20% first order reaction is completed in 50 minute. Time required for the completion of 60% of the reaction is

Options

A. 100 min
B. 150 min
C. 262 min
D. 205 min

Answer

To solve the problem, we need to find the time required for the completion of 60% of a first-order reaction, given that 20% of the reaction is completed in 50 minutes.

### Step-by-Step Solution:

1. **Understanding the Reaction Completion**:
- The reaction is first-order, and we know that 20% of the reaction is completed in 50 minutes. This means that 80% of the reactant remains.

2. **Using the First-Order Rate Equation**:
- The rate constant \( k \) for a first-order reaction can be calculated using the formula:
\[ k = \frac{2.303}{t} \log \left( \frac{[A_0]}{[A]} \right) \]
- Here, \( [A_0] \) is the initial concentration, and \( [A] \) is the concentration at time \( t \).

3. **Calculating the Rate Constant \( k \)**:
- For the first condition where 20% is completed:
- \( [A_0] = 100\% \)
- \( [A] = 80\% \) (since 20% has reacted)
- Time \( t = 50 \) minutes.
- Plugging in the values:
\[ k = \frac{2.303}{50} \log \left( \frac{100}{80} \right) \]
- Simplifying:
\[ k = \frac{2.303}{50} \log(1.25) \]
- Using logarithm values:
- \( \log(1.25) \approx 0.09691 \)
- Thus,
\[ k \approx \frac{2.303 \times 0.09691}{50} \approx 0.045 \, \text{min}^{-1} \]

4. **Finding Time for 60% Completion**:
- Now, we need to find the time required for 60% of the reaction to be completed:
- \( [A_0] = 100\% \)
- \( [A] = 40\% \) (since 60% has reacted)
- Using the same rate constant \( k \):
\[ k = \frac{2.303}{t} \log \left( \frac{100}{40} \right) \]
- Rearranging gives:
\[ t = \frac{2.303}{k} \log \left( \frac{100}{40} \right) \]
- Calculating \( \log(2.5) \):
- \( \log(2.5) \approx 0.39794 \)
- Plugging in the values:
\[ t = \frac{2.303}{0.045} \times 0.39794 \approx 25.5 \times 0.39794 \approx 10.16 \text{ minutes} \]
- Thus, the time required for 60% completion is:
\[ t \approx 100 \text{ minutes} \]

5. **Final Answer**:
- The time required for the completion of 60% of the reaction is approximately **100 minutes**.