For the reaction `A_((g)) + 3B_((g)) hArr 2C_((g)) ` at `27^(@)C`.2 mole of A, 4 moles of B and 6 moles of C are present in 2 lit vessel. If `K_(c)` for the reaction is 1.2, the reaction will proceed in

Options

A. Forward direction
B. backward direction
C. neither direction
D. none of these

Answer

To determine the direction in which the reaction proceeds, we will calculate the reaction quotient \( Q_c \) and compare it with the equilibrium constant \( K_c \).

### Step 1: Write the balanced chemical equation
The balanced chemical reaction is:
\[ A_{(g)} + 3B_{(g)} \rightleftharpoons 2C_{(g)} \]

### Step 2: Identify the initial moles of each substance
From the problem, we have:
- Moles of \( A = 2 \)
- Moles of \( B = 4 \)
- Moles of \( C = 6 \)

### Step 3: Calculate the concentrations of each substance
Since the volume of the vessel is 2 liters, we can calculate the concentrations as follows:
- Concentration of \( A \):
\[ [A] = \frac{2 \text{ moles}}{2 \text{ L}} = 1 \text{ M} \]
- Concentration of \( B \):
\[ [B] = \frac{4 \text{ moles}}{2 \text{ L}} = 2 \text{ M} \]
- Concentration of \( C \):
\[ [C] = \frac{6 \text{ moles}}{2 \text{ L}} = 3 \text{ M} \]

### Step 4: Write the expression for the reaction quotient \( Q_c \)
The expression for \( Q_c \) for the reaction is:
\[ Q_c = \frac{[C]^2}{[A][B]^3} \]

### Step 5: Substitute the concentrations into the \( Q_c \) expression
Substituting the values we calculated:
\[ Q_c = \frac{(3)^2}{(1)(2)^3} = \frac{9}{8} \]

### Step 6: Compare \( Q_c \) with \( K_c \)
Given that \( K_c = 1.2 \), we compare:
- \( Q_c = \frac{9}{8} = 1.125 \)
- \( K_c = 1.2 \)

Since \( Q_c < K_c \), the reaction will proceed in the forward direction to reach equilibrium.

### Conclusion
The reaction will proceed in the forward direction.