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

Answer

To solve the problem step by step, we need to find the volume of the flask in which the equilibrium mixture of the reaction \(3A(g) + B(g) \rightleftharpoons 2C(g)\) exists with given conditions.

### Step 1: Write the expression for the equilibrium constant \(K_c\)

The equilibrium constant \(K_c\) for the reaction is given by the formula:

\[ K_c = \frac{[C]^2}{[A]^3[B]} \]

Where:
- \([C]\) is the concentration of C,
- \([A]\) is the concentration of A,
- \([B]\) is the concentration of B.

### Step 2: Define the concentrations in terms of volume

Since we have 2 moles of each substance (A, B, and C) at equilibrium, we can express their concentrations as:

\[ [A] = \frac{n_A}{V} = \frac{2}{V} \]
\[ [B] = \frac{n_B}{V} = \frac{2}{V} \]
\[ [C] = \frac{n_C}{V} = \frac{2}{V} \]

Where \(n_A\), \(n_B\), and \(n_C\) are the number of moles of A, B, and C respectively, and \(V\) is the volume of the flask in liters.

### Step 3: Substitute the concentrations into the \(K_c\) expression

Substituting the expressions for concentrations into the \(K_c\) formula gives:

\[ K_c = \frac{\left(\frac{2}{V}\right)^2}{\left(\frac{2}{V}\right)^3 \left(\frac{2}{V}\right)} \]

### Step 4: Simplify the equation

Now, simplifying the equation:

\[ K_c = \frac{\frac{4}{V^2}}{\frac{8}{V^4}} = \frac{4}{V^2} \cdot \frac{V^4}{8} = \frac{4V^2}{8} = \frac{V^2}{2} \]

### Step 5: Set the expression equal to the given \(K_c\)

We know that \(K_c = 9.0\), so we set up the equation:

\[ \frac{V^2}{2} = 9 \]

### Step 6: Solve for \(V^2\)

Multiplying both sides by 2 gives:

\[ V^2 = 18 \]

### Step 7: Solve for \(V\)

Taking the square root of both sides:

\[ V = \sqrt{18} = 3\sqrt{2} \approx 4.24 \text{ L} \]

### Step 8: Final Calculation

Since the problem asks for the volume in liters, we round it appropriately. The volume of the flask is approximately:

\[ V \approx 4.24 \text{ L} \]

### Summary

The volume of the flask required for the equilibrium mixture is approximately **4.24 liters**.

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