To address this problem, we must first establish the connection between reaction completion time and first-order reactions. The duration for a specified reaction percentage to complete in a first-order process can be determined using the half-life formula and relevant reaction order equations.
For first-order reactions, the time \( t \) needed for a given reaction percentage to complete is defined by the equation:
\[ \ln \left(\frac{[A]_0}{[A]}\right) = kt \]
Here, \([A]_0\) denotes the initial concentration, \([A]\) is the concentration remaining at time \( t \), and \( k \) represents the rate constant.
The half-life \( t_{1/2} \) of a first-order reaction is constant and is calculated as:
\[ t_{1/2} = \frac{0.693}{k} \]
The problem states that 75% of the reaction completes in 32 minutes. This implies that 25% of the reactant remains, leading to the equation:
\[ \ln \left(\frac{100}{25}\right) = kt_{75\%} = k \times 32 \]
This simplifies to:
\[ \ln(4) = k \times 32 \]
Using the value \(\ln(4) = 1.386\), we can calculate \( k \):
\[ 1.386 = k \times 32 \]
\[ k = \frac{1.386}{32} \]
Subsequently, using \( t_{1/2} = \frac{0.693}{k} \), we substitute the value of \( k \) to find the half-life:
\[ t_{1/2} = \frac{0.693 \times 32}{1.386} \]
Simplifying this expression yields:
\[ t_{1/2} = \frac{22.176}{1.386} \approx 16 \, \text{minutes} \]
Given that the time for 50% completion (half-life) is half the time for 75% completion, the final answer is:
\[ \frac{16}{2} = 8 \, \text{minutes} \]
Therefore, the reaction achieves 50% completion in 8 minutes.