Equilibrium is a relatively less important topic in JEE Main Chemistry. However, it forms an easy chunk of the Physical Chemistry syllabus. A total of 2 questions are usually asked carrying 8 marks from this topic. Therefore, the weightage of Equilibrium in the JEE Main is around 2-3%.

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Although a low weightage compared to other topics like Atomic Structure and Electrochemistry, the questions are quite scoring. Some of the topics tested under Equilibrium are Le Chatelier's Principle, Relation between Kp and Kc, Solubility Product, and Salt Analysis. More than one question was asked from Equilibrium in the JEE Main Session on January 8th, 2020. Check JEE Main Chemistry Syllabus

Before you proceed towards studying the notes for Equilibrium, take a look at the following question -

What is the order of the basic character?

1) I > II > III > IV

2) IV > III > I > II

3) II > I > III > IV

4) IV > I > II > III

To ace Equilibrium in JEE Main Chemistry, refer to the study notes given below. 

Also Check:

• JEE Main Study Notes for States of Matter
• JEE Main Study Notes for Atomic Structure

What is Equilibrium?

Equilibrium is the state of a process in which the system's properties such as temperature, pressure, and concentration display no change over time. Two opposing processes are involved in all processes which attain equilibrium. It reaches equilibrium when the concentrations of the two competing processes are equal. The equilibrium is considered Physical Equilibrium because the competing mechanisms require only physical shifts. If chemical reactions are the opposite processes, the equilibrium is called Chemical Equilibrium

Physical Equilibrium 

Physical equilibrium can be described as equilibrium in physical processes. 

Must Read:

• JEE Main Mathematics Preparation Tips

• JEE Main Physics Preparation Tips

a. Solid - liquid equilibrium

An example of solid - liquid equilibrium is the relationship of ice and water. In a close network, ice and water achieve equilibrium at 0^oC. Around this point the rate of ice melting is equal to that of water freezing. Equilibrium is seen here as -

H2O(s) ⇔ H2O(l)

b. Liquid - gas equilibrium

Water evaporation in a closed tank is an example of liquid-gas equilibrium, where evaporation rate equals condensation rate. Equilibrium is seen here as -

H2O(l) ⇔ H2O(g)

c. Solid – Solution Equilibrium

As you add more and more salt in water that is taken in a glass jar and mixed with a glass rod, after you dissolve it a bit, you will figure out that there is no more salt going to the water and it sets down at the bottom. The solution is now reported to be saturated and in a state of equilibrium. At this point, several salt molecules from the undissolved salt enter the solution (dissolution) and the same quantity of dissolved salt is deposited back (precipitation). Therefore, the rate of dissolution at equilibrium is equal to that of precipitation.

Salt(Solid) ⇔ Salt(in solution)

d. Gas –Solution equilibrium

Within a closed vessel, the dissolution of a gas in a liquid under pressure forms a gas-liquid equilibrium. Cold drink bottles are the perfect example of this type of equilibrium. The equilibrium within the bottle is-

CO2(g) ⇔ CO2(in solution)

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Equilibrium in Chemical Process (Reversible and Irreversible Reactions)

A reaction is considered a reversible reaction when not only the reactants react to form the products under specific conditions but also the products react to form reactants under the same conditions. 

For example, 

1. 3Fe(s) + 4H2O(g) ⇔ Fe3O4(s) + 4H2(g)
2. CaCO3(s) ⇔ CaO(s) + CO2(g)
3. N2(g) + 3H2(g) ⇔ 2NH3(g)

If a reaction can not take place in the reverse direction , i.e. the products formed do not react under the same condition to give back the reactants, it is called an irreversible reaction.

For example, 

1. AgNO3(aq) + NaCl(aq) → AgCl(s) + NaNO3(g)
2. 2M(g) + O2(g) → 2MgO(s)

Typically, a chemical equilibrium is represented as - 

 where A and B are reactants, and C and D are products.

The double arrow between the left part and the right part indicates that shifts exist in both directions. 

Based on the degree of reaction, chemical reactions can be divided into three groups before equilibrium is reached. 

A) Reactions which are almost complete. 

B) Certain reactions which only continue to a very limited degree. 

C) Those reactions which continue to such an extent that the equilibrium concentrations of reactants and products are comparable. 

In fact, the state of equilibrium is fluid and not static. When the rate of forward reaction is equal to that of backward reaction, a reaction is assumed to have reached equilibrium.

If all the reactants and products of any equilibrium reaction are in the same physical state, this is known as homogeneous equilibrium. 

For example,

 N2(g) +  3H2(g) 2NH3 (g) 

Here, all the reactants and products are in the same physical state.

Heterogeneous equilibrium is the type of equilibrium in which the state of one or more reacting species varies, i.e. both reactants and products are not in the same physical condition.

For example,

2NaHCO3(s) Na2CO3(s) + CO2(g) + H2O(l)

Here, the reactant is in the solid state while the product is in the solid, liquid, and gaseous matter states. 

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Ionic Equilibrium

Chemical reactions occur mostly in solutions. Solution chemistry plays a very critical role in the process. All chemical substances consist of either polar or nonpolar units (called ions). The operation of these substances is more apparent and noticeable in solutions. The action of these substances is based on the nature and conditions of the medium in which they are applied. Consequently , it is important to understand the principles which regulate their actions in solution. 

This kind of equilibrium is found in quickly ionizing substances, or in polar compounds where ionization can be caused. Because of the ease of ionization in solvent medium, ionic and polar substances are more easily soluble in polar solvents. These solutions become rich in mobile charge carriers (ions) with the dissolution of ionic and polar substances in the solvent, and thus can conduct electricity. Substances capable of conducting electricity are called electrolytes, while non-conducting substances are called non-electrolytes.
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Characteristics of Equilibrium State

i) When the reversible reaction is conducted in a closed system, equilibrium can be achieved. 

ii) This can be done from either side of the reaction. 

iii) A catalyst can speed up the equilibrium approach but does not change the equilibrium state.

iv) It is of a dynamic nature, i.e. the reaction does not end but reactions both forward and backward take place at the same time. 

v) Changes in pressure, concentration, or temperature favor one of the reactions (forward or backward) resulting in a shift in one direction of equilibrium point.

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Law of Mass Action and Equilibrium Constant 

Guldberg and Waage developed a relationship between the rate of chemical reaction and the concentration of the reactants or, in the context of the law of mass action, their partial pressure. According to this law, “the rate at which a compound reacts is directly proportional to its active mass, and the rate of a chemical reaction is directly proportional to the sum of the active masses of reactants each elevated to a value equal to the related stoichiometric coefficient occurring in the balanced chemical equation”.

rate of reaction ∝ [A]^a.[B]^b

rate of reaction = K[A]^a[B]^b

where K is the rate constant or velocity constant of the reaction at that temperature.

Unit of rate constant (K) = [moles/lit]1–n time–1 (where n is order of reaction).

The rate of forward and backward reactions is equal in equilibrium and the concentration of reactants and products remains constant as shown in the figure after achieving equilibrium.

Hence,

Active mass = number of moles/volume in litres

Active mass of solid is taken to be unity.

Also, Active mass of reactant (a) = Conc. × activity coefficient

i.e. a = Molarity × f for dilute solution f = 1

Applying Law of mass action for general reversible reaction,

aA + bB cC + dD

Rate of forward reaction [A]a[B]b

or Rf = Kf [A]b [B]b

Similarly for backward reaction

Rb = Kb[C]c [D]d

At equilibrium Kf[A]a[B]b = Kb[C]c[D]d

The above equation is called equilibrium equation and Kc is known as equilibrium constant.

The subscript ‘c’ shows that Kc is displayed in concentrations of mol L^–1. 

At a specific temperature, the product of concentrations of the reaction products raised in the balanced chemical equation to the respective stoichiometric coefficient divided by the product of concentrations of the reactants raised to their corresponding stoichiometric coefficients has a constant value. This is known as the Law of Chemical Equilibrium or Equilibrium Law. 

In the reverse reaction, the equilibrium constant is the inverse of the equilibrium constant for the forward path reaction.

Equilibrium Constant in terms of Partial Pressures (Kp)

For reactions that involve gases,

it is more feasible to display the equilibrium constant in terms of partial pressure.

The ideal gas equation is represented as,

where c or n/v is the concentration expressed in mol/m³ (mol/L³) & R= 0.0831 bar litre/mol K.

This is an indication that at constant temperature, the pressure of any gas is directly proportional to its concentration.

For reaction in equilibrium,

We can write, With the usage of Ideal Gas Equation we get,

PA = CART = [A] RT

PB = CBRT= [B] RT

Pc = CCRT = [C] RT

Pd = CDRT= [D] RT

Substituting the above values, 

where n = (number of moles of gaseous products) – (number of moles of gaseous reactants) in the balanced chemical equation. The equation given above represents the relationship between kc and kp.

Equilibrium Constant of a Reaction that involves Condensed Phase

The representation for the equilibrium constant of a reaction between a condensed substance (solid or liquid) and a gaseous state is determined by considering only the gaseous species concentrations or partial pressures. Steady integration of condensed phases is combined with constant equilibrium.

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Facts about Equilibrium Constant

a) A huge value of KC or KP signals that the forward reaction reaches completion or very nearly so.

b) A low numerical value of KC or KP signals that the forward reaction does not reach to any significant extent.

c) A reaction will most probably reach a state of equilibrium in which both, reactants and products are present if the numerical value of KC or Kp is not very huge or low. 

d) The equilibrium constant of a forward reaction and its backward reaction are reciprocal to each other.

e) If a chemical equation is multiplied by a specific factor, its equilibrium constant should be increased to a power equal to that factor so as to avail the equilibrium constant for the new reaction.

The Reaction Quotient “Q”

Look at the following equilibrium.

PCl5 (g) PCl3(g) + Cl2 (g)

At equilibrium,

When the reaction is not in equilibrium, this ratio is known as ‘QC’. QC is the general term used for the ratio given above at any point of time. At equilibrium, QC becomes KC.

In a similar manner, PCl2 PPCl2 / PPCl2 is known as QP and at equilibrium, it turns into KP.

Suppose the reaction is at equilibrium, Q = Kc

A net reaction progresses from left to right (forward direction) if Q < KC.

A net reaction progresses from right to left (the reverse direction) if Q >Kc

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Le Chatelier’s Principle 

"When an equilibrium is exposed to a change of concentration, temperature or external pressure, the equilibrium will move in that direction where the effects of these changes are nullified." 

It can be understood by example given below. Overall, we can also foresee the direction of equilibrium with the theoretical assumption in mind.

PCl5 ——> PCl3 + Cl2

Assume that this reaction is at equilibrium and the moles of Cl2, PCl3, and PCl5 at equilibrium are a, b, and c respectively, and the total pressure is PT.

 Since,

PT  =  (a+b+c) RT / V  

KP = abRT / cV  

Suppose d moles of PCl3 is included in the system, the value of Q would be, a(b+d)RT / cV. This is evdiently more than KP. Therefore, the system will move reverse to reach equilibrium.

If the volume of the system is increased, the Q becomes abRT / cV' where V' > V. Q becomes lesser, and the system will move forward to obtain equilibrium.

If a noble gas is added at constant pressure, it leads to an increase of the system’s volume and hence, the reaction moves forward.

If the noble gas is added at constant volume, the expression of Q remains as Q = abRT / cV and the system remains in equilibrium.

Thus, for using the Le Chatlier’s principle, change the expression of KP and KC into basic terms and later notice the effect of different changes.

Solubility Product 

A solution is said to be saturated when it stays in contact with undissolved solute.

In a saturated solution of an electrolyte, two equilibria exist. This can be represented as - 

AB    AB                              

Solid unionized              ions

(dissolved)

On application of the law of action to the ionic equilibrium,

  

As the solution is saturated, the concentration of unionised molecules of the electrolyte is stable at a specific temperature, i.e., [AB] = K'= constant.

Hence,

[A+] [B-] = K[AB] = KK’ = Ks (constant)

Ks is called the solubility product. It is described as the product of the concentration of ions in a saturated solution of an electrolyte at a particular temperature.

Salt Hydrolysis 

In the scenario of salt of weak acid and weak base, the nature of medium after hydrolysis is determined in the following manner:

(i) If Ka = Kb, the medium should be neutral.

(ii) If Ka > Kb, the medium should be acidic.

(iii) If Ka < Kb, the medium should be basic.

The degree of hydrolysis of salts of weak acids and weak bases is not affected by dilution as there is no concentration term in the expression of degree of hydrolysis.

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Points to Remember in Equilibrium

1. Remember,

   • CH3COOH has Ka = 1.8*10^(–5) and pKa = 4.74;
   • For NH3, pKb = 4.77
   • Sq.rt. of 1.8 and 18 are 1.34 and 4.2 respectively.

2. Acid-base theories- Bronsted and Lewis theory

   • CAB pairs. (add H+ to get CA and remove H+ to get CB)
   • Na+, K+, NH4+ are not electrophiles or lewis acids.
   • [SbF6]-, [SiF6]2- are non-nucleophilic anions. (why?)

3. Acid-base equilibrium

1. Acid-base reactions progress towards the formation of weaker acids and weaker bases.

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Previous Year Solved Questions on Equilibrium

Q1. Find out the pH of the solution when 0.1 M CH3 COOH (50 ml) and 01. M NaOH (50 ml) are mixed, [Ka (CH3COOH)=10^-5]

Solution:

CH3 COOH CH3 COO_ + H+ …(I)

NaOH → Na+ + OH-

H+ + OH_ H2O …(II)

(I) + (II)

CH3COOH + OH- CH3COO- + H2O . (III)

0.05-X 0.05-x x

Keq of eq. (III) = Ka/Kw

Conc. of H2O remains constant

109 = x/(0.05-x)2

because the value of eq. Const. is very high

Here for x» 0.05

Let 0.05-x=a

109=0.05/a2

a = 7.0710-6

pOH= 6-log 7.07

pOH= 6 – 0.85

pH= 14-6+0.85 = 8.85

Q2. The solubility product of Pb3 (PO4)2 is 1.5 x 10^-32. Calculate the solubility in gms/litre.

Solution:

Solubility product of Pb3 (PO4)2 = 1.5 10–32

Pb3 (PO4)2 3Pb2+ + 2PO43-

Suppose x is the solubility of Pb3 (PO4)2

Then, Ksp = (3x)3 (2x)2 = 108 x5

x = 1.692 10–7 moles/lit

Molecular mass of Pb3(PO4)2 = 811

 x = 1.692 ´ 10–7 ´ 811 g/lit = 1.37 10–4 g/lit

 Solubility product is

 Ksp(SrC2O4) = [Sr2+] [C2O42–] = (5.4 10–4)2- = 2.92 10–7

Q3. Calculate the concentration of H⁺, HCO3^- and CO3^2-, in a0.01M solution of carbonic acid if the pH is 4.18.

Ka1(H2CO3) = 4.45 10–7 and Ka2 = 4.69 10–11

Solution: 

pH = – log[H+]

4.18 = – log [H+]

[H+] = 6.61 10–5

H2CO3 H+ + HCO3-

HCO3- H+ + CO32-

 [CO32-] = 4.8 10–11

Q4. Find out the molar solubility of Mg(OH)2 in 1MNH4Cl

KspMg(OH)2 = 1.8 10–11

Kb(NH3) = 1.8 10–5

Solution:

Mg(OH)2(s) Mg++ + 2OH– K1 = Ksp

2NH4+ + 2OH- 2NH4OH K2 = 1/K2b

Q5. How much of AgBr can dissolve in 1.0 L of 0.4 M NH3? Assume that [Ag (NH3)2]+ is the sole complex formed if Kf [Ag(NH3)2+]=1.0108, Ksp (AgBr)= 5.010-13

Solution: 

AgBr Ag+ + Br- 

Ag+ + 2NH3 Ag (NH2)2+

Let x= solubility ,

Then, x= [Br-]=[Ag+]+[Ag(NH3)2+]

x2=8.0´10-6 

x=2.8´10-3 M 

Revision Tips for Equilibrium

1. For the auto-protolysis of water-

• Calculate the degree of dissociation of water and its dissociation constant at 298K.
• Ionic product of water (Kw) and its reliance on temperature.

2. Use common-ion effect in Salt-analysis
3. Remember that for theory of indicators, the pH range of the buffer or indicator is pKa+1, pKa-1 i.e. two units.

Tricks to Solve Questions from Equilibrium 

Q1. Trick to solve the question: Consider the solubility to be x. Balance the equation by splitting the compounds - Ag2Cr04 and AgNO3. Enter the values found in the final equation for Ksp and find the answer to x. 

Solution: (b)

Q2. Trick to solve the question: Split HA into its elements. Correlate the rates of HA andHX to frame an equation. Equate HX to 1M. Enter the values found and derive the answer. 

Solution: (a)

Q3. Trick to solve this question: Write down the formula for pH. Substitute the values of 1 and 2 given in the formula. Use M1V1 = M2V2 to find the value of V2. 

Solution: (b)

Tips to Study Chemistry for JEE Mains 

1. Concentrate on all three sections – organic, physical and inorganic chemistry. 
2. A must-know is periodic table and chemical bonding. Grasp complete reaction and its mechanism. 
3. You ought to have a clear knowledge of basic Chemistry principles. 
4. Thoroughly learn all the key formula. 
5. Give special attention to topics such as GOC, chemical bonding, thermodynamics, electro-chemistry, and coordination compounds. 
6. Make personal and short notes for yourself.
7. Take JEE mock exams and go through the analysis to figure out what your shortcomings are. Work on your shortcomings, and increase your scores. 
8. Take chapter-wise JEE Practice tests to boost the knowledge fundamentals as and when you keep preparing for every chapter.

45 Day Chemistry Study Plan for JEE Main