Course Content
CHAPTER 3: CLASSIFICATION OF ELEMENTS
Section Name Topic Name 3 Classification of Elements and Periodicity in Properties 3.1 Why do we Need to Classify Elements ? 3.2 Genesis of Periodic Classification 3.3 Modern Periodic Law and the present form of the Periodic Table 3.4 Nomenclature of Elements with Atomic Numbers > 100 3.5 Electronic Configurations of Elements and the Periodic Table 3.6 Electronic Configurations and Types of Elements: s-, p-, d-, f – Blocks 3.7 Periodic Trends in Properties of Elements
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CHAPTER 7: EQUILIBRIUM
Section Name Topic Name 7 Equilibrium 7.1 Equilibrium in Physical Processes 7.2 Equilibrium in Chemical Processes – Dynamic Equilibrium 7.3 Law of Chemical Equilibrium and Equilibrium Constant 7.4 Homogeneous Equilibria 7.5 Heterogeneous Equilibria 7.6 Applications of Equilibrium Constants 7.7 Relationship between Equilibrium Constant K, Reaction Quotient Q and Gibbs Energy G 7.8 Factors Affecting Equilibria 7.9 Ionic Equilibrium in Solution 7.10 Acids, Bases and Salts 7.11 Ionization of Acids and Bases 7.12 Buffer Solutions 7.13 Solubility Equilibria of Sparingly Soluble Salts
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CHAPTER 10: S-BLOCK ELEMENTS
Section Name Topic Name 10 The s-Block Elements 10.1 Group 1 Elements: Alkali Metals 10.2 General Characteristics of the Compounds of the Alkali Metals 10.3 Anomalous Properties of Lithium 10.4 Some Important Compounds of Sodium 10.5 Biological Importance of Sodium and Potassium 10.6 Group 2 Elements : Alkaline Earth Metals 10.7 General Characteristics of Compounds of the Alkaline Earth Metals 10.8 Anomalous Behaviour of Beryllium 10.9 Some Important Compounds of Calcium 10.10 Biological Importance of Magnesium and Calcium
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CHAPTER 12: CHARACTERIZATION OF ORGANIC COMPOUND
Section Name Topic Name 12 Organic Chemistry – Some Basic Principles and Techniques 12.1 General Introduction 12.2 Tetravalence of Carbon: Shapes of Organic Compounds 12.3 Structural Representations of Organic Compounds 12.4 Classification of Organic Compounds 12.5 Nomenclature of Organic Compounds 12.6 Isomerism 12.7 Fundamental Concepts in Organic Reaction Mechanism 12.8 Methods of Purification of Organic Compounds 12.9 Qualitative Analysis of Organic Compounds 12.10 Quantitative Analysis
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Class 11 Chemistry Chapter 6 Thermodynamics Notes

Enthalpy: It is sum of internal energy and P-V work.

   it is denoted by H.

i.e.H= U + Pv

Class 11 Chemistry Chapter 6 Thermodynamics Notes
Class 11 Chemistry Chapter 6 Thermodynamics Notes
Thermodynamics

Heat capacity, specific heat capacity and molar heat capacity

  • Heat capacity: It is defined as amount of heat required to raise the temperature of substance by one degree Celsius.

          It is given by  :

       C=(q/∆T)

  • Specific heat capacity: It is the amount of heat required to raise the temperature of substance (1g) through one degree Celsius.q =(C x ∆t x m)q= m. C. ∆t
  • Molar heat capacity :

  It is the amount of heat required to raise the temperature of 1 mole of substance through one degree Celsius.

Molar heat capacity = c= heat capacity for 1mole.

                                            c= (c/n)

Where cm is for one mole, n=total number of moles and c is heat capacity

Types of molar heat capacities:

  • Heat capacity at constant volume.
  • Heat capacity at constant pressure.

Heat capacity at constant volume: Is defined as rate of change of internal energy with temperature at constant volume.

It is given by: C= (dU/dt)

Heat capacity at constant pressure: It is defined as rate of change of internal energy with temperature at constant pressure.

It is given by: Cp= (dH/dt)

Relation between Cp &Cv

If the volume of system is kept constant and heat is added to the system then no work is done by the system.

Thus, the heat absorbed by the system is used up completely to increase the internal energy of the system.

And if pressure is kept constant, then the energy supplied is somehow used in increasing the volume and leftover is used to increase the internal energy.

Therefore the energy required to raise the temperature at constant pressure is always more than the energy required to raise the temperature of substance at constant volume. (Cp> Cv).

We know:

Class 11 Chemistry Chapter 6 Thermodynamics Notes
Thermodynamics
Class 11 Chemistry Chapter 6 Thermodynamics Notes
Thermodynamics

The procedure consists of two steps:

  1. Combustion of know mass of substance whose heat of combustion is known: A known mass of compound is taken in platinum cup. Oxygen gas is introduced that is under high pressure in the bomb. A current is passed through the filament immersed in compound.
  2. Combustion takes place and the increase in temperature is noted, from this heat capacity can be calculated by:

Change in internal energy =q x change in temperature x m/M

Where M = molecular mass of substance and ‘m’ is mass of substance taken.

Mathematically:

Let’s say initial temperature -> t10C

In order to find temperature -> t20C

Rise in temperature -> (t2 – t1) 0C

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