1 moles of `NO_2` and 2 moles of CO are enclosed in a one litre vessel to attain the following equilibrium `NO_2 +CO hArr NO+ CO_2`. It was estimated that at the equilibrium , 25% of initial amount of CO is consumed. The equilibrium constant `K_p` is

Options

A. 1
B. `1//2`
C. `1//4`
D. `1//3`

Answer

To solve the problem, we need to find the equilibrium constant \( K_p \) for the reaction:

\[ \text{NO}_2 + \text{CO} \rightleftharpoons \text{NO} + \text{CO}_2 \]

### Step 1: Initial Moles and Changes
We start with:
- 1 mole of \( \text{NO}_2 \)
- 2 moles of \( \text{CO} \)

At equilibrium, it is given that 25% of the initial amount of \( \text{CO} \) is consumed. Since we have 2 moles of \( \text{CO} \):

\[ \text{Amount of CO consumed} = 25\% \times 2 \text{ moles} = 0.5 \text{ moles} \]

### Step 2: Calculate Moles at Equilibrium
Since 0.5 moles of \( \text{CO} \) are consumed, the change in moles of \( \text{NO}_2 \) will also be 0.5 moles (because the stoichiometry of the reaction shows a 1:1 ratio).

- Moles of \( \text{NO}_2 \) at equilibrium:
\[ 1 - 0.5 = 0.5 \text{ moles} \]

- Moles of \( \text{CO} \) at equilibrium:
\[ 2 - 0.5 = 1.5 \text{ moles} \]

- Moles of \( \text{NO} \) produced:
\[ 0 + 0.5 = 0.5 \text{ moles} \]

- Moles of \( \text{CO}_2 \) produced:
\[ 0 + 0.5 = 0.5 \text{ moles} \]

### Step 3: Total Moles at Equilibrium
At equilibrium, we have:
- \( \text{NO}_2 \): 0.5 moles
- \( \text{CO} \): 1.5 moles
- \( \text{NO} \): 0.5 moles
- \( \text{CO}_2 \): 0.5 moles

### Step 4: Calculate Concentrations
Since the volume of the vessel is 1 L, the concentrations (in mol/L) are equal to the number of moles:

- Concentration of \( \text{NO}_2 \) = 0.5 M
- Concentration of \( \text{CO} \) = 1.5 M
- Concentration of \( \text{NO} \) = 0.5 M
- Concentration of \( \text{CO}_2 \) = 0.5 M

### Step 5: Write the Expression for \( K_c \)
The equilibrium constant \( K_c \) is given by:

\[ K_c = \frac{[\text{NO}][\text{CO}_2]}{[\text{NO}_2][\text{CO}]} \]

Substituting the equilibrium concentrations:

\[ K_c = \frac{(0.5)(0.5)}{(0.5)(1.5)} = \frac{0.25}{0.75} = \frac{1}{3} \]

### Step 6: Relate \( K_c \) and \( K_p \)
Since the change in the number of moles of gas (\( \Delta n_g \)) is zero (2 moles of reactants and 2 moles of products), we can use the relationship:

\[ K_p = K_c (RT)^{\Delta n_g} \]

Since \( \Delta n_g = 0 \):

\[ K_p = K_c \cdot (RT)^0 = K_c \]

Thus,

\[ K_p = \frac{1}{3} \]

### Final Answer
The equilibrium constant \( K_p \) is:

\[ \boxed{\frac{1}{3}} \]