Chemical Equilibrium Class 11: Equilibrium Meaning, Formulas, Notes

Q: What is the equilibrium constant?

The equilibrium constant is the ratio of the concentrations of products to reactants, each raised to the power of their stoichiometric coefficients. For reversible reactions.

Q: What is Le Chatelier’s principle NCERT?

If the conditions of equilibrium are changed, it shifts to oppose the change. For example, in Haber's process, high pressure favors NH? formation.

Q: What is dynamic equilibrium?

It happens in reversible reactions when the rate of the forward reaction becomes equal to the rate of the backward reaction. Result in the same concentration of reactants and product.

Updated on Sep 16, 2025 16:14 IST

By Swayam Gupta, Associate Senior Executive

Suppose you’re watching people enter and exit a shopping mall. At some point, you notice that even though people are moving in and out, the total number of people inside the mall is still the same. This is what defines equilibrium.

Through the chemical equilibrium class 11, you will be able to predict the behavior of chemical reactions, understand medicines, and how the body maintains PH in blood, etc. While the class 11 ch6 covers the vast coursework, we have created notes of the equilibrium class 11 and provided the complex information in the easiest manner possible, which helps you to revise all the topics in less time and score higher in exams.

The topics covered are Equilibrium in Physical Processes and Chemical Processes, Dynamic Equilibrium, Equilibrium Constant, Ionic Equilibrium, Buffer Solutions, etc.

Table of content

What is Equilibrium?
Equilibrium in Physical Processes
Equilibrium in Chemical Processes – Dynamic Equilibrium
Law of Chemical Equilibrium and Equilibrium Constant
Homogeneous and Heterogeneous Equilibria
Applications of Equilibrium Constant
Relationship Between Equilibrium Constant K, Reaction Quotient Q, and Gibbs Energy G
Factors Affecting Equilibria
Ionic Equilibrium in Solution
Acids, Bases, and Salts
Ionization of Acids and Bases
Solubility Equilibria of Sparingly Soluble Salts
Chemical Equilibrium class 11 formulas
Revision Notes for Class 11 Chemistry
NCERT Solutions for Class 11 Chemistry
About the Content Reviewer
Chemical Equilibrium FAQs

What is Equilibrium?

Equilibrium is a state where opposing processes take place at the same rate, so the overall conditions don’t change. It’s like a perfectly balanced seesaw that stays in the middle.

Simply, the equilibrium meaning is - things are still happening; they’re just balanced!

For example, in an equilibrium state, water evaporates at the same rate it is condensed.

H2O (l) ⇌ H2O (vap)

Types of Equilibrium:

Physical Equilibrium
Chemical equilibrium

Equilibrium in Physical Processes

Before going into chemical reactions, let’s understand physical equilibrium.

Solid-Liquid Equilibrium (Ice and Water)

When we take ice cubes in a glass of water at 0°C. What will happen?

Some ice molecules melt into the liquid (melting)
Some water molecules freeze into the ice (freezing)

When these melting and freezing rates are equal, it is in equilibrium.

And the result will be that the amount of ice and water stays the same, even though molecules are constantly switching sides (melting and freezing).

Liquid-Vapor Equilibrium

Ever notice how water in a closed bottle seems to “disappear” from the liquid but the total amount of water stays the same?

What happens is:

Water molecules become vapor (evaporation)
Some vapor molecules return to liquid (condensation)
At equilibrium: rate of evaporation = rate of condensation

To be noted, this only works in a closed container. In an open container, vapor escapes completely.

Dissolving Sugar in Water

When you add sugar to water:

First, sugar dissolves quickly
Eventually, no more sugar can dissolve, and it becomes a saturated solution

At equilibrium, sugar dissolves at the same rate as it crystallizes out.

Equilibrium in Chemical Processes – Dynamic Equilibrium

First, understand what chemical equilibrium is.

As per the class 11 chemical equilibrium notes, chemical equilibrium can be defined as:

When the rate of forward reaction is equal to the rate of backward reaction, the concentrations of reactants and products remain the same. This state is called chemical equilibrium.

For example, formation of ammonia:

N₂(g) + 3H₂(g) ⇌ 2NH₃(g)

How chemical equilibrium develops here:

Initially, Lots of N₂ and H₂, no NH₃, the Forward reaction is fast, and no reverse reaction happens yet.

As time passes, NH₃ starts forming, the forward reaction slows down (less reactants), and the reverse reaction starts and speeds up.

Law of Chemical Equilibrium and Equilibrium Constant

The Law of Chemical Equilibrium- Mass Action Law

At equilibrium, there’s a mathematical relationship between reactant and product concentrations.

For the general reaction: aA + bB ⇌ cC + dD

Equilibrium Constant Expression:

$K_{c} = \frac{[C]^{c} [D]^{d}}{[A]^{a} [B]^{b}}$

Where:

[A], [B], [C], [D] are equilibrium concentrations
a, b, c, and d are the coefficients from the balanced equation
Kc is the equilibrium constant

Relationship Between Kc and Kp

For gas reactions, we can denote equilibrium in terms of pressure (Kp) or concentration (Kc):

$K_{p} = K_{c} {(R T)}^{∆ n}$

Where:

R = 0.0821 L·atm·mol⁻¹·K⁻¹
T = temperature in Kelvin
Δn = (moles of gaseous products) - (moles of gaseous reactants)

Reverse reaction: K’c = 1/Kc

Multiply equation by n: K”c = (Kc)ⁿ

Temperature dependent: Kc changes only with temperature

Concentration independent: Same Kc regardless of starting amounts

Homogeneous and Heterogeneous Equilibria

In the Homogeneous Equilibria, all reactants and products are in the same phase.

Gas Phase Example:

N₂(g) + 3H₂(g) ⇌ 2NH₃(g)

Kc = [NH₃]² / [N₂][H₂]³

Kp for Gas Reactions:

$K_{p} = K_{c} {(R T)}^{∆ n}$

Where:

R = 0.0821 L·atm·mol⁻¹·K⁻¹
T = temperature (K)
Δn = (moles gaseous products) - (moles gaseous reactants)

Solution Phase Example:

CH₃COOC₂H₅(aq) + H₂O(l) ⇌ CH₃COOH(aq) + C₂H₅OH(aq)

Kc = [CH₃COOH][C₂H₅OH] / [CH₃COOC₂H₅][H₂O]

Heterogeneous Equilibria

In the Heterogeneous Equilibria, Reactants and products exist in different phases.

The main rule here is: pure solids and liquids are omitted from Kc expressions

Their concentrations remain constant regardless of the amount present.

Example: CaCO₃ Decomposition

CaCO₃(s) ⇌ CaO(s) + CO₂(g)

Kc = [CO₂] (solids omitted)

Kp = PCO₂

Applications of Equilibrium Constant

The applications of the equilibrium constant to:

predict the extent of a reaction based on its magnitude,
predict the direction of the reaction, and
Calculate equilibrium concentrations.

Predicting the Extent of Reaction

Large Kc (>10³) means reaction goes nearly to completion. Example: H₂ + Cl₂ → 2HCl (Kc = 4.0 × 10³¹)
Small Kc (<10⁻³) means very little product forms. Example: N₂ + O₂ → 2NO (Kc = 4.8 × 10⁻³¹)
Moderate Kc (10⁻³ to 10³) means good amounts of both. Example: H₂ + I₂ → 2HI (Kc = 57)

Predicting the Direction of Reaction

Calculate Reaction Quotient (Qc):

$Q_{c} = \frac{[C]^{c} [D]^{d}}{[A]^{a} [B]^{b}}$ (using current concentrations)

Compare Qc with Kc:

Qc < Kc: Forward reaction occurs
Qc > Kc: Reverse reaction occurs
Qc = Kc: System at equilibrium

Calculating Equilibrium Concentrations

ICE Table Method:

Initial concentrations
Change in concentrations
Equilibrium concentrations

To calculate equilibrium concentrations start with writing down a balanced equation, set up the ICE table, write the Kc expression, then substitute and solve, at last check the answer that make sense.

Relationship Between Equilibrium Constant K, Reaction Quotient Q, and Gibbs Energy G

Gibbs Energy and Spontaneity:

- ΔG < 0: Spontaneous forward reaction

- ΔG > 0: Non-spontaneous forward reaction

- ΔG = 0: System at equilibrium

Key Relationships:

ΔG = ΔG° + RT ln Q

At equilibrium (ΔG = 0, Q = K):

ΔG° = -RT ln K

Interpretation:

- ΔG° < 0: K > 1, products favored

- ΔG° > 0: K < 1, reactants favored

- ΔG° = 0: K = 1, equal amounts

This connects thermodynamics with equilibrium. Spontaneous reactions have equilibrium constants in favor.

Factors Affecting Equilibria

Le Chatelier’s Principle states that “If you stress an equilibrium, it shifts to relieve that stress.”

Check the factors affecting equilibria:

Effect of Concentration

Adding reactant: equilibria shift toward products
Adding product: shifts toward reactants
Removing reactant: shifts toward reactants
Removing product: shifts toward products

Example: H₂ + I₂ ⇌ 2HI

Adding H₂ pushes the equilibrium right, making more HI.

Effect of Pressure (Gas Reactions)

Increasing pressure equilibria shift toward the side with fewer gas molecules

Decreasing pressure results in shifts toward the side with more gas molecules

Example: N₂ + 3H₂ ⇌ 2NH₃

Left: 4 molecules, Right: 2 molecules
High pressure favors NH₃ formation

Effect of Temperature

For exothermic reactions (ΔH < 0):

Higher temperature results in Kc decrease, and shifts left
Lower temperature results in a Kc increase, shifts right

For endothermic reactions (ΔH > 0):

With higher temperature, Kc increases, and shifts right
With lower temperature, Kc decreases, and shifts left

Effect of Catalyst

Catalyst:

Speed up both forward and reverse reactions equally
Help reach the reaction to equilibrium faster
Does not change the equilibrium position
Does not change the Kc value

Effect of Inert Gas

At constant volume, there will be no effect on the equilibrium
Constant pressure may affect the equilibrium if volume changes

Ionic Equilibrium in Solution

An equilibrium where solutions containing ions that can conduct electricity are called an ionic equilibrium.

In the ionic equilibrium class 11, the Classification of Substances is:

Strong Electrolytes: Complete ionization

- HCl, NaOH, NaCl

- α ≈ 1 (degree of ionization)

Weak Electrolytes: Partial ionization

- CH₃COOH, NH₃, H₂O

- α < 1, equilibrium between ions and molecules

Non-electrolytes: No ionization

- Sugar, ethanol, urea

Ostwald’s Dilution Law:

For weak electrolyte AB ⇌ A⁺ + B⁻

Kc = Cα² / (1-α)

For very weak electrolytes (α << 1):

Kc ≈ Cα²

α ≈ √(Kc/C)

Dilution increases the degree of ionization for weak electrolytes.

Acids, Bases, and Salts

Acids, Bases, and Salts are present everywhere, from everyday uses to something that is made by our body. Common examples of acids are- Hydrochloric acid, Acetic acid. And the basis is washing soda. When acids and bases are mixed in the right proportion, they make salt. A common example of salt is sodium chloride.

As per the Arrhenius Theory:

Acid: Produces H⁺ ions in water

HCl → H⁺ + Cl⁻

Base: Produces OH⁻ ions in water

NaOH → Na⁺ + OH⁻

As per the Brønsted-Lowry Theory:

Acid: Proton (H⁺) donor

Base: Proton (H⁺) acceptor

Example: HCl + NH₃ → NH₄⁺ + Cl⁻

Conjugate pairs: HCl/Cl⁻ and NH₄⁺/NH₃

As per the Lewis Theory:

Acid: Electron pair acceptor

Base: Electron pair donor

Example: BF₃ + NH₃ → BF₃·NH₃

Autoionization of Water:

H₂O + H₂O ⇌ H₃O⁺ + OH⁻

Kw = [H⁺][OH⁻] = 1.0 × 10⁻¹⁴ at 25°C

pH Scale:

pH = -log[H⁺]

pOH = -log[OH⁻]

pH + pOH = 14 at 25°C

pH < 7: Acidic ; pH = 7: Neutral ; pH > 7: Basic

Ionization of Acids and Bases

The Arrhenius theory of acids and bases becomes ideal for the ionization of acids and bases, as most ionizations occur in an aqueous medium.

Weak Acid Ionization:

HA + H₂O ⇌ H₃O⁺ + A⁻

$K_{a} = \frac{[H^{+}] [A^{-}]}{[HA]}$

Larger Ka = Stronger acid

Weak Base Ionization:

B + H₂O ⇌ BH⁺ + OH⁻

$K_{b} = \frac{[{BH}^{+}] [{OH}^{-}]}{[B]}$

Conjugate Acid-Base Relationship:

Ka × Kb = Kw = 1.0 × 10⁻¹⁴

pKa + pKb = 14

Polyprotic Acids

Acids that have more than one ionizable proton per molecule of the acid are called polyprotic or polybasic acids. For example, sulphuric acid, oxalic acid, etc.

H₂SO₄: Ka1 >> Ka2

First ionization is easier than the second
A higher charge makes proton removal harder

Common Ion Effect:

Adding a common ion suppresses ionization.

Example: Adding CH₃COONa to CH₃COOH solution

- Increases [CH₃COO⁻]

- Shifts the equilibrium left

- Decreases [H⁺], increases pH

Salt Hydrolysis:

Strong acid + Strong base salt: Neutral (pH = 7)

Weak acid + Strong base salt: Basic (pH > 7)

Strong acid + Weak base salt: Acidic (pH < 7)

Weak acid + Weak base salt: Depends on Ka and Kb values

Buffer Solutions

The solution that resists pH changes when small amounts of acid or base are added is called a buffer solution.

Types of Buffer Solution:

Acidic Buffer: Weak acid + its salt

CH₃COOH + CH₃COONa

Basic Buffer: Weak base + its salt

NH₃ + NH₄Cl

How Buffers Work:

For CH₃COOH/CH₃COO⁻ buffer:

Adding H⁺: CH₃COO⁻ + H⁺ → CH₃COOH

Adding OH⁻: CH₃COOH + OH⁻ → CH₃COO⁻ + H₂O

Henderson-Hasselbalch Equation:

pH = pKa + log([Salt]/[Acid])

For basic buffer:

pOH = pKb + log([Salt]/[Base])

Buffer Capacity:

Depends on the concentration of buffer components and ratio of acid to conjugate base (best when ratio = 1)

Applications of Buffer solutions:

Blood pH regulation
Laboratory procedures
Industrial processes
Pharmaceutical formulations

Solubility Equilibria of Sparingly Soluble Salts

Product of molar concentrations of ions in a saturated solution, each raised to its stoichiometric power is called Solubility Product (Ksp).

For sparingly soluble salt MₓXᵧ:

$M_{x} X_{y} (s) ⇌ x M^{n +} (aq) + y X^{m -} (aq)$

$K_{sp} = [M^{n +}]^{x} [X^{m -}]^{y}$

Examples:

AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq)

Ksp = [Ag⁺][Cl⁻]

PbI₂(s) ⇌ Pb²⁺(aq) + 2I⁻(aq)

Ksp = [Pb²⁺][I⁻]²

Predicting Precipitation:

Calculate reaction quotient for solubility (Qsp):

Qsp > Ksp: Precipitation occurs
Qsp < Ksp: No precipitation, more dissolves
Qsp = Ksp: Saturated solution

Common Ion Effect on Solubility:

Adding a common ion decreases solubility.

Example: AgCl in NaCl solution

[Cl⁻] increases from NaCl
Equilibrium shifts left
AgCl solubility decreases

Chemical Equilibrium class 11 formulas

Below, we’ve listed all the important formulas discussed in the chemical equilibrium class 11.

Concept	Formula	Explanation
Equilibrium constant (conc.)	$K_{c} = \frac{[C]^{c} [D]^{d}}{[A]^{a} [B]^{b}}$	For aA + bB ⇌ cC + dD
Equilibrium constant (pressure)	$K_{p} = \frac{(P_{C})^{c} (P_{D})^{d}}{(P_{A})^{a} (P_{B})^{b}}$	In terms of partial pressures
Relation between Kp and Kc	$K_{p} = K_{c} {(R T)}^{∆ n}$	Δn = moles of gaseous products − moles of gaseous reactants
Reaction quotient	$Q_{c} = \frac{[Products]}{[Reactants]}$	Compare Q with K to know the direction
Degree of dissociation	$α = \frac{moles dissociated}{initial moles}$	For weak electrolytes
Ostwald’s dilution law	$K_{a} = \frac{C α^{2}}{(1 - α)}$	For weak acids
Ionization constant of a weak acid	$K_{a} = \frac{[H^{+}] [A^{-}]}{[HA]}$	Applies to weak acids
Ionization constant of a weak base	$K_{b} = \frac{[{BH}^{+}] [{OH}^{-}]}{[B]}$	Applies to weak bases
Ionic product of water	$K_{w} = [H^{+}] [{OH}^{-}]$	At 25 °C, Kw = 1 × 10^−14
pH definition	$pH = - log [H^{+}]$	Acidity scale
pOH definition	$pOH = - log [{OH}^{-}]$	Basicity scale
Relation of pH and pOH	$pH + pOH = 14$	Valid at 25 °C
Salt hydrolysis constant	$K_{h} = \frac{K_{w}}{K_{a}} or \frac{K_{w}}{K_{b}}$	For the hydrolysis of salts
Buffer equation (acidic buffer)	$pH = pK a + log \frac{[}{Salt}$	Henderson–Hasselbalch equation
Buffer equation (basic buffer)	$pOH = pK b + log \frac{[}{Salt}$	Can be converted to pH = 14 − pOH
Solubility product (simple salt)	$K_{sp} = [A^{+}] [B^{-}]$	For AB ⇌ A⁺ + B⁻
Solubility product (general salt)	$K_{sp} = [A^{y +}]^{x} [B^{x -}]^{y}$	For A_xB_y ⇌ xA^y+ + yB^x−

Check also: NCERT Class 11 Chemistry Equilibrium pdf

Revision Notes for Class 11 Chemistry

Here you can check the Class 11 chemistry notes for other chapters:

Some Basic Concepts of Chemistry Class 11 Notes	Thermodynamics Class 11 Notes
Structure of Atom Class 11 Notes	Equilibrium Class 11 Notes
Classification of Elements and Periodicity in Properties Class 11 Notes	Redox Reaction Class 11 Notes
Chemical Bonding and Molecular Structure Class 11 Notes	Hydrogen Class 11 Notes
States of Matter Class 11 Notes	The s-Block Element Class 11 Notes
Organic Chemistry – Some Basic Principles and Techniques Class 11 Notes	The p-Block Element Class 11 Notes
Hydrocarbons Class 11 Notes	Environmental Chemistry Class 11 Notes

NCERT Solutions for Class 11 Chemistry

Here are the class 11 chemistry NCERT solutions.

Some Basic Concepts of Chemistry Class 11 NCERT Solutions	Thermodynamics Class 11 NCERT Solutions
Structure of Atom Class 11 NCERT Solutions	Equilibrium Class 11 NCERT Solutions
Classification of Elements and Periodicity in Properties Class 11 NCERT Solutions	Redox Reaction Class 11 NCERT Solutions
Chemical Bonding and Molecular Structure Class 11 NCERT Solutions	Hydrogen Class 11 NCERT Solutions
States of Matter Class 11 NCERT Solutions	The s-Block Element Class 11 NCERT Solutions
Organic Chemistry – Some Basic Principles and Techniques Class 11 NCERT Solutions	The p-Block Element Class 11 NCERT Solutions
Hydrocarbons Class 11 NCERT Solutions	Environmental Chemistry Class 11 NCERT Solutions

About the Content Reviewer

Praveen Kumar is a Mathematics and Chemistry Subject Matter expert. He has extensive knowledge and experience in teaching JEE Main and Class 11, and 12 students. Having cleared JEE Main with an AIR of 5000, ISI Admission Test (AIR 1), and GATE (AIR 5) allows him to leverage his personal experience with these types of exams to validate accuracy, depth and exam alignment with the Maths and Chemistry content. He ensures that the content that he has reviewed for Shiksha emphasizes unresolved issues, trust and exam relevance, as well as high syllabus standards.

Chemical Equilibrium FAQs

Here are the frequently asked questions from class 11 chemistry equilibrium.

Commonly asked questions

What is the equilibrium constant?

The equilibrium constant is the ratio of the concentrations of products to reactants, each raised to the power of their stoichiometric coefficients. For reversible reactions.

What is Le Chatelier’s principle NCERT?

If the conditions of equilibrium are changed, it shifts to oppose the change. For example, in Haber's process, high pressure favors NH? formation.

What is dynamic equilibrium?

It happens in reversible reactions when the rate of the forward reaction becomes equal to the rate of the backward reaction. Result in the same concentration of reactants and product.

Chemistry Chemical Equilibrium Exam

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News & Updates

Latest NewsPopular News

Chemical Equilibrium Class 11: Equilibrium Meaning, Formulas, Notes

What is Equilibrium?

Equilibrium in Physical Processes

Solid-Liquid Equilibrium (Ice and Water)

Liquid-Vapor Equilibrium

Dissolving Sugar in Water

Equilibrium in Chemical Processes – Dynamic Equilibrium

Law of Chemical Equilibrium and Equilibrium Constant

Relationship Between Kc and Kp

Homogeneous and Heterogeneous Equilibria

Heterogeneous Equilibria

Applications of Equilibrium Constant

Predicting the Extent of Reaction

Predicting the Direction of Reaction

Calculating Equilibrium Concentrations

Relationship Between Equilibrium Constant K, Reaction Quotient Q, and Gibbs Energy G

Factors Affecting Equilibria

Ionic Equilibrium in Solution

Acids, Bases, and Salts

Ionization of Acids and Bases

Polyprotic Acids

Common Ion Effect:

Salt Hydrolysis:

Buffer Solutions

Buffer Capacity:

Solubility Equilibria of Sparingly Soluble Salts

Common Ion Effect on Solubility:

Chemical Equilibrium class 11 formulas

Revision Notes for Class 11 Chemistry

NCERT Solutions for Class 11 Chemistry

About the Content Reviewer

Chemical Equilibrium FAQs

Commonly asked questions

Student Forum

Other Topics under this Chapter

Other Class 11th Chemistry Chapters

Popular Courses After 12th

News & Updates