To solve the problem, we will follow these steps:
### Step 1: Write the balanced chemical equation
The balanced chemical equation for the formation of ammonia from nitrogen and hydrogen is:
\[ N_2 + 3H_2 \rightleftharpoons 2NH_3 \]
### Step 2: Determine the amount of ammonia produced
We know that 100 mL of 1 M \( H_2SO_4 \) is used for neutralization. To find the moles of \( H_2SO_4 \):
\[
\text{Moles of } H_2SO_4 = \text{Volume (L)} \times \text{Molarity} = 0.1 \, \text{L} \times 1 \, \text{mol/L} = 0.1 \, \text{mol}
\]
Since \( H_2SO_4 \) reacts with ammonia in a 1:2 ratio, the moles of \( NH_3 \) produced are:
\[
\text{Moles of } NH_3 = 2 \times \text{Moles of } H_2SO_4 = 2 \times 0.1 = 0.2 \, \text{mol}
\]
### Step 3: Set up the initial and equilibrium concentrations
Initially, we have:
- \( N_2 = 1 \, \text{mol} \)
- \( H_2 = 3 \, \text{mol} \)
- \( NH_3 = 0 \)
At equilibrium, let \( x \) be the amount of \( NH_3 \) formed. We know that \( x = 0.2 \, \text{mol} \). The changes in concentration will be:
- \( N_2 \) decreases by \( \frac{x}{2} = 0.1 \, \text{mol} \)
- \( H_2 \) decreases by \( \frac{3x}{2} = 0.3 \, \text{mol} \)
Thus, at equilibrium:
- \( N_2 = 1 - 0.1 = 0.9 \, \text{mol} \)
- \( H_2 = 3 - 0.3 = 2.7 \, \text{mol} \)
- \( NH_3 = 0.2 \, \text{mol} \)
### Step 4: Calculate the equilibrium concentrations
Since the volume of the container is 1 L, the concentrations (in mol/L) are:
- \([N_2] = 0.9 \, \text{M}\)
- \([H_2] = 2.7 \, \text{M}\)
- \([NH_3] = 0.2 \, \text{M}\)
### Step 5: Write the expression for \( K_c \)
The equilibrium constant \( K_c \) for the reaction is given by:
\[
K_c = \frac{[NH_3]^2}{[N_2][H_2]^3}
\]
### Step 6: Substitute the equilibrium concentrations into the \( K_c \) expression
Substituting the values we calculated:
\[
K_c = \frac{(0.2)^2}{(0.9)(2.7)^3}
\]
### Step 7: Calculate \( K_c \)
Calculating the denominator:
\[
(2.7)^3 = 19.713
\]
Now substituting back:
\[
K_c = \frac{0.04}{0.9 \times 19.713} = \frac{0.04}{17.7417} \approx 0.00225
\]
### Final Answer
Thus, the value of \( K_c \) is approximately:
\[
K_c \approx 2.25 \times 10^{-3}
\]
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