Question:
An organic compound undergoes first order decomposition. The time taken for its decomposition to \(\frac{1}{8}\text{th}\) and \(\frac{1}{10}\text{th}\) of its initial concentration are \(t_{1/8}\) and \(t_{1/10}\) respectively. The value of \(\left[ \frac{t_{1/8}}{t_{1/10}} \right]\) is [Given \(\log_{10} 2 = 0.3\)]
An organic compound undergoes first order decomposition. The time taken for its decomposition to \(\frac{1}{8}\text{th}\) and \(\frac{1}{10}\text{th}\) of its initial concentration are \(t_{1/8}\) and \(t_{1/10}\) respectively. The value of \(\left[ \frac{t_{1/8}}{t_{1/10}} \right]\) is [Given \(\log_{10} 2 = 0.3\)]
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For first-order reactions, notice that the rate constant $k$ and the constant factor $2.303$ cancel out when taking ratios. Thus, the ratio of times is simply equal to the ratio of the log of concentration ratios:
\[ \frac{t_1}{t_2} = \frac{\log(R_1)}{\log(R_2)} \]
\[ \frac{t_1}{t_2} = \frac{\log(R_1)}{\log(R_2)} \]
Updated On: May 25, 2026
- $0.9$
- $0.6$
- $0.3$
- $0.5$
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The Correct Option is A
Solution and Explanation
Step 1: Understanding the Question:
The question asks for the ratio of time intervals ($t_{1/8}/t_{1/10}$) for a first-order reaction to decompose to $1/8\text{th}$ and $1/10\text{th}$ of its initial concentration respectively.
Step 2: Key Formula or Approach:
The integrated rate equation 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 remaining at time $t$.
Step 3: Detailed Explanation:
Let us find the expression for $t_{1/8}$:
- At $t = t_{1/8}$, the concentration remaining is $[A]_t = \frac{1}{8}[A]_0$.
\[ t_{1/8} = \frac{2.303}{k} \log_{10}\left(\frac{[A]_0}{\frac{1}{8}[A]_0}\right) = \frac{2.303}{k} \log_{10}(8) \]
Since $8 = 2^3$, we can write:
\[ t_{1/8} = \frac{2.303}{k} \times 3\log_{10}(2) \]
Now, let us find the expression for $t_{1/10}$:
- At $t = t_{1/10}$, the concentration remaining is $[A]_t = \frac{1}{10}[A]_0$.
\[ t_{1/10} = \frac{2.303}{k} \log_{10}\left(\frac{[A]_0}{\frac{1}{10}[A]_0}\right) = \frac{2.303}{k} \log_{10}(10) \]
Since $\log_{10}(10) = 1$:
\[ t_{1/10} = \frac{2.303}{k} \times 1 \]
Taking the ratio of the two times:
\[ \frac{t_{1/8}}{t_{1/10}} = \frac{\frac{2.303}{k} \times 3\log_{10}(2)}{\frac{2.303}{k} \times 1} = 3\log_{10}(2) \]
Using the given value $\log_{10}(2) = 0.3$:
\[ \frac{t_{1/8}}{t_{1/10}} = 3 \times 0.3 = 0.9 \]
Step 4: Final Answer:
The correct option is (A).
The question asks for the ratio of time intervals ($t_{1/8}/t_{1/10}$) for a first-order reaction to decompose to $1/8\text{th}$ and $1/10\text{th}$ of its initial concentration respectively.
Step 2: Key Formula or Approach:
The integrated rate equation 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 remaining at time $t$.
Step 3: Detailed Explanation:
Let us find the expression for $t_{1/8}$:
- At $t = t_{1/8}$, the concentration remaining is $[A]_t = \frac{1}{8}[A]_0$.
\[ t_{1/8} = \frac{2.303}{k} \log_{10}\left(\frac{[A]_0}{\frac{1}{8}[A]_0}\right) = \frac{2.303}{k} \log_{10}(8) \]
Since $8 = 2^3$, we can write:
\[ t_{1/8} = \frac{2.303}{k} \times 3\log_{10}(2) \]
Now, let us find the expression for $t_{1/10}$:
- At $t = t_{1/10}$, the concentration remaining is $[A]_t = \frac{1}{10}[A]_0$.
\[ t_{1/10} = \frac{2.303}{k} \log_{10}\left(\frac{[A]_0}{\frac{1}{10}[A]_0}\right) = \frac{2.303}{k} \log_{10}(10) \]
Since $\log_{10}(10) = 1$:
\[ t_{1/10} = \frac{2.303}{k} \times 1 \]
Taking the ratio of the two times:
\[ \frac{t_{1/8}}{t_{1/10}} = \frac{\frac{2.303}{k} \times 3\log_{10}(2)}{\frac{2.303}{k} \times 1} = 3\log_{10}(2) \]
Using the given value $\log_{10}(2) = 0.3$:
\[ \frac{t_{1/8}}{t_{1/10}} = 3 \times 0.3 = 0.9 \]
Step 4: Final Answer:
The correct option is (A).
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