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 for the reaction \( A + B \rightleftharpoons C + D \) given that one third of A and B are consumed at equilibrium, we can follow these steps:

### Step 1: Define Initial Concentrations
Assume we start with 1 mole of A and 1 mole of B. Since they are mixed in equimolar proportions, we have:
- \([A]_0 = 1 \, \text{mol}\)
- \([B]_0 = 1 \, \text{mol}\)
- \([C]_0 = 0 \, \text{mol}\)
- \([D]_0 = 0 \, \text{mol}\)

### Step 2: Determine Changes at Equilibrium
At equilibrium, it is given that one third of A and B are consumed. Therefore:
- Amount of A consumed = \(\frac{1}{3} \, \text{mol}\)
- Amount of B consumed = \(\frac{1}{3} \, \text{mol}\)

### Step 3: Calculate Equilibrium Concentrations
Now, we can calculate the equilibrium concentrations:
- \([A]_{eq} = [A]_0 - \text{amount consumed} = 1 - \frac{1}{3} = \frac{2}{3} \, \text{mol}\)
- \([B]_{eq} = [B]_0 - \text{amount consumed} = 1 - \frac{1}{3} = \frac{2}{3} \, \text{mol}\)
- \([C]_{eq} = [C]_0 + \text{amount formed} = 0 + \frac{1}{3} = \frac{1}{3} \, \text{mol}\)
- \([D]_{eq} = [D]_0 + \text{amount formed} = 0 + \frac{1}{3} = \frac{1}{3} \, \text{mol}\)

### Step 4: Write the Expression for the Equilibrium Constant
The equilibrium constant \( K \) for the reaction is defined as:
\[ K = \frac{[C]_{eq} \cdot [D]_{eq}}{[A]_{eq} \cdot [B]_{eq}} \]

### Step 5: Substitute the Equilibrium Concentrations into the Expression
Substituting the equilibrium concentrations we calculated:
\[ K = \frac{\left(\frac{1}{3}\right) \cdot \left(\frac{1}{3}\right)}{\left(\frac{2}{3}\right) \cdot \left(\frac{2}{3}\right)} = \frac{\frac{1}{9}}{\frac{4}{9}} = \frac{1}{4} \]

### Step 6: Final Result
Thus, the equilibrium constant \( K \) is:
\[ K = 0.25 \]