $\frac{t_{1/8}}{t_{1/10}}$ Dividing the expression for $t_{1/8}$ by the expression for $t_{1/10}$: $\frac{t_{1/8}}{t_{1/10}} = \frac{3 \log_{10}(2)}{\log_{10}(10)} = 3 \log_{10}(2)$

Given that $\log_{10}(2) = 0.3$: $\frac{t_{1/8}}{t_{1/10}} = 3 \times 0.3 = 0.9$

Step 4: Find the final required value $\frac{t_{1/8}}{t_{1/10}} \times 10 = 0.9 \times 10 = 9$