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

Explanation

To solve the problem, we need to determine the equation that describes the change in concentration of \( N_2O_5 \) with time for the given reaction:

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

Given that the rate constant \( k \) is \( 2.3 \times 10^{-2} \, s^{-1} \), we can identify that this is a first-order reaction based on the units of \( k \).

### Step-by-Step Solution:

1. **Identify the Reaction Order**:
The units of the rate constant \( k \) are \( s^{-1} \), which indicates that the reaction is first-order.

2. **Use the First-Order Reaction Equation**:
For a first-order reaction, the integrated rate law can be expressed as:
\[ \ln \left( \frac{[A_0]}{[A_t]} \right) = kt \]
where:
- \( [A_0] \) is the initial concentration of the reactant,
- \( [A_t] \) is the concentration at time \( t \),
- \( k \) is the rate constant,
- \( t \) is the time.

3. **Rearranging the Equation**:
We can rearrange the equation to express \( [A_t] \) in terms of \( [A_0] \):
\[ [A_t] = [A_0] e^{-kt} \]

4. **Substituting for \( N_2O_5 \)**:
In our case, we replace \( [A_0] \) with \( [N_2O_5]_0 \) and \( [A_t] \) with \( [N_2O_5]_t \):
\[ [N_2O_5]_t = [N_2O_5]_0 e^{-kt} \]

5. **Final Expression**:
Therefore, the equation that describes the change of \( [N_2O_5] \) with time is:
\[ [N_2O_5]_t = [N_2O_5]_0 e^{-2.3 \times 10^{-2} t} \]