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 of relating the degree of dissociation (α) of \(N_2O_4(g)\) with the equilibrium constant \(K_p\), we can follow these steps:

### Step 1: Write the balanced chemical equation
The reaction is given as:
\[ N_2O_4(g) \rightleftharpoons 2NO_2(g) \]

### Step 2: Define the initial moles and the degree of dissociation
Assume we start with 1 mole of \(N_2O_4\). If α is the degree of dissociation, then:
- Moles of \(N_2O_4\) at equilibrium = \(1 - \alpha\)
- Moles of \(NO_2\) produced = \(2\alpha\)

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

### Step 4: Calculate the partial pressures
Using the total pressure \(P_t\), we can find the partial pressures of \(N_2O_4\) and \(NO_2\):
- Partial pressure of \(N_2O_4\):
\[ P_{N_2O_4} = \left(\frac{1 - \alpha}{1 + \alpha}\right) P_t \]
- Partial pressure of \(NO_2\):
\[ P_{NO_2} = \left(\frac{2\alpha}{1 + \alpha}\right) P_t \]

### Step 5: Write the expression for the equilibrium constant \(K_p\)
The equilibrium constant \(K_p\) is given by:
\[ K_p = \frac{(P_{NO_2})^2}{P_{N_2O_4}} \]
Substituting the partial pressures:
\[ K_p = \frac{\left(\frac{2\alpha}{1 + \alpha} P_t\right)^2}{\left(\frac{1 - \alpha}{1 + \alpha} P_t\right)} \]

### Step 6: Simplify the expression
Now, simplifying the expression for \(K_p\):
\[ K_p = \frac{(2\alpha)^2 P_t^2}{(1 - \alpha)(1 + \alpha) P_t} = \frac{4\alpha^2 P_t}{1 - \alpha} \]

### Step 7: Rearranging to find α in terms of \(K_p\) and \(P_t\)
Rearranging gives:
\[ K_p(1 - \alpha) = 4\alpha^2 P_t \]
\[ K_p - K_p\alpha = 4\alpha^2 P_t \]
\[ 4\alpha^2 P_t + K_p\alpha - K_p = 0 \]

### Step 8: Solve the quadratic equation
This is a quadratic equation in terms of α:
\[ 4P_t\alpha^2 + K_p\alpha - K_p = 0 \]
Using the quadratic formula:
\[ \alpha = \frac{-K_p \pm \sqrt{(K_p)^2 + 16K_pP_t}}{8P_t} \]

### Step 9: Choose the appropriate solution
Since α must be a positive value (degree of dissociation), we take the positive root:
\[ \alpha = \frac{-K_p + \sqrt{(K_p)^2 + 16K_pP_t}}{8P_t} \]

### Final Result
Thus, the relationship between the degree of dissociation (α) and the equilibrium constant \(K_p\) is given by:
\[ \alpha = \frac{-K_p + \sqrt{(K_p)^2 + 16K_pP_t}}{8P_t} \]