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 find the equilibrium constant \( K_p \) for the reaction

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

we start with the initial amounts of the reactants and the information provided about the consumption of CO.

### Step 1: Initial Moles
We are given:
- Initial moles of \( \text{NO}_2 = 1 \, \text{mol} \)
- Initial moles of \( \text{CO} = 2 \, \text{mol} \)

### Step 2: Change in Moles
It is stated that 25% of the initial amount of CO is consumed at equilibrium. Since we have 2 moles of CO initially, the amount consumed is:

\[ \text{Amount consumed} = 0.25 \times 2 = 0.5 \, \text{mol} \]

### Step 3: Moles at Equilibrium
Now, we can determine the moles of each species at equilibrium:

- Moles of \( \text{NO}_2 \) at equilibrium:
\[ \text{NO}_2 = 1 \, \text{mol} - 0 \, \text{mol} = 1 \, \text{mol} \]

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

- Moles of \( \text{NO} \) produced (equal to the amount of CO consumed):
\[ \text{NO} = 0 + 0.5 = 0.5 \, \text{mol} \]

- Moles of \( \text{CO}_2 \) produced (equal to the amount of CO consumed):
\[ \text{CO}_2 = 0 + 0.5 = 0.5 \, \text{mol} \]

### Step 4: Equilibrium Concentrations
Since the volume of the vessel is 1 L, the concentrations (or partial pressures) of each species at equilibrium are:

- \( [\text{NO}_2] = 1 \, \text{mol/L} = 1 \, \text{atm} \)
- \( [\text{CO}] = 1.5 \, \text{mol/L} = 1.5 \, \text{atm} \)
- \( [\text{NO}] = 0.5 \, \text{mol/L} = 0.5 \, \text{atm} \)
- \( [\text{CO}_2] = 0.5 \, \text{mol/L} = 0.5 \, \text{atm} \)

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

\[ K_p = \frac{P_{\text{NO}} \cdot P_{\text{CO}_2}}{P_{\text{NO}_2} \cdot P_{\text{CO}}} \]

Substituting the equilibrium pressures:

\[ K_p = \frac{(0.5)(0.5)}{(1)(1.5)} = \frac{0.25}{1.5} = \frac{1}{6} \]

### Step 6: Final Answer
Thus, the equilibrium constant \( K_p \) is:

\[ K_p = \frac{1}{6} \]