Question

An organic compound undergoes first order decomposition. The time taken for decomposition to \(\dfrac{1}{8}\) and \(\dfrac{1}{10}\) of its initial concentration are \(t_{1/8}\) and \(t_{1/10}\) respectively. Find the value of \[ \frac{t_{1/8}}{t_{1/10}}\times10 \] (Given: \(\log 2 = 0.3\))

• A. \(3\)
• B. \(30\)
• C. \(9\)
• D. \(0.9\)

Hint:

For first order reactions, time ratios depend only on logarithms of concentration ratios—not on initial concentration.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
For a first-order reaction, the integrated rate equation is:
\[ k = \frac{2.303}{t} \log \frac{[A]_0}{[A]_t} \] where \( [A]_0 \) is the initial concentration and \( [A]_t \) is the concentration at time \( t \).
Step 2: Key Formula or Approach:
From the rate equation, time \( t \) can be expressed as:
\[ t = \frac{2.303}{k} \log \frac{[A]_0}{[A]_t} \] Step 3: Detailed Explanation:
1. Calculate \( t_{1/8} \):
Here, \( [A]_t = \frac{1}{8} [A]_0 \).
\[ t_{1/8} = \frac{2.303}{k} \log \frac{[A]_0}{\frac{1}{8}[A]_0} = \frac{2.303}{k} \log 8 = \frac{2.303}{k} \log (2^3) = \frac{3 \times 2.303 \times \log 2}{k} \]
2. Calculate \( t_{1/10} \):
Here, \( [A]_t = \frac{1}{10} [A]_0 \).
\[ t_{1/10} = \frac{2.303}{k} \log \frac{[A]_0}{\frac{1}{10}[A]_0} = \frac{2.303}{k} \log 10 = \frac{2.303}{k} \times 1 \]
3. Calculate the ratio:
\[ \frac{t_{1/8}}{t_{1/10}} = \frac{\frac{3 \times 2.303 \times 0.3}{k}}{\frac{2.303}{k}} = 3 \times 0.3 = 0.9 \]
4. Final Calculation:
\[ \text{Value} = \frac{t_{1/8}}{t_{1/10}} \times 10 = 0.9 \times 10 = 9 \]
Step 4: Final Answer:
The value is 9.