The first order rate constant for the decomposition of `N_(2)O_(5)` is `6.2 xx 10^(-2)s^(-1)`. The half-life period for this decomposition is

Options

A. 223.4 s
B. 1177.7s
C. 11.177s
D. 160.9 s

Explanation

To find the half-life period for the decomposition of \( N_2O_5 \) which follows first-order kinetics, we can use the formula for the half-life of a first-order reaction. The formula is given by:

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

where \( k \) is the first-order rate constant.

### Step-by-Step Solution:

1. **Identify the given values**:
- The rate constant \( k \) for the decomposition of \( N_2O_5 \) is given as \( 6.2 \times 10^{-2} \, s^{-1} \).

2. **Substitute the value of \( k \) into the half-life formula**:
\[ t_{1/2} = \frac{0.693}{6.2 \times 10^{-2}} \]

3. **Calculate the half-life**:
- First, calculate \( 0.693 \div 6.2 \):
\[ 0.693 \div 6.2 \approx 0.111 \]
- Now, since \( 10^{-2} \) is in the denominator, we can express the result as:
\[ t_{1/2} \approx 0.111 \times 100 \, s \approx 11.177 \, s \]

4. **Final result**:
- Therefore, the half-life period for the decomposition of \( N_2O_5 \) is approximately:
\[ t_{1/2} \approx 11.177 \, s \]

### Conclusion:
The half-life period for the decomposition of \( N_2O_5 \) is approximately **11.177 seconds**.

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