In a first order reaction, The concentration of reactant is reduced to 1/8th of the initial concentration in 75 minutea at 298 K. What is the half period of the reaction in minutes?

Options

A. 50 min
B. 15 min
C. 30 min
D. 25 min

Explanation

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

### Step 1: Understand the given information
We know that in a first-order reaction, the concentration of the reactant is reduced to 1/8th of its initial concentration in a given time. Here, the initial concentration (A) can be taken as 1 (for simplicity), and the concentration after 75 minutes (t) is A - x = 1/8.

### Step 2: Set up the equation
For a first-order reaction, the relationship between the initial concentration (A), the concentration after time t (A - x), and the rate constant (k) is given by the formula:
\[ k = \frac{2.303}{t} \log \left( \frac{A}{A - x} \right) \]

### Step 3: Substitute the known values
Here, we have:
- \( A = 1 \)
- \( A - x = \frac{1}{8} \)
- \( t = 75 \) minutes

Substituting these values into the equation gives:
\[ k = \frac{2.303}{75} \log \left( \frac{1}{\frac{1}{8}} \right) \]

### Step 4: Simplify the logarithm
The logarithm simplifies as follows:
\[ \log \left( \frac{1}{\frac{1}{8}} \right) = \log(8) = \log(2^3) = 3 \log(2) \]

Using the approximate value \( \log(2) \approx 0.301 \):
\[ \log(8) \approx 3 \times 0.301 = 0.903 \]

### Step 5: Calculate k
Now substituting back into the equation for k:
\[ k = \frac{2.303}{75} \times 0.903 \]

Calculating this gives:
\[ k \approx \frac{2.303 \times 0.903}{75} \approx \frac{2.080}{75} \approx 0.0277 \text{ min}^{-1} \]

### Step 6: Calculate the half-life (t_half)
For a first-order reaction, the half-life (t_half) is given by:
\[ t_{1/2} = \frac{0.693}{k} \]

Substituting the value of k:
\[ t_{1/2} = \frac{0.693}{0.0277} \]

Calculating this gives:
\[ t_{1/2} \approx 25 \text{ minutes} \]

### Conclusion
Thus, the half-life of the reaction is approximately **25 minutes**.

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