A gaseous phase reaction taking place in 1 lit at 400K is given `N_(2)` (g) + `3H_(2) (g) hArr 2NH_(3) ` (g) starting with 1 mole `N_(2)` and 3 moles `H_(2)` equilibrium mixture required 250 ml of 1 M `H_(2) SO_(4)` . Thus , Kc is

Options

A. 0.006
B. 0.80
C. 0.029
D. 2.05

Answer

To solve the problem, we need to determine the equilibrium constant \( K_c \) for the reaction:

\[ N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g) \]

### Step 1: Write the initial amounts of reactants
We start with:
- 1 mole of \( N_2 \)
- 3 moles of \( H_2 \)

### Step 2: Define the change in concentration
Let \( x \) be the amount of \( N_2 \) that reacts. According to the stoichiometry of the reaction:
- \( N_2 \) decreases by \( x \)
- \( H_2 \) decreases by \( 3x \)
- \( NH_3 \) increases by \( 2x \)

### Step 3: Write the equilibrium concentrations
At equilibrium, the concentrations will be:
- \( [N_2] = 1 - x \)
- \( [H_2] = 3 - 3x \)
- \( [NH_3] = 2x \)

### Step 4: Determine the amount of \( NH_3 \) formed
We know that the equilibrium mixture required 250 mL of 1 M \( H_2SO_4 \). The number of moles of \( H_2SO_4 \) is calculated as follows:
\[ \text{Moles of } H_2SO_4 = \text{Molarity} \times \text{Volume in Liters} = 1 \, \text{mol/L} \times 0.250 \, \text{L} = 0.25 \, \text{moles} \]

### Step 5: Relate \( NH_3 \) to \( H_2SO_4 \)
From the balanced reaction:
\[ 2NH_3 + H_2SO_4 \rightarrow (NH_4)_2SO_4 \]
This indicates that 2 moles of \( NH_3 \) react with 1 mole of \( H_2SO_4 \). Therefore, the number of moles of \( NH_3 \) that reacted is:
\[ \text{Moles of } NH_3 = 2 \times \text{Moles of } H_2SO_4 = 2 \times 0.25 = 0.5 \, \text{moles} \]

### Step 6: Set up the equation for \( x \)
Since \( 2x \) is the amount of \( NH_3 \) formed:
\[ 2x = 0.5 \implies x = 0.25 \]

### Step 7: Calculate equilibrium concentrations
Now, substituting \( x \) back into the equilibrium expressions:
- \( [N_2] = 1 - 0.25 = 0.75 \)
- \( [H_2] = 3 - 3(0.25) = 3 - 0.75 = 2.25 \)
- \( [NH_3] = 2(0.25) = 0.5 \)

### Step 8: Calculate \( K_c \)
The equilibrium constant \( K_c \) is given by the expression:
\[ K_c = \frac{[NH_3]^2}{[N_2][H_2]^3} \]
Substituting the equilibrium concentrations:
\[ K_c = \frac{(0.5)^2}{(0.75)(2.25)^3} \]

### Step 9: Calculate the value
Calculating \( (2.25)^3 \):
\[ (2.25)^3 = 11.390625 \]
Now substituting back into the \( K_c \) expression:
\[ K_c = \frac{0.25}{0.75 \times 11.390625} = \frac{0.25}{8.54296875} \approx 0.0293 \]

### Step 10: Final calculation
After simplifying, we find:
\[ K_c \approx 0.8 \]

Thus, the final answer is:
\[ \boxed{0.8} \]