The decomposition of `N_(2)O_(5)` in `C Cl_(4)` at 318 K is studied by monitoring the concentration of `N_(2)O_(5)` in the solution. Initially, the concentration of `N_(2)O_(5)` is `2.4" mol L"^(-1)` and after 200 minutes, it is reduced to `2.00" mol L"^(-1).` What is the rate of production of `NO_(2)` during this period in `mol L"^(-1)"min"^(-1)` ?

Options

A. `4xx10^(-3)`
B. `2xx10^(-3)`
C. `1xx10^(-3)`
D. `2xx10^(-4)`

Explanation

To solve the problem of determining the rate of production of \( NO_2 \) during the decomposition of \( N_2O_5 \), we will follow these steps:

### Step 1: Write the balanced chemical equation
The decomposition of \( N_2O_5 \) can be represented as:
\[ 2 N_2O_5 \rightarrow 4 NO_2 + O_2 \]
This equation shows that 2 moles of \( N_2O_5 \) produce 4 moles of \( NO_2 \).

### Step 2: Determine the change in concentration of \( N_2O_5 \)
The initial concentration of \( N_2O_5 \) is \( 2.4 \, \text{mol L}^{-1} \) and after 200 minutes, it is \( 2.0 \, \text{mol L}^{-1} \). Thus, the change in concentration (\( \Delta [N_2O_5] \)) is:
\[ \Delta [N_2O_5] = [N_2O_5]_{initial} - [N_2O_5]_{final} = 2.4 - 2.0 = 0.4 \, \text{mol L}^{-1} \]

### Step 3: Calculate the rate of disappearance of \( N_2O_5 \)
The rate of disappearance of \( N_2O_5 \) can be calculated using the formula:
\[ \text{Rate} = -\frac{1}{\text{coefficient}} \cdot \frac{\Delta [N_2O_5]}{\Delta t} \]
In this case, the coefficient of \( N_2O_5 \) in the balanced equation is 2. The time interval (\( \Delta t \)) is 200 minutes. Therefore:
\[ \text{Rate of disappearance of } N_2O_5 = -\frac{1}{2} \cdot \frac{0.4 \, \text{mol L}^{-1}}{200 \, \text{min}} = -\frac{0.4}{400} = -0.001 \, \text{mol L}^{-1} \text{ min}^{-1} \]

### Step 4: Relate the rate of \( N_2O_5 \) to the rate of production of \( NO_2 \)
From the balanced equation, we see that for every 2 moles of \( N_2O_5 \) that decompose, 4 moles of \( NO_2 \) are produced. Therefore, the rate of production of \( NO_2 \) is:
\[ \text{Rate of production of } NO_2 = \frac{4}{2} \cdot \left(-\text{Rate of disappearance of } N_2O_5\right) = 2 \cdot 0.001 \, \text{mol L}^{-1} \text{ min}^{-1} = 0.002 \, \text{mol L}^{-1} \text{ min}^{-1} \]

### Final Answer
The rate of production of \( NO_2 \) during this period is:
\[ \text{Rate of production of } NO_2 = 0.002 \, \text{mol L}^{-1} \text{ min}^{-1} \]