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, we will break it down into two parts: (a) calculating the specific rate constant (k) of the reaction and (b) finding the time taken for the reaction to reach 75% completion.

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

1. **Identify the given data:**
- The reaction is 20% complete in 10 minutes.
- We can assume the initial concentration of the reactant (A₀) to be 100 units (this simplifies calculations).

2. **Calculate the amount reacted (X):**
- Since the reaction is 20% complete, the amount reacted (X) is:
\[ X = 20\% \text{ of } A₀ = 20\% \text{ of } 100 = 20 \]

3. **Determine the remaining concentration:**
- The remaining concentration of the reactant (A) after 20% completion is:
\[ A = A₀ - X = 100 - 20 = 80 \]

4. **Use the first-order rate equation:**
- The formula for the rate constant (k) for a first-order reaction is given by:
\[ k = \frac{2.303}{T} \log \left( \frac{A₀}{A} \right) \]
- Substituting the values:
\[ k = \frac{2.303}{10} \log \left( \frac{100}{80} \right) \]

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

6. **Substitute and calculate k:**
- Now substituting back into the equation for k:
\[ k = \frac{2.303}{10} \times 0.09691 \approx 0.0222 \text{ min}^{-1} \]

### Part (b): Calculate the time taken for 75% completion

1. **Identify the new amount reacted (X) for 75% completion:**
- For 75% completion:
\[ X = 75\% \text{ of } A₀ = 75\% \text{ of } 100 = 75 \]

2. **Determine the remaining concentration after 75% completion:**
- The remaining concentration (A) is:
\[ A = A₀ - X = 100 - 75 = 25 \]

3. **Use the first-order rate equation to find time (T):**
- Rearranging the formula for k:
\[ k = \frac{2.303}{T} \log \left( \frac{A₀}{A} \right) \]
- We can rearrange it to solve for T:
\[ T = \frac{2.303}{k} \log \left( \frac{A₀}{A} \right) \]

4. **Substituting the known values:**
- Substitute \(k \approx 0.0222\) and calculate:
\[ T = \frac{2.303}{0.0222} \log \left( \frac{100}{25} \right) \]

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

6. **Substituting back to find T:**
- Now substituting back:
\[ T = \frac{2.303}{0.0222} \times 0.60206 \approx 63 \text{ minutes} \]

### Final Answers:
- (a) The specific rate constant \(k\) is approximately \(0.0222 \text{ min}^{-1}\).
- (b) The time taken for the reaction to reach 75% completion is approximately \(63\) minutes.