`75%` of the first order reaction completed in 3 min. `50%` of the reaction was completed in

Options

A. 24 min
B. 8 min
C. 1.5 min
D. 4 min

Answer

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

### Step 1: Understand the problem
We know that 75% of the reaction is completed in 3 minutes. This means that 25% of the reactant remains. We need to find out how long it takes for 50% of the reaction to be completed.

### Step 2: Set up the initial conditions
Let the initial concentration of the reactant be \( [A]_0 = A \). After 3 minutes, 75% of \( A \) has reacted, which means that 25% remains:
\[ [A] = A - 0.75A = 0.25A \]

### Step 3: Use the first-order reaction formula
The rate constant \( k \) for a first-order reaction can be calculated using the formula:
\[ k = \frac{2.303}{t} \log \left( \frac{[A]_0}{[A]} \right) \]
Substituting the values we have:
\[ k = \frac{2.303}{3} \log \left( \frac{A}{0.25A} \right) \]
This simplifies to:
\[ k = \frac{2.303}{3} \log (4) \]

### Step 4: Calculate \( k \)
Using the logarithm property:
\[ \log (4) = \log (2^2) = 2 \log (2) \]
Thus, we can rewrite \( k \):
\[ k = \frac{2.303}{3} \cdot 2 \log (2) = \frac{2.303 \cdot 2 \log (2)}{3} \]

### Step 5: Find the time for 50% completion
The half-life \( t_{1/2} \) for a first-order reaction is given by:
\[ t_{1/2} = \frac{0.693}{k} \]
Substituting the expression for \( k \):
\[ t_{1/2} = \frac{0.693 \cdot 3}{2.303 \cdot 2 \log (2)} \]

### Step 6: Simplify the expression
Since \( k \) was derived from \( t = 3 \) minutes, we can relate the time for 50% completion to the time for 75% completion:
\[ t_{1/2} = \frac{3}{2} = 1.5 \text{ minutes} \]

### Conclusion
Thus, the time taken for 50% of the reaction to be completed is **1.5 minutes**.