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 what percentage of the reactant remains after 120 minutes, given that the half-life of the reaction is 60 minutes.

### Step-by-Step Solution:

1. **Understanding Half-Life**:
The half-life (t₁/₂) of a first-order reaction is the time required for half of the reactant to be consumed. Here, t₁/₂ = 60 minutes.

2. **Determine the Rate Constant (k)**:
The relationship between half-life and the rate constant (k) for a first-order reaction is given by the formula:
\[ t_{1/2} = \frac{\ln(2)}{k} \]
Rearranging this, we can express k as:
\[ k = \frac{\ln(2)}{t_{1/2}} = \frac{\ln(2)}{60 \text{ minutes}} \]

3. **Calculate k**:
Using the value of ln(2) (approximately 0.693):
\[ k \approx \frac{0.693}{60} \approx 0.01155 \text{ min}^{-1} \]

4. **Determine the Time (T)**:
We are interested in the amount of reactant left after 120 minutes. We can use the integrated rate law for first-order reactions:
\[ \ln\left(\frac{[A]_0}{[A]}\right) = k \cdot t \]
Here, \([A]_0\) is the initial concentration (100% in this case), and \([A]\) is the concentration after time t.

5. **Substituting Values**:
Let \(Y\) be the percentage of the reactant left after 120 minutes. Then:
\[ \ln\left(\frac{100}{Y}\right) = k \cdot 120 \]
Substituting k:
\[ \ln\left(\frac{100}{Y}\right) = \left(\frac{0.693}{60}\right) \cdot 120 \]
Simplifying:
\[ \ln\left(\frac{100}{Y}\right) = 0.693 \cdot 2 = \ln(4) \]

6. **Solving for Y**:
Now, we can equate:
\[ \frac{100}{Y} = 4 \]
Rearranging gives:
\[ Y = \frac{100}{4} = 25 \]

7. **Conclusion**:
Therefore, after 120 minutes, 25% of the reactant will be left.

### Final Answer:
**25% of the reactant will be left after 120 minutes.**