For the reaction,
`N_(2)O_(4)(g)hArr 2NO_(2)(g)`
the reaction connecting the degree of dissociation `(alpha)` of `N_(2)O_(4)(g)` with eqilibrium constant `K_(p)` is
where `P_(tau)` is the total equilibrium pressure.

Options

A. `alpha=((K_(P)//P_(T))/(4+K_(P)//P_(T)))`
B. `alpha=((K_(P))/(4+4K_(P)))`
C. `alpha=((K_(P)//P_(T))/(4+K_(P)//P_(T)))^(1//2)`
D. `alpha=((K_(P)//P_(T))/(4+K_(P)))^(1//2)`

Answer

To solve the problem, we need to derive the relationship between the degree of dissociation (α) of \( N_2O_4(g) \) and the equilibrium constant \( K_p \) for the reaction:

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

### Step 1: Define Initial Conditions
Assume we start with 1 mole of \( N_2O_4 \) and no \( NO_2 \) at the beginning (t = 0).

- Initial moles of \( N_2O_4 \) = 1
- Initial moles of \( NO_2 \) = 0

### Step 2: Define Changes at Equilibrium
Let \( \alpha \) be the degree of dissociation of \( N_2O_4 \). At equilibrium:

- Moles of \( N_2O_4 \) = \( 1 - \alpha \)
- Moles of \( NO_2 \) = \( 2\alpha \) (since 2 moles of \( NO_2 \) are produced for every mole of \( N_2O_4 \) that dissociates)

### Step 3: Calculate Total Moles at Equilibrium
The total number of moles at equilibrium is:

\[ \text{Total moles} = (1 - \alpha) + 2\alpha = 1 + \alpha \]

### Step 4: Write the Expression for \( K_p \)
The equilibrium constant \( K_p \) for the reaction can be expressed in terms of the partial pressures of the gases. Since we are dealing with moles, we can relate the partial pressures to the total pressure \( P_t \).

The expression for \( K_p \) is:

\[ K_p = \frac{(P_{NO_2})^2}{P_{N_2O_4}} \]

### Step 5: Relate Partial Pressures to Moles
Using the mole fractions, we can express the partial pressures:

- \( P_{NO_2} = \frac{2\alpha}{1+\alpha} P_t \)
- \( P_{N_2O_4} = \frac{1 - \alpha}{1 + \alpha} P_t \)

Substituting these into the \( K_p \) expression gives:

\[ K_p = \frac{\left(\frac{2\alpha}{1+\alpha} P_t\right)^2}{\frac{1 - \alpha}{1 + \alpha} P_t} \]

### Step 6: Simplify the Expression
Simplifying the above expression:

\[ K_p = \frac{(2\alpha)^2 P_t^2}{(1 - \alpha)(1 + \alpha) P_t} \]

\[ K_p = \frac{4\alpha^2 P_t}{1 - \alpha^2} \]

### Step 7: Rearranging for \( \alpha \)
Rearranging gives:

\[ K_p (1 - \alpha^2) = 4\alpha^2 P_t \]

\[ K_p - K_p \alpha^2 = 4\alpha^2 P_t \]

\[ K_p = \alpha^2 (4P_t + K_p) \]

### Step 8: Solve for \( \alpha^2 \)
Factoring out \( \alpha^2 \):

\[ \alpha^2 = \frac{K_p}{4P_t + K_p} \]

### Step 9: Solve for \( \alpha \)
Taking the square root gives:

\[ \alpha = \sqrt{\frac{K_p}{4P_t + K_p}} \]

### Final Result
Thus, the relationship between the degree of dissociation \( \alpha \) and the equilibrium constant \( K_p \) is:

\[ \alpha = \sqrt{\frac{K_p}{4 + \frac{K_p}{P_t}}} \]