JEE Advanced 2023 Revision Notes for Chemical Equilibrium

VSAT 2022

JEE Advanced 2023 Revision Notes for Chemical Equilibrium [Free PDF Download]

Every chemical reaction maintains equilibrium. This equilibrium depends on various factors and properties of the system where the reaction is taking place. This chapter is an important part of JEE Advanced Physical Chemistry syllabus and carries a weightage of approximately 6-7%. It is very crucial to learn how chemical reactions proceed to completion and achieve equilibrium to understand advanced concepts of Chemistry. To understand these concepts better, refer to the Chemical Equilibrium JEE Advanced revision notes prepared by the subject matter experts of Vedantu.


Download this Chemical Equilibrium IIT JEE notes PDF for free from Vedantu and refer to it for your exam preparation. Follow how our subject experts have prepared these notes on every concept covered in this chapter with relevant examples and logical explanations to resolve your doubts.

Competitive Exams after 12th Science
Do you need help with your Homework? Are you preparing for Exams?
Study without Internet (Offline)

Download PDF of JEE Advanced 2023 Revision Notes for Chemical Equilibrium

Access JEE Advanced Revision Notes Chemistry Chemical Equilibrium

Reversible Reaction: The reaction which can take place in both backward and forward direction under same conditions, these are carried out in closed vessels Reactants $\underset{{{\text{Backward}}}}{\overset{{{\text{Forward}}}}{\longleftrightarrow}}$Products


Irreversible Reaction:

Under similar set of conditions, the products cannot be converted into reactants, and these reactions take place in one direction

E.g.: Thermal decomposition reaction

Evaporation of water in open vessel

$2KCl{O_{3(s)}} \to 2KC{l_{(s)}} + 3{O_2}(g)$

Precipitation reaction:  $AgN{O_{3(aq)}} + NaC{l_{(aq)}} \to AgC{l_{(s)}} \downarrow  + NaN{O_{3(aq)}}$


Equilibrium State:

When the reactants in a closed vessel at a particular temperature react to give products, the concentration of the reactants decreases, while those of products increases. After some time, there is no change in the concentration of either the reactants or products, this stage of the system is called equilibrium.

  • At equilibrium state, the rate of forward reaction and backward reaction becomes equal 

  • Equilibrium can be attained for both physical and chemical reaction

  • Equilibrium in Physical Process: Dissolution of salt

  • Equilibrium in Chemical Process: Decomposition of ammonium chloride

  • Equilibrium in Chemical Process: For general reversible reaction $A + B \to C + D$

In the beginning the rate of forward reaction is more since the concentration of reaction is high, with time, it decreases In the beginning the rate of backward reaction is zero  As time proceeds, it increases


When the rate of forward and reverse reaction is equal it is known as the equilibrium state.

  • At equilibrium both forward and the reverse reaction continue to take place with equal rates, hence equilibrium is dynamic in nature

  • By synthesising the ammonia using Haber’s process , the dynamic nature of chemical equilibrium can be demonstrated

  • Haber’s process can be conducted in following 2 ways under similar conditions

${N_{2(g)}} + 3{H_{2(g)}} \to 2N{H_{3(g)}}$

${N_{2(g)}} + 3{D_{2(g)}} \to 2N{D_{3(g)}}$


Types of Chemical Equilibria

  • Homogenous 

  • Heterogenous

Homogeneous Equilibrium: the reactants and products are in same  ${H_{2(g)}} + {I_{2(g)}}$


Heterogeneous Equilibrium: The reactants and products are not in same phase $CaC{O_{3\left( S \right)}} \to Ca{O_{\left( S \right)}} + C{O_{2\left( g \right)}}$


Law of mass action:

  • This law determines the relation between the rate of a chemical reaction and the concentration of the reactants 

  • The rate of chemical reaction at a given temperature is directly proportional to the product of active mass or molar concentration of the reactants at that instant 

  • This law is applicable to all reaction occurring in the gas phase or in the liquid phase 

  • Active mass Molar concentration for dilute solution 


Equilibrium constant with respect to Molar Concept:${K_c} = \dfrac{{{\text{Product of the concentration of product at equilibrium}}}}{{{\text{Product of the concentration of reactants at equilibrium}}}}$

This representation is known as the equilibrium law or law of chemical equilibrium

${K_c} = \dfrac{{{k_f}}}{{{k_b}}} = \dfrac{{{{\left[ C \right]}^c}{{\left[ D \right]}^d}}}{{{{\left[ A \right]}^a}{{\left[ B \right]}^b}}}$

Units of ${K_c} = {\left( {\dfrac{{moles}}{{lit}}} \right)^{\Delta n}}$

Equilibrium constant with respect to Mole Fraction:

Let us consider a reversible process

$aA + bB \to cC + dD$, the equilibrium constant

${K_x} = \dfrac{{x_C^c.x_D^d}}{{x_A^a.x_B^b}}$, where ${x_A},{x_B},{x_c}$ and ${x_D}$ are the equilibrium mole fraction of A, B, C and D.


Equilibrium constant with respect to Partial Pressure

Let us consider a reversible process

$aA + bB \to cC + dD$, the equilibrium constant

${K_p} = \dfrac{{{k_f}}}{{{k_b}}} = \dfrac{{P_C^c.P_D^d}}{{P_A^a.P_B^b}}$, where ${P_A},{P_B},{P_c}$and ${P_D}$ are the equilibrium mole fraction of A, B, C and D.

Unit of ${K_p} = {\text{ }}{\left( {atm} \right)^{\Delta n}}$; S.I unit of ${K_p} = {\text{ }}{\left( {pascal} \right)^{\Delta n}}$

$\Delta n = {n_{p(g)}} - {n_{r(g)}}$, p= products and r= reactants


Relation between KP and KC

${K_p} = {K_c}{(RT)^{\Delta n}}$

R = gas constant 

T= absolute temperature

$\Delta n$=change in number of moles


Relation between KP and Kx

${K_p} = {K_x}{({P_{total}})^{\Delta n}}$

Relation between KC and KX

${K_p} = {K_x}{({P_{total}})^{\Delta n}} = {K_c} = {\left( {RT} \right)^{\Delta n}}$


Application of Equilibrium Constant:

  • Predicting the extent of a reaction

  • Predicting the direction of the reaction


Relationship between Equilibrium constant(KC), Reaction Quotient (QC)and Gibbs energy (GC):

Vant’s Hoff isotherm equation represents the mathematical expression for equilibrium view point of thermodynamic

$\Delta G = \Delta G^\circ  + RT\ln Q$

$\Delta G^\circ $is the standard Gibbs energy

${\Delta _r}{G^\Theta } = \sum\nolimits_{\Product} {n{\Delta _f}{G^\Theta }}  - \sum\nolimits_{\operatorname{Re} ac\tan ts} {n{\Delta _f}{G^\Theta }} $

At equilibrium when $\Delta G$=0 and Q=K, then 

$\Delta G = \Delta G^\circ  + RT\ln K = 0$

$\Delta G^\circ  =  - Rt\ln K$

or, $\Delta G^\circ  =  - 2.303RT\log K$

  • If $\Delta G^\circ $<0 then $\dfrac{{ - \Delta G^\circ }}{{RT}}$ is negative and${e^{ - \Delta G^\circ /RT}} > 1$ making  K>1. Which implies that a spontaneous reaction (or) the reaction which proceeds in the forward direction to such an extent that the products are present predominantly.

  • If $\Delta G^\circ $<0 then $\dfrac{{ - \Delta G^\circ }}{{RT}}$ is negative and${e^{ - \Delta G^\circ /RT}} < 1$ making  K<1. Which implies that a non-spontaneous reaction (or) the reaction which proceeds in the forward direction to such a small degree that only a very minute quantity of product is formed.


Relation Between Vapour Density and Degree of Dissociation:

Molar density before dissociation:

$D{\text{ = }}\dfrac{{{\text{Molecular weight}}}}{{{\text{volume}}}} = \dfrac{m}{v}$

Molar density after dissociation:

$d = \dfrac{m}{{[1 + x(y - 1)]V}}$

$d\dfrac{D}{d} = [1 + x(y - 1)]1$

$x = \dfrac{{D - d}}{{d(y - 1)}}$

Y is the number of moles of products from one mole of reactant. $\dfrac{D}{d}$is also called the Van't Hoff factor.


Factor Affecting the Position of Equilibrium and Lech Atelier’s Principle:

If a system at equilibrium is subjected to a change in temperature, Pressure (or) concentration the equilibrium position is shifted to that side where the effect of change is nullified


Effect of Concentration:

The forward reaction is favourable by equilibrium shift towards right

  • Increase in concentration of reactants

  • Decrease in concentration of products


Effect of Pressure:

If $\Delta n$=0, pressure has no effect on the position of equilibrium

If $\Delta n$≠0

  • Increase of external pressure on the reaction at equilibrium favours the reaction in the direction in which the volume (or) number of moles (or) molecules decrease

  • Decrease of external pressure on the reaction at equilibrium favours the reaction in the direction in which the volume (or) number of moles (or) molecules increase.

If $\Delta n$= +ve, increase of pressure favour’s backward reaction and decrease of pressure favours forward reaction

If $\Delta n$= -ve, increase of pressure favour forward reaction and decrease of pressure favour’s backward reaction


Effect of temperature:

  • Increase of temperature favour’s endothermic reaction and decrease of temperature favour’s exothermic reaction.

  • The equilibrium constant for exothermic reaction decreases as the temperature increases.

  • The equilibrium constant for an endothermic reaction increases as the temperature increases.


Effect of Addition of Inert Gas

  • Addition of inert gas at constant volume: the total pressure of the system is increased, but the partial pressure of each reactant and product remains the same. Hence no effect on the state of equilibrium

  • Addition of inert gas at constant pressure: The total volume is increased, the number of moles per unit volume of each reactant and product is decreased. Hence equilibrium will shift to the side where number of moles are increased


Case

At constant volume

At constant pressure

$\Delta n$=0

No effect

No effect

$\Delta n$>0

No effect

Forward direction

$\Delta n$<0

No effect

Backward direction


Effect of a Catalyst:

  • A catalyst increases the rate of the chemical reaction by providing a low activation energy pathway for the reaction.

  • It increases the rate of forward and backward reactions that pass through the same transition state and does not affect equilibrium.

  • Catalyst lowers the activation energy for the forward and reverse reaction by exactly the same amount.

  • Catalyst does not affect the equilibrium composition of a reaction mixture; it does not appear in the balanced chemical equation or the equilibrium constant expression.


Equilibrium in Physical Processes

  1. Solid-liquid Equilibrium:

Ice and water are kept in a perfectly insulated thermos flask (no exchange of heat between its contents and the surroundings) at 273K and the atmospheric pressure is in equilibrium. 

Ice$ \to $Water system


  1. Liquid -Vapour Equilibrium

Water and water vapour are in equilibrium position at atmospheric pressure(1.013 bar) and at 100℃ in a closed vessel

${H_2}{O_{(l)}} \to {H_2}{O_{(vap)}}$


  1. Solid-Vapour Equilibrium

When sublimating solid is placed in a closed vessel, equilibrium is established between its solid and vapour phases

$N{H_4}Cl(solid) \to N{H_4}Cl(vapour)$

Camphor(solid) $ \to $Camphor (vapour)


Equilibrium Involving dissolution of Solids or Gases in Liquid:

  • Solids in Liquids

  • A solution in which no more solute can be dissolved at the same temperature and pressure is called saturated solution

  • In such solutions, a dynamic equilibrium exists between the solute molecules in the solid state and in the solution

  • Gases in Liquid

  • There is an equilibrium between the molecules in the gaseous state and the molecules of gas dissolved in the liquid under the pressure 


Features of Physical Equilibrium


Process

Conclusion

Liquid $\to$ Vapour

${{\text{H}}_{\text{2}}}{{\text{O}}_{{\text{(s)}}}} \to {{\text{H}}_{\text{2}}}{{\text{O}}_{{\text{(g)}}}}$

${{\text{P}}_{{{\text{H}}_{\text{2}}}{\text{O}}}}$,  Constant at given temperature

Solid $\to$ Liquid

${{\text{H}}_{\text{2}}}{{\text{O}}_{{\text{(s)}}}} \to {{\text{H}}_{\text{2}}}{{\text{O}}_{{\text{(l)}}}}$

Melting point is fixed at constant pressure

Solute(s) $ \to $ Solute(solution)

Sugar(s) $ \to $ Sugar(solution)

Concentration so solute in the solution is constant at the given temperature

Gas(g) $ \to $ Gas(a)

${\text{C}}{{\text{O}}_{{\text{2(g)}}}} \to C{{\text{O}}_{\text{2}}}_{{\text{(aq)}}}$

$\dfrac{{\left [ ga{s_{(aq)}}\right ]}}{{\left [ga{s_{(g)}}\right ]}}$is constant at given temperature

$\dfrac{{\left [ C{O_2}_{(aq)} \right ]}}{{\left [C{O_2}_{(g)} \right ]}}$is constant at a given temperature


Simultaneous Equilibrium

When more than one equilibrium is established in the same vessel, then the equilibrium concentration of the common species in all the equilibria would be the same.


Importance of Physical Chemistry Chemical Equilibrium

This chapter covered in the Physical Chemistry syllabus will explain how a chemical reaction occurs and propagates in a particular direction. It will also explain how a chemical reaction achieves equilibrium even though it has two opposing processes involved.


The chapter explains how a reaction attains equilibrium by explaining the entire concept with proper derivations. Every term in the derivation is explained in accordance with the concepts of chemical equilibrium.


All the physical and chemical aspects of a chemical reaction are explained along with their effect on the equilibrium. These explanations will help you understand how these aspects control the state of chemical equilibrium and decide the fate of a reaction.


After studying this chapter, you will be able to determine the direction in which a chemical reaction proceeds. You will also be able to analyse the products of a chemical reaction and write down an equation to represent it with this knowledge.


To understand these concepts better, download and go through the Chemical Equilibrium IIT JEE notes PDF for free from Vedantu.


Benefits of Vedantu’s Chemical Equilibrium IIT JEE Notes PDF

  • The scientific principles of chemical equilibrium are explained precisely in these revision notes to help students understand them better. These revision notes are prepared by our subject experts at Vedantu to support your exam preparation. Refer to these revision notes to prepare and revise this chapter effectively.

  • These notes will help you understand the concepts of reversible and irreversible reactions. Find out how the chemical reactions achieve equilibrium considering the factors involved in it from these notes.

  • Solve and practice the worksheet given in these notes to assess your exam preparation and find out topics on which you need to focus more. Concentrate on the sections where terms and principles are explained using mathematical expressions and derivations.

  • Resolve doubts on your own by referring to the Chemical Equilibrium IIT JEE notes PDF. Use your study time more efficiently and complete preparing this chapter in the best way possible.


Download Chemical Equilibrium JEE Advanced Revision Notes PDF

Download the free PDF of the revision notes of Chemical Equilibrium today and prepare this chapter for your JEE Advanced exam. These revision notes will help you learn how to answer fundamental questions from the explanation given by the experts. Therefore, revise the concepts of Chemical Equilibrium with these notes and score well in the JEE Advanced exam.


Important Related Links for JEE Main and JEE Advanced

FAQs on JEE Advanced 2023 Revision Notes for Chemical Equilibrium

1. How can I determine the differences in various chemical reactions?

By following the Chemical Equilibrium notes for JEE, you can easily find out the factors that determine the progress of a chemical reaction. Learn how to categorise the reactions and determine which way they will progress. Also, focus on the reactants and products and their physical and chemical properties to determine the types of chemical reactions.

2. What is homogeneous equilibrium?

When all the reactants and products are in the same physical state, the reaction attains a homogeneous equilibrium.

3. What is heterogeneous equilibrium?

When the physical states of the reactants vary from each other, it is called heterogeneous equilibrium.

4. What is a catalyst?

A catalyst is a chemical compound that does not participate chemically in a reaction but promotes it to attain equilibrium in a particular direction and speeds up the reaction.

Share this with your friends
SHARE
TWEET
SHARE
SUBSCRIBE