At a definite temperature, the equilibrium constant for a reaction, `A+BhArr2C`, was found to be 81. Starting with 1 mole A and 1 mole B, the mole fraction of C at equilibrium is :

Options

A. `(9)/(11)`
B. `(1)/(11)`
C. `(2)/(11)`
D. `(7)/(11)`

Answer

To solve the problem, we will follow these steps:

### Step 1: Write the balanced chemical equation and the expression for the equilibrium constant (K).
The reaction is:
\[ A + B \rightleftharpoons 2C \]

The equilibrium constant \( K \) is given by:
\[ K = \frac{[C]^2}{[A][B]} \]

### Step 2: Set up the initial concentrations and changes at equilibrium.
Initially, we have:
- Moles of \( A = 1 \)
- Moles of \( B = 1 \)
- Moles of \( C = 0 \)

Let \( x \) be the change in moles of \( A \) and \( B \) that react to form \( C \). At equilibrium, we have:
- Moles of \( A = 1 - x \)
- Moles of \( B = 1 - x \)
- Moles of \( C = 2x \)

### Step 3: Substitute the equilibrium concentrations into the expression for \( K \).
The equilibrium concentrations can be expressed as:
\[ K = \frac{(2x)^2}{(1 - x)(1 - x)} \]

Given that \( K = 81 \), we can write:
\[ 81 = \frac{4x^2}{(1 - x)^2} \]

### Step 4: Cross-multiply and simplify the equation.
Cross-multiplying gives:
\[ 81(1 - x)^2 = 4x^2 \]

Expanding this:
\[ 81(1 - 2x + x^2) = 4x^2 \]
\[ 81 - 162x + 81x^2 = 4x^2 \]

Rearranging the equation:
\[ 81x^2 - 162x + 81 = 0 \]

### Step 5: Solve the quadratic equation.
Dividing the entire equation by 81:
\[ x^2 - 2x + 1 = 0 \]
This factors to:
\[ (x - 1)^2 = 0 \]

Thus, \( x = 1 \).

### Step 6: Calculate the moles of \( C \) at equilibrium.
At equilibrium:
- Moles of \( C = 2x = 2(1) = 2 \)

### Step 7: Calculate the total moles at equilibrium.
Total moles at equilibrium:
\[ \text{Total moles} = (1 - x) + (1 - x) + 2x = (1 - 1) + (1 - 1) + 2(1) = 2 \]

### Step 8: Calculate the mole fraction of \( C \).
The mole fraction of \( C \) is given by:
\[ \text{Mole fraction of } C = \frac{\text{moles of } C}{\text{total moles}} = \frac{2}{2} = 1 \]

### Final Answer:
The mole fraction of \( C \) at equilibrium is \( 1 \).

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