3 moles of A and 4 moles of B are mixed together and allowed to come into equilibrium according to the following reaction.
`3A(g) + 4B(g) rarr 2C(g) + 3D(g)`
When equilibrium is reached, there is 1 mole of C. The equilibrium constant of the reaction is

Options

A. `(1/4)^(4)`
B. `(1/3)^(3)`
C. `(1/2)^(4)`
D. 1

Explanation

To find the equilibrium constant \( K_c \) for the reaction \( 3A(g) + 4B(g) \rightleftharpoons 2C(g) + 3D(g) \), we will follow these steps:

### Step 1: Write the balanced chemical equation
The balanced reaction is:
\[ 3A(g) + 4B(g) \rightleftharpoons 2C(g) + 3D(g) \]

### Step 2: Set up the initial moles of reactants and products
Initially, we have:
- Moles of \( A = 3 \)
- Moles of \( B = 4 \)
- Moles of \( C = 0 \)
- Moles of \( D = 0 \)

### Step 3: Define the change in moles at equilibrium
Let \( x \) be the change in moles of \( C \) formed at equilibrium. According to the stoichiometry of the reaction:
- For every 2 moles of \( C \) formed, 3 moles of \( A \) and 4 moles of \( B \) are consumed.

Thus, at equilibrium:
- Moles of \( A = 3 - \frac{3}{2}x \)
- Moles of \( B = 4 - 2x \)
- Moles of \( C = 2x \)
- Moles of \( D = \frac{3}{2}x \)

### Step 4: Use the information given
We are given that at equilibrium, there is 1 mole of \( C \):
\[ 2x = 1 \]
Thus, solving for \( x \):
\[ x = \frac{1}{2} \]

### Step 5: Calculate the equilibrium moles
Substituting \( x = \frac{1}{2} \) into the expressions for moles at equilibrium:
- Moles of \( A = 3 - \frac{3}{2} \times \frac{1}{2} = 3 - \frac{3}{4} = \frac{9}{4} \)
- Moles of \( B = 4 - 2 \times \frac{1}{2} = 4 - 1 = 3 \)
- Moles of \( C = 2 \times \frac{1}{2} = 1 \)
- Moles of \( D = \frac{3}{2} \times \frac{1}{2} = \frac{3}{4} \)

### Step 6: Write the expression for the equilibrium constant \( K_c \)
The equilibrium constant \( K_c \) is given by:
\[ K_c = \frac{[C]^2[D]^3}{[A]^3[B]^4} \]

### Step 7: Substitute the equilibrium concentrations
Assuming the volume of the reaction vessel is 1 liter (for simplicity):
- \([A] = \frac{9}{4} \, \text{mol/L}\)
- \([B] = 3 \, \text{mol/L}\)
- \([C] = 1 \, \text{mol/L}\)
- \([D] = \frac{3}{4} \, \text{mol/L}\)

Now substituting these values into the expression for \( K_c \):
\[ K_c = \frac{(1)^2 \left(\frac{3}{4}\right)^3}{\left(\frac{9}{4}\right)^3 (3)^4} \]

### Step 8: Simplify the expression
Calculating the numerator:
\[ (1)^2 \left(\frac{3}{4}\right)^3 = \frac{27}{64} \]

Calculating the denominator:
\[ \left(\frac{9}{4}\right)^3 = \frac{729}{64} \quad \text{and} \quad (3)^4 = 81 \]
Thus,
\[ K_c = \frac{\frac{27}{64}}{\frac{729}{64} \times 81} = \frac{27}{729 \times 81} \]

### Step 9: Final calculation
Calculating \( 729 \times 81 = 59049 \):
\[ K_c = \frac{27}{59049} \]

### Step 10: Simplifying further
Recognizing that \( 59049 = 3^{10} \) and \( 27 = 3^3 \):
\[ K_c = \frac{3^3}{3^{10}} = \frac{1}{3^7} = \frac{1}{2187} \]

### Final Answer
Thus, the equilibrium constant \( K_c \) is:
\[ K_c = \frac{1}{2187} \]