If 60 % of a first order reaction was completed in 60 minutes, 50 % of the same reaction would be completed in approximately
[log = 4 = 0.60, log 5 = 0.69].

Options

A. 50min
B. 45min
C. 60min
D. 40min

Explanation

To solve the problem, we need to determine the time required for 50% completion of a first-order reaction, given that 60% of the reaction is completed in 60 minutes.

### Step-by-Step Solution:

1. **Understanding the Reaction**:
- We have a first-order reaction where the initial concentration of reactant A is denoted as [A]₀ = A.
- After 60 minutes, 60% of A has reacted, which means 40% of A remains. Therefore, the concentration of A at t = 60 minutes is:
\[ [A] = A - 0.6A = 0.4A \]

2. **Using the First-Order Reaction Formula**:
- The integrated rate law for a first-order reaction is given by:
\[ t = \frac{2.303}{k} \log \left( \frac{[A]₀}{[A]} \right) \]
- Substituting the values for t = 60 minutes, [A]₀ = A, and [A] = 0.4A:
\[ 60 = \frac{2.303}{k} \log \left( \frac{A}{0.4A} \right) \]
- Simplifying the logarithm:
\[ \log \left( \frac{A}{0.4A} \right) = \log \left( \frac{1}{0.4} \right) = \log(2.5) = \log(10) - \log(4) = 1 - 0.6 = 0.4 \]
- Thus, we can rewrite the equation as:
\[ 60 = \frac{2.303}{k} \cdot 0.4 \]

3. **Finding the Rate Constant (k)**:
- Rearranging the equation to find k:
\[ k = \frac{2.303 \cdot 0.4}{60} \]

4. **Calculating Time for 50% Completion**:
- For 50% completion, we have:
\[ [A] = \frac{A}{2} \]
- Substituting this into the first-order equation:
\[ t_{50\%} = \frac{2.303}{k} \log \left( \frac{A}{\frac{A}{2}} \right) = \frac{2.303}{k} \log(2) \]
- Since we already have k, we can substitute it back into this equation.

5. **Relating t_{50\%} to t = 60 minutes**:
- From our earlier equation, we can set up a ratio:
\[ \frac{t_{50\%}}{60} = \frac{\log(2)}{\log(2.5)} \]
- Using the values of logarithms provided:
\[ \log(2) \approx 0.301 \quad \text{and} \quad \log(2.5) = 1 - 0.6 = 0.4 \]
- Thus:
\[ \frac{t_{50\%}}{60} = \frac{0.301}{0.4} \]
- Cross-multiplying gives:
\[ t_{50\%} = 60 \cdot \frac{0.301}{0.4} = 60 \cdot 0.7525 \approx 45.15 \text{ minutes} \]

### Final Answer:
The time required for 50% completion of the reaction is approximately **45 minutes**.