Chemical Equilibrium and Le Chatelier's Principle
Equilibrium is where chemistry stops being about "the reaction finishes" and starts being about how far it goes. It is also the chapter where a small number of formulas do a very large amount of work: the equilibrium constant, the Kp–Kc conversion, and Le Chatelier's principle. This guide covers all three with real numbers, and then walks through what actually happens to the Haber process when you change pressure, temperature or concentration.
What equilibrium really means
A reversible reaction reaches equilibrium when the forward and reverse rates become equal. The concentrations then stop changing — but the reaction has not stopped. Molecules keep converting in both directions at the same speed. That is why it is called dynamic equilibrium.
Three consequences follow immediately, and each is a standard one-mark question:
- Equilibrium is only reached in a closed system. Let a gas escape and the system can never balance.
- Equilibrium can be approached from either side — start with pure reactants or pure products and you arrive at the same constant.
- Both reactants and products are always present at equilibrium. Neither goes to zero.
The equilibrium constant
For the general reaction aA + bB ⇌ cC + dD:
Kc uses molar concentrations, Kp uses partial pressures. Pure solids and pure liquids are left out entirely, because their "concentration" — their density — does not change as the reaction proceeds. So for CaCO₃(s) ⇌ CaO(s) + CO₂(g), simply Kp = pCO₂.
The size of K tells you the position of equilibrium: K much greater than 1 means products dominate, K much less than 1 means reactants dominate, K near 1 means comparable amounts. And the only thing that changes K is temperature. Not pressure, not concentration, not a catalyst.
Converting between Kp and Kc
Use R = 0.0821 L·atm·mol⁻¹·K⁻¹ when Kp is wanted in atmospheres and concentrations are in mol/L. Count only gases in Δn — solids and liquids do not appear.
Example 1 — ammonia synthesis. N₂(g) + 3H₂(g) ⇌ 2NH₃(g) at 500 K, with Kc = 6.11 × 10⁻².
Δn = 2 − (1 + 3) = −2
RT = 0.0821 × 500 = 41.05
(RT)−2 = 1 ÷ 41.05² = 1 ÷ 1685.1 = 5.934 × 10⁻⁴
Kp = 6.11 × 10⁻² × 5.934 × 10⁻⁴ = 3.63 × 10⁻⁵
Example 2 — phosphorus pentachloride. PCl₅(g) ⇌ PCl₃(g) + Cl₂(g) at 500 K, with Kc = 1.80 × 10⁻².
Δn = 2 − 1 = +1
Kp = 1.80 × 10⁻² × 41.05 = 0.739
Example 3 — when Δn = 0. H₂(g) + I₂(g) ⇌ 2HI(g).
Δn = 2 − 2 = 0, and (RT)⁰ = 1, so Kp = Kc at every temperature. Spot this before doing any arithmetic — it saves a full minute in an exam.
Using K to find how much reacts — the ICE table
Problem. 1.00 mol of PCl₅ is placed in a 2.00 L vessel at 500 K, where Kc = 1.80 × 10⁻². Find the equilibrium concentrations and the degree of dissociation.
Initial [PCl₅] = 1.00 ÷ 2.00 = 0.500 M. Let x mol/L dissociate.
| PCl₅ | PCl₃ | Cl₂ | |
|---|---|---|---|
| Initial (M) | 0.500 | 0 | 0 |
| Change (M) | −x | +x | +x |
| Equilibrium (M) | 0.500 − x | x | x |
Kc = x² ÷ (0.500 − x) = 0.0180
x² + 0.0180x − 0.00900 = 0
x = [−0.0180 + √(0.000324 + 0.03600)] ÷ 2 = (−0.0180 + 0.19059) ÷ 2 =
0.0863 M
So [PCl₅] = 0.414 M, [PCl₃] = [Cl₂] = 0.0863 M.
Check: 0.0863² ÷ 0.414 = 0.007448 ÷ 0.414 = 0.0180 ✓
Degree of dissociation α = 0.0863 ÷ 0.500 = 0.173, i.e. 17.3%.
Note the check line. Substituting your answer back into the K expression takes ten seconds and catches almost every algebra slip.
Le Chatelier's principle
If a system at equilibrium is disturbed, it shifts in the direction that partly opposes the disturbance. "Partly" is important — the system never fully undoes the change, it only reduces it. Here is the principle applied to the Haber process, N₂(g) + 3H₂(g) ⇌ 2NH₃(g), ΔH = −92.4 kJ:
| Stress | Shift | Reason | Does K change? |
|---|---|---|---|
| Increase total pressure (smaller volume) | Right, towards NH₃ | 4 mol gas on the left, 2 on the right — fewer molecules relieve the pressure | No |
| Increase temperature | Left, back to N₂ and H₂ | The forward reaction is exothermic, so the reverse absorbs the added heat | Yes — K falls |
| Remove NH₃ as it forms | Right | Q drops below K, so the system makes more product | No |
| Add more N₂ | Right | Consumes some of the added N₂ | No |
| Add argon at constant volume | No shift | Partial pressures of N₂, H₂ and NH₃ are unchanged | No |
| Add argon at constant pressure | Left | The volume must expand, so every partial pressure falls and the side with more moles is favoured | No |
| Add an iron catalyst | No shift | Speeds both directions equally; equilibrium arrives sooner, not further along | No |
This table explains the real industrial conditions. Ammonia synthesis is run at roughly 200 atm and 700 K with an iron catalyst. High pressure is chosen because it genuinely pushes the equilibrium right. The high temperature works against yield — but at low temperature the reaction is impossibly slow, so a compromise temperature plus a catalyst is used, and unreacted gas is recycled. That trade-off between yield and rate is a favourite short-answer question.
Q versus K — the tool that predicts direction
The reaction quotient Q has exactly the same expression as K but uses whatever concentrations you have right now, equilibrium or not.
Q is the honest version of Le Chatelier: instead of arguing about which way the system "opposes" something, you calculate a number and compare. Use it whenever a question gives you a set of starting concentrations that are clearly not at equilibrium.
- Using R = 8.314 in Kp = Kc(RT)Δn and reporting atm. 8.314 belongs with pressures in pascals. For atmospheres, use 0.0821.
- Counting solids and liquids in Δn or in K. Only gases (and aqueous species for Kc) appear.
- Saying a catalyst increases yield. It never does. It changes only how fast equilibrium is reached.
- Saying pressure changes K. Pressure shifts the position of equilibrium; only temperature changes the value of K.
- Forgetting that K is defined for a specific equation. Reverse the reaction and K becomes 1/K. Double the coefficients and K becomes K². Halve them and it becomes √K.
- Dropping units of concentration into Kp. Keep the two constants separate — mixing molarities and partial pressures in one expression is a guaranteed zero.
Where equilibrium appears in exams
| Exam | Typical use |
|---|---|
| CBSE/ICSE Class 11 | Kc/Kp conversion, Le Chatelier predictions, Haber and Contact processes |
| JEE / NEET | ICE tables, degree of dissociation, simultaneous equilibria |
| IIT-JAM / CUET-PG | ΔG° = −RT ln K, temperature dependence via van 't Hoff |
| GATE / CSIR-NET | Coupled equilibria, activity coefficients, Ksp and buffer calculations |
Gas equilibria start with PV = nRT. Before you can build a Kp expression you usually need partial pressures from moles, volume and temperature — and that is exactly what the Ideal Gas Law calculator does, solving for whichever variable you leave blank.
Open the Ideal Gas Law (PV = nRT) Calculator →Physical chemistry is where most students lose marks, and equilibrium is where it begins. ABC Chemistry runs Class 11–12 coaching at its Gurugram centre and online across India, plus dedicated IIT-JAM, GATE and CSIR-NET batches — details at abcchemistry.in.