For a first order reaction, if the time taken for completion of `50%` of the reaction is `t` second, the time required for completion of `99.9%` of the reaction is `nt`. Find the value of n?

Answer

To solve the problem, we need to use the first-order reaction kinetics and the integrated rate law for first-order reactions.

### Step-by-Step Solution:

1. **Understanding the First-Order Reaction:**
For a first-order reaction, the rate constant \( k \) can be expressed using the formula:
\[ k = \frac{2.303}{t} \log\left(\frac{[A_0]}{[A_t]}\right) \]
where:
- \( [A_0] \) is the initial concentration,
- \( [A_t] \) is the concentration at time \( t \),
- \( t \) is the time taken for the reaction.

2. **Data for 50% Completion:**
When 50% of the reaction is completed, the concentration at time \( t \) can be expressed as:
\[ [A_t] = 100 - 50 = 50 \]
Therefore, substituting into the equation:
\[ k = \frac{2.303}{t} \log\left(\frac{100}{50}\right) = \frac{2.303}{t} \log(2) \]

3. **Data for 99.9% Completion:**
For 99.9% completion, the concentration at time \( nt \) is:
\[ [A_t] = 100 - 99.9 = 0.1 \]
Substituting into the equation gives:
\[ k = \frac{2.303}{nt} \log\left(\frac{100}{0.1}\right) = \frac{2.303}{nt} \log(1000) \]
Since \( \log(1000) = 3 \), we can simplify this to:
\[ k = \frac{2.303 \cdot 3}{nt} \]

4. **Equating the Two Expressions for k:**
Since the rate constant \( k \) is the same for both times, we can set the two expressions equal to each other:
\[ \frac{2.303}{t} \log(2) = \frac{2.303 \cdot 3}{nt} \]

5. **Canceling Common Terms:**
We can cancel \( 2.303 \) and \( t \) from both sides:
\[ \log(2) = \frac{3}{n} \]

6. **Solving for n:**
Rearranging gives:
\[ n = \frac{3}{\log(2)} \]
Using the approximate value \( \log(2) \approx 0.301 \):
\[ n = \frac{3}{0.301} \approx 9.965 \]
Rounding this gives \( n \approx 10 \).

### Final Answer:
The value of \( n \) is \( 10 \).

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