If 75% of a first order reaction was completed in 90 minutes, 60% of the same reaction would be completed in approximately (in minutes ) ..............
(Take : log 2 = 0.30 , log 2.5 = 0.40)

Answer

To solve the problem, we will use the first-order reaction kinetics formula and the information given about the completion of the reaction.

### Step-by-Step Solution:

1. **Understanding the Reaction Completion**:
- We know that 75% of the reaction is completed in 90 minutes. This means that 25% of the reactant remains.
- If we denote the initial concentration of the reactant as \( [A_0] \), then after 75% completion, the concentration left is:
\[ [A] = [A_0] \times (1 - 0.75) = [A_0] \times 0.25 = \frac{[A_0]}{4} \]

2. **Using the First-Order Rate Equation**:
- The first-order rate equation is given by:
\[ k = \frac{2.303}{t} \log \left( \frac{[A_0]}{[A]} \right) \]
- For 75% completion, substituting the values:
\[ k = \frac{2.303}{90} \log \left( \frac{[A_0]}{\frac{[A_0]}{4}} \right) = \frac{2.303}{90} \log(4) \]
- Since \( \log(4) = 2 \log(2) \) and using \( \log(2) = 0.30 \):
\[ \log(4) = 2 \times 0.30 = 0.60 \]
- Therefore, we can express \( k \) as:
\[ k = \frac{2.303}{90} \times 0.60 \]

3. **Calculating the Rate Constant \( k \)**:
- Now, calculating \( k \):
\[ k = \frac{2.303 \times 0.60}{90} = \frac{1.3818}{90} \approx 0.0153 \text{ min}^{-1} \]

4. **Finding Time for 60% Completion**:
- For 60% completion, 40% of the reactant remains:
\[ [A] = [A_0] \times (1 - 0.60) = [A_0] \times 0.40 = \frac{[A_0]}{2.5} \]
- Now, substituting this into the first-order rate equation:
\[ k = \frac{2.303}{t} \log \left( \frac{[A_0]}{\frac{[A_0]}{2.5}} \right) = \frac{2.303}{t} \log(2.5) \]
- Using \( \log(2.5) = 0.40 \):
\[ k = \frac{2.303}{t} \times 0.40 \]

5. **Equating the Two Expressions for \( k \)**:
- Since \( k \) is constant, we can set the two equations for \( k \) equal to each other:
\[ \frac{2.303 \times 0.60}{90} = \frac{2.303 \times 0.40}{t} \]
- Cancel \( 2.303 \) from both sides:
\[ \frac{0.60}{90} = \frac{0.40}{t} \]

6. **Solving for \( t \)**:
- Cross-multiplying gives:
\[ 0.60t = 0.40 \times 90 \]
- Thus:
\[ 0.60t = 36 \implies t = \frac{36}{0.60} = 60 \text{ minutes} \]

### Final Answer:
The time required for 60% completion of the reaction is approximately **60 minutes**.