`9.2` grams of `N_(2)O_(4(g))` is taken in a closed one litre vessel and heated till the following equilibrium is reached `N_(2)O_(4(g))hArr2NO_(2(g))`. At equilibrium, `50% N_(2)O_(4(g))` is dissociated. What is the equilibrium constant (in mol `litre^(-1)`) (Molecular weight of `N_(2)O_(4) = 92`) ?

Options

A. `0.1`
B. `0.4`
C. `0.2`
D. `2`

Answer

To solve the problem, we need to determine the equilibrium constant \( K_c \) for the reaction:

\[ N_2O_4(g) \rightleftharpoons 2 NO_2(g) \]

### Step 1: Calculate the number of moles of \( N_2O_4 \)

Given:
- Mass of \( N_2O_4 = 9.2 \) grams
- Molecular weight of \( N_2O_4 = 92 \) g/mol

Using the formula for number of moles:

\[ \text{Number of moles} = \frac{\text{mass}}{\text{molecular weight}} = \frac{9.2 \text{ g}}{92 \text{ g/mol}} = 0.1 \text{ moles} \]

### Step 2: Determine the initial concentration of \( N_2O_4 \)

Since the volume of the vessel is 1 liter, the concentration \( C \) of \( N_2O_4 \) is:

\[ C = \frac{\text{Number of moles}}{\text{Volume}} = \frac{0.1 \text{ moles}}{1 \text{ L}} = 0.1 \text{ mol/L} \]

### Step 3: Set up the equilibrium expression

At equilibrium, we know that 50% of \( N_2O_4 \) is dissociated. Thus, the change in concentration can be expressed as follows:

- Initial concentration of \( N_2O_4 = 0.1 \) mol/L
- Change in concentration of \( N_2O_4 = -0.5 \times 0.1 = -0.05 \) mol/L
- Concentration of \( N_2O_4 \) at equilibrium = \( 0.1 - 0.05 = 0.05 \) mol/L
- Concentration of \( NO_2 \) formed = \( 2 \times 0.05 = 0.1 \) mol/L

### Step 4: Write the equilibrium constant expression

The equilibrium constant \( K_c \) is given by:

\[ K_c = \frac{[\text{NO}_2]^2}{[\text{N}_2O_4]} \]

Substituting the equilibrium concentrations:

\[ K_c = \frac{(0.1)^2}{0.05} \]

### Step 5: Calculate \( K_c \)

Calculating the value:

\[ K_c = \frac{0.01}{0.05} = 0.2 \text{ mol/L} \]

### Final Answer

The equilibrium constant \( K_c \) is:

\[ \boxed{0.2 \text{ mol/L}} \]

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