consider the given reaction, 3A(g) + B(g) hArr 2C(g) at a given temperature if a mixture of 2 mol each of A, B and C exist at equilibrium and `K_c =9` then volume of the flask will be

Options

A. 3L
B. 6L
C. 9L
D. 36L

Answer

To solve the problem, we need to determine the volume of the flask given the equilibrium concentrations and the equilibrium constant \( K_c \).

### Step-by-Step Solution:

1. **Write the balanced chemical equation**:
\[ 3A(g) + B(g) \rightleftharpoons 2C(g) \]

2. **Identify the equilibrium conditions**:
We are given that at equilibrium, there are 2 moles of A, 2 moles of B, and 2 moles of C.

3. **Define the volume of the flask**:
Let the volume of the flask be \( V \) liters.

4. **Calculate the equilibrium concentrations**:
The equilibrium concentration of a substance is given by the formula:
\[ \text{Concentration} = \frac{\text{Number of moles}}{\text{Volume}} \]
Therefore, the equilibrium concentrations are:
- For A:
\[ [A] = \frac{2}{V} \]
- For B:
\[ [B] = \frac{2}{V} \]
- For C:
\[ [C] = \frac{2}{V} \]

5. **Write the expression for the equilibrium constant \( K_c \)**:
The equilibrium constant \( K_c \) is defined as:
\[ K_c = \frac{[\text{Products}]^{\text{stoichiometric coefficient}}}{[\text{Reactants}]^{\text{stoichiometric coefficient}}} \]
For the given reaction:
\[ K_c = \frac{[C]^2}{[A]^3[B]} \]
Substituting the equilibrium concentrations:
\[ K_c = \frac{\left(\frac{2}{V}\right)^2}{\left(\frac{2}{V}\right)^3 \left(\frac{2}{V}\right)} \]

6. **Simplify the expression**:
\[ K_c = \frac{\frac{4}{V^2}}{\frac{8}{V^4}} = \frac{4}{V^2} \cdot \frac{V^4}{8} = \frac{4V^2}{8} = \frac{V^2}{2} \]

7. **Set the expression equal to the given \( K_c \)**:
We know from the problem that \( K_c = 9 \):
\[ \frac{V^2}{2} = 9 \]

8. **Solve for \( V^2 \)**:
Multiply both sides by 2:
\[ V^2 = 18 \]

9. **Take the square root to find \( V \)**:
\[ V = \sqrt{18} = 3\sqrt{2} \approx 4.24 \text{ liters} \]

10. **Final answer**:
The volume of the flask is approximately \( 4.24 \) liters.