A vessel (A) contains I mole each of `N_(2) & O_(2)` and another vesser (B) contains 2 mole each of `N_(2) & O_(2)` . Both vessels are heated to same temperature till equilibrium is established in-both cases. Then, correct statement is

Options

A. `K_(C) "for " N_(2) + O_(2) hArr ` 2NO in A & B are in the ratio 1:2
B. `K_(P) "for " N_(2) + O_(2) hArr ` 2NO in A & B are in the ratio 1:3
C. `K_(C) "for" N_(2) + O_(2) hArr ` 2NO in A & B are equal
D. `K_(p) "for " N_(2) + O_(2) hArr ` 2NO in A &B are in the ratio 2:1

Answer

To solve the problem, we will analyze the equilibrium established in both vessels A and B containing nitrogen (N₂) and oxygen (O₂) gases.

### Step-by-Step Solution:

1. **Identify the Initial Conditions:**
- Vessel A contains 1 mole of N₂ and 1 mole of O₂.
- Vessel B contains 2 moles of N₂ and 2 moles of O₂.

2. **Write the Balanced Chemical Equation:**
The reaction taking place is:
\[ N_2(g) + O_2(g) \rightleftharpoons 2NO(g) \]

3. **Set Up the Equilibrium Expressions:**
For both vessels, we can set up the equilibrium expressions for the reaction. Let \( x \) be the moles of N₂ that react in vessel A, and \( y \) be the moles of N₂ that react in vessel B.

- For vessel A:
- Initial moles: N₂ = 1, O₂ = 1
- Change: N₂ = -x, O₂ = -x, NO = +2x
- At equilibrium:
- N₂ = \( 1 - x \)
- O₂ = \( 1 - x \)
- NO = \( 2x \)

The equilibrium constant \( K_c \) for vessel A is:
\[ K_c = \frac{(2x)^2}{(1-x)(1-x)} = \frac{4x^2}{(1-x)^2} \]

- For vessel B:
- Initial moles: N₂ = 2, O₂ = 2
- Change: N₂ = -y, O₂ = -y, NO = +2y
- At equilibrium:
- N₂ = \( 2 - y \)
- O₂ = \( 2 - y \)
- NO = \( 2y \)

The equilibrium constant \( K_c \) for vessel B is:
\[ K_c = \frac{(2y)^2}{(2-y)(2-y)} = \frac{4y^2}{(2-y)^2} \]

4. **Equate the Equilibrium Constants:**
Since both reactions are the same and occur at the same temperature, we can equate the two expressions for \( K_c \):
\[ \frac{4x^2}{(1-x)^2} = \frac{4y^2}{(2-y)^2} \]
Simplifying gives:
\[ \frac{x^2}{(1-x)^2} = \frac{y^2}{(2-y)^2} \]

5. **Relate \( x \) and \( y \):**
Taking the square root of both sides, we have:
\[ \frac{x}{1-x} = \frac{y}{2-y} \]
Cross-multiplying gives:
\[ x(2-y) = y(1-x) \]
Rearranging leads to:
\[ 2x - xy = y - xy \implies 2x = y \]
Thus, we find:
\[ x = \frac{y}{2} \]

6. **Conclusion on \( K_c \) and \( K_p \):**
Since both vessels are at the same temperature and involve the same reaction, the equilibrium constants \( K_c \) for both vessels will be equal. Additionally, since the change in moles of gas (\( \Delta n \)) is zero (2 moles of reactants yield 2 moles of products), we have:
\[ K_p = K_c \]

### Final Statement:
The correct statement is that \( K_c \) for the reaction in both vessels A and B is equal, and since \( K_p = K_c \) under these conditions, both \( K_c \) and \( K_p \) are the same for both vessels.