The rate constant `k`, for the reaction `N_(2)O_(5)(g) rarr 2NO_(2) (g) + (1)/(2) O_(2)(g)` is `2.3 xx 10^(-2) s^(-1)`. Which equation given below describes the change of `[N_(2)O_(5)]` with time ? `[N_(2)O_(5)]_(0)` and `[N_(2)O_(5)]_(t)` correspond to concentration of `N_(2)O_(5)` initially and at time, `t` ?

Options

A. `[N_(2)O_(5)]t=[N_(2)O_(5)]_(0)+kt`
B. `[N_(2)O_(5)]_(0)=[N_(2)O_(5)]_(t)^(e^(kt))`
C. `log[N_(2)O_(5)]_(t)=log[N_(2)O_(5)]_(0)+kt`
D. In`([N_(2)O_(5)]_(0))/([N_(2)O_(5)]_(t))`=kt.

Explanation

To solve the problem, we need to derive the equation that describes the change in concentration of \( N_2O_5 \) with time for a first-order reaction. The reaction given is:

\[ N_2O_5(g) \rightarrow 2NO_2(g) + \frac{1}{2}O_2(g) \]

### Step-by-Step Solution:

1. **Identify the Order of the Reaction**:
The rate constant \( k \) is given as \( 2.3 \times 10^{-2} \, s^{-1} \). The unit \( s^{-1} \) indicates that this is a first-order reaction.

2. **Write the Integrated Rate Law for a First-Order Reaction**:
The integrated rate law for a first-order reaction is given by the equation:
\[ \ln \left( \frac{[A]_t}{[A]_0} \right) = -kt \]
where:
- \( [A]_t \) is the concentration of the reactant at time \( t \),
- \( [A]_0 \) is the initial concentration of the reactant,
- \( k \) is the rate constant,
- \( t \) is time.

3. **Rearranging the Equation**:
We can rearrange the equation to express \( [A]_t \):
\[ [A]_t = [A]_0 e^{-kt} \]
For our specific case, substituting \( [A] \) with \( [N_2O_5] \):
\[ [N_2O_5]_t = [N_2O_5]_0 e^{-kt} \]

4. **Expressing in Terms of Natural Logarithm**:
Alternatively, we can express it in logarithmic form:
\[ \ln [N_2O_5]_t = \ln [N_2O_5]_0 - kt \]

5. **Final Equation**:
Thus, the equation that describes the change of \( [N_2O_5] \) with time is:
\[ [N_2O_5]_t = [N_2O_5]_0 e^{-kt} \]

### Conclusion:
The correct equation that describes the change in concentration of \( N_2O_5 \) with time is:
\[ [N_2O_5]_t = [N_2O_5]_0 e^{-kt} \]