`N_(2)O_(3)` dissociates into `NO` and `NO_(2)`. At equilibrium pressure of `3 atm` , all three gases were found to have equal number of moles in a vessel. In another vessel, equimolormixture of `N_(2)O_(3), NO` and `NO_(2)` are taken at the same temperature but at an initial pressure of `9 atm` then find the partial pressure of `NO_(2)` (in `atm`) at equilibrium in second vessel!
`N_(2)O_(3)(g) hArr NO(g) + NO_(2) (g)`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the Reaction
The dissociation reaction is given as:
\[ N_2O_3(g) \rightleftharpoons NO(g) + NO_2(g) \]

### Step 2: Analyze the First Vessel
In the first vessel, at equilibrium, the total pressure is given as \( 3 \, \text{atm} \) and all three gases have equal moles. Let's denote the number of moles of each gas at equilibrium as \( n \). Therefore, we can write:
- Moles of \( N_2O_3 = n \)
- Moles of \( NO = n \)
- Moles of \( NO_2 = n \)

The total number of moles at equilibrium is:
\[ n + n + n = 3n \]

Using the ideal gas law, the total pressure \( P \) is related to the total number of moles \( n \) and volume \( V \) by:
\[ P = \frac{nRT}{V} \]
Given that the total pressure is \( 3 \, \text{atm} \):
\[ P = \frac{3nRT}{V} = 3 \, \text{atm} \]

### Step 3: Calculate the Equilibrium Constant \( K_p \)
At equilibrium, the partial pressures can be expressed as:
- \( P_{N_2O_3} = \frac{nRT}{V} \)
- \( P_{NO} = \frac{nRT}{V} \)
- \( P_{NO_2} = \frac{nRT}{V} \)

Since all gases have equal moles, the partial pressures are equal:
\[ P_{N_2O_3} = P_{NO} = P_{NO_2} = 1 \, \text{atm} \]

Now, we can find \( K_p \):
\[ K_p = \frac{P_{NO} \cdot P_{NO_2}}{P_{N_2O_3}} = \frac{1 \cdot 1}{1} = 1 \]

### Step 4: Analyze the Second Vessel
In the second vessel, we start with an equimolar mixture of \( N_2O_3 \), \( NO \), and \( NO_2 \) at an initial pressure of \( 9 \, \text{atm} \).

Let the initial moles of each gas be \( x \). Thus, the initial pressures are:
- \( P_{N_2O_3} = 3 \, \text{atm} \)
- \( P_{NO} = 3 \, \text{atm} \)
- \( P_{NO_2} = 3 \, \text{atm} \)

### Step 5: Set Up the Equilibrium Expression
At equilibrium, let \( p \) be the change in pressure due to the dissociation of \( N_2O_3 \):
- The pressure of \( N_2O_3 \) will decrease by \( p \): \( P_{N_2O_3} = 3 - p \)
- The pressures of \( NO \) and \( NO_2 \) will increase by \( p \): \( P_{NO} = 3 + p \) and \( P_{NO_2} = 3 + p \)

### Step 6: Write the Equation for \( K_p \)
Using the equilibrium constant from the first vessel:
\[ K_p = \frac{(3 + p)(3 + p)}{3 - p} = 1 \]

### Step 7: Solve for \( p \)
Expanding the equation:
\[ (3 + p)^2 = 1(3 - p) \]
\[ 9 + 6p + p^2 = 3 - p \]
Rearranging gives:
\[ p^2 + 7p + 6 = 0 \]

### Step 8: Factor the Quadratic Equation
Factoring:
\[ (p + 6)(p + 1) = 0 \]
Thus, \( p = -6 \) or \( p = -1 \). Since pressure cannot be negative, we take \( p = 1 \).

### Step 9: Calculate the Partial Pressure of \( NO_2 \)
Now, substituting \( p \) back to find the partial pressure of \( NO_2 \):
\[ P_{NO_2} = 3 + p = 3 + 1 = 4 \, \text{atm} \]

### Final Answer
The partial pressure of \( NO_2 \) at equilibrium in the second vessel is:
\[ \boxed{4 \, \text{atm}} \]