For the reaction N2(g)+3H2(g)⇌2NH3(g), Kc=0.5 at 400 K. Initially, 1 mol of N2 and 3 mol of H2 are placed in a 2 L container. Find the equilibrium concentrations and the percentage conversion of N2.
Solution — Step by Step
Initial concentrations:
[N2]0=1/2=0.5 M
[H2]0=3/2=1.5 M
[NH3]0=0
Let x = mol/L of N2 that reacts. By stoichiometry, 3x of H2 reacts and 2x of NH3 forms.
This is a high-order polynomial — generally messy. We try a smart simplification: notice that 1.5−3x=3(0.5−x). Substitute:
Kc=(0.5−x)⋅27(0.5−x)34x2=27(0.5−x)44x2
27(0.5−x)44x2=0.5⟹4x2=13.5(0.5−x)4
Take square roots (positive on both sides):
2x=13.5(0.5−x)2≈3.674(0.5−x)2
Let u=0.5−x. Then x=0.5−u, so 2(0.5−u)=3.674u2, i.e. 1−2u=3.674u2, or 3.674u2+2u−1=0.
Quadratic formula:
u=2×3.674−2+4+14.696=7.348−2+4.324≈0.316
So x=0.5−0.316≈0.184 M.
[N2]eq=0.5−0.184=0.316 M
[H2]eq=1.5−3(0.184)=0.948 M
[NH3]eq=2(0.184)=0.368 M
Percent conversion of N2: 0.184/0.5×100%=36.8%.
Quick sanity check on Kc: (0.368)2/[(0.316)(0.948)3]=0.1354/[0.316×0.853]=0.1354/0.270≈0.50 ✓
Why This Works
The ICE table organises stoichiometry: how much each species changes is locked together by the balanced equation. The trick we used — recognising 1.5=3×0.5 which means the reactants are in stoichiometric ratio — collapses the cubic and quartic terms into one variable.
Whenever the problem gives initial moles in the exact stoichiometric ratio, you can always factor like this. Otherwise you’re stuck with a messy polynomial that needs numerical methods.
Alternative Method
For approximate problems, the “small x” approximation works when K is small: assume x≪0.5, simplify denominators to 0.5 and 1.5, solve for x, then check whether the assumption holds. Here x≈0.18 which is 36% of 0.5 — too large for the approximation to hold accurately. The full quadratic was needed.
For ammonia synthesis problems specifically, NEET often gives stoichiometric initial ratios. Recognising this and substituting [H2]=3[N2] early collapses the algebra dramatically.
Common Mistake
The two killer errors:
Forgetting the stoichiometric coefficients in the change row. If x mol/L of N2 reacts, then 3x of H2 reacts (not x) and 2x of NH3 forms (not x). Always read the balanced equation carefully.
Mixing Kc and Kp.Kc uses concentrations; Kp uses partial pressures. They differ by a factor of (RT)Δn. For this reaction Δn=2−4=−2. Don’t substitute partial pressures into a Kc expression.
Final answer:[N2]=0.316 M, [H2]=0.948 M, [NH3]=0.368 M, 36.8% conversion.
Want to master this topic?
Read the complete guide with more examples and exam tips.