For the first order reaction `A to 2 B ,1` mole of reactant A gives 0.2 moles of B after 100 minutes. The half life of the reaction is ...... min. (Round off to the nearest integer).
[Use `ln 2 =0. 69, ln 10 = 2. 3`
Properties of logarithms : ln `x ^(y) = y ln x, ln ((x)/(y)) = ln x - ln y]`
Round off to the nearest integer)

Answer

To solve the problem, we need to find the half-life of a first-order reaction where 1 mole of reactant A produces 0.2 moles of product B after 100 minutes.

### Step-by-Step Solution:

1. **Identify the Change in Concentration**:
- Initially, we have 1 mole of A.
- After 100 minutes, 0.2 moles of B are formed, which means 0.2 moles of A have reacted.
- Therefore, the amount of A remaining = 1 - 0.2 = 0.8 moles.

2. **Use the First-Order Reaction Formula**:
The integrated rate law for a first-order reaction is given by:
\[ \ln \left( \frac{[A]_0}{[A]} \right) = k t \]
where:
- \([A]_0\) = initial concentration of A = 1 mole
- \([A]\) = concentration of A after time t = 0.8 moles
- \(t\) = time = 100 minutes
- \(k\) = rate constant

3. **Substituting Values**:
\[ \ln \left( \frac{1}{0.8} \right) = k \cdot 100 \]

4. **Calculate the Natural Logarithm**:
\[ \ln(1) - \ln(0.8) = 0 - \ln(0.8) \]
Using the property of logarithms:
\[ \ln(0.8) = \ln \left( \frac{8}{10} \right) = \ln(8) - \ln(10) \]
We know that:
\[ \ln(10) \approx 2.3 \quad \text{and} \quad \ln(8) = 3 \ln(2) \approx 3 \times 0.69 = 2.07 \]
Therefore:
\[ \ln(0.8) \approx 2.07 - 2.3 = -0.23 \]
Thus:
\[ \ln(0.8) \approx -0.223 \]

5. **Finding the Rate Constant \(k\)**:
\[ -\ln(0.8) = k \cdot 100 \]
\[ k = \frac{-(-0.223)}{100} = 0.00223 \, \text{min}^{-1} \]

6. **Calculate the Half-Life \(t_{1/2}\)**:
The half-life of a first-order reaction is given by:
\[ t_{1/2} = \frac{0.693}{k} \]
Substituting the value of \(k\):
\[ t_{1/2} = \frac{0.693}{0.00223} \approx 310.3 \, \text{minutes} \]

7. **Round Off to the Nearest Integer**:
Rounding 310.3 gives us:
\[ t_{1/2} \approx 310 \, \text{minutes} \]

### Final Answer:
The half-life of the reaction is approximately **310 minutes**.