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:
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- \(\frac{4}{3}\)
- \(\frac{3}{4}\)
- \(\frac{2}{3}\)
- \(\frac{3}{2}\)
The Correct Option is C
Approach Solution - 1
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:
- The reactant decomposes to one fourth (\(\frac{1}{4}\)) of its initial concentration in time \(t_1\).
- 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:
- For \(t_1\):
- 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:
- \(\frac{2}{3}\)
Approach Solution -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: \[ \frac{t_1}{t_2} = \frac{\log 4}{\log 8} = \frac{2 \log 2}{3 \log 2} = \frac{2}{3} \]
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