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

To solve this problem, we need to understand the concept of a first-order decomposition reaction. In a first-order reaction, the time taken for a reactant to reduce to a fraction of its initial concentration is given by the formula for the half-life:

\(t = \frac{\ln(n)}{k}\)

where:

• \(n\) is the fraction of the initial concentration remaining.
• \(k\) is the rate constant.

According to the problem, we have two scenarios:

1. The reactant decomposes to one fourth (\(\frac{1}{4}\)) of its initial concentration in time \(t_1\).
2. The reactant decomposes to one eighth (\(\frac{1}{8}\)) of its initial concentration in time \(t_2\).

We can apply the first-order reaction formula to these cases:

1. For \(t_1\):
2. For \(t_2\):

The ratio of \(t_1\) to \(t_2\) is given by:

\(\frac{t_1}{t_2} = \frac{\ln(4)}{\ln(8)}\)

We know that:

• \(\ln(4) = 2 \ln(2)\)
• \(\ln(8) = 3 \ln(2)\)

Substituting these values, we have:

\(\frac{t_1}{t_2} = \frac{2 \ln(2)}{3 \ln(2)} = \frac{2}{3}\)

Thus, the ratio \(\frac{t_1}{t_2}\) is \(\frac{2}{3}\). Therefore, the correct option is:

• \(\frac{2}{3}\)

Answer 2

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: \[ \frac{t_1}{t_2} = \frac{\log 4}{\log 8} = \frac{2 \log 2}{3 \log 2} = \frac{2}{3} \]