20% deocmposition of `H_(2)O_(2)` in presence of an acid requires 5 min. The time required for 50% decomposition in minutes is

Options

A. 15.52
B. 1.552
C. 0.1552
D. 7.76

Explanation

To solve the problem of determining the time required for 50% decomposition of \( H_2O_2 \) given that 20% decomposition takes 5 minutes, we will use the principles of first-order kinetics.

### Step-by-Step Solution:

1. **Understanding the Problem**: We know that 20% of \( H_2O_2 \) decomposes in 5 minutes. We need to find the time required for 50% decomposition.

2. **Using the First-Order Kinetics Formula**: The formula for the rate of a first-order reaction is given by:
\[ t = \frac{2.303}{k} \log \left( \frac{A}{A - x} \right) \]
where:
- \( t \) is the time,
- \( k \) is the rate constant,
- \( A \) is the initial concentration,
- \( x \) is the amount decomposed.

3. **Calculating the Rate Constant \( k \)**: For the 20% decomposition:
- Initial concentration \( A = 100 \) (assuming 100 units),
- Amount decomposed \( x = 20 \),
- Remaining concentration \( A - x = 100 - 20 = 80 \).

Plugging these values into the formula:
\[ 5 = \frac{2.303}{k} \log \left( \frac{100}{80} \right) \]
Rearranging to find \( k \):
\[ k = \frac{2.303 \cdot \log \left( \frac{100}{80} \right)}{5} \]

4. **Calculating the Logarithm**:
\[ \log \left( \frac{100}{80} \right) = \log(1.25) \approx 0.09691 \]
Thus,
\[ k = \frac{2.303 \cdot 0.09691}{5} \approx 0.04463 \text{ min}^{-1} \]

5. **Finding Time for 50% Decomposition**: For 50% decomposition, \( x = 50 \):
- Remaining concentration \( A - x = 100 - 50 = 50 \).

Using the half-life formula for first-order reactions:
\[ t_{1/2} = \frac{0.693}{k} \]
Substituting the value of \( k \):
\[ t_{1/2} = \frac{0.693}{0.04463} \approx 15.52 \text{ minutes} \]

6. **Conclusion**: The time required for 50% decomposition of \( H_2O_2 \) is approximately 15.52 minutes.