For the reaction `A+B hArr 3. C` at `25^(@)C`, a `3 L` vessel contains 1, 2, and 4 moles of A, B and C respectively. Predict the direction of reaction if:
a. `K_(c)` for the reaction is `10`.
b. `K_(c)` for the reaction is `15`.
c. `K_(c)` for the reaction is `10.66`

Explanation

To solve the problem, we need to analyze the reaction \( A + B \rightleftharpoons 3C \) and determine the direction of the reaction based on the given equilibrium constants \( K_c \) and the concentrations of the reactants and products.

### Step-by-Step Solution:

1. **Calculate the Concentrations:**
- The number of moles of A, B, and C are given as 1, 2, and 4 respectively, and the volume of the vessel is 3 L.
- Concentration of A:
\[ [A] = \frac{\text{moles of A}}{\text{volume}} = \frac{1}{3} \, \text{mol/L} \]
- Concentration of B:
\[ [B] = \frac{\text{moles of B}}{\text{volume}} = \frac{2}{3} \, \text{mol/L} \]
- Concentration of C:
\[ [C] = \frac{\text{moles of C}}{\text{volume}} = \frac{4}{3} \, \text{mol/L} \]

2. **Calculate the Reaction Quotient \( Q_c \):**
- The reaction quotient \( Q_c \) is calculated using the formula:
\[ Q_c = \frac{[C]^3}{[A][B]} \]
- Substituting the concentrations:
\[ Q_c = \frac{\left(\frac{4}{3}\right)^3}{\left(\frac{1}{3}\right)\left(\frac{2}{3}\right)} = \frac{\frac{64}{27}}{\frac{2}{9}} = \frac{64}{27} \times \frac{9}{2} = \frac{288}{54} = 10.66 \]

3. **Compare \( Q_c \) with \( K_c \) for each case:**
- **Case a:** \( K_c = 10 \)
- Since \( Q_c (10.66) > K_c (10) \), the reaction will shift to the left (backward direction) to reach equilibrium.

- **Case b:** \( K_c = 15 \)
- Since \( Q_c (10.66) < K_c (15) \), the reaction will shift to the right (forward direction) to reach equilibrium.

- **Case c:** \( K_c = 10.66 \)
- Since \( Q_c (10.66) = K_c (10.66) \), the reaction is already at equilibrium and will not shift in either direction.

### Summary of Results:
- a. The reaction shifts **backward** (to the left).
- b. The reaction shifts **forward** (to the right).
- c. The reaction is **at equilibrium**.