Question

A first-order reaction is 75 percent completed in 6000s. What is the half-life?

• A. 50 minutes
• B. 45 minutes
• C. 60 minutes
• D. 20 minutes

Hint:

For first-order reactions, the half-life is independent of the initial concentration and is given by \( t_{1/2} = \frac{0.693}{k} \).

Answer 1

The Correct Option is D

For a first-order reaction, the fraction of reactant remaining at time \( t \) is given by the equation: \[ \ln \left( \frac{[A]}{[A_0]} \right) = -kt \] where \( [A] \) is the concentration at time \( t \), \( [A_0] \) is the initial concentration, and \( k \) is the rate constant. Given that 75 percent of the reaction is completed, 25% of the reactant remains. Thus: \[ \frac{[A]}{[A_0]} = 0.25 \] Substituting into the equation: \[ \ln(0.25) = -k \times 6000 \] Solving for \( k \), we get: \[ k = \frac{-\ln(0.25)}{6000} = 0.000578 \, \text{s}^{-1} \] The half-life for a first-order reaction is given by: \[ t_{1/2} = \frac{0.693}{k} \] Substituting the value of \( k \): \[ t_{1/2} = \frac{0.693}{0.000578} \approx 1200 \, \text{s} \approx 20 \, \text{minutes} \]