Question

For a first order reaction, the ratio of the time for \(75%\) completion of a reaction to the time for \(50%\) completion is \(\dots\dots\dots\). (Integer answer)

Hint:

For first-order reactions, \(t_{75%} = 2 \times t_{50%}\) and \(t_{87.5%} = 3 \times t_{50%}\). These power-of-two relationships are very common in exams.

Answer 1

Correct Answer: 2

Step 1: Understanding the Concept:
In a first-order reaction, the time required for a certain percentage of completion depends only on the rate constant and the fraction remaining. A useful property is that the time for \(75%\) completion is exactly twice the half-life.
Step 2: Key Formula or Approach:
\[ t = \frac{2.303}{k} \log \left( \frac{[A]_0}{[A]_t} \right) \]
Step 3: Detailed Explanation:
1. Time for \(50%\) completion (\(t_{1/2}\)):
Amount remaining is \(50%\) of \([A]_0\).
\[ t_{50%} = \frac{2.303}{k} \log \left( \frac{100}{50} \right) = \frac{2.303 \log 2}{k} = \frac{0.693}{k} \]
2. Time for \(75%\) completion (\(t_{75%}\)):
Amount remaining is \(25%\) of \([A]_0\).
\[ t_{75%} = \frac{2.303}{k} \log \left( \frac{100}{25} \right) = \frac{2.303 \log 4}{k} = \frac{2.303 \times 2 \log 2}{k} \]
3. Calculation of Ratio:
\[ \text{Ratio} = \frac{t_{75%}}{t_{50%}} = \frac{(2 \times 2.303 \log 2) / k}{(2.303 \log 2) / k} = 2 \]
Step 4: Final Answer:
The ratio is 2.