For the reaction `A+B hArr C+D`, the initial concentrations of A and B are equal. The equilibrium concentration of C is two times the equilibrium concentration of A. The value of equilibrium constant is ………..

Options

A. `1`
B. `2`
C. `3`
D. `4`

Explanation

To solve the problem, we need to analyze the given chemical reaction and the information provided about the concentrations at equilibrium.

### Step-by-Step Solution:

1. **Identify the Reaction**:
The reaction is given as:
\[ A + B \rightleftharpoons C + D \]

2. **Initial Concentrations**:
Let the initial concentrations of A and B be \( [A]_0 = [B]_0 = 1 \, \text{mol/L} \) (since they are equal).

3. **Equilibrium Concentrations**:
- Let \( x \) be the amount of A and B that reacts at equilibrium.
- Therefore, at equilibrium:
- Concentration of A: \( [A] = 1 - x \)
- Concentration of B: \( [B] = 1 - x \)
- Concentration of C: \( [C] = x \)
- Concentration of D: \( [D] = x \)

4. **Given Condition**:
The problem states that the equilibrium concentration of C is two times the equilibrium concentration of A:
\[ [C] = 2[A] \]
Substituting the expressions we have:
\[ x = 2(1 - x) \]

5. **Solve for \( x \)**:
Rearranging the equation:
\[ x = 2 - 2x \]
Adding \( 2x \) to both sides:
\[ 3x = 2 \]
Dividing by 3:
\[ x = \frac{2}{3} \]

6. **Calculate Equilibrium Concentrations**:
Now we can find the equilibrium concentrations:
- \( [C] = x = \frac{2}{3} \)
- \( [A] = 1 - x = 1 - \frac{2}{3} = \frac{1}{3} \)
- \( [B] = 1 - x = \frac{1}{3} \)
- \( [D] = x = \frac{2}{3} \)

7. **Equilibrium Constant Expression**:
The equilibrium constant \( K_c \) is given by:
\[ K_c = \frac{[C][D]}{[A][B]} \]
Substituting the equilibrium concentrations:
\[ K_c = \frac{\left(\frac{2}{3}\right)\left(\frac{2}{3}\right)}{\left(\frac{1}{3}\right)\left(\frac{1}{3}\right)} \]

8. **Calculate \( K_c \)**:
Simplifying the expression:
\[ K_c = \frac{\frac{4}{9}}{\frac{1}{9}} = 4 \]

### Final Answer:
The value of the equilibrium constant \( K_c \) is **4**.

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