Question

A first order reaction is 75% completed in 6000 s. What is its half life period at the same temperature? ________.

• A. 15 min
• B. 25 min
• C. 75 min
• D. 50 min
• E. 60 min

Hint:

$t_{75\%} = 2 t_{1/2}$; $t_{87.5\%} = 3 t_{1/2}$; $t_{93.75\%} = 4 t_{1/2}$.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept
This problem involves the kinetics of a first-order reaction. A key characteristic of first-order reactions is that their half-life is constant, and there is a direct relationship between the time taken for any fraction of completion and the half-life.
Step 2: Key Formula or Approach
Method 1: Conceptual understanding of half-lives.
- After one half-life (\(t_{1/2}\)), 50% of the reaction is complete, and 50% of the reactant remains. - After a second half-life (total time \(2 \times t_{1/2}\)), 50% of the \textit{remaining} reactant reacts. This is \(0.5 \times 50% = 25%\) of the original amount. - The total percentage completed after two half-lives is \(50% + 25% = 75%\). - Therefore, for a first-order reaction, the time required for 75% completion is exactly twice the half-life: \(t_{75%} = 2 \times t_{1/2}\). Method 2: Using the integrated rate law.
The integrated rate law for a first-order reaction is \(k = \frac{2.303}{t} \log\left(\frac{A_0}{A_t}\right)\), and the half-life is \(t_{1/2} = \frac{0.693}{k}\). We can use the first equation to find \(k\) and then find \(t_{1/2}\).
Step 3: Detailed Explanation
Using Method 1 (Conceptual):
We are given that the time for 75% completion is 6000 s. \[ t_{75%} = 6000 \text{ s} \] Using the relationship \(t_{75%} = 2 \times t_{1/2}\): \[ 6000 \text{ s} = 2 \times t_{1/2} \] Solve for \(t_{1/2}\): \[ t_{1/2} = \frac{6000 \text{ s}}{2} = 3000 \text{ s} \] The question asks for the half-life in minutes. We need to convert seconds to minutes. \[ t_{1/2} \text{ in minutes} = \frac{3000 \text{ s}}{60 \text{ s/min}} = 50 \text{ min} \] Step 4: Final Answer
The half-life period of the reaction is 50 minutes.