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

Options

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

Answer

To find the equilibrium constant \( K_c \) for the reaction

\[ N_2O_4 \rightleftharpoons 2NO_2, \]

we will follow these steps:

### Step 1: Write the expression for \( K_c \)

The equilibrium constant \( K_c \) is defined as:

\[ K_c = \frac{[NO_2]^2}{[N_2O_4]} \]

where \([NO_2]\) and \([N_2O_4]\) are the molar concentrations of \( NO_2 \) and \( N_2O_4 \) at equilibrium, respectively.

### Step 2: Calculate the concentrations

Given:
- 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

We can calculate the concentrations:

1. Concentration of \( N_2O_4 \):

\[ [N_2O_4] = \frac{\text{moles of } N_2O_4}{\text{volume}} = \frac{0.2 \, \text{mol}}{2 \, \text{L}} = 0.1 \, \text{mol/L} \]

2. Concentration of \( NO_2 \):

\[ [NO_2] = \frac{\text{moles of } NO_2}{\text{volume}} = \frac{2 \times 10^{-3} \, \text{mol}}{2 \, \text{L}} = 1 \times 10^{-3} \, \text{mol/L} \]

### Step 3: Substitute the concentrations into the \( K_c \) expression

Now we substitute the concentrations into the \( K_c \) expression:

\[ K_c = \frac{(1 \times 10^{-3})^2}{0.1} \]

### Step 4: 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 \( K_c \) for the reaction is:

\[ K_c = 1 \times 10^{-5} \]

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