In a gaseous reaction `A+2B iff 2C+D` the initial concentration of B was 1.5 times that of A. At equilibrium the concentration of A and D were equal. Calculate the equilibrium constant `K_(C)`.

Answer

To solve the problem step by step, let’s analyze the reaction and the information given:

**Reaction:**
\[ A + 2B \rightleftharpoons 2C + D \]

**Given Information:**
1. The initial concentration of B is 1.5 times that of A.
2. At equilibrium, the concentration of A and D are equal.

### Step 1: Define Initial Concentrations
Let the initial concentration of A be \( C \).
Then, the initial concentration of B will be \( 1.5C \).

### Step 2: Define Change in Concentrations at Equilibrium
Let \( x \) be the amount of A that reacts at equilibrium.
From the stoichiometry of the reaction:
- A decreases by \( x \): \( C - x \)
- B decreases by \( 2x \): \( 1.5C - 2x \)
- C increases by \( 2x \): \( 2x \)
- D increases by \( x \): \( x \)

### Step 3: Set Up the Equilibrium Condition
According to the problem, at equilibrium, the concentration of A and D are equal:
\[ C - x = x \]

### Step 4: Solve for \( x \)
From the equation \( C - x = x \):
\[ C = 2x \implies x = \frac{C}{2} \]

### Step 5: Substitute \( x \) Back into Concentrations
Now substitute \( x \) back into the equilibrium expressions:
- Concentration of A at equilibrium:
\[ C - x = C - \frac{C}{2} = \frac{C}{2} \]
- Concentration of B at equilibrium:
\[ 1.5C - 2x = 1.5C - 2\left(\frac{C}{2}\right) = 1.5C - C = 0.5C \]
- Concentration of C at equilibrium:
\[ 2x = 2\left(\frac{C}{2}\right) = C \]
- Concentration of D at equilibrium:
\[ x = \frac{C}{2} \]

### Step 6: Write the Expression for the Equilibrium Constant \( K_c \)
The equilibrium constant \( K_c \) is given by:
\[ K_c = \frac{[C]^2[D]}{[A][B]^2} \]

### Step 7: Substitute the Equilibrium Concentrations into the \( K_c \) Expression
Substituting the equilibrium concentrations:
\[ K_c = \frac{(C)^2\left(\frac{C}{2}\right)}{\left(\frac{C}{2}\right)\left(0.5C\right)^2} \]

### Step 8: Simplify the Expression
Calculating the numerator:
\[ (C)^2 \left(\frac{C}{2}\right) = \frac{C^3}{2} \]
Calculating the denominator:
\[ \left(\frac{C}{2}\right) \left(0.5C\right)^2 = \left(\frac{C}{2}\right) \left(\frac{C^2}{4}\right) = \frac{C^3}{8} \]

Now substituting back into the \( K_c \) expression:
\[ K_c = \frac{\frac{C^3}{2}}{\frac{C^3}{8}} = \frac{C^3}{2} \times \frac{8}{C^3} = 4 \]

### Final Answer
Thus, the equilibrium constant \( K_c \) is:
\[ \boxed{4} \]