The compounds A and B are mixed in equimolar proportion to form the products, `A+B hArr C+D`
At equilibrium, one third of A and B are consumed. The equilibrium constant for the reaction is

Options

A. 0.5
B. 4
C. 2.5
D. 0.25

Explanation

To find the equilibrium constant (Kc) for the reaction \( A + B \rightleftharpoons C + D \), we will follow these steps:

### Step 1: Define Initial Conditions
Let the initial moles of both A and B be \( A \) (since they are mixed in equimolar proportions). At time \( t = 0 \):
- Moles of A = \( A \)
- Moles of B = \( A \)
- Moles of C = 0
- Moles of D = 0

### Step 2: Define Change at Equilibrium
According to the problem, one third of A and B are consumed at equilibrium. Therefore:
- Moles of A consumed = \( \frac{A}{3} \)
- Moles of B consumed = \( \frac{A}{3} \)

### Step 3: Calculate Moles at Equilibrium
At equilibrium, the moles of each substance will be:
- Moles of A = \( A - \frac{A}{3} = \frac{2A}{3} \)
- Moles of B = \( A - \frac{A}{3} = \frac{2A}{3} \)
- Moles of C = \( 0 + \frac{A}{3} = \frac{A}{3} \)
- Moles of D = \( 0 + \frac{A}{3} = \frac{A}{3} \)

### Step 4: Write the Expression for Kc
The equilibrium constant \( K_c \) is given by the formula:
\[ K_c = \frac{[C][D]}{[A][B]} \]
Substituting the equilibrium concentrations:
\[ K_c = \frac{\left(\frac{A}{3}\right)\left(\frac{A}{3}\right)}{\left(\frac{2A}{3}\right)\left(\frac{2A}{3}\right)} \]

### Step 5: Simplify the Expression
Now, simplify the expression:
\[ K_c = \frac{\frac{A^2}{9}}{\frac{4A^2}{9}} = \frac{A^2}{9} \times \frac{9}{4A^2} = \frac{1}{4} \]
Thus, \( K_c = 0.25 \).

### Conclusion
The equilibrium constant for the reaction is \( K_c = 0.25 \).

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