In a chemical equilibrium `A + B hArr` C + D, when one mole each of the two reactants are mixed, `0.6` mole each of the products are formed. The equilibrium constant calculated is

Options

A. 1
B. `0.36`
C. `2.25`
D. `4//9`

Explanation

To calculate the equilibrium constant (K) for the reaction \( A + B \rightleftharpoons C + D \), we can follow these steps:

### Step-by-Step Solution:

1. **Initial Moles of Reactants**:
- We start with 1 mole of A and 1 mole of B.
- Initial concentrations:
\[ [A] = 1 \text{ mole}, \quad [B] = 1 \text{ mole} \]

2. **Moles of Products at Equilibrium**:
- At equilibrium, we are given that 0.6 moles of C and 0.6 moles of D are formed.
- Therefore, the equilibrium concentrations of the products are:
\[ [C] = 0.6 \text{ moles}, \quad [D] = 0.6 \text{ moles} \]

3. **Change in Moles of Reactants**:
- Since 0.6 moles of products are formed, the change in moles of A and B can be calculated as:
\[ \text{Change in } A = 1 - 0.6 = 0.4 \text{ moles} \]
\[ \text{Change in } B = 1 - 0.6 = 0.4 \text{ moles} \]

4. **Equilibrium Concentrations of Reactants**:
- The equilibrium concentrations of A and B will be:
\[ [A] = 1 - 0.6 = 0.4 \text{ moles} \]
\[ [B] = 1 - 0.6 = 0.4 \text{ moles} \]

5. **Expression for Equilibrium Constant (K)**:
- The equilibrium constant \( K \) is defined as:
\[ K = \frac{[C][D]}{[A][B]} \]

6. **Substituting the Values into the K Expression**:
- Now substituting the equilibrium concentrations into the expression:
\[ K = \frac{(0.6)(0.6)}{(0.4)(0.4)} \]

7. **Calculating K**:
- Performing the multiplication:
\[ K = \frac{0.36}{0.16} = 2.25 \]

### Final Answer:
The equilibrium constant \( K \) for the reaction is **2.25**.

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