Question

In a first order decomposition reaction, the time taken for the decomposition of reactant to one fourth and one eighth of its initial concentration are $ t_1 $ and $ t_2 $ (s), respectively. The ratio $ t_1 / t_2 $ will be:

• A. \(\frac{4}{3}\)
• B. \(\frac{3}{4}\)
• C. \(\frac{2}{3}\)
• D. \(\frac{3}{2}\)

Hint:

- For first-order reactions, time depends on the fraction remaining, not the absolute concentration - Each half-life period is equal for first-order reactions - The time to reach 1/4 is less than time to reach 1/8 (\(\frac{2}{3}\) ratio)

Answer 1

The Correct Option is C

Step 1: Recall first-order kinetics equation \[ t = \frac{2.303}{k} \log \frac{[A]_0}{[A]} \] where \(k\) is the rate constant, \([A]_0\) is initial concentration, and \([A]\) is concentration at time \(t\). 
Step 2: Calculate \(t_1\) for 1/4th decomposition \[ t_1 = \frac{2.303}{k} \log \frac{1}{1/4} = \frac{2.303}{k} \log 4 = \frac{2.303}{k} \times 2 \log 2 \] 
Step 3: Calculate \(t_2\) for 1/8th decomposition \[ t_2 = \frac{2.303}{k} \log \frac{1}{1/8} = \frac{2.303}{k} \log 8 = \frac{2.303}{k} \times 3 \log 2 \] 
Step 4: For decomposition to: 
- 1/4 remaining: \( t_1 = \frac{2.303}{k} \log 4 \) - 1/8 remaining: \( t_2 = \frac{2.303}{k} \log 8 \) Thus: \[ \frac{t_1}{t_2} = \frac{\log 4}{\log 8} = \frac{2 \log 2}{3 \log 2} = \frac{2}{3} \]