Time required for 50% completion of the first order reaction in one hr. What is the time required for 99% completion of the same reaction ?

Answer

To solve the problem of finding the time required for 99% completion of a first-order reaction when the time for 50% completion is given as 1 hour, we can follow these steps:

### Step 1: Understand the relationship for first-order reactions
For a first-order reaction, the rate constant \( k \) can be calculated using the formula:
\[ k = \frac{2.303}{t_{1/2}} \log \left( \frac{[A_0]}{[A]} \right) \]
where:
- \( t_{1/2} \) is the half-life (time for 50% completion),
- \( [A_0] \) is the initial concentration,
- \( [A] \) is the concentration at time \( t \).

### Step 2: Calculate the rate constant \( k \)
Given that the time for 50% completion (\( t_{1/2} \)) is 1 hour (or 60 minutes), and the initial concentration \( [A_0] \) is 100% and the concentration at \( t_{1/2} \) is 50%, we can substitute these values into the equation:
\[ k = \frac{2.303}{60} \log \left( \frac{100}{50} \right) \]
Calculating this gives:
\[ k = \frac{2.303}{60} \log(2) \]
Using \( \log(2) \approx 0.301 \):
\[ k = \frac{2.303 \times 0.301}{60} \approx 0.0115 \, \text{min}^{-1} \]

### Step 3: Calculate the time for 99% completion
For 99% completion, only 1% of the reactant remains. Thus, we can use the same formula:
\[ t = \frac{2.303}{k} \log \left( \frac{[A_0]}{[A]} \right) \]
Here, \( [A_0] = 100\% \) and \( [A] = 1\% \), so:
\[ t = \frac{2.303}{0.0115} \log \left( \frac{100}{1} \right) \]
Calculating this gives:
\[ t = \frac{2.303}{0.0115} \log(100) = \frac{2.303}{0.0115} \times 2 \]
Using \( \log(100) = 2 \):
\[ t = \frac{2.303 \times 2}{0.0115} \approx 400.6 \, \text{minutes} \]

### Step 4: Convert minutes to hours
To convert minutes to hours, we divide by 60:
\[ t \approx \frac{400.6}{60} \approx 6.68 \, \text{hours} \]

### Final Answer
The time required for 99% completion of the reaction is approximately **6.68 hours**.

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