The degree of dissociation of `N_(2)O_(4)(1)` obeying the equilibrium,
`N_(2)O_(4) (g) hArr 2NO_(2)(g)`, is approximately related to the pressure at equilibrium by `:`

Options

A. `alpha prop P`
B. `alpha prop (1)/(sqrt(P))`
C. `alpha prop (1)/(P^(2))`
D. `alpha prop (1)/(P^(4))`

Answer

To solve the problem regarding the degree of dissociation of \( N_2O_4 \) at equilibrium, we will follow these steps:

### Step 1: Write the equilibrium reaction
The equilibrium reaction is:
\[ N_2O_4 (g) \rightleftharpoons 2 NO_2 (g) \]

### Step 2: Define the degree of dissociation
Let \( \alpha \) be the degree of dissociation of \( N_2O_4 \). This means that if we start with 1 mole of \( N_2O_4 \), then at equilibrium:
- The amount of \( N_2O_4 \) remaining will be \( 1 - \alpha \).
- The amount of \( NO_2 \) produced will be \( 2\alpha \).

### Step 3: Calculate total moles at equilibrium
The total number of moles at equilibrium can be expressed as:
\[ \text{Total moles} = (1 - \alpha) + 2\alpha = 1 + \alpha \]

### Step 4: Calculate partial pressures
To find the partial pressures, we need to use the mole fractions:
- The partial pressure of \( N_2O_4 \):
\[ P_{N_2O_4} = \left( \frac{1 - \alpha}{1 + \alpha} \right) P \]
- The partial pressure of \( NO_2 \):
\[ P_{NO_2} = \left( \frac{2\alpha}{1 + \alpha} \right) P \]

### Step 5: Write the expression for \( K_p \)
The equilibrium constant \( K_p \) for the reaction can be expressed as:
\[ K_p = \frac{(P_{NO_2})^2}{P_{N_2O_4}} = \frac{\left( \frac{2\alpha}{1 + \alpha} P \right)^2}{\left( \frac{1 - \alpha}{1 + \alpha} P \right)} \]

### Step 6: Simplify the expression for \( K_p \)
Substituting the expressions for partial pressures into the \( K_p \) expression:
\[ K_p = \frac{(2\alpha)^2 P^2}{(1 - \alpha)(1 + \alpha)} = \frac{4\alpha^2 P^2}{(1 - \alpha)(1 + \alpha)} \]

### Step 7: Assume \( \alpha \) is small
Since \( \alpha \) is small, we can approximate \( 1 - \alpha \approx 1 \):
\[ K_p \approx \frac{4\alpha^2 P^2}{1} = 4\alpha^2 P^2 \]

### Step 8: Solve for \( \alpha \)
Rearranging gives:
\[ \alpha^2 = \frac{K_p}{4P^2} \]
Taking the square root:
\[ \alpha = \sqrt{\frac{K_p}{4P^2}} = \frac{1}{2} \sqrt{\frac{K_p}{P^2}} = \frac{\sqrt{K_p}}{2P} \]

### Step 9: Determine the relationship between \( \alpha \) and \( P \)
From the final expression, we can see that:
\[ \alpha \propto \frac{1}{P} \]
This indicates that the degree of dissociation \( \alpha \) is inversely proportional to the square root of the pressure \( P \).

### Conclusion
Thus, the degree of dissociation of \( N_2O_4 \) is inversely proportional to the square root of the pressure at equilibrium.

---