Equilibrium Class 11 Chemistry Notes PDF Download
Chapter 6 Class 11 Chemistry Equilibrium gives insight into the dynamic balance between the forward and reverse reactions in a chemical process. Equilibrium is a fundamental concept in chemistry that describes a state where the concentrations of reactants and products remain constant over time. This chapter explains equilibrium principles, including the conditions required for a system to reach equilibrium, the factors affecting equilibrium position, and the quantitative measures used to express equilibrium states.
Table of Content
Chapter 6 Equilibrium Class 11 Notes lets you quickly access and review the chapter content. For a comprehensive study experience, check out the Class 11 Chemistry Revision Notes FREE PDF here and refer to the CBSE Class 11 Chemistry Syllabus for detailed coverage. Vedantu's notes offer a focused, student-friendly approach, setting them apart from other resources and providing you with the best tools for success.
Introduction:
Chemical equilibrium is an important aspect of chemical and biological processes. When a liquid evaporates in a closed container, molecules with relatively higher kinetic energy escape the liquid surface into the vapour phase and a number of liquid molecules from the vapour phase strike the liquid surface and are retained in the liquid phase. It gives rise to a constant vapour pressure because of an equilibrium in which the number of molecules leaving the liquid equals the number returning to liquid from the vapour. Then the system has reached an equilibrium state at this stage. Thus, at equilibrium, the rate of evaporation is equal to the rate of condensation. It may be represented by
${{H}_{2}}O(l)\rightleftharpoons {{H}_{2}}O(vap)$
The above double arrow indicates that the process is simultaneously going in both directions. The equilibrium mixture means the mixture of reactants and products in the equilibrium state is called an equilibrium mixture.
Equilibrium in the Physical Process:
The most important and familiar examples are the phase transformation process. Eg.
\[solid\rightleftharpoons liquid\]
\[liquid\rightleftharpoons gas\]
\[solid\rightleftharpoons gas\]
a. Solid-liquid Equilibrium:
At a particular temperature and pressure, the ice and water are in equilibrium. The normal melting point or normal freezing point of the substance is the temperature at which the solid phase and liquid phases are equilibrium for any pure substance at atmospheric pressure. When the system at dynamic equilibrium will follow the conditions:
Both the opposing processes occur simultaneously
Both the process occurs at the same rate, so the amount of ice and water remains constant.
b. Liquid-vapour equilibrium:
The number of water molecules from the gaseous state into the liquid state increases until the equilibrium is reached.
i.e rate of evaporation = rate of condensation
\[{{H}_{2}}O(l)\rightleftharpoons {{H}_{2}}O(vap)\]
The equilibrium water pressure means the pressure at which the water molecules remain constant at a given temperature and the vapour pressure of water increases with temperature. The evaporation of liquid depends on,
The nature of the liquid
The amount of liquid
The applied temperature
For example, in a closed vessel water and water vapour are in equilibrium at atmospheric pressure (1.031 bar) and at ${{100}^{0}}C$. This means that for any pure liquid at one atmospheric pressure (1.0301 bar), then the normal boiling point is said to be the temperature at which liquid and vapours are at equilibrium.
c. Solid-vapour Equilibrium:
The examples of this type of system are considered when solids sublime to vapour phase.
\[{{I}_{2}}(solid)\rightleftharpoons {{I}_{2}}(vapour)\]
\[Camphor(solid)\rightleftharpoons Camphor(vapour)\]
\[N{{H}_{4}}Cl(solid)\rightleftharpoons N{{H}_{4}}Cl(vapour)\]
General Characteristics of Equilibrium Involving Physical Processes:
Following characteristics are common to the system at equilibrium for which the physical processes are discussed above:
At a given temperature, the equilibrium is possible only in closed systems
If both the opposing processes occur at the same rate then the system is dynamic at stable conditions.
In a system, all measurable properties remain constant.
Equilibrium is characterized by the constant value of one of its parameters given, and then it is attained for a physical process.
Equilibrium in Chemical Processes:
Compared to physical systems chemical reactions also attain a state of equilibrium. Those chemical reactions can occur both in forward and backward directions. The chemical equilibrium is dynamic in nature when the rate of forward reaction is equal to the rate of backward reaction.
Reversible Reactions
Consider a general case of a reversible reaction
\[A+B\rightleftharpoons C+D\]
With passing time, there is a decrease in the concentration of reactants A and B, the increase in the concentration of products C and D which leads to a decrease in the rate of forward reaction and an increase in the rate of backward reaction.
When the two reactions occur at the same rate the system will reach a state of equilibrium.
Irreversible Reactions
The products formed will not react back to form reactants under the same conditions; it is called the Irreversible reaction. These reactions cannot take place in the reverse direction.
\[A+B\to C+D\]
It is a state of minimum Gibb’s energy
$dG=0 and \Delta G=0$ at this state
Rate of forward reaction = rate of backward reaction
This equilibrium is dynamic and stable in nature
Dynamic Nature of Chemical Equilibrium:
This dynamic nature of chemical equilibrium can be explained by the synthesis of ammonia by Haber’s process. This process starts with definite amounts of ${{N}_{2}}\ and {{H}_{2}}$ and carries out a reaction when equilibrium is attained at a particular temperature. At equilibrium the concentrations of ${{N}_{2}},{{H}_{2}}\ and N{{H}_{3}}$ are constant.
Characteristics of Chemical Equilibrium:
The concentration of each of the reactants and the products become constant at equilibrium.
The rate of forward reaction becomes equal to the rate of backward reaction at equilibrium and hence equilibrium is dynamic in nature.
None of the products is allowed to escape out or separate out as a solid then only chemical equilibrium can be established.
Equilibrium Constant:
An equilibrium mixture is a mixture of reactants and products in the equilibrium state.
Consider a general reversible reaction,
\[A+B\rightleftharpoons C+D\]
Where A and B are reactants, C and D are products in the above-balanced equation.
\[{{K}_{c}}=\frac{[C][D]}{[A][B]}\]
Where ${{K}_{c}}$ is Equilibrium constant and expressed in concentrations of $mol{{L}^{-1}}$.
At a given temperature, the product of concentrations of the reaction products raised to the respective stoichiometric coefficient in the balanced chemical equation divided by the product of concentrations of the reactants raised to their individual stoichiometric coefficients has a constant value. This is known as the Equilibrium Law or Law of Chemical Equilibrium.
The equilibrium constant for the general reaction,
\[aA+bB\rightleftharpoons cC+dD\]
Equilibrium constant expressed as,
\[{{K}_{c}}=\dfrac{{{[C]}^{c}}{{[D]}^{d}}}{{{[A]}^{a}}{{[B]}^{b}}}\]
Where [A], [B], [C] and [D] are the equilibrium concentrations of the reactants and products.
The product of the molar concentrations of the products, each raised to the power equal to its stoichiometric coefficient is divided by the product of the molar concentrations of the reactants, each raised to the power equal to its stoichiometric coefficient is constant at constant temperature is called Equilibrium Constant.
Characteristics of Equilibrium Constant:
The value of the equilibrium constant for a particular reaction is always constant depending only upon the temperature of the reaction and is independent of the concentrations of the reactants with which we start or the direction from which the equilibrium approached.
The value of the equilibrium constant is inverted when the reaction is reversed.
The equilibrium constant for the new equation is the square root of K. (i.e, $\sqrt{K}$) when the equation is divided by 2.
The equilibrium constant for the new equation is the square of K (i.e, ${{K}^{2}}$) when the equation is multiplied by 2.
By the addition of a catalyst to the reaction will not affect the value of the equilibrium.
Predicting the Extent of Reaction:
If the value of ${{K}_{c}}>{{10}^{3}}$ will be high, then the equilibrium reaction is forward dominant.
If the value of ${{K}_{c}}<{{10}^{-3}}$, then the equilibrium reaction is backward dominant.
Moderate value of ${{K}_{c}}$ (between ${{10}^{3}}$ and ${{10}^{-3}}$ ) dominates equilibrium neither directions.
Predicting the Direction of the Reaction- Reaction Quotient:
We can predict the direction in which the reaction will proceed at any stage with the help of the equilibrium constant. Reaction Quotient Q will calculate for this purpose. In the same way as the equilibrium constant ${{K}_{c}}$, Reaction Quotient will not necessarily equilibrium values.
\[aA+bB\rightleftharpoons cC+dD\]
\[{{Q}_{c}}=\dfrac{{{[C]}^{c}}{{[D]}^{d}}}{{{[A]}^{a}}{{[B]}^{b}}}\]
the reaction proceeds in the direction of reactants when ${{Q}_{c}}>{{K}_{c}}$
the reaction proceeds in the direction of products when ${{Q}_{c}}<{{K}_{c}}$
the reaction mixture is already at equilibrium when ${{Q}_{c}}={{K}_{c}}$
Calculating Equilibrium Concentrations:
In case of a problem in which we know the initial concentrations but do not know any of the equilibrium concentrations, the following three steps shall be followed:
Step 1. Write the balanced equation for the reaction.
Step 2. Under the balanced equation, make a table that lists for each substance involved in the reaction:
(a) The initial concentration,
(b) The change in concentration on going to equilibrium, and
(c) The equilibrium concentration.
In constructing the table, define x as the concentration (mol/L) of one of the substances that react on going to equilibrium, then use the stoichiometry of the reaction to determine the concentrations of the other substances in terms of x.
Step 3. Substitute the equilibrium concentrations into the equilibrium equation for the reaction and solve for x.
Step 4. Calculate the equilibrium concentrations from the calculated value of x.
Step 5. Check your results by substituting them into the equilibrium equation.
Relationship Between Equilibrium Constant and Gibbs Free Energy:
The value of the equilibrium constant for a reaction does not depend on the rate of reaction.
If a change in Gibbs free energy $\Delta G$ is negative, then the reaction proceeds in the forward reaction with spontaneity.
If a change in Gibbs free energy $\Delta G$ is positive, the products of the forward reaction shall be converted to reactants with a non-spontaneous process.
If the change in Gibbs free energy is zero, then the reaction is at equilibrium and there is no longer any free energy left to drive the reaction.
A mathematical expression of this thermodynamic view of equilibrium can be described by the following equation:
\[\Delta G=\Delta {{G}^{o}}+RT\ln Q\]
Where Q is Reaction Quotient. And ${{G}^{o}}$ is standard Gibbs free energy.
At equilibrium, when $\Delta {{G}^{o}}=0$ and ${{Q}_{c}}={{K}_{c}}$
\[\Delta G=\Delta {{G}^{o}}+RT\ln {{K}_{c}}=0\]
\[\Delta {{G}^{o}}=-RT\ln {{K}_{c}}\]
\[\ln {{K}_{c}}=-\frac{\Delta {{G}^{o}}}{RT}\]
\[K={{e}^{-\frac{\Delta {{G}^{o}}}{RT}}}\]
from the above equation, from the values of $\Delta {{G}^{o}}$ , the reaction spontaneity can be interpreted.
Homogeneous Equilibria:
In an equilibrium system, all the reactants and products are in the same phase is known as a Homogenous system.
For example,
\[{{N}_{2}}(g)+3{{H}_{2}}(g)\rightleftharpoons 2N{{H}_{3}}(g)\]
In the above reaction, all reactants and products are in the gaseous phase.
\[C{{H}_{3}}COO{{C}_{2}}{{H}_{5}}(aq)+{{H}_{2}}O(l)\rightleftharpoons C{{H}_{3}}COOH(aq)+{{C}_{2}}{{H}_{5}}COOH(aq)\]
In the above reaction, all reactants and products are homogeneous solution phases.
Heterogeneous Equilibria:
An equilibrium system having more than one phase is called heterogeneous equilibrium. A familiar example of this type of system is the equilibrium between water vapour and liquid water in a closed container.
\[{{H}_{2}}O(l)\rightleftharpoons {{H}_{2}}O(vap)\]
Similarly, there is an equilibrium between a solid and its saturated solution is a heterogeneous equilibrium.
\[Ca{{(OH)}_{2}}(s)\rightleftharpoons C{{a}^{+2}}(aq)+2O{{H}^{-}}(aq)\]
Le Chatelier’s Principle:
This principle helps to decide what course of reaction adopts and make a qualitative prediction about the effect of changes in conditions on equilibrium.
It states that “a change in any of the factors that determine the equilibrium conditions of a system will cause the system to change in such a manner so as to reduce or to counteract the effect of the change.”
This principle is applicable to all physical and chemical equilibrium systems.
a. Effect of Concentration Change:
Generally, equilibrium is disturbed by the addition or removal of any reactant or product. Le Chatelier’s principle predicts that:
The concentration of added reactant or product is relieved by the direction of the net reaction that consumes the added substance.
The concentration of removed reactant or product is relieved by the direction of net reaction that replenishes the added substance.
“When the concentration of any of the reactants or products in a reaction at equilibrium is changed, the composition of the equilibrium mixture changes so as to minimize the effect of concentration changes”.
b. Effect of Temperature Change
Because the Reaction Quotient Q no longer equals to the equilibrium constant when equilibrium is disturbed by the change in the concentrations, pressure, or volume and further the composition of the equilibrium mixture changes.
The value of the equilibrium constant changed when a change in the temperature was observed.
Generally, the temperature change dependence of the equilibrium constant depends on the sign of the $\Delta H$ for the reaction.
If $\Delta H$is negative, then the equilibrium constant of the exothermic reaction decreases because of temperature increases.
If $\Delta H$ is positive, then the equilibrium constant of the endothermic reaction increases because temperature decreases.
c. Effect of Pressure Change
When the total number of moles of gaseous products is different, then a pressure change obtained changing the volume can affect the yield of products in case of a gaseous reaction. Le-Chatlier’s principle is applied to heterogeneous equilibrium the effect of pressure changes on solids and liquids can be ignored because the volume of a solution or liquid is nearly independent of pressure.
d. Effect of Volume Change
The effect of the increase in pressure is equivalent to the effect of a decrease in volume.
So the effect of a decrease in volume will be shifted towards the equilibrium in the direction in which the number of moles decreases.
e. Effect a Catalyst
The chemical substance increases the rate of a chemical reaction by making available a new low energy pathway for the conversion of reactants to products.
A catalyst increases the rate of forward and reverse reactions that pass through the transition state which does not affect the equilibrium.
For example: contact process
\[2S{{O}_{2}}(g)+{{O}_{2}}(g)\rightleftharpoons 2S{{O}_{3}}(g);{{K}_{c}}=1.7\times {{10}^{26}}\]
The oxidation of sulphur dioxide to sulphur trioxide is partially very slow.
So, Platinum or divanidium pentaoxide (${{V}_{2}}{{O}_{5}}$) is used as a catalyst to increase the rate of reaction.
f. Effect of Inert Gas Addition
If an inert gas such as argon is added and the volume kept constant this does not take any part in the reaction. Thus, the equilibrium remains undisturbed. Because the addition of inert gas does not change the partial pressures or the molar concentrations of substances involved in the reaction at constant volume.
Ionic Equilibrium in Solution:
An example of equilibrium reaction that involves ions is given below:
\[F{{e}^{+3}}(aq)+SC{{N}^{-}}(aq)\rightleftharpoons [Fe{{(SCN]}^{2+}}(aq)\]
Micheal Faraday classified the substances into two categories based on their ability to conduct electricity. Those are electrolytes and non-electrolytes. Because of the presence of the ions, the solution of electrolytes conducts electricity.
For example, an aqueous solution of NaCl is dissociated completely into $N{{a}^{+}}$ and $C{{l}^{-}}$ ions, because of almost 100% ionization in the case of sodium chloride. But in the case of acetic acid which is a weak electrolyte, 5% ionization takes place. Hence, equilibrium is established between ions and unionized molecules in weak electrolytes. This type of equilibrium involving ions in an aqueous solution is known as Ionic Equilibrium.
Classification of Electrolytes:
Based on the strength of electrolytes are classified into two categories:
a. Strong Electrolytes
The electrolytes which are easily broken into ions with complete dissociation are known as strong electrolytes.
Eg: $HCl,NaCl,NaOH,HN{{O}_{3}},HCl{{O}_{4}},{{H}_{2}}S{{O}_{4}}$
b. Weak Electrolytes
The electrolytes which are partially dissociated into ions are known as weak electrolytes.
Eg:$C{{H}_{3}}COOH,N{{H}_{4}}OH,HCN,{{H}_{2}}{{C}_{2}}{{O}_{4}}$ , and all organic acids and bases.
Acids, Bases and Salts:
There are various theories from the explanation of acids and bases.
Arrhenius Concept of Acids and Bases:
According to this theory, acids are the substances that give ${{H}^{+}}$ ions and bases which produce hydroxyl ions $O{{H}^{-}}$ in water.
Arrhenius Acids:
According to Arrhenius, acids increases the ${{H}^{+}}$ ion concentration in water.
The ionization of acids can be represented by the following equations for HX (aq):
\[HX(aq)\to {{H}^{+}}(aq)+{{X}^{-}}(aq)\]
Or
\[HX(aq)+{{H}_{2}}O(l)\to {{H}_{3}}{{O}^{+}}(aq)+{{X}^{-}}(aq)\]
Example: $HCl,{{H}_{2}}S{{O}_{4}},C{{H}_{3}}COOH$ etc.
Arrhenius Bases:
According to Arrhenius, bases increases the $O{{H}^{-}}$ ions concentration in water.
For example: $NaOH,Ca{{(OH)}_{2}},N{{H}_{4}}OH$ etc.
MOH like base molecule ionizes in aqueous solution from the following equation:
\[MOH(aq)\to {{M}^{+}}(aq)+O{{H}^{-}}(aq)\]
The Bronsted-lowry Acids and Bases:
According to Brönsted-Lowry theory, acid is a substance that is capable of donating a hydrogen ion H+ and bases are substances capable of accepting a hydrogen ion, H+. In short, acids are proton donors and bases are proton acceptors.
For example, consider the dissolution of $N{{H}_{3}}$ in ${{H}_{2}}O$ represented by the following equation:
In this reaction, a water molecule that acts as a proton donor is known as Lowry-Bronsted acid and an ammonia molecule that acts as a proton acceptor is known as Lowry-Bronsted base.
Lewis Acids and Bases:
According to this theory, acid is a species that accepts electron pairs and a base that donates electron pairs.
A typical example is a reaction of electron-deficient species $B{{F}_{3}}$ with $N{{H}_{3}}$.
\[B{{F}_{3}}+:N{{H}_{3}}\to B{{F}_{3}}:N{{H}_{3}}\]
Lewis acids are electron deficient species like $AlC{{l}_{3}},C{{o}^{+3}},M{{g}^{+2}}$
Lewis bases are electron pair donating species like ${{H}_{2}}O,N{{H}_{3}},O{{H}^{-}}$
The Ionization Constant of Water and its Ionic Product:
Some substances have the unique ability to act as acid and base like water molecules.
\[{{H}_{2}}O(l)(acid)+{{H}_{2}}O(l)(base)\rightleftharpoons {{H}_{3}}{{O}^{+}}(aq)+O{{H}^{-}}(aq)\]
The dissociation constant is represented by:
\[K=\frac{[{{H}_{3}}{{O}^{+}}][O{{H}^{-}}]}{[{{H}_{2}}O]}\]
The concentration of water is neglected from the denominator as water is pure liquid and its concentration remains constant. Then the equilibrium constant is known as the Ionic Product Of Water, ${{K}_{w}}$.
\[{{K}_{w}}=[{{H}^{+}}][O{{H}^{-}}]\]
The concentration of ${{H}^{+}}$ = $1.0\times {{10}^{-7}}M$ at 298K
As dissociation of water produces equal number of $[{{H}^{+}}]=[O{{H}^{-}}]=1\times {{10}^{-7}}M$
\[{{K}_{w}}=[{{H}_{3}}{{O}^{+}}][O{{H}^{-}}]={{(1\times{{10}^{-7}})}^{2}}=1\times {{10}^{-14}}{{M}^{2}}\]
We can distinguish acidic, neutral, and basic solutions by the relative values of the ${{H}_{3}}{{O}^{+}}$ and $O{{H}^{-}}$ concentrations:
Acidic: $[{{H}_{3}}{{O}^{+}}]>[O{{H}^{-}}]$
Neutral: $[{{H}_{3}}{{O}^{+}}]=[O{{H}^{-}}]$
Basic: $[{{H}_{3}}{{O}^{+}}]<[O{{H}^{-}}]$
The pH Scale:
Hydronium ion concentration in molarity is more conveniently expressed on a logarithmic scale known as the pH Scale.
From the definition of pH,
\[pH=-\log [{{H}^{+}}]\]
at ${{25}^{o}}C$ pure water has a concentration of hydrogen ions, $[{{H}^{+}}]={{10}^{-7}}M$
Hence, the pH of pure water is given as:
$pH=-\log ({{10}^{-7}})=7$
Acidic solutions posses a concentration of hydrogen ions, $[O{{H}^{-}}]>{{10}^{-7}}M$ , while basic solution posses a concentration of hydrogen ions, $[O{{H}^{-}}]<{{10}^{-7}}M$
Thus, acidic solution has pH < 7
Basic solution has pH < 7
Neutral solution has pH = 7
Then the ionic product of water, ${{K}_{w}}=[{{H}_{3}}{{O}^{+}}][O{{H}^{-}}]={{10}^{-14}}$
Taking negative logarithms on both sides of equation, we obtain
\[-\log {{K}_{w}}=-\log \{[{{H}_{3}}{{O}^{+}}][O{{H}^{-}}]\}=-\log {{10}^{-14}}\]
\[p{{K}_{w}}=pH+pOH=14\]
Ionization Constants of Weak Acids:
Consider a weak acid HX which is partially ionized in the aqueous solution. The equilibrium can be expressed by:
\[\underset{\text{Initial concentration}}{\mathop{{}}}\,\underset{0}{\mathop{HX(aq)}}\,+{{H}_{2}}O(l)\rightleftharpoons \underset{0}{\mathop{{{H}_{3}}{{O}^{+}}(aq)}}\,+\underset{0}{\mathop{{{X}^{-}}(aq)}}\,\]
Let $\alpha$ be the extent of ionization change (M)
\[\underset{-C\alpha }{\mathop{HX(aq)}}\,+{{H}_{2}}O(l)\rightleftharpoons \underset{\alpha }{\mathop{{{H}_{3}}{{O}^{+}}(aq)}}\,+\underset{\alpha }{\mathop{{{X}^{-}}(aq)}}\,\]
\[\underset{\text{Equilibrium concentration}}{\mathop{{}}}\,\underset{C-C\alpha }{\mathop{HX(aq)}}\,+{{H}_{2}}O(l)\rightleftharpoons \underset{C\alpha }{\mathop{{{H}_{3}}{{O}^{+}}(aq)}}\,+\underset{C\alpha }{\mathop{{{X}^{-}}(aq)}}\,\]
Here C is the initial concentration of the undissociated acid HX. Then the equilibrium constant for the above-discussed reaction is,
\[{{K}_{a}}=\frac{{{C}^{2}}{{\alpha }^{2}}}{C(1-\alpha )}=\frac{C{{\alpha }^{2}}}{1-\alpha }\]
${{K}_{a}}$ is called acid dissociation constant or ionization constant.
the pH scale for the hydrogen ion concentration has been so useful that besides $p{{K}_{w}}$
\[p{{K}_{a}}=-\log ({{K}_{a}})\]
Ionization Constants of Weak Bases:
The ionization of base MOH can be represented by the following equation:
\[MOH\rightleftharpoons {{M}^{+}}(aq)+O{{H}^{-}}(aq)\]
In a weak base there is partial ionization of MOH into M+ and OH–, the case is similar to that of acid-dissociation equilibrium. The equilibrium constant for base ionization is called the base ionization constant and is represented by Kb. It can be expressed in terms of concentration in molarity of various species
in equilibrium by the following equation:
\[{{K}_{b}}=\frac{[{{M}^{+}}][O{{H}^{-}}]}{[MOH]}\]
Alternatively, if c = initial concentration of base and $\alpha $ = degree of ionization of base i.e. the extent to which the base ionizes. When equilibrium is reached, the equilibrium constant can be written as:
\[{{K}_{b}}=\frac{{{C}^{2}}{{\alpha }^{2}}}{C(1-\alpha )}=\frac{C{{\alpha }^{2}}}{1-\alpha }\]
The pH scale for the hydrogen ion concentration has been extended to get:
\[p{{K}_{b}}=-\log ({{K}_{b}})\]
Hydrolysis of Salts:
The reactions between acids and bases in definite proportions form salts that undergo ionization in water. The cations or anions formed on ionization of salts either exit as hydrated ions in aqueous solutions or interact with water to reform corresponding acids or bases depending upon the natural salts.
Consider three types of hydrolysis of the salts:
Salts of weak acid and strong base. Eg:$C{{H}_{3}}COONa$
\[pH=7+\frac{1}{2}p{{K}_{a}}-\frac{1}{2}\log C\]
Salts of strong acid and weak base. Eg: $N{{H}_{4}}Cl$
\[pH=7-\frac{1}{2}p{{K}_{b}}-\frac{1}{2}\log C\]
Salts of weak acid and weak base. Eg: $C{{H}_{3}}COON{{H}_{4}}$
\[pH=7+\frac{1}{2}p{{K}_{b}}-\frac{1}{2}p{{K}_{b}}\]
Buffer Solutions:
Many body fluids e.g., blood or urine have definite pH and any deviation in their pH indicates malfunctioning of the body. The control of pH is also very important in many chemical and biochemical processes. Many medical and cosmetic formulations require that these be kept and administered at a
particular pH. The solutions which resist change in pH on dilution are called Buffer Solutions.
Solubility Equilibria of Sparingly Soluble Salts:
Each salt has its characteristic solubility which depends on temperature. We classify salts on the basis of their solubility in the following three categories
For example,
\[AgCl(s)\leftrightarrow A{{g}^{+}}(aq)+C{{l}^{-}}(aq)\]
Solubility product, ${{K}_{sp}}=[A{{g}^{+}}][C{{l}^{-}}]=I.P$ = Ionic product
If ionic product < ${{K}_{sp}}$ , more salt can be dissolved through forward reaction.
If ionic product > ${{K}_{sp}}$, precipitation of solid salt takes place as a backward reaction.
If ionic product = ${{K}_{sp}}$ no more salt can be dissolved. This stage is the saturation point.
Class 11 Chemistry Chapter 6 Important Topics and Subtopics Covered
Class 11 Chemistry Chapters 6 Formulas
1. Equilibrium Constant 𝐾𝑐= $\dfrac{[C]c \cdot [D]d}{[A]a \cdot [B]b}$
Le Chatelier's Principle:
States that if a system at equilibrium is subjected to a change in concentration, temperature, or pressure, the system will adjust itself to counteract the effect of the change and establish a new equilibrium.
Equilibrium Shifts:
Effect of Concentration: Adding reactants or products will shift the equilibrium in the direction that reduces the concentration change.
Effect of Temperature: Increasing temperature favors endothermic reactions, while decreasing temperature favors exothermic reactions.
Effect of Pressure (for gases): Increasing pressure favors the side with fewer moles of gas.
Equilibrium Expressions for Heterogeneous Equilibria:
For 𝑎𝐴+𝑏𝐵⇌𝑐𝐶+𝑑𝐷
aA+bB⇌cC+dD involving solids or liquids, their concentrations are considered constant and not included in the equilibrium expression.
Importance of Revision Notes for Class 11 Chemistry Chapter 6
Summarises Key Points: Condenses important concepts for quick review.
Saves Time: Provides a fast way to revise before exams.
Highlights Essentials: Focuses on crucial topics and definitions.
Improves Memory: Helps in better retention of information.
Enhances Exam Prep: Targets weak areas for more effective study.
Clarifies Concepts: Simplifies complex ideas for easier understanding.
Includes Visuals: Uses diagrams and charts for better grasp.
Boosts Confidence: Prepares students thoroughly for exams.
Tips For Learning the Class 11 Chapter 6
Focus on core processes with illustrations and examples.
Draw and label diagrams for clarity.
Create summaries of each process.
Connect concepts to everyday examples.
Solve past exam questions to test understanding.
Explain concepts to others to reinforce learning.
Revisit material frequently to retain information.
Conclusion
Chapter 6 on Equilibrium is crucial for understanding how chemical systems behave under various conditions. Students gain valuable insights into both theoretical and practical aspects of chemistry by learning about equilibrium constants, Le Chatelier's Principle, and the factors affecting equilibrium. This knowledge is foundational for further studies in chemistry and for applying these concepts to real-world scenarios, such as industrial reactions and environmental processes.
Related Study Materials for Class 11 Chapter 6
Revision Notes Links For Class 11 Chemistry Revision Notes
Related Study Materials Links for Class 11 Chemistry
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FAQs on Equilibrium Class 11 Notes: CBSE Chemistry Chapter 6
1. How can I quickly summarise the concept of chemical equilibrium for revision?
A quick summary is that chemical equilibrium is a dynamic state in a reversible reaction where the rate of the forward reaction equals the rate of the reverse reaction. This results in the concentrations of reactants and products remaining constant over time, even though the reactions are still occurring. It's a state of balance, not inactivity. For a deeper dive, you can explore the Law of Chemical Equilibrium in detail.
2. What are the essential topics to cover in a quick revision of the Class 11 Equilibrium chapter?
For an effective revision of the Equilibrium chapter, you should create a concept map covering these key areas:
Equilibrium in Physical and Chemical Processes: Understanding the basic dynamic nature.
Law of Mass Action and the Equilibrium Constant: The relationship between Kp and Kc.
Le Chatelier's Principle: How changes in concentration, temperature, and pressure affect equilibrium.
Ionic Equilibrium: Concepts of acids, bases, pH, and hydrolysis.
Buffer Solutions and Solubility Product (Ksp): Their definitions and applications.
After revising, test your knowledge with Important Questions for Class 11 Chemistry Chapter 6.
3. How does Le Chatelier's principle help in predicting the outcome of a reaction, and what's a common mistake to avoid?
Le Chatelier's principle states that if a change is applied to a system at equilibrium, the system will shift to counteract that change. This helps predict how a reaction will respond to changes in concentration, temperature, or pressure. A common mistake is thinking that adding a catalyst will shift the equilibrium. A catalyst speeds up both the forward and reverse reactions equally, allowing equilibrium to be reached faster, but it does not change the position of the equilibrium itself.
4. Why are there two different equilibrium constants, Kp and Kc, and how do I decide which to use in a summary?
The two constants exist to handle different states of matter. Kc uses the molar concentrations of reactants and products, making it suitable for aqueous solutions and gases. Kp uses the partial pressures of gaseous reactants and products, so it is used exclusively for reactions involving gases. For a quick summary, remember: if the reaction involves gases and pressure data is given, use Kp. Otherwise, or for solutions, use Kc. They are related by the formula Kp = Kc(RT)Δng.
5. What is the core difference to remember between physical and chemical equilibrium for a quick recap?
The core difference is in the substances involved. Physical equilibrium is a balance between different phases of the same substance, such as ice melting into water (H₂O(s) ⇌ H₂O(l)). Chemical equilibrium, however, is a balance in a reversible reaction where reactants form new chemical substances (products), such as the formation of ammonia from nitrogen and hydrogen (N₂(g) + 3H₂(g) ⇌ 2NH₃(g)).
6. Can you provide a concise summary of ionic equilibrium for Class 11?
Ionic equilibrium is the part of the chapter that deals with the balance of ions in a solution, mainly in water. The key concepts to summarise are:
Acids and Bases: Definitions according to Arrhenius, Brønsted-Lowry, and Lewis theories.
Ionization: The extent to which strong and weak electrolytes dissociate into ions.
pH Scale: A measure of acidity or alkalinity in a solution.
Salt Hydrolysis: How salts of weak acids or bases react with water to change the solution's pH.
For practice, you can refer to the NCERT Solutions for Class 11 Equilibrium.
7. Beyond definitions, how do buffer solutions and the common ion effect practically work to resist pH changes?
Practically, a buffer solution (a weak acid and its conjugate base, or a weak base and its conjugate acid) resists pH change by having a component to neutralize added acid and another to neutralize added base. The common ion effect is the underlying mechanism. When an ion common to the weak electrolyte is added, it suppresses the electrolyte's ionization, shifting the equilibrium. In a buffer, this effect keeps the concentrations of the acid/base and its conjugate partner stable, ready to absorb pH shocks.
8. What is the key takeaway about the solubility product (Ksp) for revision purposes?
The key takeaway is that the Solubility Product (Ksp) is an equilibrium constant that quantifies the solubility of a sparingly soluble salt. For a salt like AgCl, Ksp = [Ag⁺][Cl⁻]. A higher Ksp value means the salt is more soluble. It is crucial for predicting whether a precipitate will form when two solutions are mixed: if the ionic product (Qsp) exceeds the Ksp, precipitation occurs.
