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. 6 L
B. 9 L
C. 36 L
D. None of these

Answer

To solve the problem, we need to determine the volume of the flask (denoted as \( V \)) in which the equilibrium concentrations of the gases A, B, and C are established. The reaction is given as:

\[ 3A(g) + B(g) \rightleftharpoons 2C(g) \]

We know that at equilibrium, there are 2.0 moles of each gas A, B, and C. The equilibrium constant \( K_c \) is given as 9.0.

### Step-by-Step Solution:

1. **Define the Equilibrium Concentrations:**
- Let the volume of the flask be \( V \) liters.
- The equilibrium concentrations can be expressed as:
- \([A] = \frac{2.0 \, \text{mol}}{V}\)
- \([B] = \frac{2.0 \, \text{mol}}{V}\)
- \([C] = \frac{2.0 \, \text{mol}}{V}\)

2. **Write the Expression for \( K_c \):**
- The expression for the equilibrium constant \( K_c \) for the reaction is:
\[ K_c = \frac{[C]^2}{[A]^3[B]} \]
- Substituting the equilibrium concentrations into the expression:
\[ K_c = \frac{\left(\frac{2.0}{V}\right)^2}{\left(\frac{2.0}{V}\right)^3 \left(\frac{2.0}{V}\right)} \]

3. **Simplify the Expression:**
- This simplifies to:
\[ K_c = \frac{\frac{4.0}{V^2}}{\frac{8.0}{V^4}} = \frac{4.0 \cdot V^4}{8.0 \cdot V^2} = \frac{4V^2}{8} = \frac{V^2}{2} \]

4. **Set Up the Equation with Given \( K_c \):**
- We know \( K_c = 9.0 \), so we set up the equation:
\[ \frac{V^2}{2} = 9 \]

5. **Solve for \( V^2 \):**
- Multiplying both sides by 2 gives:
\[ V^2 = 18 \]

6. **Calculate \( V \):**
- Taking the square root of both sides:
\[ V = \sqrt{18} = 6 \, \text{liters} \]

### Conclusion:
The volume of the flask is \( 6 \, \text{liters} \).