Two moles of each reactant A and B are taken in a reaction flask. They react in the following manner,
`A(g)+B(g)hArr C(g)+D(g)`
At equilibrium, it was found that the concentration of C is triple to that of B the equilibrium constant for the reaction is

Options

A. 4.5
B. 6
C. 9
D. `1//6`

Answer

To solve the problem, we need to determine the equilibrium constant \( K_c \) for the reaction \( A(g) + B(g) \rightleftharpoons C(g) + D(g) \) given the initial conditions and the relationship between the concentrations of the reactants and products at equilibrium.

### Step-by-Step Solution:

1. **Identify Initial Moles**:
- We start with 2 moles of \( A \) and 2 moles of \( B \).
- Therefore, the initial moles are:
- \( [A]_0 = 2 \)
- \( [B]_0 = 2 \)
- \( [C]_0 = 0 \)
- \( [D]_0 = 0 \)

2. **Define Change in Moles**:
- Let \( x \) be the number of moles of \( A \) and \( B \) that react to form products \( C \) and \( D \).
- At equilibrium, the moles will be:
- \( [A] = 2 - x \)
- \( [B] = 2 - x \)
- \( [C] = x \)
- \( [D] = x \)

3. **Use the Given Relationship**:
- We know that the concentration of \( C \) is triple that of \( B \):
\[ [C] = 3[B] \]
- Substituting the expressions for \( [C] \) and \( [B] \):
\[ x = 3(2 - x) \]

4. **Solve for \( x \)**:
- Expanding the equation:
\[ x = 6 - 3x \]
- Rearranging gives:
\[ 4x = 6 \implies x = \frac{6}{4} = 1.5 \]

5. **Determine Equilibrium Concentrations**:
- Now substituting \( x \) back into the expressions for the equilibrium concentrations:
- \( [A] = 2 - 1.5 = 0.5 \)
- \( [B] = 2 - 1.5 = 0.5 \)
- \( [C] = 1.5 \)
- \( [D] = 1.5 \)

6. **Calculate the Equilibrium Constant \( K_c \)**:
- The equilibrium constant \( K_c \) is given by:
\[ K_c = \frac{[C][D]}{[A][B]} \]
- Substituting the equilibrium concentrations:
\[ K_c = \frac{(1.5)(1.5)}{(0.5)(0.5)} = \frac{2.25}{0.25} = 9 \]

### Final Answer:
The equilibrium constant \( K_c \) for the reaction is **9**.