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. 40 minutes
B. 50 minutes
C. 45 minutes
D. 60 minutes

Explanation

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

### Step-by-Step Solution:

1. **Understanding the Reaction Order**:
- We are dealing with a first-order reaction. The rate of a first-order reaction can be described by the equation:
\[ k = \frac{2.303}{T} \log \left( \frac{A_0}{A_0 - x} \right) \]
where \( k \) is the rate constant, \( T \) is the time taken, \( A_0 \) is the initial concentration, and \( x \) is the amount reacted.

2. **Finding the Rate Constant (k)**:
- Given that 60% of the reaction is completed in 60 minutes, we can express this mathematically:
- \( x = 0.6 A_0 \) (60% reacted)
- \( A_0 - x = 0.4 A_0 \) (40% remaining)
- Substituting into the equation:
\[ k = \frac{2.303}{60} \log \left( \frac{A_0}{0.4 A_0} \right) \]
- This simplifies to:
\[ k = \frac{2.303}{60} \log \left( \frac{1}{0.4} \right) = \frac{2.303}{60} \log(2.5) \]

3. **Calculating log(2.5)**:
- We can express \( \log(2.5) \) using the properties of logarithms:
\[ \log(2.5) = \log\left(\frac{10}{4}\right) = \log(10) - \log(4) \]
- Given that \( \log(4) = 0.60 \):
\[ \log(2.5) = 1 - 0.60 = 0.40 \]

4. **Substituting Back to Find k**:
- Now substituting \( \log(2.5) \) back into the equation for \( k \):
\[ k = \frac{2.303}{60} \times 0.40 \]
- Calculating this gives:
\[ k = \frac{2.303 \times 0.40}{60} \approx \frac{0.9212}{60} \approx 0.01535 \, \text{min}^{-1} \]

5. **Finding the Half-Life (T_half)**:
- The half-life for a first-order reaction is given by:
\[ T_{1/2} = \frac{0.693}{k} \]
- Substituting the value of \( k \):
\[ T_{1/2} = \frac{0.693}{0.01535} \approx 45.2 \, \text{minutes} \]
- Rounding this gives approximately 45 minutes.

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

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