In a vessel of 5L,26 moles of A and 4 moles of B were placed. At equilibrium 1 mole of C was present. The `K_(C)` for the reaction :
`A+2BhArrC` is

Options

A. 0.25
B. 0. 50
C. 2.5
D. 4.8

Answer

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

### Step 1: Write the Initial and Equilibrium Concentrations
- **Initial Moles:**
- Moles of A = 26
- Moles of B = 4
- Moles of C = 0 (since it is not present initially)

- **At Equilibrium:**
- Moles of C = 1 (given)

Using the stoichiometry of the reaction, we can find the changes in the moles of A and B:
- For every 1 mole of C formed, 1 mole of A is consumed and 2 moles of B are consumed.

Thus:
- Moles of A at equilibrium = \( 26 - 1 = 25 \)
- Moles of B at equilibrium = \( 4 - 2(1) = 2 \)

### Step 2: Calculate the Concentrations
The concentration \( C \) is calculated using the formula:
\[ C = \frac{\text{Number of moles}}{\text{Volume in liters}} \]
Given that the volume of the vessel is 5 L, we can calculate the concentrations:

- Concentration of A:
\[ [C_A] = \frac{25 \text{ moles}}{5 \text{ L}} = 5 \text{ M} \]

- Concentration of B:
\[ [C_B] = \frac{2 \text{ moles}}{5 \text{ L}} = 0.4 \text{ M} \]

- Concentration of C:
\[ [C_C] = \frac{1 \text{ mole}}{5 \text{ L}} = 0.2 \text{ M} \]

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

### Step 4: Substitute the Concentrations into the \( K_c \) Expression
Substituting the values we calculated:
\[ K_c = \frac{[C]}{[A][B]^2} = \frac{0.2}{(5)(0.4)^2} \]

### Step 5: Calculate \( K_c \)
Calculating \( K_c \):
\[ K_c = \frac{0.2}{5 \times 0.16} = \frac{0.2}{0.8} = 0.25 \]

### Final Answer
The value of \( K_c \) for the reaction is \( 0.25 \).

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