Question

A first-order reaction is 25% complete in 30 minutes. How much time will it take for the reaction to be 75% complete?

• A. 90 min
• B. 60 min
• C. 120 min
• D. 150 min

Hint:

A useful shortcut for first-order reactions: The time for 75% completion (\(t_{75\%}\)) is exactly two half-lives (\(2 \times t_{1/2}\)). However, since 25% is not a half-life, you must use the ratio of logs as shown above.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept
For a first-order reaction, the time required for a certain percentage of completion depends on the rate constant \(k\), which remains constant throughout the reaction.

Step 2: Key Formula or Approach
The integrated rate law for a first-order reaction is: \[ k = \frac{2.303}{t} \log\left(\frac{a}{a-x}\right) \] where \(a\) is the initial concentration and \(x\) is the amount reacted.

Step 3: Detailed Explanation
1. Find \(k\) from 25% completion: - \(t = 30\), \(x = 0.25a\), so \(a-x = 0.75a\). - \(k = \frac{2.303}{30} \log\left(\frac{1}{0.75}\right) = \frac{2.303}{30} \log(1.33)\) 2. Calculate \(t\) for 75% completion: - \(x = 0.75a\), so \(a-x = 0.25a\). - \(t_{75\%} = \frac{2.303}{k} \log\left(\frac{1}{0.25}\right) = \frac{2.303}{k} \log(4)\) 3. Substitute \(k\): - \(t_{75\%} = 30 \times \frac{\log(4)}{\log(1.33)}\) - Using \(\log(4) \approx 0.602\) and \(\log(1.33) \approx 0.124\): - \(t_{75\%} = 30 \times \frac{0.602}{0.124} \approx 30 \times 4.85 \approx 145.5 \text{ min}\). - The closest standard option provided in such exam sets is 150 min.

Step 4: Final Answer
The reaction will be 75% complete in approximately 150 minutes.