Question
For the equilibrium N2(g)+3H2(g)⇌2NH3(g), Kc=0.5 at 700 K. A 1 L vessel initially contains 2 mol N₂, 3 mol H₂, and no NH₃. Find the equilibrium concentrations and the equilibrium yield of ammonia.
Solution — Step by Step
Let x mol/L of N₂ react.
| Species | Initial | Change | Equilibrium |
|---|
| N₂ | 2 | −x | 2−x |
| H₂ | 3 | −3x | 3−3x |
| NH₃ | 0 | +2x | 2x |
Kc=[N2][H2]3[NH3]2=(2−x)(3−3x)3(2x)2=0.5
Factor: (3−3x)3=27(1−x)3. So
27(2−x)(1−x)34x2=0.5⇒27(2−x)(1−x)38x2=1
This is a quartic. Trial: x=0.55. LHS denom: 27×1.45×(0.45)3=27×1.45×0.0911≈3.567. LHS num: 8×0.3025=2.42. Ratio ≈0.679. Try x=0.6: denom 27×1.4×0.064=2.42. Num 8×0.36=2.88. Ratio ≈1.19. Interpolate: x≈0.58.
With x≈0.58:
[N2]≈1.42 mol/L, [H2]≈1.26 mol/L, [NH3]≈1.16 mol/L.
Yield: 20.58×100%≈29% of N₂ converted.
[N2]≈1.42, [H2]≈1.26, [NH3]≈1.16 mol/L; yield ≈29%.
Why This Works
The ICE-table approach is the universal recipe for equilibrium concentration problems. Define the extent of reaction (one variable), express every species in terms of it, plug into Kc, solve. The polynomial that results may need numerical iteration when an exact algebraic solution is messy.
For ammonia synthesis specifically, the Kc value is moderate at this temperature — partial conversion is realistic. At lower T and higher pressure (Le Chatelier), conversion improves, which is why industrial Haber process operates at ~450°C and 200 atm.
Alternative Method
For small x (large Kc would not justify this here), you could neglect x relative to initial values and solve a simpler equation. With Kc=0.5, the approximation is poor — you’d get 4x2/27⋅2⋅1=0.5⇒x≈0.82, which is too far from the iterated answer. Use approximations only when x≪ initial concentrations.
Common Mistake
Students treat Kp and Kc interchangeably. They are related by Kp=Kc(RT)Δn where Δn is the change in moles of gas. For this reaction Δn=2−4=−2, so Kp=Kc. If a problem gives Kp, convert before using ICE concentrations.