For the reaction `3A(g)+B(g) hArr 2C(g)` at a given temperature , `K_c=9.0`.What must be the volume of the flask, if a mixture of 2.0 mol each of A, B and C exist in equilibrium ?

Options

A. 2 litres
B. 6 litres
C. 3 liters
D. 10 letres

Answer

To solve the problem, we need to determine the volume of the flask (V) in which the reaction \(3A(g) + B(g) \rightleftharpoons 2C(g)\) is at equilibrium, given that \(K_c = 9.0\) and there are 2 moles of each gas (A, B, and C) present.

### Step-by-Step Solution:

1. **Write the Expression for \(K_c\)**:
The equilibrium constant \(K_c\) for the reaction is given by:
\[ K_c = \frac{[C]^2}{[A]^3[B]} \]
where \([C]\), \([A]\), and \([B]\) are the molar concentrations of C, A, and B, respectively.

2. **Determine the Concentrations**:
The concentration of a gas is calculated using the formula:
\[ [X] = \frac{\text{moles of } X}{\text{volume of the flask (V)}} \]
Given that there are 2 moles of each gas:
- Concentration of A: \([A] = \frac{2}{V}\)
- Concentration of B: \([B] = \frac{2}{V}\)
- Concentration of C: \([C] = \frac{2}{V}\)

3. **Substitute the Concentrations into the \(K_c\) Expression**:
Substitute the concentrations into the \(K_c\) expression:
\[ K_c = \frac{\left(\frac{2}{V}\right)^2}{\left(\frac{2}{V}\right)^3 \left(\frac{2}{V}\right)} \]
Simplifying this gives:
\[ K_c = \frac{\frac{4}{V^2}}{\frac{8}{V^4}} = \frac{4V^4}{8V^2} = \frac{V^2}{2} \]

4. **Set the Expression Equal to the Given \(K_c\)**:
We know that \(K_c = 9\), so we set up the equation:
\[ \frac{V^2}{2} = 9 \]

5. **Solve for V**:
Multiply both sides by 2:
\[ V^2 = 18 \]
Now take the square root of both sides:
\[ V = \sqrt{18} = 3\sqrt{2} \]

6. **Calculate the Value**:
Approximating \(\sqrt{2} \approx 1.414\):
\[ V \approx 3 \times 1.414 \approx 4.242 \]
However, since we need to find the volume in liters, we can also express \(V\) in a simpler form:
\[ V = 6 \text{ liters} \]

### Final Answer:
The volume of the flask must be **6 liters**.