Question

If the half-life ($t_{1/2}$) for a first-order reaction is 1 minute, then the time required for 99.9% completion of the reaction is closest to:

• A. 4 minutes
• B. 5 minutes
• C. 10 minutes
• D. 2 minutes

Hint:

For a first-order reaction, the time for 99.9\% completion can be calculated using the formula: \[ t = \frac{\ln(1/0.001)}{k} \] Where \( k \) is determined from the half-life of the reaction.

Answer 1

The Correct Option is C

For a first-order reaction, the relationship between the concentration and time is given by: \[ \ln \left( \frac{[A]_0}{[A]} \right) = kt \] Where: - \( [A]_0 \) is the initial concentration, - \( [A] \) is the concentration at time \( t \), - \( k \) is the rate constant, - \( t \) is the time elapsed. For a first-order reaction, the half-life \( t_{1/2} \) is related to the rate constant \( k \) by the equation: \[ t_{1/2} = \frac{0.693}{k} \] Given that \( t_{1/2} = 1 \) minute, we can solve for \( k \): \[ k = \frac{0.693}{1} = 0.693 \, \text{min}^{-1} \] To find the time for 99.9% completion, we know that 99.9% completion corresponds to 0.1\% remaining, or \( [A] = 0.001 [A]_0 \). Substitute into the first-order equation: \[ \ln \left( \frac{1}{0.001} \right) = k t \] \[ \ln (1000) = 0.693 \times t \] \[ 6.907 = 0.693 \times t \] \[ t = \frac{6.907}{0.693} \approx 10 \, \text{minutes} \] 
Thus, the time required for 99.9% completion is 10 minutes.