A first order reaction is `20%` complete in `10 min`. Calculate (a) the specific rate constant of the reaction and (b) the time taken for the reaction to reach `75%` completion.

Explanation

To solve the problem step by step, we will address both parts: (a) calculating the specific rate constant (k) and (b) finding the time taken for the reaction to reach 75% completion.

### Step-by-Step Solution:

#### (a) Calculate the specific rate constant (k)

1. **Identify the initial conditions**:
- Let the initial concentration of the reactant be \( [A]_0 = 100 \) (arbitrary units).
- Since the reaction is 20% complete, the concentration of the reactant that has reacted is \( x = 20 \).
- Therefore, the remaining concentration is \( [A] = [A]_0 - x = 100 - 20 = 80 \).

2. **Use the first-order reaction formula**:
The integrated rate law for a first-order reaction is given by:
\[ k = \frac{2.303}{t} \log \left( \frac{[A]_0}{[A]} \right) \]
where:
- \( t = 10 \) minutes (time taken for 20% completion),
- \( [A]_0 = 100 \),
- \( [A] = 80 \).

3. **Substitute the values into the formula**:
\[ k = \frac{2.303}{10} \log \left( \frac{100}{80} \right) \]

4. **Calculate the logarithm**:
\[ \log \left( \frac{100}{80} \right) = \log(1.25) \approx 0.09691 \]

5. **Calculate k**:
\[ k = \frac{2.303}{10} \times 0.09691 \approx 0.0223 \, \text{min}^{-1} \]

#### (b) Calculate the time taken for the reaction to reach 75% completion

1. **Determine the remaining concentration at 75% completion**:
- If 75% of the reactant has reacted, then \( x = 75 \).
- The remaining concentration is \( [A] = [A]_0 - x = 100 - 75 = 25 \).

2. **Use the first-order reaction formula again**:
\[ k = \frac{2.303}{t} \log \left( \frac{[A]_0}{[A]} \right) \]
Rearranging gives:
\[ t = \frac{2.303}{k} \log \left( \frac{[A]_0}{[A]} \right) \]

3. **Substitute the known values**:
\[ t = \frac{2.303}{0.0223} \log \left( \frac{100}{25} \right) \]

4. **Calculate the logarithm**:
\[ \log \left( \frac{100}{25} \right) = \log(4) \approx 0.60206 \]

5. **Calculate t**:
\[ t = \frac{2.303}{0.0223} \times 0.60206 \approx 62.16 \, \text{minutes} \]

### Final Answers:
(a) The specific rate constant \( k \) is approximately \( 0.0223 \, \text{min}^{-1} \).

(b) The time taken for the reaction to reach 75% completion is approximately \( 62.16 \, \text{minutes} \).