Problem Analysis
A first-order reaction achieves 50% completion in 45 minutes. The objective is to determine the time necessary for 99.9% reaction completion.
First-Order Half-Life Application
The half-life (\( t_{1/2} \)) for a first-order reaction is defined by: \[ t_{1/2} = \frac{0.693}{k}, \] where \( k \) represents the rate constant.
Rate Constant Calculation
Given \( t_{1/2} = 45 \) minutes: \[ k = \frac{0.693}{45} \, \text{min}^{1}. \]
First-Order Kinetics Equation
The governing equation for a first-order reaction is: \[ \ln \left( \frac{[A]_0}{[A]} \right) = kt, \] with the following definitions:
\( [A]_0 \) denotes the initial concentration.
\( [A] \) represents the concentration at time \( t \).
\( k \) is the rate constant.
\( t \) signifies the time elapsed.
Time for 99.9% Completion Calculation
For 99.9% completion, the remaining concentration is \( [A] = 0.001 [A]_0 \).
Substitution into the first-order equation yields: \[ \ln \left( \frac{[A]_0}{0.001 [A]_0} \right) = kt. \] \[ \ln (1000) = kt. \] Therefore, \[ t = \frac{\ln (1000)}{k}. \]
Substituting \( k = \frac{0.693}{45} \): \[ t = \frac{\ln (1000)}{\frac{0.693}{45}} = \frac{6.908}{0.693} \times 45 \approx 450 \, \text{minutes}. \]
Conversion to hours: \[ t = \frac{450}{60} = 7.5 \, \text{hours}. \]
Option Alignment
The computed duration of 7.5 hours aligns with option (B). Final Answer: The time required for 99.9% of the reaction to be completed is (B) 7.5 hours.