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 find the mole fraction of C at equilibrium for the reaction \( A + B \rightleftharpoons 2C \) with an equilibrium constant \( K_c = 81 \), we can follow these steps:

### Step 1: Set up the initial conditions
We start with 1 mole of A and 1 mole of B. Therefore, the initial moles are:
- \( [A]_0 = 1 \)
- \( [B]_0 = 1 \)
- \( [C]_0 = 0 \)

### Step 2: Define the change in moles
Let \( x \) be the amount of A and B that reacts. At equilibrium, the moles will be:
- \( [A] = 1 - x \)
- \( [B] = 1 - x \)
- \( [C] = 2x \)

### Step 3: Write the expression for the equilibrium constant
The equilibrium constant \( K_c \) is given by:
\[ K_c = \frac{[C]^2}{[A][B]} \]
Substituting the equilibrium concentrations:
\[ K_c = \frac{(2x)^2}{(1-x)(1-x)} = \frac{4x^2}{(1-x)^2} \]
Given that \( K_c = 81 \), we can set up the equation:
\[ \frac{4x^2}{(1-x)^2} = 81 \]

### Step 4: Solve for \( x \)
Cross-multiplying gives:
\[ 4x^2 = 81(1 - x)^2 \]
Expanding the right side:
\[ 4x^2 = 81(1 - 2x + x^2) = 81 - 162x + 81x^2 \]
Rearranging the equation:
\[ 0 = 81x^2 - 162x + 81 - 4x^2 \]
\[ 0 = 77x^2 - 162x + 81 \]

### Step 5: Use the quadratic formula
We can apply the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 77 \), \( b = -162 \), and \( c = 81 \).
\[ x = \frac{162 \pm \sqrt{(-162)^2 - 4 \cdot 77 \cdot 81}}{2 \cdot 77} \]
Calculating the discriminant:
\[ (-162)^2 = 26244 \]
\[ 4 \cdot 77 \cdot 81 = 24924 \]
\[ \sqrt{26244 - 24924} = \sqrt{320} = 17.888 \]
Now substituting back into the formula:
\[ x = \frac{162 \pm 17.888}{154} \]
Calculating the two possible values for \( x \):
1. \( x_1 = \frac{179.888}{154} \approx 1.167 \) (not possible since \( x \) cannot exceed 1)
2. \( x_2 = \frac{144.112}{154} \approx 0.935 \)

### Step 6: Calculate the mole fraction of C
At equilibrium, the moles of C are:
\[ [C] = 2x = 2 \cdot 0.935 = 1.87 \]
Total moles at equilibrium:
\[ \text{Total moles} = (1 - x) + (1 - x) + 2x = 2 - 2x + 2x = 2 \]
Thus, the mole fraction of C is:
\[ \text{Mole fraction of C} = \frac{1.87}{2} = 0.935 \]

### Final Answer
The mole fraction of C at equilibrium is approximately \( \frac{9}{11} \).