In the decomposition reaction `AB_(5)(g)hArrAB_(3)(g)+B_(2)(g)`, at equilibrium in a 10 litre closed vessel at `227^(@)C`, 2 moles of `AB_(3)`, 5 moles of `B_(2)` and 4 moles of `AB_(5)`, are present. The equilibrium contstant `K_(c)` for the formation of `AB_(5)(g)` is

Options

A. 0.25
B. `4.0`
C. 0.04
D. 2.5

Explanation

To find the equilibrium constant \( K_c \) for the formation of \( AB_5(g) \) from the reaction:

\[ AB_5(g) \rightleftharpoons AB_3(g) + B_2(g) \]

we start by writing the expression for the equilibrium constant \( K_c \):

\[ K_c = \frac{[AB_3][B_2]}{[AB_5]} \]

### Step 1: Calculate the concentrations of the species at equilibrium.

Given:
- Volume of the vessel = 10 L
- Moles of \( AB_3 \) = 2 moles
- Moles of \( B_2 \) = 5 moles
- Moles of \( AB_5 \) = 4 moles

The concentration of each species can be calculated using the formula:

\[ \text{Concentration} = \frac{\text{Number of moles}}{\text{Volume in liters}} \]

Calculating the concentrations:

- Concentration of \( AB_3 \):
\[ [AB_3] = \frac{2 \text{ moles}}{10 \text{ L}} = 0.2 \text{ M} \]

- Concentration of \( B_2 \):
\[ [B_2] = \frac{5 \text{ moles}}{10 \text{ L}} = 0.5 \text{ M} \]

- Concentration of \( AB_5 \):
\[ [AB_5] = \frac{4 \text{ moles}}{10 \text{ L}} = 0.4 \text{ M} \]

### Step 2: Substitute the concentrations into the \( K_c \) expression.

Now we can substitute the concentrations into the \( K_c \) expression:

\[ K_c = \frac{[AB_3][B_2]}{[AB_5]} = \frac{(0.2)(0.5)}{0.4} \]

### Step 3: Calculate \( K_c \).

Calculating the value:

\[ K_c = \frac{0.1}{0.4} = 0.25 \]

### Final Answer:

The equilibrium constant \( K_c \) for the formation of \( AB_5(g) \) is:

\[ K_c = 0.25 \]

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