What Chemical Equilibrium Actually Means
Picture this: you’re making tea and the colour stops getting darker even though the tea bag is still in. That’s not equilibrium — but it’s a useful analogy for apparent stillness with actual activity. In chemistry, chemical equilibrium is when a reversible reaction’s forward and backward rates become equal, so concentrations stop changing.
The key word is reversible. Not all reactions are. When you burn paper, that’s one-way — no amount of wishful thinking brings the ash back. But when nitrogen and hydrogen react to form ammonia, the ammonia simultaneously breaks back down. At some point, the rate of making ammonia equals the rate of it decomposing. That point is equilibrium.
Here’s what students miss: at equilibrium, the reaction hasn’t stopped. Both forward and backward reactions are happening continuously — just at the same rate. Concentrations are constant, not because nothing’s happening, but because the two processes cancel out. This is called dynamic equilibrium.
Key Terms & Definitions
Reversible reaction — A reaction that can proceed in both forward and backward directions under the same conditions. Shown with ⇌ (double arrow).
Dynamic equilibrium — The state where forward and backward reaction rates are equal and macroscopic properties (concentration, colour, pressure) remain constant, though molecular-level changes continue.
Equilibrium constant (K) — A number that tells you where equilibrium lies — whether mostly products, mostly reactants, or somewhere in between.
Degree of dissociation (α) — Fraction of original moles that have reacted. If α = 0.5, half the molecules have dissociated.
Homogeneous equilibrium — All species in the same phase (e.g., all gases or all in aqueous solution).
Heterogeneous equilibrium — Species in different phases (e.g., solid + gas, solid + liquid).
Equilibrium Constants: Kc and Kp
This is the most calculation-heavy part of Chapter 7, and it’s also the highest-scoring. Let’s get precise.
Kc — The Concentration-Based Constant
For a general reaction:
where square brackets denote molar concentrations at equilibrium. Concentrations of pure solids and pure liquids are not included — they’re constant and are absorbed into K.
Students write Kc expressions including solid or liquid reactants. If CaCO₃(s) ⇌ CaO(s) + CO₂(g), then Kc = [CO₂] only. Pure solids have activity = 1.
Kp — The Pressure-Based Constant
For reactions involving gases, we express equilibrium in terms of partial pressures:
where:
- R = 0.0821 L·atm·mol⁻¹·K⁻¹ (use this when pressure is in atm)
- T = temperature in Kelvin
- Δnₘg = moles of gaseous products − moles of gaseous reactants
When Δnₘg = 0, Kp = Kc. This happens for reactions like H₂ + I₂ ⇌ 2HI (1+1 → 2, so Δnₘg = 0).
What Does the Value of K Tell Us?
| K value | Interpretation |
|---|---|
| K >> 1 (say, 10³ or more) | Equilibrium heavily favours products |
| K << 1 (say, 10⁻³ or less) | Equilibrium heavily favours reactants |
| K ≈ 1 | Significant concentrations of both |
| K is very large (10¹⁰⁺) | Reaction goes nearly to completion |
Reaction Quotient (Qc) — The Direction Predictor
Before reaching equilibrium, a reaction mixture has a reaction quotient Qc, calculated using the same expression as Kc but with current concentrations (not equilibrium ones).
- If Qc < Kc: reaction proceeds forward (more products needed)
- If Qc > Kc: reaction proceeds backward (excess products decompose)
- If Qc = Kc: system is already at equilibrium
This is heavily tested in JEE Main — expect a numerical where you’re given initial concentrations and asked which direction the reaction shifts.
Le Chatelier’s Principle
This is the conceptual heart of the chapter. Le Chatelier’s Principle states:
When a system at equilibrium is subjected to a stress, it shifts in the direction that partially relieves that stress.
“Stress” means any change: concentration, pressure, temperature, or addition of catalyst.
Effect of Concentration Change
Adding more of a reactant → equilibrium shifts forward (toward products). Removing a product → equilibrium shifts forward. Adding more of a product → equilibrium shifts backward.
In industrial synthesis (Haber process for NH₃, Contact process for SO₃), products are continuously removed from the reaction vessel. This keeps shifting equilibrium forward, increasing yield without changing K.
Effect of Pressure Change
Only affects equilibria involving gases. More pressure → equilibrium shifts toward the side with fewer moles of gas.
For N₂ + 3H₂ ⇌ 2NH₃:
- Left side: 1 + 3 = 4 moles gas
- Right side: 2 moles gas
Increasing pressure → shifts right → more NH₃ formed. This is why Haber process runs at 200 atm.
For H₂ + I₂ ⇌ 2HI (Δnₘg = 0): pressure has no effect on position of equilibrium.
Effect of Temperature Change
This is different from concentration and pressure changes — temperature change actually changes the value of K.
| Reaction type | Increase temperature | K value |
|---|---|---|
| Exothermic (ΔH < 0) | Shifts backward | Decreases |
| Endothermic (ΔH > 0) | Shifts forward | Increases |
Think of it this way: adding heat to an exothermic reaction is like adding a product (heat is a product). By Le Chatelier, the system shifts backward to absorb excess heat.
Effect of Catalyst
A catalyst speeds up both forward and backward reactions equally. It does not shift the equilibrium position and does not change K. It only helps the system reach equilibrium faster.
JEE Main frequently asks: “Which of the following does NOT change the equilibrium constant K?” Answer: concentration, pressure, catalyst. Only temperature changes K.
Solved Examples
Example 1 — Easy (CBSE Board Level)
For the reaction: N₂(g) + 3H₂(g) ⇌ 2NH₃(g), Kc = 0.5 mol⁻² L² at 400°C. Calculate Kp.
Solution:
Δnₘg = 2 − (1 + 3) = −2
T = 400 + 273 = 673 K
Example 2 — Medium (JEE Main Level)
At 500 K, Kp for the reaction PCl₅(g) ⇌ PCl₃(g) + Cl₂(g) is 1.78 atm. If the initial pressure of PCl₅ is 2 atm, find the degree of dissociation α.
Solution:
Let α = degree of dissociation. At equilibrium:
| Species | PCl₅ | PCl₃ | Cl₂ |
|---|---|---|---|
| Initial moles | 1 | 0 | 0 |
| Equilibrium moles | 1−α | α | α |
| Total moles = 1 + α |
Partial pressures (total pressure P = 2 atm initially, at equilibrium P_total = 2(1+α)… wait, let me use mole fractions):
Actually, at equilibrium, total moles = 1 + α. Total pressure at equilibrium:
By stoichiometry, if initial pressure of PCl₅ = P₀ = 2 atm:
Example 3 — Hard (JEE Advanced Level)
At 700 K, for the reaction 2SO₂(g) + O₂(g) ⇌ 2SO₃(g), Kp = 1.8 × 10³ atm⁻¹. Calculate Kc at 700 K.
Solution:
Δnₘg = 2 − (2 + 1) = −1
When Δnₘg is negative, Kc > Kp. When Δnₘg is positive, Kc < Kp. This quick check catches arithmetic errors before you finalise your answer.
Exam-Specific Tips
CBSE Board (Class 11 & 12 Overlap)
The Chapter 7 weightage in CBSE Class 11 finals is typically 5-7 marks. Focus areas:
- Writing correct Kc and Kp expressions (1-2 marks)
- Kp = Kc(RT)^Δn calculation (2-3 marks)
- Le Chatelier’s principle application (1-2 marks, usually 1-mark MCQ or short answer)
CBSE marking scheme: always show the formula first, then substitution, then answer with units. Units of K depend on Δn — many students lose marks here.
JEE Main
Chemical Equilibrium appears in 1-2 questions per shift, typically 4-8 marks. Recent PYQ focus (2022-2024):
- Calculating Kc given equilibrium concentrations
- Direction of shift when Q is given
- Effect of conditions on K vs. equilibrium position (conceptual trap)
- JEE Main 2024 Shift 1 had a question on degree of dissociation of PCl₅ — exactly the type in Example 2 above.
NEET
NEET tests Le Chatelier’s principle heavily — typically 1 question per paper. The 2023 NEET paper had a question on the Haber process conditions. Buffer solutions and ionic equilibrium (Henderson-Hasselbalch equation) also appear under this chapter’s umbrella in NEET.
For JEE Main, memorise: K depends only on temperature. Catalyst changes rate, not K. Pressure and concentration change shift direction, not K. This single concept appears in multiple forms.
Common Mistakes to Avoid
Mistake 1: Including solids/liquids in Kc or Kp expression. CaCO₃(s) ⇌ CaO(s) + CO₂(g) → Kp = P(CO₂). Only gases and aqueous species appear in K expressions.
Mistake 2: Confusing “equilibrium shifts” with “K changes.” Adding a reactant shifts equilibrium forward — but K stays the same. Students write “K increases.” Wrong. K changes only with temperature.
Mistake 3: Using wrong R value in Kp = Kc(RT)^Δn. Use R = 0.0821 L·atm·mol⁻¹·K⁻¹ when pressure is in atm. Use R = 8.314 J·mol⁻¹·K⁻¹ only when pressure is in Pa. Mixing these is a guaranteed wrong answer.
Mistake 4: Forgetting to convert temperature to Kelvin. Every thermodynamic formula uses T in Kelvin. T = 400°C is 673 K, not 400 K.
Mistake 5: Wrong sign of Δnₘg. Δnₘg = (moles of gaseous products) − (moles of gaseous reactants). For N₂ + 3H₂ ⇌ 2NH₃, it’s 2 − 4 = −2. A positive Δn means Kp > Kc; negative means Kp < Kc.
Practice Questions
Q1. Write the Kc expression for: Fe₃O₄(s) + 4H₂(g) ⇌ 3Fe(s) + 4H₂O(g)
Kc = [H₂O]⁴ / [H₂]⁴
Since Fe₃O₄ and Fe are pure solids, they are not included. Only gaseous/aqueous species appear.
Q2. For A(g) ⇌ 2B(g), Kp = 4 atm at 300 K. What is Kc?
Δnₘg = 2 − 1 = 1
Kp = Kc(RT)^1
Kc = Kp / (RT) = 4 / (0.0821 × 300) = 4 / 24.63 = 0.162 mol L⁻¹
Q3. At equilibrium, [H₂] = 0.5 M, [I₂] = 0.5 M, [HI] = 4 M for H₂ + I₂ ⇌ 2HI. Calculate Kc.
Kc = [HI]² / ([H₂][I₂]) = (4)² / (0.5 × 0.5) = 16 / 0.25 = 64
No units since Δnₘg = 0.
Q4. Qc for a reaction is 0.02 and Kc is 0.5. Which direction does the reaction proceed?
Qc (0.02) < Kc (0.5), so the reaction proceeds in the forward direction to form more products until equilibrium is reached.
Q5. For N₂O₄(g) ⇌ 2NO₂(g), ΔH = +57 kJ/mol. What happens to K when temperature is increased?
The forward reaction is endothermic (ΔH > 0). Increasing temperature shifts equilibrium forward (toward products), so K increases. This is one of the few cases where K actually changes — because temperature changed.
Q6. At 1000 K, Kp for CO(g) + ½O₂(g) ⇌ CO₂(g) is 3.3 × 10⁴ atm⁻¹/². Find Kc.
Δnₘg = 1 − (1 + 0.5) = −0.5
Kc = Kp / (RT)^(Δng) = Kp × (RT)^(0.5)
RT = 0.0821 × 1000 = 82.1
Kc = 3.3 × 10⁴ × √82.1 = 3.3 × 10⁴ × 9.06 = 2.99 × 10⁵ mol⁻¹/² L¹/²
Q7. In the synthesis of ammonia, what happens to the yield of NH₃ if (a) pressure is increased, (b) temperature is increased, (c) catalyst is added?
N₂ + 3H₂ ⇌ 2NH₃, ΔH = −92 kJ/mol
(a) Pressure increased: Δnₘg = 2−4 = −2, so more pressure shifts toward fewer moles of gas (right) → yield of NH₃ increases.
(b) Temperature increased: Reaction is exothermic, so increasing temp shifts equilibrium left → yield of NH₃ decreases. K decreases.
(c) Catalyst added: No effect on yield or K. Only reaches equilibrium faster.
Q8. For PCl₃(g) + Cl₂(g) ⇌ PCl₅(g), Kc = 2 × 10³ L mol⁻¹ at 298 K. If [PCl₃] = [Cl₂] = 0.01 M initially and [PCl₅] = 0.1 M, is the system at equilibrium?
Calculate Qc: Qc = [PCl₅] / ([PCl₃][Cl₂]) = 0.1 / (0.01 × 0.01) = 0.1 / 0.0001 = 1000
Kc = 2000
Qc (1000) < Kc (2000), so the system is not at equilibrium. It will shift forward, forming more PCl₅.
Q9 (Challenge). α = 0.4 for PCl₅ ⇌ PCl₃ + Cl₂ at total pressure 2 atm. Calculate Kp.
Using the formula derived earlier:
Kp = α²P / (1 − α²)
= (0.4)² × 2 / (1 − 0.16)
= (0.16 × 2) / 0.84
= 0.32 / 0.84
= 0.381 atm
FAQs
Why does K have no units sometimes?
Kc and Kp are technically dimensionless — we express concentrations/pressures relative to a standard state (1 mol/L or 1 atm). But at the CBSE and JEE level, we assign units based on Δn for practical calculations. If Δnₘg = 0, K is unitless. If Δnₘg = +1, Kp has units of atm, and so on.
Can K ever be negative?
No. K is always positive because it’s a ratio of product concentrations/pressures to reactant concentrations/pressures, all of which are positive numbers raised to positive powers.
What is the difference between Kc and Kp — when do I use which?
Use Kc for reactions in solution (aqueous) or when concentrations are given. Use Kp for gas-phase reactions when partial pressures are given. Both describe the same equilibrium — they’re just different ways to express it. They’re related by Kp = Kc(RT)^Δn.
Does adding an inert gas affect equilibrium?
At constant volume: no effect (partial pressures of reactants/products don’t change). At constant pressure: adding inert gas increases total pressure, which decreases mole fractions, effectively reducing partial pressures of reactants/products → equilibrium shifts toward more moles of gas (if Δnₘg ≠ 0).
Why does the Haber process use 450-500°C if higher temperature decreases NH₃ yield?
This is a classic industrial compromise. At low temperatures, K is high (more NH₃ favoured), but the rate is too slow to be economical. At 450°C with an iron catalyst, the rate is acceptable and yield is workable (~15-25%). Removing NH₃ continuously as it forms compensates for the lower K at higher temperature.
What is the relationship between equilibrium constant and Gibbs free energy?
At the JEE Advanced level: ΔG° = −RT ln K. If K > 1, ΔG° < 0 (spontaneous in forward direction). If K < 1, ΔG° > 0 (non-spontaneous). This connects thermodynamics with equilibrium — a topic that appears in JEE Advanced physical chemistry.
How is ionic equilibrium related to chemical equilibrium?
Ionic equilibrium is a special case of chemical equilibrium applied to weak acids, weak bases, and sparingly soluble salts in water. Ka, Kb, Kw, and Ksp are all equilibrium constants with the same mathematical structure as Kc — just applied to different systems.
Can the same reaction have different K values at the same temperature if written differently?
Yes. If you write N₂ + 3H₂ ⇌ 2NH₃ with Kc = K, then for ½N₂ + 3/2 H₂ ⇌ NH₃, Kc’ = √K. And for 2NH₃ ⇌ N₂ + 3H₂ (reverse), Kc” = 1/K. This is a favourite board question and also appears in JEE Main numericals.