The first order rate constant for the decomposition of `N_(2)O_(5)` is `6.2 xx 10^(-4) sec^(-1)`. The `t_(1//2)` of decomposition is

Options

A. 1117.7
B. 111.7
C. 223.4
D. 160.9

Answer

To find the half-life (\( t_{1/2} \)) of the decomposition of \( N_2O_5 \) for a first-order reaction, we can use the formula for the half-life of a first-order reaction, which 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 data**:
- The first-order rate constant \( k = 6.2 \times 10^{-4} \, \text{sec}^{-1} \).

2. **Write the half-life formula for a first-order reaction**:
- The formula is \( t_{1/2} = \frac{0.693}{k} \).

3. **Substitute the value of \( k \) into the formula**:
- \( t_{1/2} = \frac{0.693}{6.2 \times 10^{-4}} \).

4. **Calculate the half-life**:
- First, calculate the denominator:
\[ 6.2 \times 10^{-4} = 0.00062 \]
- Now perform the division:
\[ t_{1/2} = \frac{0.693}{0.00062} \approx 1117.74 \, \text{seconds} \]

5. **Round the answer to an appropriate number of significant figures**:
- The half-life \( t_{1/2} \approx 1172.74 \, \text{seconds} \).

### Final Answer:
The half-life of the decomposition of \( N_2O_5 \) is approximately **1172.74 seconds**.