Unit 8
General and Physical Chemistry
Class 11 Chemistry
Chemical Equilibrium
Nepal eNotes source title: Equilibria
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NEB/CDC syllabus scope: Unit 8 – Chemical Equilibrium is a 3-teaching-hour General and Physical Chemistry unit. It covers physical and chemical equilibrium, the dynamic nature of equilibrium, law of mass action, equilibrium expressions and equilibrium constants, the relationship between Kp and Kc, and Le Chatelier’s principle applied to concentration, pressure, temperature and catalysts. The curriculum explicitly states that numericals are not required for this unit.
1. Equilibrium: Basic Idea
Equilibrium
A state in which opposing processes occur at equal rates so that the observable macroscopic properties of the system remain constant with time.
A state in which opposing processes occur at equal rates so that the observable macroscopic properties of the system remain constant with time.
Equilibrium can be physical or chemical. Chemical equilibrium is reached in a reversible reaction when the forward and reverse reactions continue at equal rates.
Important
Equilibrium does not mean that the forward and reverse reactions stop. It means their rates become equal.
Equilibrium does not mean that the forward and reverse reactions stop. It means their rates become equal.
Diagram 1: Dynamic balance between forward and reverse reactions
2. Reversible Reactions
Reversible reaction
A reaction that can proceed in both forward and reverse directions under the same set of conditions.
A reaction that can proceed in both forward and reverse directions under the same set of conditions.
A reversible reaction is written using a double arrow:
A + B ⇌ C + D
Initially, if only reactants are present, the forward rate is high and the reverse rate is very low. As products accumulate, the reverse rate increases until both rates become equal.
Condition for equilibrium
A chemical equilibrium must be established in a closed system so that reactants and products are not continually lost from the reaction mixture.
A chemical equilibrium must be established in a closed system so that reactants and products are not continually lost from the reaction mixture.
3. Physical Equilibrium
Physical equilibrium involves opposite physical processes occurring at equal rates without a change in chemical identity.
Examples
- Liquid ⇌ vapour in a closed container.
- Solid ⇌ liquid at the melting point under suitable conditions.
- Solute dissolving ⇌ solute crystallizing in a saturated solution.
Liquid–Vapour Equilibrium
H₂O(l) ⇌ H₂O(g)At equilibrium in a closed container, the rate of evaporation equals the rate of condensation. The amount of liquid and vapour may remain constant even though molecules continuously cross the phase boundary.
Diagram 2: Physical equilibrium is also dynamic
4. Chemical Equilibrium
Chemical equilibrium
The state of a reversible chemical reaction in a closed system when the forward and reverse reaction rates are equal and reactant/product concentrations remain constant with time.
The state of a reversible chemical reaction in a closed system when the forward and reverse reaction rates are equal and reactant/product concentrations remain constant with time.
Example: Haber Equilibrium
N₂(g) + 3H₂(g) ⇌ 2NH₃(g)
At equilibrium, N₂ and H₂ continue forming NH₃ while NH₃ simultaneously decomposes to N₂ and H₂ at exactly the same overall rate.
Constant does not mean equal
At equilibrium, reactant and product concentrations are constant, but they are not necessarily equal.
At equilibrium, reactant and product concentrations are constant, but they are not necessarily equal.
5. Dynamic Nature of Chemical Equilibrium
Consider a reversible reaction:
A ⇌ B- At the beginning, [A] is high and [B] may be zero; the forward reaction dominates.
- As B forms, the reverse reaction becomes faster.
- Eventually, forward and reverse rates become equal.
- After this point, concentrations remain constant although molecular reactions continue.
Diagram 3: Dynamic equilibrium from equal reaction rates
6. Law of Mass Action
For the general reversible reaction:
aA + bB ⇌ cC + dD
At a fixed temperature, the equilibrium relationship between the activities of products and reactants can be written in the concentration-based introductory form:
Kc = [C]c[D]d / ([A]a[B]b)
The exponents are the stoichiometric coefficients in the balanced chemical equation.
Pure solids and pure liquids
In standard equilibrium expressions, pure solids and pure liquids are omitted because their activities are treated as constant.
In standard equilibrium expressions, pure solids and pure liquids are omitted because their activities are treated as constant.
Examples
For:
N₂(g) + 3H₂(g) ⇌ 2NH₃(g)Kc = [NH₃]² / ([N₂][H₂]³)
For:
CaCO₃(s) ⇌ CaO(s) + CO₂(g)Kc = [CO₂]
Diagram 4: Law-of-mass-action equilibrium expression
7. Equilibrium Constant Kc
Kc
The equilibrium constant written in terms of equilibrium molar concentrations of relevant species at a specified temperature.
The equilibrium constant written in terms of equilibrium molar concentrations of relevant species at a specified temperature.
For:
aA(g) + bB(g) ⇌ cC(g) + dD(g)Kc = [C]c[D]d / [A]a[B]b
Important Points
- Kc has one definite value for a given reaction at a fixed temperature.
- K changes if the temperature changes.
- Changing initial concentrations does not change K at the same temperature; the equilibrium composition shifts until the same K is restored.
- A catalyst does not change K.
8. Equilibrium Constant Kp
Kp
The equilibrium constant for a gaseous reaction written using equilibrium partial pressures.
The equilibrium constant for a gaseous reaction written using equilibrium partial pressures.
For:
aA(g) + bB(g) ⇌ cC(g) + dD(g)
Kp = (PC)c(PD)d / [(PA)a(PB)b]
Gas phases only
Kp is useful when the equilibrium is expressed using the partial pressures of gaseous species.
Kp is useful when the equilibrium is expressed using the partial pressures of gaseous species.
9. Significance / Importance of the Equilibrium Constant
The magnitude of K indicates the relative extent of reaction at equilibrium.
| Magnitude of K | General interpretation |
|---|---|
| K ≫ 1 | Products are strongly favored at equilibrium |
| K ≈ 1 | Appreciable amounts of reactants and products are present |
| K ≪ 1 | Reactants are strongly favored at equilibrium |
K does not tell reaction speed
A very large K means the equilibrium position favors products; it does not mean the reaction is fast. Equilibrium thermodynamics and reaction kinetics are different ideas.
A very large K means the equilibrium position favors products; it does not mean the reaction is fast. Equilibrium thermodynamics and reaction kinetics are different ideas.
Diagram 5: Interpreting equilibrium-constant magnitude
10. Relationship between Kp and Kc
For a gaseous equilibrium:
aA(g) + bB(g) ⇌ cC(g) + dD(g)
Kp = Kc(RT)Δn
where:
Δn = total gaseous product coefficients − total gaseous reactant coefficients
Δn = (c + d) − (a + b)
| Δn | Relationship |
|---|---|
| 0 | Kp = Kc |
| > 0 | Kp = Kc(RT)positive |
| < 0 | Kp = Kc/(RT)|Δn| |
Count only gaseous species in Δn
Pure solids and liquids do not contribute to Δn in the Kp–Kc relation.
Pure solids and liquids do not contribute to Δn in the Kp–Kc relation.
11. Derivation of Kp = Kc(RT)Δn
For each ideal gaseous component i:
PᵢV = nᵢRTSince concentration [i] = nᵢ/V:
Pᵢ = [i]RT
For:
aA + bB ⇌ cC + dDWrite Kp:
Kp = (PC)^c(PD)^d / [(PA)^a(PB)^b]Replace each partial pressure by concentration × RT:
Kp = ([C]RT)^c([D]RT)^d / [([A]RT)^a([B]RT)^b]Separate the concentration terms and RT powers:
Kp = {[C]^c[D]^d / [A]^a[B]^b} × (RT)^(c+d−a−b)The bracketed concentration ratio is Kc, therefore:
Kp = Kc(RT)Δn
Diagram 6: Kp–Kc derivation using the ideal gas equation
12. Le Chatelier’s Principle
Le Chatelier’s principle
When a system at equilibrium is disturbed by a change in concentration, pressure or temperature, the equilibrium shifts in the direction that tends to oppose the imposed change and establishes a new equilibrium.
When a system at equilibrium is disturbed by a change in concentration, pressure or temperature, the equilibrium shifts in the direction that tends to oppose the imposed change and establishes a new equilibrium.
The principle helps predict the direction of shift; it does not replace the equilibrium-constant expression.
Syllabus note
The current Grade 11 curriculum explicitly says numericals are not required for Le Chatelier’s principle. Focus on qualitative prediction and explanation.
The current Grade 11 curriculum explicitly says numericals are not required for Le Chatelier’s principle. Focus on qualitative prediction and explanation.
13. Effect of Concentration
For:
N₂(g) + 3H₂(g) ⇌ 2NH₃(g)- Adding N₂ or H₂ shifts equilibrium toward NH₃.
- Removing NH₃ shifts equilibrium toward NH₃ to replace some removed product.
- Adding NH₃ shifts equilibrium toward N₂ and H₂.
- Removing a reactant shifts equilibrium toward the reactant side.
K remains unchanged
At constant temperature, changing concentration changes the equilibrium position but does not change the value of the equilibrium constant.
At constant temperature, changing concentration changes the equilibrium position but does not change the value of the equilibrium constant.
Diagram 7: Qualitative concentration response
14. Effect of Pressure / Volume on Gaseous Equilibria
Changing pressure matters mainly when gases are present and the two sides contain different total numbers of gaseous moles.
Pressure Increase by Decreasing Volume
- Equilibrium shifts toward the side with fewer moles of gas.
Pressure Decrease by Increasing Volume
- Equilibrium shifts toward the side with more moles of gas.
Example: Haber Equilibrium
N₂(g) + 3H₂(g) ⇌ 2NH₃(g)Left side: 4 mol gaseous coefficients. Right side: 2. Increasing pressure favors the right side.
If gas moles are equal
If the total gaseous stoichiometric coefficients are the same on both sides, a simple pressure/volume change does not favor either side by this mole-count criterion.
If the total gaseous stoichiometric coefficients are the same on both sides, a simple pressure/volume change does not favor either side by this mole-count criterion.
Pure solids and liquids
Do not count solids or liquids when applying the gaseous-mole pressure rule.
Do not count solids or liquids when applying the gaseous-mole pressure rule.
15. Effect of Temperature
Temperature changes are treated by considering heat as a reactant or product.
Exothermic Forward Reaction
Reactants ⇌ Products + heat- Increasing temperature adds “heat” → equilibrium shifts toward reactants.
- Decreasing temperature removes heat → equilibrium shifts toward products.
Endothermic Forward Reaction
Reactants + heat ⇌ Products- Increasing temperature shifts toward products.
- Decreasing temperature shifts toward reactants.
Temperature is special
Unlike concentration and pressure changes, changing temperature changes the actual value of the equilibrium constant K.
Unlike concentration and pressure changes, changing temperature changes the actual value of the equilibrium constant K.
Diagram 8: Heat behaves like a reactant or product for qualitative prediction
16. Effect of a Catalyst
A catalyst lowers the activation-energy barrier for both forward and reverse reactions.
- It increases both forward and reverse rates.
- It allows equilibrium to be reached faster.
- It does not change the equilibrium composition.
- It does not change K.
- It does not shift equilibrium to either side.
High-yield sentence
A catalyst changes the rate of attaining equilibrium, not the position of equilibrium.
A catalyst changes the rate of attaining equilibrium, not the position of equilibrium.
Diagram 9: Catalyst changes kinetics, not equilibrium position
17. Concept Examples
Example 1: Write Kc
For:
H₂(g) + I₂(g) ⇌ 2HI(g) Kc = [HI]² / ([H₂][I₂])
Example 2: Omit Pure Solids
For:
CaCO₃(s) ⇌ CaO(s) + CO₂(g) Kc = [CO₂]CaCO₃ and CaO are pure solids and do not appear in the equilibrium expression.
Example 3: Determine Δn
For:
N₂(g) + 3H₂(g) ⇌ 2NH₃(g) Δn = 2 − (1+3) = −2 Kp = Kc(RT)⁻² = Kc/(RT)²
Example 4: When Kp = Kc
For:
H₂(g) + I₂(g) ⇌ 2HI(g)Δn = 2 − 2 = 0.
Kp = Kc
Example 5: Pressure Change
For N₂ + 3H₂ ⇌ 2NH₃, increasing pressure shifts equilibrium to the right because the right side has fewer gaseous moles.
Example 6: Catalyst
Adding an iron catalyst to the Haber equilibrium does not increase the equilibrium yield by shifting the reaction. It helps the system reach the same equilibrium more rapidly.
Example 7: Exothermic Reaction
If the forward reaction releases heat, raising temperature shifts equilibrium toward reactants; lowering temperature favors products.
18. High-Yield Comparison Tables
Physical vs Chemical Equilibrium
| Feature | Physical Equilibrium | Chemical Equilibrium |
|---|---|---|
| Identity of substance | No new chemical species required | Reactants and products differ chemically |
| Examples | Liquid ⇌ vapour, dissolution ⇌ crystallization | N₂ + 3H₂ ⇌ 2NH₃ |
| Dynamic? | Yes | Yes |
| Macroscopic state | Constant at equilibrium | Constant composition at equilibrium |
Effect of Equilibrium Stresses
| Change | Equilibrium response | Does K change? |
|---|---|---|
| Concentration | Shifts to oppose concentration change | No, if T constant |
| Pressure / volume | For gaseous systems, favors side according to gas-mole count | No, if T constant |
| Temperature | Shifts according to exothermic/endothermic direction | Yes |
| Catalyst | No shift; equilibrium reached faster | No |
19. Common Exam Mistakes
- Saying equilibrium means the reaction stops.
- Saying forward and reverse concentrations become equal. It is the rates that become equal.
- Forgetting that equilibrium requires a reversible process in a closed system.
- Confusing physical equilibrium with a static condition.
- Writing reactants in the numerator and products in the denominator of K.
- Forgetting to raise concentration/pressure terms to stoichiometric coefficients.
- Including pure solids or pure liquids in standard equilibrium expressions.
- Using initial concentrations instead of equilibrium concentrations when defining Kc.
- Thinking a large K means a fast reaction.
- Thinking K changes when concentration or pressure is changed at constant temperature.
- Forgetting that K depends on temperature.
- Counting solid/liquid coefficients in Δn for Kp = Kc(RT)^Δn.
- Calculating Δn as reactant moles − product moles. Correct: gaseous products − gaseous reactants.
- Applying the pressure rule when no gases are involved.
- Applying the pressure rule when gaseous mole counts are the same on both sides and claiming a preferred side.
- For an exothermic reaction, saying increasing temperature favors products; it favors the endothermic reverse direction.
- Saying a catalyst shifts equilibrium toward products.
- Saying a catalyst changes K.
- Doing unnecessary equilibrium numericals even though the current Grade 11 curriculum marks numericals as not required for this unit.
20. Important Exam Questions
Very Short / Short Questions
- Define equilibrium.
- Define reversible reaction.
- Define physical equilibrium with an example.
- Define chemical equilibrium.
- Why is chemical equilibrium called dynamic?
- What condition must be satisfied by forward and reverse rates at equilibrium?
- State the law of mass action.
- Write the equilibrium-constant expression for a general reversible reaction.
- Write Kc for N₂ + 3H₂ ⇌ 2NH₃.
- Why are pure solids and liquids omitted from equilibrium expressions?
- Define Kc.
- Define Kp.
- What is the significance of a very large K?
- Does K indicate reaction speed? Explain.
- Derive or state Kp = Kc(RT)^Δn.
- Define Δn in the Kp–Kc relation.
- When does Kp = Kc?
- State Le Chatelier’s principle.
- Explain the effect of increasing reactant concentration.
- Explain the effect of increasing pressure on N₂ + 3H₂ ⇌ 2NH₃.
- Explain the effect of temperature on an exothermic equilibrium.
- Explain the effect of temperature on an endothermic equilibrium.
- What is the effect of a catalyst on equilibrium position and K?
- Which factor among concentration, pressure, temperature and catalyst changes K?
Long / Descriptive Questions
- Explain physical and chemical equilibrium with examples.
- Explain the dynamic nature of chemical equilibrium with a rate–time diagram.
- State and explain the law of mass action.
- Explain the equilibrium constant and its significance.
- Derive the relationship between Kp and Kc.
- State Le Chatelier’s principle and explain the effects of concentration, pressure and temperature.
- Explain why a catalyst does not change the position of equilibrium.
- Apply Le Chatelier’s principle qualitatively to the Haber equilibrium.
Exam Strategy
This is a short **3-hour unit**. Prioritize: dynamic equilibrium → Kc expression → Kp/Kc derivation → Le Chatelier effects. The official content table marks numericals not required, so focus on correct equations, derivations and qualitative reasoning.
This is a short **3-hour unit**. Prioritize: dynamic equilibrium → Kc expression → Kp/Kc derivation → Le Chatelier effects. The official content table marks numericals not required, so focus on correct equations, derivations and qualitative reasoning.
21. One-Minute Revision
- Unit 8: Chemical Equilibrium — 3 teaching hours.
- Nepal eNotes source page uses the title “Equilibria.”
- A reversible reaction proceeds in both directions.
- Equilibrium is dynamic, not static.
- At equilibrium: forward rate = reverse rate.
- Reactant/product concentrations become constant, not necessarily equal.
- A closed system is required to establish equilibrium properly.
- Physical equilibrium involves opposing physical processes.
- Chemical equilibrium involves reversible chemical reactions.
- Law of mass action gives the equilibrium expression.
- For aA+bB⇌cC+dD, Kc = [C]^c[D]^d/[A]^a[B]^b.
- Pure solids and liquids are omitted from standard K expressions.
- Kc uses equilibrium concentrations.
- Kp uses equilibrium gas partial pressures.
- Large K favors products; small K favors reactants.
- K magnitude does not tell reaction speed.
- Kp = Kc(RT)^Δn.
- Δn = gaseous product coefficients − gaseous reactant coefficients.
- If Δn=0, Kp=Kc.
- Le Chatelier: equilibrium shifts to oppose a disturbance.
- Adding reactant generally shifts toward products.
- Removing product generally shifts toward products.
- Increasing pressure favors fewer gaseous moles.
- Decreasing pressure favors more gaseous moles.
- For an exothermic forward reaction, raising T shifts left.
- For an endothermic forward reaction, raising T shifts right.
- Only temperature changes K among the standard Le Chatelier factors.
- A catalyst does not shift equilibrium or change K.
- A catalyst only makes equilibrium establish faster.
- Current syllabus: numericals not required for Unit 8.
22. Diagram Practice
- Forward and reverse arrows at equilibrium.
- Liquid–vapour physical equilibrium.
- Forward/reverse rate versus time graph.
- Law-of-mass-action expression map.
- Meaning of small, medium and large K.
- Kp–Kc derivation flow.
- Concentration stress diagram.
- Temperature effect on exothermic/endothermic equilibria.
- Catalyzed vs uncatalyzed energy profiles.
Source handling: The original Nepal eNotes Equilibria PDF remains embedded above using the Google Drive file directly linked from the Nepal eNotes source page. The current NEB/CDC Grade 11 Chemistry curriculum names this topic Unit 8 – Chemical Equilibrium and assigns 3 teaching hours. The typed section follows that current scope and emphasizes qualitative Le Chatelier reasoning because the official content table states numericals not required. It is a searchable, responsive study companion and is not claimed to be a word-for-word transcription of the handwritten PDF.
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