The rate constant of the production of 2B (g) by the reaction , `A(g)overset(Delta)rarr2B(g)` is `2.48xx10^(-4)s^(-1)` A 1 : 1 molar ratio of A to B in the reaction mixture is attained after

Options

A. 26.25 minute
B. 27.25 minute
C. 28.25 minute
D. 0 minute

Answer

To solve the problem, we need to determine the time at which the molar ratio of A to B becomes 1:1 in the reaction \( A(g) \overset{\Delta}{\rightarrow} 2B(g) \) given the rate constant \( k = 2.48 \times 10^{-4} \, s^{-1} \).

### Step-by-Step Solution:

1. **Understanding the Reaction**:
The reaction shows that 1 mole of A produces 2 moles of B. Therefore, if we let \( x \) be the amount of A that reacts, then \( 2x \) moles of B will be formed.

2. **Setting Up Initial Conditions**:
At time \( t = 0 \):
- Moles of A = \( A \) (initial amount)
- Moles of B = 0

3. **Expressing Moles at Time \( t \)**:
After time \( t \):
- Moles of A = \( A - x \)
- Moles of B = \( 2x \)

4. **Establishing the Molar Ratio**:
We want the molar ratio of A to B to be 1:1:
\[ A - x = 2x \]
Rearranging gives:
\[ A = 3x \quad \Rightarrow \quad x = \frac{A}{3} \]

5. **Applying the First Order Rate Law**:
The first-order rate law is given by:
\[ kt = 2.303 \log \left( \frac{[A]_0}{[A]_t} \right) \]
Where:
- \( [A]_0 = A \) (initial concentration)
- \( [A]_t = A - x = A - \frac{A}{3} = \frac{2A}{3} \)

6. **Substituting Values into the Rate Law**:
\[ kt = 2.303 \log \left( \frac{A}{\frac{2A}{3}} \right) = 2.303 \log \left( \frac{3}{2} \right) \]
Simplifying gives:
\[ kt = 2.303 \log \left( 1.5 \right) \]

7. **Calculating the Logarithm**:
The logarithm \( \log(1.5) \approx 0.176 \).

8. **Rearranging for Time \( t \)**:
\[ t = \frac{2.303 \cdot 0.176}{k} \]
Substituting \( k = 2.48 \times 10^{-4} \, s^{-1} \):
\[ t = \frac{2.303 \cdot 0.176}{2.48 \times 10^{-4}} \approx 1634.33 \, \text{seconds} \]

9. **Converting Seconds to Minutes**:
\[ t \approx \frac{1634.33}{60} \approx 27.24 \, \text{minutes} \]

10. **Final Answer**:
The closest option is **27.25 minutes** (Option B).