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

Options

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

Explanation

To solve the problem step by step, we will follow the principles of first-order kinetics.

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

We know that in a first-order reaction, the half-life (t₁/₂) is given by the formula:

\[ t_{1/2} = \frac{0.693}{k} \]

In the problem, the concentration of the reactant decreases from 0.8 M to 0.4 M in 15 minutes. This means that the time taken for the concentration to halve (from 0.8 M to 0.4 M) is 15 minutes. Therefore, we can set:

\[ t_{1/2} = 15 \text{ minutes} \]

Now, we can rearrange the half-life formula to find k:

\[ k = \frac{0.693}{t_{1/2}} = \frac{0.693}{15} \]

Calculating this gives:

\[ k \approx 0.0462 \text{ min}^{-1} \]

### Step 2: Use the rate constant to find the time for the concentration change from 0.1 M to 0.025 M

For a first-order reaction, we can use the integrated rate law:

\[ \ln \left( \frac{A_0}{A_t} \right) = kt \]

Where:
- \(A_0\) is the initial concentration (0.1 M)
- \(A_t\) is the concentration at time t (0.025 M)
- \(k\) is the rate constant (0.0462 min⁻¹)
- \(t\) is the time we want to find

Substituting the values into the equation:

\[ \ln \left( \frac{0.1}{0.025} \right) = 0.0462 \cdot t \]

Calculating the left side:

\[ \frac{0.1}{0.025} = 4 \quad \Rightarrow \quad \ln(4) \approx 1.3863 \]

Now substituting back into the equation:

\[ 1.3863 = 0.0462 \cdot t \]

### Step 3: Solve for t

Now, we can solve for t:

\[ t = \frac{1.3863}{0.0462} \approx 30 \text{ minutes} \]

### Final Answer

The time taken for the concentration of the reactant to change from 0.1 M to 0.025 M is approximately **30 minutes**.

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