10 moles of `A_(2), 15` moles of `B_(2)` and 5 moles of AB are placed in a 2 litre vessel and allowed the come to equilibrium. The final concentration of AB is 10.5 M, `A_(2)(g)+B_(2)(g)hArr 2AB(g)`
Determine the value of equilibrium constant `(K_(C))` for the reaction.

Options

A. 25.3
B. 31.5
C. 36.3
D. 40.5

Answer

To determine the equilibrium constant \( K_c \) for the reaction \( A_2(g) + B_2(g) \rightleftharpoons 2AB(g) \), we will follow these steps:

### Step 1: Calculate Initial Concentrations
We start with the initial moles of each substance:
- Moles of \( A_2 \) = 10 moles
- Moles of \( B_2 \) = 15 moles
- Moles of \( AB \) = 5 moles

Since the volume of the vessel is 2 liters, we can calculate the initial concentrations:
\[ \text{Initial concentration of } A_2 = \frac{10 \text{ moles}}{2 \text{ L}} = 5 \text{ M} \]
\[ \text{Initial concentration of } B_2 = \frac{15 \text{ moles}}{2 \text{ L}} = 7.5 \text{ M} \]
\[ \text{Initial concentration of } AB = \frac{5 \text{ moles}}{2 \text{ L}} = 2.5 \text{ M} \]

### Step 2: Set Up the Change in Concentration
Let \( x \) be the amount of \( A_2 \) and \( B_2 \) that reacts at equilibrium. The changes in concentrations will be:
- For \( A_2 \): \( 5 - x \)
- For \( B_2 \): \( 7.5 - x \)
- For \( AB \): \( 2.5 + 2x \)

### Step 3: Use the Final Concentration of \( AB \)
We know the final concentration of \( AB \) is 10.5 M. Therefore:
\[ 2.5 + 2x = 10.5 \]
Solving for \( x \):
\[ 2x = 10.5 - 2.5 = 8 \quad \Rightarrow \quad x = 4 \]

### Step 4: Calculate Equilibrium Concentrations
Now we can calculate the equilibrium concentrations:
- Concentration of \( A_2 \):
\[ \text{Equilibrium concentration of } A_2 = 5 - x = 5 - 4 = 1 \text{ M} \]
- Concentration of \( B_2 \):
\[ \text{Equilibrium concentration of } B_2 = 7.5 - x = 7.5 - 4 = 3.5 \text{ M} \]
- Concentration of \( AB \):
\[ \text{Equilibrium concentration of } AB = 10.5 \text{ M} \]

### Step 5: Write the Expression for \( K_c \)
The expression for the equilibrium constant \( K_c \) for the reaction is given by:
\[ K_c = \frac{[AB]^2}{[A_2][B_2]} \]
Substituting the equilibrium concentrations:
\[ K_c = \frac{(10.5)^2}{(1)(3.5)} \]

### Step 6: Calculate \( K_c \)
Now we calculate \( K_c \):
\[ K_c = \frac{110.25}{3.5} = 31.5 \]

### Final Answer
Thus, the equilibrium constant \( K_c \) for the reaction is:
\[ \boxed{31.5} \]

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