`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. `1//2`
D. `1//4`

Explanation

To solve the problem, we will follow these steps:

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

### Step 2: Set up the initial moles of reactants and products
At the start (T=0), we have:
- Moles of A = 4
- Moles of B = 4
- Moles of C = 0
- Moles of D = 0

### Step 3: Determine the change in moles at equilibrium
According to the problem, 2 moles of C are formed at equilibrium. Since the stoichiometric coefficients of C and D are the same, we can assume that 2 moles of D are also formed. Therefore, the changes in moles can be represented as follows:
- Change in A = -2 (2 moles of A are consumed)
- Change in B = -2 (2 moles of B are consumed)
- Change in C = +2 (2 moles of C are formed)
- Change in D = +2 (2 moles of D are formed)

### Step 4: Calculate the equilibrium moles
At equilibrium, the moles of each substance will be:
- Moles of A = 4 - 2 = 2
- Moles of B = 4 - 2 = 2
- Moles of C = 0 + 2 = 2
- Moles of D = 0 + 2 = 2

### Step 5: Write the expression for the equilibrium constant (K)
The equilibrium constant \( K \) for the reaction is given by:
\[ K = \frac{[C][D]}{[A][B]} \]

### Step 6: Substitute the equilibrium concentrations into the expression
Since we are dealing with moles and the volume cancels out, we can directly substitute the equilibrium moles:
\[ K = \frac{(2)(2)}{(2)(2)} = \frac{4}{4} = 1 \]

### Final Answer
Thus, the value of the equilibrium constant \( K \) is:
\[ K = 1 \]

---