The half life period of a first order reaction is 60 minutes. What percentage of the reactant will be left behind after 120 minutes ?

Explanation

To solve the problem, we need to determine the percentage of the reactant left after 120 minutes for a first-order reaction with a half-life of 60 minutes.

### Step-by-Step Solution:

1. **Understand the Half-Life of a First-Order Reaction:**
The half-life (T½) 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 60 minutes, we can rearrange the formula to find \( k \):
\[ k = \frac{0.693}{T_{1/2}} = \frac{0.693}{60 \text{ min}} \approx 0.01155 \text{ min}^{-1} \]

3. **Determine the Total Time and Number of Half-Lives:**
We need to find out how many half-lives fit into 120 minutes:
\[ \text{Number of half-lives} = \frac{120 \text{ min}}{60 \text{ min}} = 2 \]

4. **Calculate the Remaining Percentage of Reactant:**
For a first-order reaction, the amount of reactant remaining after \( n \) half-lives can be calculated using the formula:
\[ \text{Remaining percentage} = \left( \frac{1}{2} \right)^n \times 100\% \]
Substituting \( n = 2 \):
\[ \text{Remaining percentage} = \left( \frac{1}{2} \right)^2 \times 100\% = \frac{1}{4} \times 100\% = 25\% \]

5. **Conclusion:**
After 120 minutes, 25% of the reactant will be left.

### Final Answer:
The percentage of the reactant left after 120 minutes is **25%**.