`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)`) (Moleculatr weight of `N_(2)O_(4) = 92`) ?

Options

A. `0.1`
B. `0.4`
C. `0.01`
D. `2`

Explanation

To solve the problem step-by-step, we will calculate 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

To find the number of moles:
\[ \text{Moles of } N_2O_4 = \frac{\text{Mass}}{\text{Molecular Weight}} = \frac{9.2 \, \text{g}}{92 \, \text{g/mol}} = 0.1 \, \text{mol} \]

### Step 2: Calculate the concentration of \( N_2O_4 \)
Since the volume of the vessel is 1 litre:
\[ \text{Concentration of } N_2O_4 = \frac{\text{Moles}}{\text{Volume}} = \frac{0.1 \, \text{mol}}{1 \, \text{L}} = 0.1 \, \text{mol/L} \]

### Step 3: Determine the change in concentration at equilibrium
Given that 50% of \( N_2O_4 \) dissociates:
- Initial concentration of \( N_2O_4 = 0.1 \, \text{mol/L} \)
- Change in concentration of \( N_2O_4 \) at equilibrium:
\[ \text{Dissociated } N_2O_4 = 0.1 \, \text{mol/L} \times 0.5 = 0.05 \, \text{mol/L} \]
- Concentration of \( N_2O_4 \) at equilibrium:
\[ [N_2O_4]_{eq} = 0.1 - 0.05 = 0.05 \, \text{mol/L} \]

### Step 4: Calculate the concentration of \( NO_2 \) at equilibrium
From the stoichiometry of the reaction, for every mole of \( N_2O_4 \) that dissociates, 2 moles of \( NO_2 \) are produced:
\[ [NO_2]_{eq} = 2 \times 0.05 = 0.1 \, \text{mol/L} \]

### Step 5: Write the expression for the equilibrium constant \( K_c \)
The expression for \( K_c \) for the reaction is:
\[ K_c = \frac{[NO_2]^2}{[N_2O_4]} \]

### Step 6: Substitute the equilibrium concentrations into the \( K_c \) expression
Substituting the equilibrium concentrations:
\[ K_c = \frac{(0.1)^2}{0.05} = \frac{0.01}{0.05} = 0.2 \, \text{mol/L} \]

### Final Answer
The equilibrium constant \( K_c \) is \( 0.2 \, \text{mol/L} \).

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