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 solve the problem, we will follow these steps:

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

### Step 2: Set up the initial concentrations
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 A and B that react. According to the stoichiometry of the reaction:
- For every 3 moles of A that react, 4 moles of B react, producing 2 moles of C and 3 moles of D.

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

### Step 4: Use the information given to find \( x \)
We are given that at equilibrium, there is 1 mole of C:
\[ 2 \times \frac{2}{3}x = 1 \]
This simplifies to:
\[ \frac{4}{3}x = 1 \]
\[ x = \frac{3}{4} \]

### Step 5: Calculate the equilibrium moles of A, B, C, and D
Now we can find the moles of each substance at equilibrium:
- Moles of A = \( 3 - \frac{3}{4} = \frac{9}{4} \)
- Moles of B = \( 4 - \frac{4}{3} \times \frac{3}{4} = 4 - 1 = 3 \)
- Moles of C = 1 (given)
- Moles of D = \( 3 \times \frac{3}{4} = \frac{9}{4} \)

### Step 6: Write the expression for the equilibrium constant \( K_c \)
The equilibrium constant \( K_c \) is given by:
\[ K_c = \frac{[C]^2[D]^3}{[A]^3[B]^4} \]
Substituting the equilibrium concentrations (assuming a volume of 1 L for simplicity):
\[ K_c = \frac{(1)^2 \left(\frac{9}{4}\right)^3}{\left(\frac{9}{4}\right)^3 \cdot (3)^4} \]

### Step 7: Calculate \( K_c \)
Calculating the values:
\[ K_c = \frac{1 \cdot \left(\frac{9}{4}\right)^3}{\left(\frac{9}{4}\right)^3 \cdot 81} \]
This simplifies to:
\[ K_c = \frac{1}{81} \]

### Final Answer
The equilibrium constant \( K_c \) is:
\[ K_c = \frac{1}{81} \]

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