A first order reaction is carried out with an initial concentration of 10 moles per litre and 75 % of the reactant changing to the product if t 50% is 20 min.

Options

A. 40
B. 80
C. 160
D. cannot be calculated

Answer

To solve the problem, we need to determine the time required for 75% of the reactant to convert into the product in a first-order reaction. We know the following:

- Initial concentration (\( [A]_0 \)) = 10 moles per liter
- 75% of the reactant changes to product
- Half-life (\( t_{1/2} \)) = 20 minutes

### Step-by-Step Solution:

1. **Understanding the Reaction Completion**:
- If 75% of the reactant has reacted, then 25% of the reactant remains.
- This means that the concentration of the reactant after 75% conversion is:
\[ [A] = [A]_0 \times (1 - 0.75) = 10 \, \text{moles/liter} \times 0.25 = 2.5 \, \text{moles/liter} \]

2. **Using the First-Order Kinetics Equation**:
- The first-order kinetics can be described by the equation:
\[ k = \frac{0.693}{t_{1/2}} \]
- Substituting the half-life into the equation:
\[ k = \frac{0.693}{20 \, \text{min}} = 0.03465 \, \text{min}^{-1} \]

3. **Applying the Integrated Rate Law**:
- The integrated rate law for a first-order reaction is given by:
\[ \ln \left( \frac{[A]_0}{[A]} \right) = kt \]
- Substituting the known values:
\[ \ln \left( \frac{10 \, \text{moles/liter}}{2.5 \, \text{moles/liter}} \right) = 0.03465 \, \text{min}^{-1} \times t \]

4. **Calculating the Natural Logarithm**:
- Calculate the left-hand side:
\[ \ln(4) = 1.3863 \]
- Therefore, we have:
\[ 1.3863 = 0.03465 \, t \]

5. **Solving for Time (t)**:
- Rearranging the equation to solve for \( t \):
\[ t = \frac{1.3863}{0.03465} \approx 40 \, \text{minutes} \]

### Final Answer:
The time required for 75% of the reactant to convert into product is approximately **40 minutes**.