Question

The rate constant for a first-order reaction is \( 0.0693 \, \text{min}^{-1} \). What is the half-life of the reaction?

• A. \( 5 \, \text{min} \)
• B. \( 10 \, \text{min} \)
• C. \( 15 \, \text{min} \)
• D. \( 20 \, \text{min} \)

Hint:

For first-order reactions, the half-life is independent of the initial concentration and can be quickly calculated using \( t_{1/2} = \frac{\ln 2}{k} \). Memorize \( \ln 2 \approx 0.693 \) for efficiency.

Answer 1

The Correct Option is B

For a first-order reaction, the half-life (\( t_{1/2} \)) is given by the formula:
\[ t_{1/2} = \frac{\ln 2}{k} \] Where:
- \( k = 0.0693 \, \text{min}^{-1} \) (rate constant),
- \( \ln 2 \approx 0.693 \).
Substitute the values:
\[ t_{1/2} = \frac{0.693}{0.0693} \] \[ t_{1/2} \approx \frac{0.693}{0.0693} \approx 10 \, \text{min} \] To confirm:
\[ 0.693 \div 0.0693 = \frac{693}{69.3} = 10 \] Thus, the half-life of the reaction is \( 10 \, \text{min} \).