Sunday, 26 January 2020

Equilibrium Chemistry Class 11 Notes

Equilibrium Chemistry Class 11 Notes

Equilibrium Chemistry Class 11 Notes

These are Chemical Equilibrium revision notes of Equilibrium Chemistry Class 11 Notes. These notes provided for study purpose by website.


Chemical reaction occur in one direction from reactant to products are termed as irreversible chemical reactions.
There direction is indicated by an arrow (→) pointing from reactant to products.
  1. C(s)   +   O2(s)   →   CO2(S)
  2. AgNO3(aq)   +   NaCl(aq)   →   AgCl(S)   ↓   +   NaNO3(aq)
  3. 2KClO3(S)   →  2KCl(S)   +   3O2(S)   ↑
Chemical reactions that do not go to completion in any direction and that can simultaneously occur in both forward (from reactant to product) and reverse (from product to reactant) direction are called reversible chemical reactions.
There direction is indicated by double arrows, one pointing in forward direction and other in reverse direction.


Physical equilibrium is defined as the equilibrium between two phase of the same substance when the rates of forward and reverse processes are equal.
The evaporation of liquid water in closed container is an example of physical equilibrium as the rate of evaporation and condensation are equal.
Another example of physical equilibriums are-
Physical Equilibrium
Chemical equilibrium is defined as the state of chemical reaction at which the rates of forward and reverse reactions are equal and the concentration of reactants and products reach constant values.
Consider a reversible reaction-
As reaction proceeds in forward direction with time, the concentration of A and B molecules decreases and concentration of C and D molecules increases. As the concentration of reactant molecules A and B decreases, the rate of forward reaction decreases and as the concentration of product molecules C and D increases the rate of reverse reaction increases.
After a certain time the rate of forward and reverse reaction gets equal and concentration of reactant and product reach constant values. At this state the chemical system is at equilibrium. As both forward and reverse reaction proceeds at the same rate therefore the chemical equilibrium is dynamic in nature.
Chemical Equilibrium – Equilibrium Chemistry Class 11 Notes by


The law of mass action states that the rate chemical reaction is directly proportional to the product of concentrations (active mass) of the reactants, with each concentration is raised to a power equal to its stoichiometric coefficient in the balance equation.
Consider the general reaction
aA   +   bB   →   Products
According to the law of mass of action,
Rate   α   [A]a [B]b
 Rate   =  K [A]a [B]b
The above equation represents rate equation and K is proportionality constant called rate constant.


The equilibrium constant (Kc) is defined as the ratio of the product of equilibrium concentrations (mol / L) of products to the product of equilibrium concentration of reactants, with the concentration of each substance raised to the power equal to its stoichiometric coefficient in the balanced chemical equation.
Consider the reversible chemical reaction
According to the law of mass action,
                     Rateforward   α   [A]a [B]b
                     Rateforward   =   K[A]a [B]b
Similarly       Ratereverse   α   [C]c [D]d
                     Ratereverse   =   K[C]c [D]d
At equilibrium the rates of forward and reverse reaction are equal. Thus,
Rateforward   =   Ratereverse
K[A]a [B]b   =   K[C]c [D]d
  •  Kf / Kr   =    [C]c [D]d /  [A]a [B]b     =    Kc
i.e      Kc   =    Kf / Kr   =    [C]c [D]d /  [A]a [B]b    
The above equation represents equation for equilibrium constant Kc. Where Kf and Kr are the rate constant of forward and reverse reactions respectively.
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Equilibrium constant with respect to partial pressure (Kp)

For reactions involving gases, the equilibrium constant is expressed in terms of partial pressure.
For the reaction
The equilibrium constant in terms of partial pressure (Kp) can be expressed by replacing the molar concentrations by partial pressures.
Kp    =       [PC]c [PD]d   /  [PA]a [PB]b
Where PA, PB, PC, PD are equilibrium partial pressures of A, B, C and D respectively.
Relationship between Kp and Kc
The partial pressure of each component in a mixture of ideal gaseous is directly proportional to its concentration at constant temperature.
For component A,
PA  =  (nA/V) RT
Where    (nA/V)   =   [A]  is molar concentration in mol dm-3
               PA  =  [A] RT
Similarly   PB  =  [B] RT
                 PC  =  [C] RT
                 PD  =  [D] RT
Now substituting equation for PA, PB, PC, and PD in equation of equilibrium constant in terms of partial pressure we get,
Where  Δn  =  (c+d) – (a+b) =  number of moles of gaseous product – number of moles gaseous reactant
                R  =  gas constant = 0.08206 L atm K-1 mol-1

Characteristics of equilibrium constant (Kc)

  1. The value of equilibrium constant (Kc) depends on the temperature.
  2. The value of equilibrium constant (Kc) does not depend upon original concentration of reactants or products, pressure, volume and catalyst.
  3. The values of equilibrium constant (Kc) are generally expressed without units.
  4. Higher values of Kc or Kp means more concentration of products is formed and the equilibrium point is more towards right hand side and vice versa.

Application of equilibrium constant

  • To predict direction of reaction
If Qc < Kc, the reaction will proceed from left to right, in forward direction generating more product to attain the equilibrium.
If Qc = Kc, the reaction is at equilibrium and hence no net reaction occurs.
If Qc > Kc, the reaction will proceed from right to left, in reverse direction generating more reactant to attain the equilibrium.
  • To know the extent of reaction
If Kc > 103, then the reaction is in favor of products and nearly goes to completion.
If Kc < 10-3, then the concentration of reactant is much greater than that of product. This shows that reaction does not take place.
If 103 > Kc > 10-3, then significant concentration of both reactants and products is present at equilibrium.
  • To calculate equilibrium concentration
By knowing the value of equilibrium constant, the unknown concentration can be calculated.

Le – chatelier’s principal

It states that, when a system at equilibrium is subjected to a change in any of the factors determining the equilibrium conditions of a system, system will respond in such a way as to minimize the effect of change.

Factors affecting equilibrium

  • Effect of change in concentration
If concentration of one or all of the reactant species is increased, the equilibrium shifts in the forward direction and more of the product are formed. Alternatively, if the product species is increased, the equilibrium shifts in the backward direction forming more reactant.
  • Effect of temperature
Increase in temperature shifts the equilibrium in endothermic direction. And decrease in temperature shifts the equilibrium in exothermic direction.
  • Effect of pressure
According to Le – chatelier’s principal, increase of pressure on a system at equilibrium shifts the equilibrium in the direction in which pressure reduced i.e. in direction where number of gaseous species is less.
  • Effect of volume
Increase in pressure means decrease in volume, so the effect of change of volume is exactly reverse to that of pressure.
  • Effect of catalyst
Catalyst has no effect on the equilibrium point because it speeds up both the forward and the backward reactions to the same extent.


The equilibrium between ionic species in solution is called ionic equilibrium.
Water soluble compounds are classified as either electrolytes or non – electrolytes.
The electrolytes are compound that ionize (or dissociate) into their ions and conduct on electric current in aqueous solution.
The non – electrolytes are compound that do not ionize into their ions and hence do not conduct electric current in aqueous solution.

Strong and weak electrolytes

Electrolytes are classified as strong and weak electrolytes.
  • Strong electrolytes
The substances which ionize almost completely into ions in aqueous solution are called strong electrolytes. For example HCl, H2SO4, HNO3, NaOH, KOH, NaCl, KNO3 etc. are strong electrolytes.
HCl   +   H2O   →   H3O  +   Cl
HNO  +   H2O   →   H3O+   +   NO3
  • Weak electrolytes
The substances which ionize to a small extent in aqueous solution are called weak electrolytes. For example, CH3COOH, NH4OH, HCN etc. are weak electrolytes.
In such cases, the molecules are in equilibrium with their ions.

Degree of dissociation

The degree of dissociation of an electrolyte is defined as the fraction of the total number of moles of the electrolytes that dissociates into its ion when an equilibrium is reached. It is given by expression,
 α   =    Number of moles of electrolyte ionized as ion /  Total number of moles of electrolytes
Percent dissociation = α x 100

Ostwald’s dilution law

If an electrolyte AB having concentration C moles per liter dissociates in aqueous solution and α is the degree of dissociation, then
The above mathematical relation is Ostwald’s dilution law.
  • For weak acids
  • Similarly, for weak base


Arrhenius Theory

According to this theory,
Acid is a substance which contains hydrogen and produces H+ ions in aqueous solution.
For example   :   HCl(aq)  →  H+(aq)   +  Cl(aq)
Base is a substance that contain OH group and produce OH ions in aqueous solution.
For example   :  NaOH(aq)  →  Na+(aq)  +  OH(aq)
This theory cannot be applied to compound which do not contains free H+ or OH ions.

Bronsted – Lowery Theory

According to Bronsted – Lowery theory an Acid is any substance that can donate a proton (H+) to another substance and Base is any substance that can accept a proton (H+) from another substance. i.e. acid is proton donor and base is proton acceptor. For example
The pair of acids and bases which are formed from each other by the gain or loss of a proton are called conjugate acid – base pairs.

Lewis Theory

According to Lewis theory an acid is a substance which can accept a pair of electrons while a base is a substance which can donate a pair of electrons.


K is the equilibrium constant for the ionization equilibrium.
Since the dissociation of water takes place to a small extent, the concentration of the undissociated water is nearly constant. Thus [H2O] = constant = K’
Therefore   Kw =  [H+] [OH]
Where Kw is constant and is known as ionic product of water. At 298K, the value of
Kw is 1 x 10-14 mol2 L-2
                 Kw   =   [H+] [OH]
                       =   1 x 10-14 at 298K 
As the concentration of [H+] and [OH] ions are equal in pure water,If  [H+]  =  [OH]  :  the solution is neutral
    [H+]  >  [OH]  :  the solution is acidic     [H+]  <  [OH]  :  the solution is basic


 The PH of a solution is defined as the negative logarithm, to the base I0, of the molar concentration of H+ ions in solution.
Mathematically, the pH of a solution is expressed as
     PH = – log10[H+]
 Hence [H+] = 10-pH
Similarly, poH of a solution is defined as the negative logarithm, to the base 10, of the molar concentration of OH ions in solution.
Thus mathematically
       POH = log10 [OH]
Relationship between PH and POH at 298K (25ᵒc)
      PH + POH = 14


Hydrolysis is defined as reaction in which cations or anions or both ions of salt react with ions of water to produce acidity or alkalinity (basicity)  or sometimes even neutrality.
  1. The aqueous solution of the salt of weak acid and strong base is alkaline.
  2. The aqueous solution of salt of weak base and strong acid is acidic.
  3. The aqueous solution of salt of strong acid and strong base is neutral.
  4. The aqueous solution of salt of weak base and weak acid may be acidic, basic or neutral depending upon relative strengths of acids and base.


Buffer solution is defined as a solution which resists change in pH on addition of small amounts of acids or bases or on dilution of the solution.
The ability of buffer solution to resist changes in pH value on addition of small amounts of either acid or base or on is called buffer action.
  1. A solution containing equimolar amounts of CH3COOH and CH3COONa is acidic buffer solution.
  2. A solution containing equimolar amounts of NH4OH and NH4Cl is basic buffer solution.
The pH of acidic and basic solution is given by Henderson-Hasselbalch equation for acidic buffer,
Buffer capacity is defined as the number of moles acid or base required to add one litre of solution to change the pH of unity.
Buffer capacity =   Number of moles of acid or base added per liter / Change in pH


When a saturated solution of a sparingly soluble salt is put into water, some of the salt dissolve in water and dissociates into its ion and most of the salt remains undissolved. Thus there is equilibrium between undissolved solid salt and its dissolved ions. This equilibrium is called solubility equilibrium.
For example AgCl, dissolves as-
As the concentration of [AgCl] is constant


The common ion effects state that the ionization of weak electrolyte is suppressed in presence of strong electrolyte containing an ion common to the weak electrolyte.
 For example, in a solution containing weak base NH4OH and its soluble ion salt NH4Cl. In a weak base  NH4OH  there exist equilibrium and salt NH4Cl dissociates completely into its ion.
As the concentration of NH4ions increases then according to the Le Chatelier’s principal the equilibrium shifts towards left i.e. towards NHOH this produce non ionized NHOH molecule the result is that the ionization of NHOH is suppressed due to the presence of NH4CI containing a common NH4+ ion .
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Sunday, 5 January 2020

Amines Class 12 Notes

Amines Class 12 Notes

In this revision lecture notes, we will learn about Amines Class 12 Chemistry Notes. We will cover all basic scientific knowledge about Amines in this article. So, enjoy learning chemistry with Chemistry Notes Info @

Amines Class 12 Chemistry Notes


            Amines are the organic derivatives of ammonia (NH3) in which one, two, or all three hydrogen atoms attached to nitrogen are replaced by equivalent number of same or different alkyl and/ or aryl groups.
Like ammonia, amines have pyramidal geometry and the nitrogen atom in amines, the nitrogen is attached to sp3 hybridized carbon of alkyl group and in aromatic amines to sp2 hybridized carbon of aryl group. H – N – H , C – N – H or C – N – C bond angle is less than 109ᵒ28’.


            Depending upon the number of hydrogen atoms replaced by alkyl or aryl groups attached to nitrogen atom in ammonia molecule, amines are classified as –

1) Primary (1ᵒ) Amines

The functional group present is – NH2 (Amino group)
e.g.     CH3 – NH2 Methylamine

2) Secondary (2ᵒ) Amines

The functional group present is – NH-  (Amino group)
e.g.     CH3 – NH– CHDimethylamine

3) Tertiary (3ᵒ) Amines                

The functional group present is -N= (Tertiary nitrogen atom)

Secondary and tertiary amines are further classified as

Secondary and Tertiary amines of Amines Class 12 Notes are further classified into two groups i.e.

a) Simple / Symmetrical amines

In simple amines same alkyl or aryl groups are attached to the nitrogen.

b) Mixed / Unsymmetrical amines

In mixed amines different alkyl or aryl groups are attached to the nitrogen.


1) By ammonolysis of alkyl halides

            When ethyl bromide is heated with alcoholic ammonia at 373K in sealed copper tube, it gives mixture of ethylamine, diethyl amine and triethyl amine along with tetraethyl ammonium bromide.
C2H– Br + NH3     Δ → C2H– NH2 + HBr
C2H– NH2 + C2H– Br    Δ  → (C2H5)2NH + HBr
(C2H5)2NH + C2H– Br    Δ  → (C2H5)3N + HBr
(C2H5)3 + C2H-Br    Δ  → (C2H5)4N+Br
NH3     RX→ R – NH2    RX→ R2NH  RX→ R3N    RX→ R4N+X

2) By reduction of Nitro Compounds

            Both aliphatic and aromatic primary amines can be prepared by the reduction of nitro compounds by either catalytically with H2 in the presence of Raney Ni, Pt or Pd or chemically with active metals.
e.g.       CH3NO2    +     3H  Raney Ni/Pt→     CH3NH2     +     2H2O         Nitromethane                   Ethanol          Methylamine

3) By Gabriel phthalimide synthesis

            Phthalmide reacts with ethanolic potassium hydroxide to give potassium salt of phthalmide. In this step N – H proton is removed to give imide ion. It is then heated to give N – alkyl phthalimide, which on alkaline hydrolysis give a primary amine. Share these Amines Class 12 Notes of chemistry with your friends.

4) By Reduction of Alkyl nitrites / cyanides

            Nitriles on reduction with lithium aluminium hydride (LiAlH4) or catalytic hydrogenation produce primary amines.
   R – CN (Alkyl cyanide)     — H2 / Ni & Na(Hg) / C2H5OH  →   R – CH2 – NH2  (1ᵒ Amine)                  
5) By reduction of amides
            Acid amides on reduction with lithium aluminium hydride give corresponding amines.

6) By Hoffmann bromamide degradation

            The conversion of amides into amines in the presence of bromine and alkali is known as Hoffmann degradation of amides.
R – CO – NH2 (Amides) +   Br2  +   4NaOH       →   R – NH (Amine) +  Na2CO3  +  2NaBr  +  2H2O


1) Aliphatic amines with low molecular weight are colourless, gaseous compounds with fishy odour. High molecular weight aliphatic amines are solid.
2) Pure aromatic amines such as aniline are colourless liquid. Arylamines are toxic in nature.
3) Aliphatic amines are soluble in water. As the molar mass increase solubility decreases.
4) Amines are less polar than the corresponding alcohol but more polar than corresponding alkanes.
5) Amines are higher boiling points than corresponding alkanes but lower than corresponding alcohols or carboxylic acids. The order or boiling points of isomeric amines is 1ᵒ > 2ᵒ > 3ᵒ.


            The lone pair of electron on nitrogen makes amines both basic and nucleophilic. They react with acids to form salts and react with nucleophiles in many reactions.

1) Basic nature of Amines

            Nitrogen atom of amines contains a lone pair of electron which can be donated. Thus, amines act as Lewis bases and are Lowery – Bronsted bases as they accept a proton.
The parent amine is regenerated when alkyl ammonium halide is treated with NaOH.
R – NH3+X (Alkyl Ammonium Halide) + NaOH → R – NH2 (Amine)  + NaX  + H2O
Amines are weak bases. They dissolve in water and produce OH ions. Aqueous Solutions of amines turn the color of litmus paper from red to blue.
  R – NH2     +     H – O – H     ↔     R – NH3+    +     OH
Weak Base                                                           Strong base
            The equilibrium lies far to the left as OH is a stronger base than amines. The expressions for the equilibrium constant K, basicity constant Kb and pKb values are as follows.
Amines class XII notes - Amines Class 12 Notes - ChemistryNotesInfo
PKb = -logKb
Strong bases have high values of Kb and low values of PKb.

2) Action of nitrous acid

  • On primary (1ᵒ) amines
Except methylamine, primary amines react with nitrous acid, in cold condition to give alcohol and nitrogen gas.
  R – NH2  (1ᵒ amines) +   NaNO +  2HCl   273 – 278K→   R – OH  +  HCl  +  N2  ↑
  • On secondary (2ᵒ) amines
Secondary amines react with nitrous acid to give the N – nitrosoamines which are generally pale yellow oils.
 R2 – NH (2ᵒ amine) +  HNO2  —NaNO2 + dil. HCl (at 273 – 278K)
→ R2N – N = O  +  H2O
  • On tertiary (3ᵒ) amines
Tertiary amines react with nitrous acid to form water soluble nitrite salt. As visible change observed, it is said that there is no reaction.
     R3N (3ᵒ amine) + HNO2  NaNO2 / dil. HCl ( at 273 – 278K )
→    [ R3NH ]+NO2   (No change)

3) Acylation of amines

            Acylation is the replacement of a hydrogen atom of amino group by acyl group (R – CO). It is nucleophilic substitution reaction. Acetylation of primary / secondary amines with acetyl chloride or acetic anhydride gives corresponding amine.

4) Carbylamine reaction  

            Aliphatic and aromatic primary amines on heating with chloroform and ethanolic potassium hydroxide form isocyanides or carbylamines. This reaction is known as carbylamines reaction or isocyanides test.
R – NH2  +  CHCl3  +  3KOH  Heat→   R – NC   +   3KCl  +   3H2O

5) Reaction with arylsulphonyl Chloride (Hinsberg’s test)

  • Primary amines when reacts with benzene Sulphonyl chloride ( Hinsberg’s reagent),
it yields N – ethylbenzene sulphonamide.
  • Secondary amines reacts with benzene sulphonyl chloride, it yields N, N – dimethylbenzene sulphonamides.
  • Tertiary amines do not react with benzene sulphonyl chloride.

6) Electrophilic aromatic substitution

  • Bromination
Reaction of aniline with bromine water at room temperature results in the formation of white precipitate of 2, 4, 6 – tribromoaniline.
  • Nitration
In direct nitration of aniline with concentrated nitric acid at 288K in the presence of sulphuric acid, a mixture of ortho, meta and para isomers of nitroaniline are obtained as product. Also, dark coloured tars are obtained due to oxidation.
  • Sulphonation
Reaction of aniline with cold concentrated Sulphuric acid gives anilinium hydrogen sulphate. Heating anilinium hydrogen sulphate with Sulphuric acid at 453K to 473K gives p – aminobenzene sulfonic acid (sulphonic acid) as major product.


            The diazonium salts have the general formula RN2+X where R stands for aryl group and X ion may be Cl, Br, HSO4, BF4, etc.
The N2+ group is called diazonium group.
e.g.  C6H5N2+Cl IS named as benzenediazonium chloride.
The conversion of primary aliphatic or aromatic amines into diazonium salts is known as diazotization.


            Diazonium salts are prepared by the reaction of nitrous acid in cold condition with alkyl/aryl primary amines (other than methylamine).


Reaction involving displacement of nitrogen (Diazonium group)
  • Replacement by -Cl , -Br and –CN
The reaction in which copper (1) salts are used to replace nitrogen in diazonium salt is called sandmeyer reaction
Yield in sandmeyer reaction is better than yield in Gattermann reaction.
  •  Replacement by – I
Ar – N+2X   +   KI       Δ  →    Ar – I   +   KX   +   N2    ↑
  • Replacement by – F
Ar – N+2X       HBr  →  Ar – N+2BF4     Δ  →    Ar – F   +   BF3   +   N2    ↑
  • Replacement by – H
Ar – N+2X  + H3PO2  +  H2O      CuCl  →  ArH   +   H3PO3   +   HX   +   N2    ↑
  • Replacement by – OH
Ar – N+2X  + H2O   Dil. H2SO4  →  Ar – OH    +   HX   +   N2    ↑
  • Replacement by – NO2
Ar – N+2X      HBF4  →  Ar – N+2BF4     NaNO2 / Cu &  Δ  →    Ar – NO2   +  NaBF4   +   N2    ↑


1) Arene diazonium salts can be prepared from nearly all primary amines and are helpful in the synthesis of variety of organic compounds.
2) Arenes diazonium salts are used as useful intermediates to introduce various groups into aromatic ring. Such as – F, – Cl, -Br, – I, -CN, -NO2, -OH, -H, etc.
3) The halide or Cyano group can be easily introduced in aromatic ring through diazonium salts.
4) The compounds which cannot be prepared by direct electrophilic aromatic substitution can be prepared by replacement of diazo group.
5) Azo compounds obtained from diazonium salts are strongly coloured and are used as dyes. 
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