`8` mol of gas `AB_(3)` are introduced into a `1.0 dm^(3)` vessel. It dissociates as `2AB_(3)(g) hArr A_(2)(g)+3B_(2)(g)`
At equilibrium, 2 mol of `A_(2)` is found to be present. The equilibrium constant for the reaction is

Options

A. (a) `2 mol^(2) L^(-2)`
B. (b) `3 mol^(2) L^(-2)`
C. (c) `27 mol^(2) L^(-2)`
D. (d) `36 mol^(2) L^(-2)`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Write the balanced chemical equation
The dissociation reaction is given as:
\[ 2AB_3(g) \rightleftharpoons A_2(g) + 3B_2(g) \]

### Step 2: Determine the initial moles and changes at equilibrium
Initially, we have 8 moles of \( AB_3 \). Let \( a \) be the amount of \( AB_3 \) that dissociates. The changes in moles can be represented as follows:
- Initial moles of \( AB_3 \): \( 8 \)
- Change in moles of \( AB_3 \): \( -2a \) (since 2 moles of \( AB_3 \) dissociate to produce 1 mole of \( A_2 \) and 3 moles of \( B_2 \))
- Moles of \( A_2 \) produced: \( a \)
- Moles of \( B_2 \) produced: \( 3a \)

At equilibrium, the moles of each species will be:
- \( AB_3 \): \( 8 - 2a \)
- \( A_2 \): \( a \)
- \( B_2 \): \( 3a \)

### Step 3: Use the information given about \( A_2 \)
We are told that at equilibrium, there are 2 moles of \( A_2 \):
\[ a = 2 \]

### Step 4: Substitute \( a \) into the expressions for moles at equilibrium
Now substituting \( a = 2 \):
- Moles of \( AB_3 \) at equilibrium:
\[ 8 - 2(2) = 8 - 4 = 4 \]
- Moles of \( A_2 \) at equilibrium:
\[ a = 2 \]
- Moles of \( B_2 \) at equilibrium:
\[ 3a = 3(2) = 6 \]

### Step 5: Write the expression for the equilibrium constant \( K_c \)
The equilibrium constant \( K_c \) is given by the formula:
\[ K_c = \frac{[A_2]^1[B_2]^3}{[AB_3]^2} \]

### Step 6: Calculate the concentrations
Since the volume of the vessel is \( 1.0 \, \text{dm}^3 \) (or 1 liter), the concentrations are equal to the number of moles:
- Concentration of \( A_2 \):
\[ [A_2] = \frac{2}{1} = 2 \, \text{mol/L} \]
- Concentration of \( B_2 \):
\[ [B_2] = \frac{6}{1} = 6 \, \text{mol/L} \]
- Concentration of \( AB_3 \):
\[ [AB_3] = \frac{4}{1} = 4 \, \text{mol/L} \]

### Step 7: Substitute the concentrations into the equilibrium constant expression
Now substituting the concentrations into the expression for \( K_c \):
\[ K_c = \frac{(2)^1(6)^3}{(4)^2} \]

### Step 8: Calculate \( K_c \)
Calculating the values:
\[ K_c = \frac{2 \cdot 216}{16} = \frac{432}{16} = 27 \]

### Final Answer
The equilibrium constant \( K_c \) for the reaction is:
\[ K_c = 27 \]

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