`4` moles of A are mixed with `4` moles of B, when `2` moles of C are formed at equilibrium according to the reaction `A+B hArr C+D`.
The value of equilibrium constant is

Options

A. 4
B. 1
C. `sqrt2`
D. `sqrt4`

Answer

To find the equilibrium constant for the reaction \( A + B \rightleftharpoons C + D \) given the initial moles and the amount formed at equilibrium, we can follow these steps:

### Step 1: Write the balanced equation
The balanced chemical equation is:
\[ A + B \rightleftharpoons C + D \]

### Step 2: Identify initial moles
Initially, we have:
- Moles of A = 4
- Moles of B = 4
- Moles of C = 0
- Moles of D = 0

### Step 3: Determine changes at equilibrium
At equilibrium, it is given that 2 moles of C are formed. Therefore:
- Moles of C = 2
- Moles of D = 2 (since the stoichiometry of the reaction indicates that for every mole of C formed, one mole of D is also formed)
- Moles of A = 4 - 2 = 2 (2 moles of A reacted)
- Moles of B = 4 - 2 = 2 (2 moles of B reacted)

### Step 4: Write the equilibrium moles
At equilibrium, we have:
- Moles of A = 2
- Moles of B = 2
- Moles of C = 2
- Moles of D = 2

### Step 5: Calculate concentrations
Assuming the volume of the container is \( V \), the concentrations at equilibrium will be:
- Concentration of A = \( \frac{2}{V} \)
- Concentration of B = \( \frac{2}{V} \)
- Concentration of C = \( \frac{2}{V} \)
- Concentration of D = \( \frac{2}{V} \)

### Step 6: Write the expression for the equilibrium constant
The equilibrium constant \( K_c \) is given by the formula:
\[ K_c = \frac{[C][D]}{[A][B]} \]

### Step 7: Substitute the concentrations into the equilibrium constant expression
Substituting the equilibrium concentrations into the expression:
\[ K_c = \frac{\left(\frac{2}{V}\right) \left(\frac{2}{V}\right)}{\left(\frac{2}{V}\right) \left(\frac{2}{V}\right)} \]

### Step 8: Simplify the expression
\[ K_c = \frac{\frac{4}{V^2}}{\frac{4}{V^2}} = 1 \]

### Conclusion
The value of the equilibrium constant \( K_c \) is:
\[ K_c = 1 \]