For the reaction `:` `" "2A+B hArr 3C` at `298 K, K_(c)=49`
A `3L` vessel contains 2,1 and 3 molesof `A, B` and `C` respectively.
The reaction at the same temperature

Options

A. must proceed in forward direction
B. must proceed in backward direction
C. must be in equilibrium
D. cannot be predicted

Explanation

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

### Step-by-Step Solution:

1. **Identify the Reaction and Given Data:**
The reaction is:
\[ 2A + B \rightleftharpoons 3C \]
Given:
- \( K_c = 49 \)
- Moles of \( A = 2 \)
- Moles of \( B = 1 \)
- Moles of \( C = 3 \)
- Volume of the vessel = 3 L

2. **Calculate the Concentrations:**
To find the concentrations of each species, we use the formula:
\[ \text{Concentration} = \frac{\text{Number of moles}}{\text{Volume in liters}} \]
- Concentration of \( A \):
\[ [A] = \frac{2 \text{ moles}}{3 \text{ L}} = \frac{2}{3} \text{ M} \]
- Concentration of \( B \):
\[ [B] = \frac{1 \text{ mole}}{3 \text{ L}} = \frac{1}{3} \text{ M} \]
- Concentration of \( C \):
\[ [C] = \frac{3 \text{ moles}}{3 \text{ L}} = 1 \text{ M} \]

3. **Calculate the Reaction Quotient \( Q_c \):**
The expression for \( Q_c \) for the reaction is:
\[ Q_c = \frac{[C]^3}{[A]^2[B]} \]
Substituting the concentrations:
\[ Q_c = \frac{(1)^3}{\left(\frac{2}{3}\right)^2 \left(\frac{1}{3}\right)} \]
Simplifying:
\[ Q_c = \frac{1}{\left(\frac{4}{9}\right) \left(\frac{1}{3}\right)} = \frac{1}{\frac{4}{27}} = \frac{27}{4} = 6.75 \]

4. **Compare \( Q_c \) with \( K_c \):**
We have:
- \( Q_c = 6.75 \)
- \( K_c = 49 \)

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

5. **Conclusion:**
The reaction will shift to the right (forward direction) to form more products.

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

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