`NO_(2)` required for a reaction is produced by the decomposition of `N_(2)O_(5)` is `"CC"l_(4)` as per the equatiion `2N_(2)O_(5)(g)to4NO_(2)(g)+O_(2)(g)`. The initial concentration fo `N_(2)O_(5)` is `3.00 "mol"L^(-1)` and it is `2.75 "mol"L^(-1)` after 30 minutes. The rate of formation of `NO_(2)` is `4.67xx10^(-x) "mol" L^(-1)"min"^(-1)`. the numerical value of x is __________.

Options

A. `2.083xx10^(-3)" mol L"^(-1)" m in"^(-1)`
B. `4.167xx10^(-3)" mol L"^(-1)" m in"^(-1)`
C. `1.667xx10^(-2)" mol L"^(-1)" m in"^(-1)`
D. `8.33xx10^(-3)" mol L"^(-1)" m in"^(-1)`

Explanation

To solve the problem, we need to calculate the rate of formation of \( NO_2 \) based on the decomposition of \( N_2O_5 \) using the provided reaction equation and concentration data.

### Step-by-Step Solution:

1. **Write the balanced chemical equation**:
\[ 2N_2O_5(g) \rightarrow 4NO_2(g) + O_2(g) \]

2. **Identify the initial and final concentrations of \( N_2O_5 \)**:
- Initial concentration of \( N_2O_5 \) = \( 3.00 \, \text{mol L}^{-1} \)
- Final concentration of \( N_2O_5 \) after 30 minutes = \( 2.75 \, \text{mol L}^{-1} \)

3. **Calculate the change in concentration of \( N_2O_5 \)**:
\[ \Delta [N_2O_5] = [N_2O_5]_{initial} - [N_2O_5]_{final} = 3.00 - 2.75 = 0.25 \, \text{mol L}^{-1} \]

4. **Calculate the rate of disappearance of \( N_2O_5 \)**:
The rate of disappearance is given by:
\[ \text{Rate} = -\frac{\Delta [N_2O_5]}{\Delta t} = -\frac{0.25 \, \text{mol L}^{-1}}{30 \, \text{min}} = -\frac{0.25}{30} = -0.00833 \, \text{mol L}^{-1} \text{ min}^{-1} \]

5. **Relate the rate of disappearance of \( N_2O_5 \) to the rate of formation of \( NO_2 \)**:
According to the stoichiometry of the reaction:
\[ -\frac{1}{2} \frac{d[N_2O_5]}{dt} = \frac{1}{4} \frac{d[NO_2]}{dt} \]
Therefore, we can express the rate of formation of \( NO_2 \) as:
\[ \frac{d[NO_2]}{dt} = -2 \left(-\frac{d[N_2O_5]}{dt}\right) \]

6. **Substitute the rate of disappearance of \( N_2O_5 \)**:
\[ \frac{d[NO_2]}{dt} = 2 \times 0.00833 \, \text{mol L}^{-1} \text{ min}^{-1} = 0.01667 \, \text{mol L}^{-1} \text{ min}^{-1} \]

7. **Express the rate of formation of \( NO_2 \) in scientific notation**:
\[ 0.01667 \, \text{mol L}^{-1} \text{ min}^{-1} = 1.67 \times 10^{-2} \, \text{mol L}^{-1} \text{ min}^{-1} \]

8. **Compare with the given rate**:
The problem states that the rate of formation of \( NO_2 \) is \( 4.67 \times 10^{-x} \, \text{mol L}^{-1} \text{ min}^{-1} \). From our calculation:
\[ 1.67 \times 10^{-2} = 4.67 \times 10^{-x} \]

9. **Solve for \( x \)**:
To find \( x \), we can equate the powers of ten:
\[ 10^{-2} = 10^{-x} \implies x = 2 \]

### Final Answer:
The numerical value of \( x \) is \( \boxed{2} \).