In a first order reaction, the concentration of the reactant decreases form `0.8 M` to `0.4 M` in `15 min`. The time taken for the concentration to change form `0.1 M` to `0.025 M` is

Options

A. `60 min`
B. `15 min`
C. `7.5 min`
D. `30 min`

Explanation

To solve the problem step by step, we will use the first-order reaction kinetics formula.

### Step 1: Determine the rate constant (k)

Given:
- Initial concentration (C₀) = 0.8 M
- Final concentration (C) = 0.4 M
- Time (t) = 15 minutes

For a first-order reaction, the rate constant (k) can be calculated using the formula:

\[ k = \frac{2.303}{t} \log \left( \frac{C_0}{C} \right) \]

Substituting the values:

\[ k = \frac{2.303}{15} \log \left( \frac{0.8}{0.4} \right) \]

Calculating the logarithm:

\[ \frac{0.8}{0.4} = 2 \quad \text{and} \quad \log(2) \approx 0.3010 \]

Now substituting this back into the equation:

\[ k = \frac{2.303}{15} \times 0.3010 \]

Calculating \( k \):

\[ k \approx \frac{2.303 \times 0.3010}{15} \approx 0.0462 \, \text{min}^{-1} \]

### Step 2: Calculate the time taken for the concentration to change from 0.1 M to 0.025 M

Now we need to find the time taken for the concentration to change from:
- Initial concentration (C₀) = 0.1 M
- Final concentration (C) = 0.025 M

Using the same first-order reaction formula:

\[ t = \frac{2.303}{k} \log \left( \frac{C_0}{C} \right) \]

Substituting the values:

\[ t = \frac{2.303}{0.0462} \log \left( \frac{0.1}{0.025} \right) \]

Calculating the logarithm:

\[ \frac{0.1}{0.025} = 4 \quad \text{and} \quad \log(4) = 2 \log(2) \approx 2 \times 0.3010 = 0.6020 \]

Now substituting this back into the equation:

\[ t = \frac{2.303}{0.0462} \times 0.6020 \]

Calculating \( t \):

\[ t \approx \frac{2.303 \times 0.6020}{0.0462} \approx 30 \, \text{minutes} \]

### Final Answer:
The time taken for the concentration to change from 0.1 M to 0.025 M is **30 minutes**.

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