The reaction's equilibrium expression is:
\[
K_c = \frac{[\text{NH}_3]^2}{[\text{N}_2][\text{H}_2]^3}
\]
Given:
- Equilibrium constant, \( K_c = 0.5 \)
- Initial concentrations: \( [\text{N}_2] = 1.0 \, \text{mol/L} \), \( [\text{H}_2] = 1.0 \, \text{mol/L} \), \( [\text{NH}_3] = 0 \, \text{mol/L} \).
Let \( -2x \) represent the change in \( \text{N}_2 \) concentration and \( -3x \) represent the change in \( \text{H}_2 \) concentration. The concentration of \( \text{NH}_3 \) increases by \( +2x \). At equilibrium, the concentrations are:
- \( [\text{N}_2] = 1.0 - 2x \)
- \( [\text{H}_2] = 1.0 - 3x \)
- \( [\text{NH}_3] = 2x \)
Substituting these into the equilibrium expression yields:
\[
0.5 = \frac{(2x)^2}{(1.0 - 2x)(1.0 - 3x)^3}
\]
Solving this equation for \( x \), after simplification, results in \( x = 0.25 \).
The equilibrium concentrations are calculated as follows:
- \( [\text{N}_2] = 1.0 - 2(0.25) = 0.75 \, \text{mol/L} \)
- \( [\text{H}_2] = 1.0 - 3(0.25) = 0.75 \, \text{mol/L} \)
- \( [\text{NH}_3] = 2(0.25) = 0.5 \, \text{mol/L} \)
Therefore, the final equilibrium concentrations are \( [\text{N}_2] = 0.75 \, \text{mol/L}, [\text{H}_2] = 0.75 \, \text{mol/L}, [\text{NH}_3] = 0.5 \, \text{mol/L} \).