Question

For a first-order reaction, the rate constant \( k \) is \( 1.386 \times 10^{-3} \, \text{s}^{-1} \). What is the half-life \( t_{1/2} \) of the reaction? (Use \( \ln 2 = 0.693 \))

• A. 300 s
• B. 500 s
• C. 200 s
• D. 1000 s

Hint:

For a first-order reaction, the half-life is independent of the initial concentration and is given by \( t_{1/2} = \frac{\ln 2}{k} \).

Answer 1

The Correct Option is B

- For a first-order reaction, the half-life \( t_{1/2} \) is given by: \[ t_{1/2} = \frac{\ln 2}{k} \]
- Given \( k = 1.386 \times 10^{-3} \, \text{s}^{-1} \), \( \ln 2 = 0.693 \): \[ t_{1/2} = \frac{0.693}{1.386 \times 10^{-3}} = \frac{0.693}{1.386} \times 10^3 = 0.5 \times 10^3 = 500 \, \text{s} \]
- This matches option (B).