for the reaction `2A_((g)) hArr B_((g)) + 3C_((g))`, at a given temperature, `K_(c)` = 16. What must be the volume of the falsk, If a mixture of 2 mole cach A, B and C exist in equilibrium ?

Options

A. `(1)/(4)`
B. `(1)/(2)`
C. 1
D. 2

Answer

To solve the problem step-by-step, we will analyze the given equilibrium reaction and use the equilibrium constant expression to find the required volume of the flask.

### Step 1: Write the balanced chemical equation
The reaction is given as:
\[ 2A_{(g)} \rightleftharpoons B_{(g)} + 3C_{(g)} \]

### Step 2: Identify the equilibrium concentrations
According to the problem, we have:
- 2 moles of A
- 2 moles of B
- 2 moles of C

Let the volume of the flask be \( V \) liters. The molar concentrations of A, B, and C can be expressed as:
- Concentration of A, \([A] = \frac{2}{V}\)
- Concentration of B, \([B] = \frac{2}{V}\)
- Concentration of C, \([C] = \frac{2}{V}\)

### Step 3: Write the expression for the equilibrium constant \( K_c \)
The equilibrium constant \( K_c \) for the reaction is given by:
\[ K_c = \frac{[B][C]^3}{[A]^2} \]

Substituting the concentrations into the expression:
\[ K_c = \frac{\left(\frac{2}{V}\right) \left(\frac{2}{V}\right)^3}{\left(\frac{2}{V}\right)^2} \]

### Step 4: Simplify the expression
Calculating the numerator:
\[ [B] = \frac{2}{V}, \quad [C]^3 = \left(\frac{2}{V}\right)^3 = \frac{8}{V^3} \]
Thus, the numerator becomes:
\[ [B][C]^3 = \frac{2}{V} \cdot \frac{8}{V^3} = \frac{16}{V^4} \]

Calculating the denominator:
\[ [A]^2 = \left(\frac{2}{V}\right)^2 = \frac{4}{V^2} \]

Now substituting back into the \( K_c \) expression:
\[ K_c = \frac{\frac{16}{V^4}}{\frac{4}{V^2}} = \frac{16}{V^4} \cdot \frac{V^2}{4} = \frac{16V^2}{4V^4} = \frac{4}{V^2} \]

### Step 5: Set the expression equal to the given \( K_c \)
We are given that \( K_c = 16 \):
\[ \frac{4}{V^2} = 16 \]

### Step 6: Solve for \( V^2 \)
Cross-multiplying gives:
\[ 4 = 16V^2 \]
Dividing both sides by 16:
\[ V^2 = \frac{4}{16} = \frac{1}{4} \]

### Step 7: Take the square root to find \( V \)
Taking the square root of both sides:
\[ V = \sqrt{\frac{1}{4}} = \frac{1}{2} \text{ liters} \]

### Conclusion
The volume of the flask must be \( \frac{1}{2} \) liters.

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