For a given reaction of first order, it takes `20` minutes for the concentration to drop from `1.0 "mol liter"^(-1)` to `0.6 "mol litre"^(-1)`. The time required for the concentration to drop from `0.6 "mol litre"^(-1)` to `0.36 "mol litre"^(-1)` will be

Options

A. more than 20 minutes
B. less than 20 minutes
C. equal to 20 minutes
D. infinity

Explanation

To solve the problem, we need to use the first-order kinetics formula, which relates the concentration of a reactant to time. The formula is given by:

\[ t = \frac{2.303}{k} \log \left( \frac{[A]_0}{[A]} \right) \]

Where:
- \( t \) is the time taken for the reaction,
- \( k \) is the rate constant,
- \([A]_0\) is the initial concentration,
- \([A]\) is the final concentration.

### Step 1: Calculate the rate constant \( k \)

From the first part of the problem, we know that it takes 20 minutes for the concentration to drop from \( 1.0 \, \text{mol L}^{-1} \) to \( 0.6 \, \text{mol L}^{-1} \).

Using the formula:

\[ 20 = \frac{2.303}{k} \log \left( \frac{1.0}{0.6} \right) \]

Calculating the logarithm:

\[ \log \left( \frac{1.0}{0.6} \right) = \log(1.6667) \approx 0.2218 \]

Now substituting this value into the equation:

\[ 20 = \frac{2.303}{k} \cdot 0.2218 \]

Rearranging to solve for \( k \):

\[ k = \frac{2.303 \cdot 0.2218}{20} \]

Calculating \( k \):

\[ k \approx \frac{0.5112}{20} \approx 0.02556 \, \text{min}^{-1} \]

### Step 2: Calculate the time for the concentration to drop from \( 0.6 \, \text{mol L}^{-1} \) to \( 0.36 \, \text{mol L}^{-1} \)

Now we will use the same formula to find the time required for the concentration to drop from \( 0.6 \, \text{mol L}^{-1} \) to \( 0.36 \, \text{mol L}^{-1} \):

\[ t = \frac{2.303}{k} \log \left( \frac{0.6}{0.36} \right) \]

Calculating the logarithm:

\[ \log \left( \frac{0.6}{0.36} \right) = \log(1.6667) \approx 0.2218 \]

Substituting \( k \) into the equation:

\[ t = \frac{2.303}{0.02556} \cdot 0.2218 \]

Calculating \( t \):

\[ t \approx \frac{2.303 \cdot 0.2218}{0.02556} \approx 20 \, \text{minutes} \]

### Conclusion

The time required for the concentration to drop from \( 0.6 \, \text{mol L}^{-1} \) to \( 0.36 \, \text{mol L}^{-1} \) is also **20 minutes**.