`60%` of a first order reaction was completed in `60 min`. The time taken for reactants to decompose to half of their original amount will be

Options

A. `~~ 30 min`
B. `~~ 45 min`
C. `~~ 20 min`
D. `~~ 40 min`

Explanation

To solve the problem, we need to find the half-life (T1/2) of a first-order reaction given that 60% of the reaction is completed in 60 minutes.

### Step-by-Step Solution:

1. **Understand the Reaction**:
- We have a first-order reaction where 60% of the reactant has decomposed in 60 minutes. This means that 40% of the reactant remains.

2. **Set Up the First-Order Kinetics Equation**:
- The integrated rate law for a first-order reaction is given by:
\[ t = \frac{2.303}{k} \log \left( \frac{[A_0]}{[A]} \right) \]
- Here, \( [A_0] \) is the initial concentration and \( [A] \) is the concentration at time \( t \).

3. **Apply the Given Data**:
- Since 60% has decomposed, we have:
\[ [A] = 0.4 [A_0] \]
- Substituting this into the equation for \( t = 60 \) minutes:
\[ 60 = \frac{2.303}{k} \log \left( \frac{[A_0]}{0.4 [A_0]} \right) \]
- This simplifies to:
\[ 60 = \frac{2.303}{k} \log \left( \frac{1}{0.4} \right) \]
- The logarithm can be rewritten as:
\[ \log \left( \frac{1}{0.4} \right) = \log(2.5) = \log(10/4) = \log(10) - \log(4) \]

4. **Calculate the Logarithms**:
- Using known values:
\[ \log(10) = 1 \quad \text{and} \quad \log(4) \approx 0.602 \]
- Thus:
\[ \log(2.5) \approx 1 - 0.602 = 0.398 \]

5. **Substitute Back to Find k**:
- Now substituting back into the equation:
\[ 60 = \frac{2.303}{k} \times 0.398 \]
- Rearranging gives:
\[ k = \frac{2.303 \times 0.398}{60} \]

6. **Calculate the Half-Life (T1/2)**:
- The half-life for a first-order reaction is given by:
\[ T_{1/2} = \frac{0.693}{k} \]
- Substitute \( k \) from the previous step:
\[ T_{1/2} = \frac{0.693 \times 60}{2.303 \times 0.398} \]

7. **Final Calculation**:
- Calculate \( T_{1/2} \):
\[ T_{1/2} \approx 45 \text{ minutes} \]

### Conclusion:
The time taken for the reactants to decompose to half of their original amount (T1/2) is approximately **45 minutes**.