`99%` at a first order reaction was completed in `32 min`. When will `99.9%` of the reaction complete.

Options

A. (a) 50 minute
B. (b) 46 minutes
C. (c) 49 minute
D. (d) 48 minute

Answer

To solve the problem of determining when 99.9% of a first-order reaction will be completed, given that 99% was completed in 32 minutes, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Reaction**:
- We start with an initial concentration of A (let's assume 100 grams).
- After 32 minutes, 99% of A has reacted, meaning only 1 gram of A remains.

2. **Using the First-Order Rate Equation**:
- The rate constant \( k \) for a first-order reaction is given by the equation:
\[ k = \frac{1}{t} \ln\left(\frac{A_0}{A_t}\right) \]
- Here, \( A_0 \) is the initial concentration (100 grams) and \( A_t \) is the concentration at time \( t \) (1 gram after 32 minutes).

3. **Calculating the Rate Constant \( k \)**:
- Plugging in the values:
\[ k = \frac{1}{32} \ln\left(\frac{100}{1}\right) \]
- This simplifies to:
\[ k = \frac{1}{32} \ln(100) \]
- Calculating \( \ln(100) \):
\[ \ln(100) = 4.605 \]
- Therefore:
\[ k = \frac{1}{32} \times 4.605 \approx 0.1439 \text{ min}^{-1} \]

4. **Finding Time for 99.9% Completion**:
- For 99.9% completion, only 0.1 gram of A remains.
- Thus, \( A_t = 0.1 \) grams.
- We can use the same rate equation to find the new time \( t \):
\[ k = \frac{1}{t} \ln\left(\frac{100}{0.1}\right) \]
- This simplifies to:
\[ k = \frac{1}{t} \ln(1000) \]
- Calculating \( \ln(1000) \):
\[ \ln(1000) = 6.907 \]
- Rearranging to solve for \( t \):
\[ t = \frac{1}{k} \ln(1000) \]
- Substituting \( k \):
\[ t = \frac{1}{0.1439} \times 6.907 \]
- Calculating \( t \):
\[ t \approx 48 \text{ minutes} \]

### Final Answer:
The time required for 99.9% completion of the reaction is approximately **48 minutes**.