When 3 moles of A and 1 mole of B are mixed in 1 litre vessel, the following reaction takes place `A_((g)) + B_((g))hArr2C_((g)). 1.5` moles of C are formed. The equilibrium constant for the reaction is

Options

A. `0.12`
B. `0.25`
C. `0.50`
D. `4.0`

Explanation

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

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

### Step 2: Determine initial moles
Initially, we have:
- Moles of A = 3
- Moles of B = 1
- Moles of C = 0 (since it hasn't formed yet)

### Step 3: Determine moles at equilibrium
At equilibrium, we are given that 1.5 moles of C are formed. Since the stoichiometry of the reaction shows that 1 mole of A and 1 mole of B produce 2 moles of C, we can determine how many moles of A and B are consumed.

From the formation of 1.5 moles of C:
- Moles of A consumed = \( \frac{1.5}{2} = 0.75 \) moles
- Moles of B consumed = \( \frac{1.5}{2} = 0.75 \) moles

### Step 4: Calculate moles of A and B at equilibrium
Now we can find the moles of A and B remaining at equilibrium:
- Moles of A at equilibrium = Initial moles of A - Moles of A consumed
\[ = 3 - 0.75 = 2.25 \]
- Moles of B at equilibrium = Initial moles of B - Moles of B consumed
\[ = 1 - 0.75 = 0.25 \]

### Step 5: Write the expression for the equilibrium constant
The equilibrium constant \( K_c \) for the reaction is given by:
\[ K_c = \frac{[C]^2}{[A][B]} \]
Where \([C]\), \([A]\), and \([B]\) are the molar concentrations of C, A, and B at equilibrium.

### Step 6: Calculate the concentrations
Since the reaction occurs in a 1-liter vessel, the concentrations are equal to the number of moles:
- \([C] = 1.5 \, \text{mol/L}\)
- \([A] = 2.25 \, \text{mol/L}\)
- \([B] = 0.25 \, \text{mol/L}\)

### Step 7: Substitute the concentrations into the equilibrium expression
Now we can substitute these values into the equilibrium constant expression:
\[ K_c = \frac{(1.5)^2}{(2.25)(0.25)} \]

### Step 8: Calculate \( K_c \)
Calculating the numerator:
\[ (1.5)^2 = 2.25 \]

Calculating the denominator:
\[ (2.25)(0.25) = 0.5625 \]

Now substituting back into the equation:
\[ K_c = \frac{2.25}{0.5625} = 4 \]

Thus, the equilibrium constant \( K_c \) for the reaction is:
\[ K_c = 4 \]