For the reaction `a+bhArrc+d`, initially concentrations of a and b are equal and at equilibrium the concentration of will be twice of that of a. What will be equilibrium constant for the reaction ?

Options

A. 2
B. 9
C. 4
D. 3

Explanation

To solve the problem step by step, we will analyze the reaction and the given conditions to find the equilibrium constant (K).

### Step-by-Step Solution:

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

2. **Define Initial Concentrations**:
Let the initial concentrations of A and B be equal, say \( [A]_0 = [B]_0 = a \).
The initial concentrations of C and D are zero:
\[ [C]_0 = [D]_0 = 0 \]

3. **Define Changes at Equilibrium**:
Let \( x \) be the change in concentration of A and B that reacts to form C and D.
At equilibrium, the concentrations will be:
\[ [A] = a - x \]
\[ [B] = a - x \]
\[ [C] = x \]
\[ [D] = x \]

4. **Use the Given Condition**:
According to the problem, the concentration of D at equilibrium is twice that of A:
\[ [D] = 2[A] \]
Substituting the expressions we have:
\[ x = 2(a - x) \]

5. **Solve for x**:
Rearranging the equation:
\[ x = 2a - 2x \]
\[ 3x = 2a \]
\[ x = \frac{2}{3}a \]

6. **Substitute x Back into Concentrations**:
Now we can find the equilibrium concentrations:
\[ [A] = a - x = a - \frac{2}{3}a = \frac{1}{3}a \]
\[ [B] = a - x = \frac{1}{3}a \]
\[ [C] = x = \frac{2}{3}a \]
\[ [D] = x = \frac{2}{3}a \]

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

8. **Simplify the Expression**:
\[ K = \frac{\frac{4}{9}a^2}{\frac{1}{9}a^2} \]
The \( a^2 \) terms cancel out:
\[ K = \frac{4}{9} \div \frac{1}{9} = 4 \]

9. **Final Answer**:
The equilibrium constant \( K \) for the reaction is:
\[ K = 4 \]