For the reaction `A(g) +3B(g) hArr 2C(g) `at `27^(@)C`, 2 moles of `A`, 4 moles of B and 6 moles of C are present in 2 litre 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

Explanation

To solve the problem, we need to determine the direction in which the reaction will proceed based on the given concentrations and the equilibrium constant \( K_c \).

### Step-by-Step Solution:

1. **Write the Reaction and Given Information:**
The reaction is:
\[ A(g) + 3B(g) \rightleftharpoons 2C(g) \]
Given:
- Moles of \( A = 2 \)
- Moles of \( B = 4 \)
- Moles of \( C = 6 \)
- Volume of the vessel = 2 L
- \( K_c = 1.2 \)

2. **Calculate the Concentrations:**
Concentration is calculated using the formula:
\[ \text{Concentration} = \frac{\text{Number of moles}}{\text{Volume in liters}} \]
- 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 \( C \):
\[ [C] = \frac{6 \text{ moles}}{2 \text{ L}} = 3 \text{ M} \]

3. **Calculate \( Q_c \):**
The reaction quotient \( Q_c \) is calculated using the formula:
\[ Q_c = \frac{[C]^2}{[A][B]^3} \]
Substituting the values:
\[ Q_c = \frac{(3)^2}{(1)(2)^3} = \frac{9}{8} = 1.125 \]

4. **Compare \( Q_c \) and \( K_c \):**
- We have \( K_c = 1.2 \) and \( Q_c = 1.125 \).
- Since \( K_c > Q_c \) (1.2 > 1.125), the reaction will proceed in the forward direction to reach equilibrium.

5. **Conclusion:**
The reaction will proceed in the forward direction to increase the concentration of products until \( Q_c \) equals \( K_c \).

### Final Answer:
The reaction will proceed in the **forward direction**.

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