The rate equation for the decomposition of `N_(2)O_(5)` in `"CC"l_(4)` is rate `=K[N_(2)O_(4)]` where `K=6.3x10^(-4)s^(-1)` at 320 K. what would be the initial rate of decompositioni of `N_(2)O_(5)` in a .10 M solution of `N_(2)O_(4)`?

Options

A. `6.3xx10^(-5)` mol `"litre"^(-1)s^(-1)`
B. `0.63xx10^(-6)` mol `"litre"^(-1)s^(-1)`
C. `6.3xx10^(-5)` mol `"litre"^(-1)s^(-1)`
D. `0.63xx10^(-4)` mol `"litre"^(-1)s^(-1)`

Explanation

To solve the problem, we need to calculate the initial rate of decomposition of \( N_2O_5 \) using the given rate equation and the concentration of \( N_2O_4 \).

### Step-by-Step Solution:

1. **Identify the Rate Equation**:
The rate of the reaction is given by the equation:
\[ \text{Rate} = k [N_2O_4] \]
where \( k = 6.3 \times 10^{-4} \, s^{-1} \) and \( [N_2O_4] = 0.10 \, M \).

2. **Substitute the Values into the Rate Equation**:
We can substitute the values of \( k \) and \( [N_2O_4] \) into the rate equation:
\[ \text{Rate} = (6.3 \times 10^{-4} \, s^{-1}) \times (0.10 \, M) \]

3. **Calculate the Rate**:
Now, we perform the multiplication:
\[ \text{Rate} = 6.3 \times 10^{-4} \times 0.10 = 6.3 \times 10^{-5} \, \text{mol L}^{-1} \text{s}^{-1} \]

4. **Final Result**:
Therefore, the initial rate of decomposition of \( N_2O_5 \) is:
\[ \text{Rate} = 6.3 \times 10^{-5} \, \text{mol L}^{-1} \text{s}^{-1} \]

### Conclusion:
The initial rate of decomposition of \( N_2O_5 \) in a 0.10 M solution of \( N_2O_4 \) is \( 6.3 \times 10^{-5} \, \text{mol L}^{-1} \text{s}^{-1} \).