In an equilibrium `A+B hArr C+D`, A and B are mixed in vesel 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 proceed step by step to find the equilibrium constant \( K_c \) for the reaction \( A + B \rightleftharpoons C + D \).

### Step 1: Define Initial Concentrations
Let the initial concentration of \( B \) be \( a \) M. According to the problem, the initial concentration of \( A \) is twice that of \( B \):
- Initial concentration of \( A = 2a \) M
- Initial concentration of \( B = a \) M
- Initial concentration of \( C = 0 \) M
- Initial concentration of \( D = 0 \) M

### Step 2: Set Up the Change in Concentrations
Let \( x \) be the change in concentration of \( C \) and \( D \) at equilibrium. Therefore, the changes for \( A \) and \( B \) will be:
- Change in concentration of \( A = -x \)
- Change in concentration of \( B = -x \)
- Change in concentration of \( C = +x \)
- Change in concentration of \( D = +x \)

### Step 3: Write Equilibrium Concentrations
At equilibrium, the concentrations will be:
- Concentration of \( A = 2a - x \)
- Concentration of \( B = a - x \)
- Concentration of \( C = x \)
- Concentration of \( D = x \)

### Step 4: Use Given Information
It is given that the concentration of \( C \) at equilibrium is three times the equilibrium concentration of \( B \):
\[ x = 3(a - x) \]
Solving for \( x \):
\[ x = 3a - 3x \\ x + 3x = 3a \\ 4x = 3a \\ x = \frac{3a}{4} \]

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

### Step 6: Write the Expression for \( K_c \)
The equilibrium constant \( K_c \) is given by the expression:
\[ K_c = \frac{[C][D]}{[A][B]} \]
Substituting the equilibrium concentrations:
\[ K_c = \frac{\left(\frac{3a}{4}\right)\left(\frac{3a}{4}\right)}{\left(\frac{5a}{4}\right)\left(\frac{a}{4}\right)} = \frac{\frac{9a^2}{16}}{\frac{5a^2}{16}} = \frac{9}{5} \]

### Step 7: Final Calculation
Calculating \( K_c \):
\[ K_c = \frac{9}{5} = 1.8 \]

### Conclusion
The final value of the equilibrium constant \( K_c \) is approximately 2 when rounded to the nearest integer.