Time for half change for a first order reaction is 25 minutes. What time will be required for `99%` reaction ?

Answer

To solve the problem of determining the time required for a first-order reaction to reach 99% completion when the half-life is given as 25 minutes, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the relationship of half-life and rate constant (k)**:
For a first-order reaction, the half-life (t_half) is related to the rate constant (k) by the formula:
\[ t_{1/2} = \frac{0.693}{k} \]
Given that \( t_{1/2} = 25 \) minutes, we can rearrange this formula to find k:
\[ k = \frac{0.693}{t_{1/2}} = \frac{0.693}{25} \text{ min}^{-1} \]

2. **Calculate the rate constant (k)**:
\[ k = \frac{0.693}{25} = 0.02772 \text{ min}^{-1} \]

3. **Set up the equation for the first-order reaction**:
The integrated rate law for a first-order reaction is given by:
\[ \ln \left( \frac{[A_0]}{[A]} \right) = kt \]
Where:
- \([A_0]\) is the initial concentration,
- \([A]\) is the concentration at time t,
- \(t\) is the time elapsed.

4. **Determine the concentrations**:
If 99% of the reaction has occurred, then 1% of the reactant remains. If we assume the initial concentration \([A_0] = 100\), then:
\[ [A] = 100 - 99 = 1 \]

5. **Substituting values into the equation**:
Now substituting the values into the integrated rate law:
\[ \ln \left( \frac{100}{1} \right) = kt \]
This simplifies to:
\[ \ln(100) = kt \]

6. **Calculate \(\ln(100)\)**:
\[ \ln(100) = 4.605 \]

7. **Substituting k into the equation**:
Now substituting the value of k:
\[ 4.605 = 0.02772 \cdot t \]

8. **Solving for time (t)**:
\[ t = \frac{4.605}{0.02772} \approx 166.16 \text{ minutes} \]

### Final Answer:
The time required for the reaction to reach 99% completion is approximately **166.16 minutes**.

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