In an equilibrium `A+B hArr C+D`, A and B are mixed in vessel at temperature T. The initial concentration of A was twice the initial concentration of B. After the equilibrium has reaches, concentration of C was thrice the equilibrium concentration of B. Calculate `K_(c)`.

Explanation

To solve the problem, we will follow these steps:

### Step 1: Define Initial Concentrations
Let the initial concentration of B be \( x \). According to the problem, the initial concentration of A is twice that of B, so:
- Initial concentration of A = \( 2x \)
- Initial concentration of B = \( x \)

### Step 2: Write the Change in Concentrations at Equilibrium
The reaction is:
\[ A + B \rightleftharpoons C + D \]

Let \( \alpha \) be the change in concentration of A and B at equilibrium. Therefore, at equilibrium:
- Concentration of A = \( 2x - \alpha \)
- Concentration of B = \( x - \alpha \)
- Concentration of C = \( \alpha \)
- Concentration of D = \( \alpha \)

### Step 3: Use Given Information About Concentrations
According to the problem, the concentration of C at equilibrium is three times the equilibrium concentration of B:
\[ \alpha = 3(x - \alpha) \]

### Step 4: Solve for \( \alpha \)
Rearranging the equation:
\[ \alpha = 3x - 3\alpha \]
\[ \alpha + 3\alpha = 3x \]
\[ 4\alpha = 3x \]
\[ \alpha = \frac{3x}{4} \]

### Step 5: Substitute \( \alpha \) Back to Find Equilibrium Concentrations
Now substituting \( \alpha \) back into the equilibrium concentrations:
- Concentration of A = \( 2x - \frac{3x}{4} = \frac{8x}{4} - \frac{3x}{4} = \frac{5x}{4} \)
- Concentration of B = \( x - \frac{3x}{4} = \frac{4x}{4} - \frac{3x}{4} = \frac{x}{4} \)
- Concentration of C = \( \alpha = \frac{3x}{4} \)
- Concentration of D = \( \alpha = \frac{3x}{4} \)

### Step 6: Write the Expression for \( K_c \)
The equilibrium constant \( K_c \) is given by:
\[ K_c = \frac{[C][D]}{[A][B]} \]

Substituting the equilibrium concentrations:
\[ K_c = \frac{\left(\frac{3x}{4}\right)\left(\frac{3x}{4}\right)}{\left(\frac{5x}{4}\right)\left(\frac{x}{4}\right)} \]

### Step 7: Simplify the Expression
Calculating the numerator:
\[ K_c = \frac{\frac{9x^2}{16}}{\frac{5x^2}{16}} \]
Canceling \( x^2 \) and simplifying:
\[ K_c = \frac{9}{5} \]

### Final Answer
Thus, the equilibrium constant \( K_c \) is:
\[ K_c = \frac{9}{5} \]

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