The half-life of a first order reaction is 30 min.
How long would be required for 25% of the reactant to be decomposed ?

Explanation

To solve the problem, we need to determine how long it takes for 25% of the reactant to decompose in a first-order reaction, given that the half-life (t₁/₂) is 30 minutes.

### Step-by-Step Solution:

1. **Understand the Half-Life of a First-Order Reaction**:
The half-life (t₁/₂) of a first-order reaction is given by the formula:
\[ t_{1/2} = \frac{0.693}{k} \]
where \( k \) is the rate constant.

2. **Calculate the Rate Constant (k)**:
Given that the half-life is 30 minutes, we can rearrange the formula to find \( k \):
\[ k = \frac{0.693}{t_{1/2}} = \frac{0.693}{30 \text{ min}} \approx 0.0231 \text{ min}^{-1} \]

3. **Determine the Remaining Concentration**:
If 25% of the reactant is decomposed, then 75% of the reactant remains. We can denote the initial concentration as \( A_0 \) and the concentration at time \( t \) as \( A_t \):
\[ A_t = 0.75 A_0 \]

4. **Use the First-Order Kinetics Equation**:
The integrated rate law for a first-order reaction is:
\[ k = \frac{1}{t} \ln\left(\frac{A_0}{A_t}\right) \]
Plugging in the values we have:
\[ k = \frac{1}{t} \ln\left(\frac{A_0}{0.75 A_0}\right) = \frac{1}{t} \ln\left(\frac{1}{0.75}\right) \]

5. **Substituting the Value of k**:
We already calculated \( k \):
\[ 0.0231 = \frac{1}{t} \ln\left(\frac{4}{3}\right) \]
Here, \( \frac{1}{0.75} = \frac{4}{3} \).

6. **Calculate \( \ln\left(\frac{4}{3}\right) \)**:
We can calculate \( \ln\left(\frac{4}{3}\right) \):
\[ \ln\left(\frac{4}{3}\right) \approx 0.2877 \]

7. **Rearranging to Find Time (t)**:
Rearranging the equation to solve for \( t \):
\[ t = \frac{\ln\left(\frac{4}{3}\right)}{k} \]
Substituting the values:
\[ t = \frac{0.2877}{0.0231} \approx 12.43 \text{ minutes} \]

### Final Answer:
It would take approximately **12.43 minutes** for 25% of the reactant to decompose.