18.4 g of `N_(2)O_(4)` is taken in a 1 L closed vessel and heated till the equilibrium is reached.
`N_(2)O_(4(g))rArr2NO_(2(g))`
At equilibrium it is found that 50% of `N_(2)O_(4)` is dissociated . What will be the value of equilibrium constant?

Options

A. 0.2
B. 2
C. 0.4
D. 0.8

Answer

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

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

we will follow these steps:

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

Given:
- Mass of \( N_2O_4 = 18.4 \, \text{g} \)
- Molar mass of \( N_2O_4 = 92 \, \text{g/mol} \)

Using the formula for moles:

\[ \text{moles} = \frac{\text{mass}}{\text{molar mass}} = \frac{18.4 \, \text{g}}{92 \, \text{g/mol}} = 0.2 \, \text{mol} \]

### Step 2: Determine the initial and equilibrium concentrations

Initially, in a 1 L vessel:
- Moles of \( N_2O_4 \) = 0.2 mol
- Moles of \( NO_2 \) = 0 mol

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

\[ \text{Amount dissociated} = 0.5 \times 0.2 \, \text{mol} = 0.1 \, \text{mol} \]

Thus, at equilibrium:
- Moles of \( N_2O_4 \) remaining = \( 0.2 - 0.1 = 0.1 \, \text{mol} \)
- Moles of \( NO_2 \) formed = \( 2 \times 0.1 = 0.2 \, \text{mol} \)

### Step 3: Write the expression for the equilibrium constant \( K_c \)

The equilibrium constant expression for the reaction is given by:

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

### Step 4: Calculate the equilibrium concentrations

Since the volume of the vessel is 1 L, the concentrations are equal to the number of moles:

- Concentration of \( N_2O_4 \) at equilibrium = \( [N_2O_4] = 0.1 \, \text{mol/L} \)
- Concentration of \( NO_2 \) at equilibrium = \( [NO_2] = 0.2 \, \text{mol/L} \)

### Step 5: Substitute the concentrations into the \( K_c \) expression

Now substitute the values into the equilibrium constant expression:

\[ K_c = \frac{(0.2)^2}{0.1} = \frac{0.04}{0.1} = 0.4 \]

### Final Answer

Thus, the value of the equilibrium constant \( K_c \) is:

\[ \boxed{0.4} \]

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