A reaction, `A(g)+2B(g)hArr 2C(g)+D(g)` was studied using an initial concentraction of A and B were found to be equal. The value 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 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 \( [A]_0 = A \) and the initial concentration of \( B \) be \( [B]_0 = 1.5A \).

### Step 2: Set Up the Change in Concentrations
As the reaction proceeds to equilibrium, let \( x \) be the amount of \( A \) that reacts. Therefore, at equilibrium:
- The concentration of \( A \) will be \( [A] = A - x \)
- The concentration of \( B \) will be \( [B] = 1.5A - 2x \)
- The concentration of \( C \) will be \( [C] = 2x \)
- The concentration of \( D \) will be \( [D] = x \)

### Step 3: Apply the Equilibrium Condition
We are given that at equilibrium, the concentrations of \( A \) and \( B \) are equal:
\[ A - x = 1.5A - 2x \]

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

### Step 5: Substitute \( x \) Back to Find Concentrations
Now substituting \( x \) back into the equilibrium concentrations:
- \( [A] = A - 0.5A = 0.5A \)
- \( [B] = 1.5A - 2(0.5A) = 0.5A \)
- \( [C] = 2(0.5A) = A \)
- \( [D] = 0.5A \)

### Step 6: 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:
\[ K_c = \frac{(A)^2(0.5A)}{(0.5A)(0.5A)^2} \]
\[ K_c = \frac{A^2 \cdot 0.5A}{0.5A \cdot 0.25A^2} = \frac{0.5A^3}{0.125A^3} = 4 \]

### Step 7: Relate \( K_p \) to \( K_c \)
The relationship between \( K_p \) and \( K_c \) is given by:
\[ K_p = K_c (RT)^{\Delta n} \]
Where \( \Delta n \) is the change in moles of gas:
- Moles of products: \( 2 + 1 = 3 \)
- Moles of reactants: \( 1 + 2 = 3 \)
Thus, \( \Delta n = 3 - 3 = 0 \).

### Step 8: Calculate \( K_p \)
Since \( \Delta n = 0 \):
\[ K_p = K_c (RT)^0 = K_c \cdot 1 = 4 \]

### Final Answer
The value of \( K_p \) is \( 4 \).

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