`A+2 B hArr 2C"
2 mole each A and B present in 10 litre so that C form is 1 mole, Calculate `K_(c)`

Options

A. `1.5`
B. `6.67`
C. `0.15`
D. `2.3`

Explanation

To calculate the equilibrium constant \( K_c \) for the reaction \( A + 2B \rightleftharpoons 2C \), we will follow these steps:

### Step 1: Write the balanced equation
The balanced equation is given as:
\[ A + 2B \rightleftharpoons 2C \]

### Step 2: Identify initial moles and changes
Initially, we have:
- 2 moles of \( A \)
- 2 moles of \( B \)
- 0 moles of \( C \)

At equilibrium, it is given that 1 mole of \( C \) is formed.

### Step 3: Determine the change in moles
Let \( x \) be the change in moles of \( C \) formed. Since 2 moles of \( C \) are produced from 1 mole of \( A \) and 2 moles of \( B \), we can say:
\[ 2x = 1 \]
Thus, \( x = 0.5 \).

### Step 4: Calculate moles at equilibrium
Now, we can calculate the moles of \( A \) and \( B \) at equilibrium:
- Moles of \( C \) at equilibrium = \( 1 \) mole (given)
- Moles of \( A \) at equilibrium = \( 2 - 0.5 = 1.5 \) moles
- Moles of \( B \) at equilibrium = \( 2 - 2(0.5) = 1 \) mole

### Step 5: Calculate concentrations
We need to convert these moles into concentrations. The volume of the solution is given as 10 liters.
- Concentration of \( A \) = \( \frac{1.5 \text{ moles}}{10 \text{ L}} = 0.15 \, \text{M} \)
- Concentration of \( B \) = \( \frac{1 \text{ mole}}{10 \text{ L}} = 0.1 \, \text{M} \)
- Concentration of \( C \) = \( \frac{1 \text{ mole}}{10 \text{ L}} = 0.1 \, \text{M} \)

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

### Step 7: Substitute the concentrations into the expression
Now, substituting the concentrations we calculated:
\[ K_c = \frac{(0.1)^2}{(0.15)(0.1)^2} \]

### Step 8: Simplify the expression
\[ K_c = \frac{0.01}{0.15 \times 0.01} = \frac{0.01}{0.0015} \]

### Step 9: Calculate \( K_c \)
\[ K_c = \frac{0.01}{0.0015} = \frac{10}{1.5} \approx 6.67 \]

### Final Answer
Thus, the equilibrium constant \( K_c \) is approximately \( 6.67 \).

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