`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. `0.2`
C. `0.4`
D. 2

Answer

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

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

### 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 moles:

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

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

Since the volume of the vessel is 1 L:

\[ \text{Initial concentration of } N_2O_4 = \frac{0.1 \, \text{mol}}{1 \, \text{L}} = 0.1 \, \text{M} \]

### Step 3: Set up the equilibrium expression

At equilibrium, it is given that 50% of \( N_2O_4 \) is dissociated. Therefore:

- Initial concentration of \( N_2O_4 = 0.1 \, \text{M} \)
- Change in concentration of \( N_2O_4 = -0.05 \, \text{M} \) (50% of 0.1 M)
- Equilibrium concentration of \( N_2O_4 = 0.1 - 0.05 = 0.05 \, \text{M} \)

For \( NO_2 \), since 1 mole of \( N_2O_4 \) produces 2 moles of \( NO_2 \):

- Change in concentration of \( NO_2 = +0.1 \, \text{M} \) (2 times the change in \( N_2O_4 \))
- Equilibrium concentration of \( NO_2 = 0 + 0.1 = 0.1 \, \text{M} \)

### Step 4: Write the expression for the equilibrium constant \( K_c \)

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

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

Substituting the equilibrium concentrations:

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

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

\[ K_c = \frac{0.01}{0.05} = 0.2 \]

### Final Answer

The equilibrium constant \( K_c \) is \( 0.2 \, \text{mol L}^{-1} \).

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