The decomposition of `N_(2)O_(4)` to `NO_(2)` is carried out at `280^(@)C` in chloroform. When equilibrium is reached, `0.2` mol of `N_(2)O_(4)` and `2xx10^(-3)` mol of `NO_(2)` are present in a `2L` solution. The equilibrium constant for the reaction
`N_(2)O_(4) hArr 2NO_(2)` is

Options

A. `1xx10^(-2)`
B. `2xx10^(-3)`
C. `1xx10^(-5)`
D. `2xx10^(-5)`

Explanation

To find the equilibrium constant \( K_c \) for the decomposition of \( N_2O_4 \) to \( NO_2 \), we can follow these steps:

### Step 1: Write the balanced chemical equation
The decomposition reaction is:
\[ N_2O_4 \rightleftharpoons 2 NO_2 \]

### Step 2: Identify the equilibrium concentrations
From the problem, we know:
- Moles of \( N_2O_4 \) at equilibrium = 0.2 mol
- Moles of \( NO_2 \) at equilibrium = \( 2 \times 10^{-3} \) mol
- Volume of the solution = 2 L

### Step 3: Calculate the equilibrium concentrations
The concentration \( C \) is calculated using the formula:
\[ C = \frac{\text{moles}}{\text{volume (L)}} \]

For \( N_2O_4 \):
\[ C_{N_2O_4} = \frac{0.2 \text{ mol}}{2 \text{ L}} = 0.1 \text{ M} \]

For \( NO_2 \):
\[ C_{NO_2} = \frac{2 \times 10^{-3} \text{ mol}}{2 \text{ L}} = 1 \times 10^{-3} \text{ M} \]

### Step 4: Write the expression for the equilibrium constant \( K_c \)
The expression for the equilibrium constant \( K_c \) for the reaction is:
\[ K_c = \frac{[NO_2]^2}{[N_2O_4]} \]

### Step 5: Substitute the equilibrium concentrations into the expression
Substituting the values we calculated:
\[ K_c = \frac{(1 \times 10^{-3})^2}{0.1} \]

### Step 6: Calculate \( K_c \)
Calculating the numerator:
\[ (1 \times 10^{-3})^2 = 1 \times 10^{-6} \]

Now substituting back into the equation:
\[ K_c = \frac{1 \times 10^{-6}}{0.1} = \frac{1 \times 10^{-6}}{1 \times 10^{-1}} = 1 \times 10^{-5} \]

### Final Answer
Thus, the equilibrium constant \( K_c \) is:
\[ K_c = 1 \times 10^{-5} \]

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