The methods employed for the determination of equilibrium constant are different for different equilibria depending on the nature of the system and the type of reaction. The task becomes comparatively easy when we can measure some physical property such as vapour density, refractive index, optical rotation, electrical conductivity etc. In simple gaseous dissociation such as `N_(2)O_(4) rarr 2NO_(2) `. 0.1 mole of `N_(2)O_(4(g))` is decomposed under atmospheric condition at `25^(@)C`.
calculate Value of `K_(p)` when degree of dissociation becomes equal to `50%`

Options

A. 1.5 times
B. Remains constant
C. 2 times
D. Decreases

Answer

To calculate the equilibrium constant \( K_p \) for the dissociation of \( N_2O_4 \) into \( 2NO_2 \) when the degree of dissociation is 50%, we can follow these steps:

### Step 1: Write the reaction and initial conditions
The reaction is:
\[ N_2O_4(g) \rightleftharpoons 2NO_2(g) \]
Initially, we have 0.1 moles of \( N_2O_4 \).

### Step 2: Define the degree of dissociation
Let the degree of dissociation be \( \alpha \). Given that \( \alpha = 50\% = 0.5 \).

### Step 3: Calculate the moles at equilibrium
Initially, we have:
- Moles of \( N_2O_4 = 0.1 \)
- Moles of \( NO_2 = 0 \)

At equilibrium, the moles will be:
- Moles of \( N_2O_4 = 0.1 - 0.1 \cdot \alpha = 0.1 - 0.05 = 0.05 \)
- Moles of \( NO_2 = 2 \cdot (0.1 \cdot \alpha) = 2 \cdot 0.05 = 0.1 \)

### Step 4: Calculate total moles at equilibrium
Total moles at equilibrium:
\[ N_{total} = \text{Moles of } N_2O_4 + \text{Moles of } NO_2 = 0.05 + 0.1 = 0.15 \]

### Step 5: Calculate \( K_p \)
The expression for \( K_p \) is given by:
\[ K_p = \frac{(P_{NO_2})^2}{P_{N_2O_4}} \]
Using the ideal gas law, we can relate moles to partial pressures. Assuming the total pressure \( P_{total} = 1 \, \text{atm} \):
- Partial pressure of \( N_2O_4 = \frac{n_{N_2O_4}}{N_{total}} \cdot P_{total} = \frac{0.05}{0.15} \cdot 1 = \frac{1}{3} \, \text{atm} \)
- Partial pressure of \( NO_2 = \frac{n_{NO_2}}{N_{total}} \cdot P_{total} = \frac{0.1}{0.15} \cdot 1 = \frac{2}{3} \, \text{atm} \)

Now substituting into the \( K_p \) expression:
\[ K_p = \frac{\left(\frac{2}{3}\right)^2}{\frac{1}{3}} = \frac{\frac{4}{9}}{\frac{1}{3}} = \frac{4}{9} \cdot 3 = \frac{4}{3} \]

### Step 6: Final calculation
Thus, the value of \( K_p \) is:
\[ K_p = \frac{4}{3} \approx 1.33 \]