4 moles of A are mixed with 4 moles of B. At equilibrium for the raction `A+BhArrC+D`, 2 moles of C and D are formed. The equilibrium constant for the reaction will be

Options

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

Answer

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

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

### Step 2: Set up the initial moles
Initially, 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
At equilibrium, it is given that 2 moles of C and 2 moles of D are formed. Therefore, the change in moles can be represented as:
- Change in moles of A = -2 (since 2 moles of A are consumed)
- Change in moles of B = -2 (since 2 moles of B are consumed)
- Change in moles of C = +2 (since 2 moles of C are formed)
- Change in moles of D = +2 (since 2 moles of D are formed)

### Step 4: Calculate the equilibrium moles
Now, we can calculate the moles of each substance at equilibrium:
- Moles of A at equilibrium = \( 4 - 2 = 2 \)
- Moles of B at equilibrium = \( 4 - 2 = 2 \)
- Moles of C at equilibrium = \( 0 + 2 = 2 \)
- Moles of D at equilibrium = \( 0 + 2 = 2 \)

### Step 5: Write the expression for the equilibrium constant \( K_c \)
The equilibrium constant \( K_c \) for the reaction is given by:
\[ K_c = \frac{[C][D]}{[A][B]} \]
where \([X]\) represents the concentration of substance X.

### Step 6: Calculate the concentrations
Assuming the volume of the reaction mixture is 1 liter, the concentrations at equilibrium are:
- \([A] = \frac{2 \text{ moles}}{1 \text{ L}} = 2 \, \text{M}\)
- \([B] = \frac{2 \text{ moles}}{1 \text{ L}} = 2 \, \text{M}\)
- \([C] = \frac{2 \text{ moles}}{1 \text{ L}} = 2 \, \text{M}\)
- \([D] = \frac{2 \text{ moles}}{1 \text{ L}} = 2 \, \text{M}\)

### Step 7: Substitute the concentrations into the equilibrium constant expression
Now substitute the concentrations into the expression for \( K_c \):
\[ K_c = \frac{(2)(2)}{(2)(2)} = \frac{4}{4} = 1 \]

### Final Answer
The equilibrium constant \( K_c \) for the reaction is:
\[ \boxed{1} \]