Read the following paragraph and answer the questions given below,
for a general reaction, `aA+bB hArr cC=dD`, equilibrium constant `K_c` is given by `K_c=([C]^c[D]^d)/([A]^a[B]^b)` However, when all reactants and products are gases the equilibrium constant is generally expressed in terms of partial pressure.
`K_p=(P_C^cxx P _D^d)/(P_A^axxP_B^b)`
The value of `K_p` for the equilibrium of the reaction, `N_2O_4 (g) hArr 2NO_2(g)` is 2 then the % dissociation of `N_2O_4` at 0.5 atm pressure will be

Options

A. 25
B. 68
C. 71
D. 82

Answer

To solve the problem of finding the percentage dissociation of \( N_2O_4 \) at a pressure of 0.5 atm, we will follow these steps:

### Step 1: Write the reaction and equilibrium expression
The reaction is:
\[ N_2O_4 (g) \rightleftharpoons 2NO_2 (g) \]
The equilibrium constant in terms of partial pressures is given by:
\[ K_p = \frac{(P_{NO_2})^2}{P_{N_2O_4}} \]
We know that \( K_p = 2 \).

### Step 2: Define initial conditions and changes
Let the initial amount of \( N_2O_4 \) be 1 mole. At equilibrium, let \( \alpha \) be the degree of dissociation of \( N_2O_4 \). Therefore:
- Initial moles of \( N_2O_4 = 1 \)
- Moles of \( NO_2 \) formed = \( 2\alpha \)
- At equilibrium:
- Moles of \( N_2O_4 = 1 - \alpha \)
- Moles of \( NO_2 = 2\alpha \)

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

### Step 4: Calculate partial pressures
The partial pressures can be expressed as:
\[ P_{N_2O_4} = \frac{(1 - \alpha)}{(1 + \alpha)} \times P \]
\[ P_{NO_2} = \frac{(2\alpha)}{(1 + \alpha)} \times P \]
Given \( P = 0.5 \) atm, we can substitute this into the equations.

### Step 5: Substitute into the equilibrium expression
Substituting the partial pressures into the \( K_p \) expression:
\[ K_p = \frac{\left(\frac{(2\alpha)}{(1 + \alpha)} \times 0.5\right)^2}{\frac{(1 - \alpha)}{(1 + \alpha)} \times 0.5} \]
This simplifies to:
\[ K_p = \frac{(2\alpha)^2 \times 0.5}{(1 - \alpha) \times 0.5} = \frac{4\alpha^2}{1 - \alpha} \]
Setting this equal to 2 (the given \( K_p \)):
\[ \frac{4\alpha^2}{1 - \alpha} = 2 \]

### Step 6: Solve for \( \alpha \)
Cross-multiplying gives:
\[ 4\alpha^2 = 2(1 - \alpha) \]
\[ 4\alpha^2 + 2\alpha - 2 = 0 \]
Dividing through by 2:
\[ 2\alpha^2 + \alpha - 1 = 0 \]
Using the quadratic formula:
\[ \alpha = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-1 \pm \sqrt{1 + 8}}{4} = \frac{-1 \pm 3}{4} \]
Calculating the roots:
\[ \alpha = \frac{2}{4} = 0.5 \quad \text{(valid solution)} \]
\[ \alpha = \frac{-4}{4} = -1 \quad \text{(not valid)} \]

### Step 7: Calculate percentage dissociation
The percentage dissociation is given by:
\[ \text{Percentage dissociation} = \alpha \times 100 = 0.5 \times 100 = 50\% \]

### Final Answer
The percentage dissociation of \( N_2O_4 \) at 0.5 atm pressure is **50%**.

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