`K_c" for "A+B Leftrightarrow 3C` is 20 at `25^@C`. If a 2-litre contains 1,2 and 4 moles of A, B and C respectively, the reaction at `25^@C` shall

Options

A. proceed from left to right
B. proceed from right to left
C. be at equilibrium
D. not occur

Answer

To solve the problem, we need to determine the direction of the reaction based on the given equilibrium constant \( K_c \) and the concentrations of the reactants and products.

### Step-by-Step Solution:

1. **Write the Balanced Chemical Equation**:
The reaction is given as:
\[ A + B \leftrightarrow 3C \]

2. **Calculate the Initial Concentrations**:
We are given that in a 2-liter container, there are:
- 1 mole of A
- 2 moles of B
- 4 moles of C

The concentrations can be calculated as:
\[ [A] = \frac{1 \text{ mole}}{2 \text{ L}} = 0.5 \text{ M} \]
\[ [B] = \frac{2 \text{ moles}}{2 \text{ L}} = 1 \text{ M} \]
\[ [C] = \frac{4 \text{ moles}}{2 \text{ L}} = 2 \text{ M} \]

3. **Write the Expression for the Equilibrium Constant \( K_c \)**:
The expression for \( K_c \) for the reaction is:
\[ K_c = \frac{[C]^3}{[A][B]} \]
Substituting the concentrations:
\[ K_c = \frac{(2)^3}{(0.5)(1)} = \frac{8}{0.5} = 16 \]

4. **Determine the Reaction Quotient \( Q_c \)**:
The reaction quotient \( Q_c \) is calculated using the same expression as \( K_c \):
\[ Q_c = \frac{[C]^3}{[A][B]} = \frac{(2)^3}{(0.5)(1)} = 16 \]

5. **Compare \( Q_c \) with \( K_c \)**:
We know that:
\[ K_c = 20 \quad \text{and} \quad Q_c = 16 \]
Since \( K_c > Q_c \) (20 > 16), the system is not at equilibrium.

6. **Determine the Direction of the Reaction**:
When \( K_c > Q_c \), the reaction will shift to the right (towards the products) to reach equilibrium. Therefore, the reaction will proceed in the forward direction.

### Conclusion:
The reaction will move in the forward direction (from left to right).

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