Question

If 75$\%$ of a first-order reaction gets completed in 32 minutes, the time taken for 50$\%$ completion of this reaction is:

• A. 16 minutes
• B. 78 minutes
• C. 8 minutes
• D. 4 minutes

Answer 1

The Correct Option is C

To solve this problem, we need to understand the relationship between reaction completion time for a first-order reaction. For a first-order reaction, the time taken for a certain percentage of the reaction to complete can be determined using the half-life formula and the reaction order equations.

For first-order reactions, the time \( t \) required for a given percentage of the reaction to complete can be described by the equation:

\[ \ln \left(\frac{[A]_0}{[A]}\right) = kt \]

where \([A]_0\) is the initial concentration, \([A]\) is the remaining concentration at time \( t \), and \( k \) is the rate constant.

The half-life \( t_{1/2} \) for a first-order reaction is a constant and is given by:

\[ t_{1/2} = \frac{0.693}{k} \]

According to the problem, 75% of the reaction gets completed in 32 minutes. This means 25% of the reactant remains, which gives us:

\[ \ln \left(\frac{100}{25}\right) = kt_{75\%} = k \times 32 \]

Simplifying this, we have:

\[ \ln(4) = k \times 32 \]

Using the natural logarithm value, \(\ln(4) = 1.386\), we can solve for \( k \):

\[ 1.386 = k \times 32 \]

\[ k = \frac{1.386}{32} \]

Now, since \( t_{1/2} = \frac{0.693}{k} \), substituting for \( k \) gives:

\[ t_{1/2} = \frac{0.693 \times 32}{1.386} \]

Simplifying this equation, we find:

\[ t_{1/2} = \frac{22.176}{1.386} \approx 16 \, \text{minutes} \]

Since the 50% completion time (half-life) is half of the \( 75\% \) completion time, the answer is:

\[ \frac{16}{2} = 8 \, \text{minutes} \]

Hence, the time taken for 50% completion of the reaction is 8 minutes.

Answer 2

\(\text{For a first-order reaction, the relationship between time and completion is given by:} \\ t = \frac{0.693}{k} \\ \text{For 75\% completion, the time is related to } k \text{ as follows:} \\ t_{75\%} = \frac{t_{50\%}}{0.693 \times 11} \\ \text{Given that the 75\% completion time is 32 minutes, the time for 50\% completion is:} \\ t_{50\%} = \frac{32}{0.693} \approx 8 \, \text{minutes}\)