Question

A first order reaction has a rate constant 0.00813 min\(^{-1}\). How long will it take for 60% completion?

• A. 98.7 min
• B. 56.35 min
• C. 112.7 min
• D. 62.77 min

Hint:

For first-order reactions, the time for a given percentage of completion can be calculated using the formula \( \ln \left( \frac{A_0}{A_t} \right) = kt \).

Answer 1

The Correct Option is C

Step 1: First order reaction formula.
For a first-order reaction, the relation between the concentration at any time \( t \) and the initial concentration is: \[ \ln \left( \frac{A_0}{A_t} \right) = kt \] Where: - \( A_0 \) is the initial concentration, - \( A_t \) is the concentration at time \( t \), - \( k \) is the rate constant, and - \( t \) is the time.

Step 2: Calculation.
For 60% completion, 40% of the reactant remains. Using the first-order rate equation: \[ \ln \left( \frac{1}{0.4} \right) = 0.00813 \times t \] Solving for \( t \): \[ \ln(2.5) = 0.00813 \times t \quad \Rightarrow \quad 0.9163 = 0.00813 \times t \] \[ t = \frac{0.9163}{0.00813} \approx 112.7 \, \text{min} \]
Step 3: Conclusion.
The time required for 60% completion is approximately 112.7 minutes, so the correct answer is (C) 112.7 min.