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 in which the reaction will proceed based on the given equilibrium constant \( K_c \) and the reaction quotient \( Q_c \).

### Step-by-Step Solution:

1. **Identify the Reaction and Given Data:**
The reaction is:
\[ 2A + B \rightleftharpoons 3C \]
Given:
- \( K_c = 49 \) at \( 298 \, K \)
- Moles in a \( 3 \, L \) vessel: \( A = 2 \, \text{moles}, B = 1 \, \text{mole}, C = 3 \, \text{moles} \)

2. **Calculate Concentrations:**
The concentration of each component can be calculated using the formula:
\[ \text{Concentration} = \frac{\text{Number of moles}}{\text{Volume (L)}} \]
For \( A \):
\[ [A] = \frac{2 \, \text{moles}}{3 \, \text{L}} = \frac{2}{3} \, \text{M} \]
For \( B \):
\[ [B] = \frac{1 \, \text{mole}}{3 \, \text{L}} = \frac{1}{3} \, \text{M} \]
For \( C \):
\[ [C] = \frac{3 \, \text{moles}}{3 \, \text{L}} = 1 \, \text{M} \]

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

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

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

5. **Conclusion:**
The reaction must proceed in the forward direction.

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

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