`N_2+3H_2 hArr 2NH_3, K_c =1.2`
At the start of a reaction, there are 0.249 mol `N_2 , 3.21 xx 10^(-2) mol H_2 and 6.42 xx 10^(-4) mol NH_3` in a 3.50 L reaction vessel at `375^@C` . Hence reaction will proceed in

Options

A. forward direction
B. Backward direction
C. At equilibrium
D. stops

Answer

To determine the direction in which the reaction will proceed, we need to calculate the reaction quotient \( Q_c \) and compare it with the equilibrium constant \( K_c \).

### Step 1: Write the balanced chemical equation
The balanced chemical equation is:
\[ N_2 + 3H_2 \rightleftharpoons 2NH_3 \]

### Step 2: Write the expression for \( K_c \)
The equilibrium constant \( K_c \) is given by:
\[ K_c = \frac{[NH_3]^2}{[N_2][H_2]^3} \]

### Step 3: Calculate the initial concentrations
We need to calculate the concentrations of each species using the formula:
\[ \text{Concentration} = \frac{\text{moles}}{\text{volume (L)}} \]

Given:
- Moles of \( N_2 = 0.249 \) mol
- Moles of \( H_2 = 3.21 \times 10^{-2} \) mol
- Moles of \( NH_3 = 6.42 \times 10^{-4} \) mol
- Volume of the vessel = 3.50 L

Calculating the concentrations:
- \([N_2] = \frac{0.249}{3.50} = 0.0714 \, \text{M}\)
- \([H_2] = \frac{3.21 \times 10^{-2}}{3.50} = 0.00917 \, \text{M}\)
- \([NH_3] = \frac{6.42 \times 10^{-4}}{3.50} = 1.83 \times 10^{-4} \, \text{M}\)

### Step 4: Calculate \( Q_c \)
Now we can calculate \( Q_c \) using the initial concentrations:
\[ Q_c = \frac{[NH_3]^2}{[N_2][H_2]^3} = \frac{(1.83 \times 10^{-4})^2}{(0.0714)(0.00917)^3} \]

Calculating \( Q_c \):
\[ Q_c = \frac{3.35 \times 10^{-8}}{(0.0714)(7.74 \times 10^{-7})} = \frac{3.35 \times 10^{-8}}{5.53 \times 10^{-8}} \approx 0.606 \]

### Step 5: Compare \( Q_c \) with \( K_c \)
Given \( K_c = 1.2 \):
- Since \( Q_c (0.606) < K_c (1.2) \), the reaction will proceed in the forward direction to produce more products.

### Conclusion
The reaction will proceed in the forward direction.

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