A reaction, `A(g)+2B(g)hArr 2C(g)+D(g)` was studied using an initial concentraction of B which was 1.5 times that of A. But the equilibrium concentrations of A and B were found to be equal. The vlue of `K_(p)` for the equilibrium is

Options

A. 4
B. 6
C. 8
D. 12

Answer

To find the value of \( K_p \) for the equilibrium reaction \( A(g) + 2B(g) \rightleftharpoons 2C(g) + D(g) \), we will follow these steps:

### Step 1: Define Initial Concentrations
Let the initial concentration of \( A \) be \( a \). Since the initial concentration of \( B \) is 1.5 times that of \( A \), we can express it as:
\[ [B]_{initial} = 1.5a \]

### Step 2: Set Up Change in Concentrations
At equilibrium, let \( x \) be the amount of \( A \) that reacts. The changes in concentrations will be:
- For \( A \): \( [A] = a - x \)
- For \( B \): \( [B] = 1.5a - 2x \)
- For \( C \): \( [C] = 2x \)
- For \( D \): \( [D] = x \)

### Step 3: Use Given Information
We are told that at equilibrium, the concentrations of \( A \) and \( B \) are equal:
\[ a - x = 1.5a - 2x \]

### Step 4: Solve for \( x \)
Rearranging the equation:
\[ a - x = 1.5a - 2x \implies 2x - x = 1.5a - a \implies x = 0.5a \]

### Step 5: Substitute \( x \) Back
Now we can substitute \( x \) back into the expressions for \( [A] \) and \( [B] \):
- \( [A] = a - 0.5a = 0.5a \)
- \( [B] = 1.5a - 2(0.5a) = 1.5a - a = 0.5a \)

### Step 6: Calculate Equilibrium Concentrations
At equilibrium:
- \( [A] = 0.5a \)
- \( [B] = 0.5a \)
- \( [C] = 2(0.5a) = a \)
- \( [D] = 0.5a \)

### Step 7: Write the Expression for \( K_c \)
The equilibrium constant \( K_c \) is given by:
\[ K_c = \frac{[C]^2[D]}{[A][B]^2} \]
Substituting the equilibrium concentrations:
\[ K_c = \frac{(a)^2(0.5a)}{(0.5a)(0.5a)^2} \]

### Step 8: Simplify the Expression
\[ K_c = \frac{a^2 \cdot 0.5a}{0.5a \cdot (0.25a^2)} = \frac{0.5a^3}{0.125a^3} = 4 \]

### Step 9: Relate \( K_c \) and \( K_p \)
Since the change in the number of moles (\( \Delta n \)) is 0 (3 moles of products and 3 moles of reactants), we have:
\[ K_p = K_c \]
Thus,
\[ K_p = 4 \]

### Final Answer
The value of \( K_p \) for the equilibrium is:
\[ \boxed{4} \]