When 3.00 mole of A and 1.00 mole of B are mixed in a 1,00 litre vessel , the following reaction takes place
`A(g) +B(g) hArr 2C(g)`
the equilibrium mixture contains 0.5 mole of C. What is the value of equilibrium constant for the reaction ?

Options

A. 0.12
B. 6
C. 1.5
D. 3

Answer

To find the equilibrium constant (Kc) for the reaction \( A(g) + B(g) \rightleftharpoons 2C(g) \) given the initial moles of A and B and the moles of C at equilibrium, we can follow these steps:

### Step 1: Write the balanced equation
The balanced equation for the reaction is:
\[ A(g) + B(g) \rightleftharpoons 2C(g) \]

### Step 2: Set up the initial moles
From the problem, we know:
- Initial moles of A = 3.00 moles
- Initial moles of B = 1.00 mole
- Initial moles of C = 0.00 moles (since it is produced in the reaction)

### Step 3: Define the change in moles
Let \( x \) be the amount of A and B that react to form C. According to the stoichiometry of the reaction:
- For every 1 mole of A that reacts, 1 mole of B reacts, and 2 moles of C are produced.

### Step 4: Write the equilibrium expression
At equilibrium, we have:
- Moles of A = \( 3.00 - x \)
- Moles of B = \( 1.00 - x \)
- Moles of C = \( 2x \)

Given that the equilibrium mixture contains 0.5 moles of C, we can set up the equation:
\[ 2x = 0.5 \]
From this, we can solve for \( x \):
\[ x = \frac{0.5}{2} = 0.25 \]

### Step 5: Calculate the equilibrium moles
Now we can find the equilibrium moles of A and B:
- Moles of A at equilibrium = \( 3.00 - 0.25 = 2.75 \) moles
- Moles of B at equilibrium = \( 1.00 - 0.25 = 0.75 \) moles
- Moles of C at equilibrium = \( 0.5 \) moles (as given)

### Step 6: Calculate concentrations
Since the reaction occurs in a 1.00 litre vessel, the concentrations (in moles per litre) are the same as the number of moles:
- Concentration of A = \( [A] = 2.75 \, \text{mol/L} \)
- Concentration of B = \( [B] = 0.75 \, \text{mol/L} \)
- Concentration of C = \( [C] = 0.5 \, \text{mol/L} \)

### Step 7: Write the expression for Kc
The equilibrium constant \( K_c \) is given by:
\[ K_c = \frac{[C]^2}{[A][B]} \]

### Step 8: Substitute the values into the Kc expression
Substituting the equilibrium concentrations into the expression:
\[ K_c = \frac{(0.5)^2}{(2.75)(0.75)} \]

### Step 9: Calculate Kc
Calculating the values:
\[ K_c = \frac{0.25}{2.0625} \]
\[ K_c = 0.1219 \]

### Step 10: Round the answer
Rounding to two decimal places:
\[ K_c \approx 0.12 \]

Thus, the value of the equilibrium constant \( K_c \) for the reaction is approximately **0.12**.

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