Energy can take many forms, including kinetic energy produced by an object’s movement, potential energy produced by an object’s position, heat energy transferred from one object to another due to a temperature difference, radiant energy associated with sunlight, the electrical energy produced in galvanic cells, the chemical energy stored in chemical substances, and so on.

All of these different types of energy may be transformed from one form to the other.

For example, as water in a dam reservoir falls, its potential energy is turned into kinetic energy, and if the falling water is utilized to power turbines, the kinetic energy of the water is converted into electrical energy. However, if the water collides with rocks near the dam’s base. The kinetic energy of the object is transformed into thermal energy. 

As a result, the various kinds of energy are quantitatively tied to one another. Thermodynamics is the study of such quantitative relationships between various kinds of energy. Energy shifts occur as a result of physical and chemical processes. The study of energy transitions in these processes is the main focus of thermodynamics.

Thermodynamics is the branch of science that studies the many kinds of energy, their quantitative connections, and the energy changes that occur in physical and chemical processes. 

Table of Contents

Gibbs Energy

J.W. Gibbs was an American theoretician. He introduced a new thermodynamic function called Gibbs energy denoted as G. The second law of thermodynamics states that ΔS_total = (ΔS_system + ΔS_Surrounding) must be positive for all spontaneous processes. To assess the spontaneity of a process, two entropy changes, ΔS_System and ΔS_Surrounding must be determined. As a result, it is more straightforward to describe the criteria of spontaneity in terms of the system’s thermodynamic features alone, without respect for the surroundings and this problem was solved by Gibbs.

Gibbs energy (G) is defined as,

G = H – TS

where:

• S is the entropy of the system,
• H is Enthalpy, and
• T is the Temperature.

G is also a state function because, H, T, and S are state functions. 

The change in Gibbs Energy (ΔG) on the initial and final state of the system and not on the path connecting the two states. The change in Gibbs energy at constant temperature and pressure is defined as:

ΔG = ΔH – T ΔS

where:

• ΔS is the change in entropy of the system and
• ΔH is the change in Enthalpy.

Gibbs Energy and Spontaneity

The total entropy change can be written as,

ΔS_total = ΔS_system + ΔS_surrounding = ΔS + ΔS_surr

According to the second law of thermodynamics, ΔS_total > 0 at constant temperature and pressure for the process to be spontaneous. If ΔH is the enthalpy change accompanying the reaction, that is the enthalpy change of the system then the change in enthalpy of the surroundings is (-ΔH)

Therefore,

ΔS_surr = -ΔH / T

Hence the total entropy is given by,

ΔS_total = ΔS – ΔH / T   

This equation shows that ΔS_total is expressed in terms of the properties of the system only.

Rearranging the equation we get,

-T ΔS_total = -T ΔS + ΔH

or

-T ΔStotal = ΔH -T ΔS 

Combining the above two equations, we get,

ΔG = -T ΔS_total    

This equation indicates that ΔG and ΔS_total have opposite signs because T is always positive. Thus, for a spontaneous process carried out at a constant temperature and pressure ΔS_total  > 0 and hence ΔG < 0.

Gibbs energy of a system decreases in a spontaneous change that takes place at constant temperature and pressure. On contrary, for a non-spontaneous reaction ΔS_total and hence ΔG > 0.

Gibbs energy of a system increases in a non-spontaneous change that takes place at constant temperature and pressure. The end of the spontaneous process is an equilibrium that corresponds to a minimum in G. Hence the change in Gibbs energy is:

• ΔG < 0, the process is spontaneous.
• ΔG > 0, the process is non-spontaneous.
• ΔG = 0, the process is at equilibrium.

Factor affecting the Spontaneity

Consider the equation,

ΔG = ΔH – T ΔS

The value ΔG determines whether a physical or chemical change will occur spontaneously. The equations ΔH and ΔS correspond to the values of the system alone. The equation states that two elements influence the spontaneity of reactions

• ΔH is the amount of heat transmitted at constant pressure and temperature, and
• ΔS is the rise or reduction in molecular disorder.

The spontaneous process is favoured by a decrease in enthalpy (-ΔH) and increase in entropy (ΔS) On the other hand non-spontaneous reaction is favoured by an increase in enthalpy (+ΔH) and decrease in entropy (-ΔS). The term temperature in the equation is an important component in determining the relative relevance of the enthalpy and entropy contributions to ΔG. If ΔH and ΔS in the equation are both positive or both negative, the sign of ΔG and hence the spontaneity of the reaction, depends on temperature. 

ΔH	ΔS	ΔG	Spontaneity of reaction
Negative   (exothermic)	Positive	Negative	Reactions are spontaneous at all temperatures.
Negative   (exothermic)	Negative	Negative or Positive	Reactions become spontaneous at low temperatures when |T. ΔS| < |ΔH|.
Positive   (endothermic)	Positive	Negative or Positive	Reactions become spontaneous at low temperatures when T.ΔS < ΔH.
Positive   (endothermic)	Negative	Positive	Reactions are non-spontaneous at all temperatures.

Temperature of Equilibrium

At equilibrium, i.e., ΔG = 0, the process is neither spontaneous nor non-spontaneous because it is balanced between spontaneous and non-spontaneous behavior. (+ΔH)

So, 

ΔG = ΔH – T ΔS = 0

Hence,

ΔH = TΔS or T = ΔH / ΔS

T is the temperature at which the transition from spontaneous to non-spontaneous behavior happens. T is calculated on the assumption that ΔH and ΔS are temperature independent. In reality, ΔH and ΔS change with temperature. However, for modest temperature changes, the variance in them will not add considerable mistakes. 

ΔG and Equilibrium constant

All of the substances (reactants and products) in a chemical reaction may not be in their normal forms. As a result of the connection, the change in Gibbs energy of a reaction is related to the change in standard Gibbs energy. 

ΔG = ΔG° + RT ln Q

where:

• ΔG° is the standard Gibbs energy change (change in Gibbs energy when all the substances are in their standard state).
• Q is the reaction quotient.

The expression of the reaction quotient is similar to that of the equilibrium constant, but there is one single difference between them, i.e., Equilibrium concentrations or partial pressures of products and reactants are included in the equilibrium constant. Whereas Q is expressed in terms of reactant beginning concentration partial pressures and product final concentrations or pressures. 

For Example, consider the below example:

aA +bB ⇢ cC + dD

For the above reaction, the reaction Quotient is given by 

or

When the values of concentration or partial pressure are other than equilibrium values. When the reaction reaches equilibrium, the concentrations and partial pressure reach their equilibrium values and at this stage, Q = K.

At equilibrium, ΔG = 0 and Q = K, then the standard Gibbs energy equation becomes,

0 = ΔG° + RT ln K

Hence,

ΔG° = -RT ln K = -2.303RT log₁₀K

This equation gives the relationship between standard Gibbs energy change for the reaction and its equilibrium constant.

Solved Problems

Problem 1: Determine whether the reaction is spontaneous or non-spontaneous for the given value of ΔH and ΔS. Also, state whether they are exothermic or endothermic.

• ΔH = – 40 kJ and ΔS = +135 JK^-1 at 300K
• ΔH = – 60 kJ and ΔS = -160 JK^-1 at 400K

Solution:

 ΔH = – 40 kJ,  ΔS = +135 J K^-1 = 0.135 kJ K^-1 and T = 300K

 ΔG = -40 (kJ) – 0.135(kJ K^-1) × 300(K)

= 80.5 kJ

Because ΔG is negative, the reaction is spontaneous. The negative ΔH value indicates that the reaction is exothermic.

ΔH = – 60 kJ,  ΔS = – 160 J K^-1 = – 0.160 kJ K^-1 and T = 400K

 ΔG = -60 (kJ) – 0.160(kJ K^-1) × 400(K)

 = -60 kJ + 64 kJ = 4kJ

The reaction is non-spontaneous because ΔG is positive and exothermic as ΔH is negative.

Problem 2: For a certain reaction ΔH = -25kJ and ΔS = -40J K^-1. At what temperature will it change from spontaneous to non-spontaneous.

Solution;

T = ΔH / ΔS

 ΔH = – 25 kJ,  ΔS = -40 J K^-1  = 0.04 kJ K-1 

Hence, T = -25(kJ) / -0.04 kJ K^-1  = 625K

Because both ΔH and ΔS are negative, the reaction will occur spontaneously at lower temperatures. As a result, the reaction will be spontaneous below 625K and non-spontaneous beyond 625K.

At 625K, the transition from spontaneous to non-spontaneous occurs.

Problem 3: Determine  ΔS_total and decide whether the following reaction is spontaneous at 298K.

ΔH° = -24.8 kJ, ΔS° = 15 J K^-1

Solution: 

The heat evolved in the reaction is 24.8 kJ. The same quantity of heat is absorbed by the surroundings. 

Hence, Entropy change of the surrounding will be,

ΔS_surr = ΔH° / T

= – [(-24.8 (kJ)) / 298 (K)]

= + 83.2 J K^-1

ΔS_total = ΔS_System + ΔS_Surr 

ΔS_Sys = ΔS° = 15 J K^-1

 = 15(J K^-1) + 83.2 (J K^-1)

= 98.2 J K^-1

As ΔS_total is positive, the reaction is spontaneous at 298 K.

Problem 4: Determine whether the reaction,

N₂​O₄ ​(g)⟶2NO₂​ (g)

is spontaneous at 298 K from the following data.

Δ_fH° (N₂O₄) = 9.16 kJ mol^-1, Δ_fH° (NO₂) = 33.2 kJ mol^-1

Solution: 

ΔH=∑Δf​H°(products)−∑Δf​H°(reactants

= 2 × Δ_fH° (NO₂) – Δ_fH° (N₂O₄)

= 2(mol) × 33.2(kJ mol^-1) -1(mol) × 9.16(kJ mol^-1)

= +57.24 kJ

ΔG° = ΔH° – TΔS°

57.24(kJ) – 298(K) × 175.8 × 10^-3 (kJ K^-1)

= +4.85 kJ.

Because ΔG° is positive, the reaction is non-spontaneous at 298 K. 

The temperature at which the reaction changes from spontaneous to non-spontaneous is given by,

T = ΔH° / ΔS°

= 57.24(kJ) / 0.1758(kJ K^-1)

= 325.6 K

Because ΔH° and ΔS° are both positive, the reaction will be spontaneous at high temperature.

The reaction will be spontaneous above 325.6 K.

Problem 5: Determine K_p for the reaction,

2SO₂​(g) + O₂​(g) ⟶ 2SO_3​(g)

is 7.1 × 10²⁴ at 298 K. Calculate ΔG° for the reaction (R = 8.314 JK^-1 mol^-1).

Solution:

ΔG° = -2.303RT log₁₀ K_p

K_p = 7.1 × 10²⁴ 

R = 8.314 JK^-1 mol^-1

T = 298K

Hence,

  ΔG°  = -2.303 × 8.314 × 10^-3 (kJ K^-1 mol^-1) ×  log₁₀ (7.1 ×  10²⁴)

          = -141.8 kJ mol^-1

Problem statement 6: Calculate K_p for the reaction at 513 K,

2NOCl (g) ⟶ 2NO(g) + Cl₂​(g)

with ΔG° = 17.38 kJ mol^-1.

Solution:

ΔG° = -2.303 RT log₁₀ K_p

ΔG° = 17.38 kJ mol^-1

R = 8.314 J K^-1 mol^-1

T = 513K

Hence,

log₁₀ k_p = – ΔG° / 2.303 RT

             = – (17380(J mol^-1) / 2.303 × 8.314 ( J K^-1 mol^-1) × 513(K))

             = -1.769

Hence, K_p = antilog(-1.769)

                = 0.017          

Neeraj Anand, Param Anand

Er. Neeraj K.Anand is a freelance mentor and writer who specializes in Engineering & Science subjects. Neeraj Anand received a B.Tech degree in Electronics and Communication Engineering from N.I.T Warangal & M.Tech Post Graduation from IETE, New Delhi. He has over 30 years of teaching experience and serves as the Head of Department of ANAND CLASSES. He concentrated all his energy and experiences in academics and subsequently grew up as one of the best mentors in the country for students aspiring for success in competitive examinations. In parallel, he started a Technical Publication "ANAND TECHNICAL PUBLISHERS" in 2002 and Educational Newspaper "NATIONAL EDUCATION NEWS" in 2014 at Jalandhar. Now he is a Director of leading publication "ANAND TECHNICAL PUBLISHERS", "ANAND CLASSES" and "NATIONAL EDUCATION NEWS". He has published more than hundred books in the field of Physics, Mathematics, Computers and Information Technology. Besides this he has written many books to help students prepare for IIT-JEE and AIPMT entrance exams. He is an executive member of the IEEE (Institute of Electrical & Electronics Engineers. USA) and honorary member of many Indian scientific societies such as Institution of Electronics & Telecommunication Engineers, Aeronautical Society of India, Bioinformatics Institute of India, Institution of Engineers. He has got award from American Biographical Institute Board of International Research in the year 2005.

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CBSE Class 11 Chemistry Syllabus

CBSE Class 11 Chemistry Syllabus is a vast which needs a clear understanding of the concepts and topics. Knowing CBSE Class 11 Chemistry syllabus helps students to understand the course structure of Chemistry.

Unit-wise CBSE Class 11 Syllabus for Chemistry

Below is a list of detailed information on each unit for Class 11 Students.

UNIT I – Some Basic Concepts of Chemistry

General Introduction: Importance and scope of Chemistry.

Nature of matter, laws of chemical combination, Dalton’s atomic theory: concept of elements,
atoms and molecules.

Atomic and molecular masses, mole concept and molar mass, percentage composition, empirical and molecular formula, chemical reactions, stoichiometry and calculations based on stoichiometry.

UNIT II – Structure of Atom

Discovery of Electron, Proton and Neutron, atomic number, isotopes and isobars. Thomson’s model and its limitations. Rutherford’s model and its limitations, Bohr’s model and its limitations, concept of shells and subshells, dual nature of matter and light, de Broglie’s relationship, Heisenberg uncertainty principle, concept of orbitals, quantum numbers, shapes of s, p and d orbitals, rules for filling electrons in orbitals – Aufbau principle, Pauli’s exclusion principle and Hund’s rule, electronic configuration of atoms, stability of half-filled and completely filled orbitals.

UNIT III – Classification of Elements and Periodicity in Properties

Significance of classification, brief history of the development of periodic table, modern periodic law and the present form of periodic table, periodic trends in properties of elements -atomic radii, ionic radii, inert gas radii, Ionization enthalpy, electron gain enthalpy, electronegativity, valency. Nomenclature of elements with atomic number greater than 100.

UNIT IV – Chemical Bonding and Molecular Structure

Valence electrons, ionic bond, covalent bond, bond parameters, Lewis structure, polar character of covalent bond, covalent character of ionic bond, valence bond theory, resonance, geometry of covalent molecules, VSEPR theory, concept of hybridization, involving s, p and d orbitals and shapes of some simple molecules, molecular orbital theory of homonuclear diatomic molecules(qualitative idea only), Hydrogen bond.

UNIT V – Chemical Thermodynamics

Concepts of System and types of systems, surroundings, work, heat, energy, extensive and intensive properties, state functions. First law of thermodynamics – internal energy and enthalpy, measurement of U and H, Hess’s law of constant heat summation, enthalpy of bond dissociation, combustion, formation, atomization, sublimation, phase transition, ionization, solution and dilution. Second law of Thermodynamics (brief introduction)
Introduction of entropy as a state function, Gibb’s energy change for spontaneous and nonspontaneous processes.
Third law of thermodynamics (brief introduction).

UNIT VI – Equilibrium

Equilibrium in physical and chemical processes, dynamic nature of equilibrium, law of mass action, equilibrium constant, factors affecting equilibrium – Le Chatelier’s principle, ionic equilibrium- ionization of acids and bases, strong and weak electrolytes, degree of ionization,
ionization of poly basic acids, acid strength, concept of pH, hydrolysis of salts (elementary idea), buffer solution, Henderson Equation, solubility product, common ion effect (with illustrative examples).

UNIT VII – Redox Reactions

Concept of oxidation and reduction, redox reactions, oxidation number, balancing redox reactions, in terms of loss and gain of electrons and change in oxidation number, applications of redox reactions.

UNIT VIII – Organic Chemistry: Some basic Principles and Techniques

General introduction, classification and IUPAC nomenclature of organic compounds. Electronic displacements in a covalent bond: inductive effect, electromeric effect, resonance and hyper conjugation. Homolytic and heterolytic fission of a covalent bond: free radicals, carbocations, carbanions, electrophiles and nucleophiles, types of organic reactions.

UNIT IX – Hydrocarbons

Classification of Hydrocarbons
Aliphatic Hydrocarbons:
Alkanes – Nomenclature, isomerism, conformation (ethane only), physical properties, chemical reactions.
Alkenes – Nomenclature, structure of double bond (ethene), geometrical isomerism, physical properties, methods of preparation, chemical reactions: addition of hydrogen, halogen, water, hydrogen halides (Markovnikov’s addition and peroxide effect), ozonolysis, oxidation, mechanism of electrophilic addition.
Alkynes – Nomenclature, structure of triple bond (ethyne), physical properties, methods of preparation, chemical reactions: acidic character of alkynes, addition reaction of – hydrogen, halogens, hydrogen halides and water.

Aromatic Hydrocarbons:

Introduction, IUPAC nomenclature, benzene: resonance, aromaticity, chemical properties: mechanism of electrophilic substitution. Nitration, sulphonation, halogenation, Friedel Craft’s alkylation and acylation, directive influence of functional group in monosubstituted benzene. Carcinogenicity and toxicity.

To know the CBSE Syllabus for all the classes from 1 to 12, visit the Syllabus page of CBSE. Meanwhile, to get the Practical Syllabus of Class 11 Chemistry, read on to find out more about the syllabus and related information in this page.

CBSE Class 11 Chemistry Practical Syllabus with Marking Scheme

In Chemistry subject, practical also plays a vital role in improving their academic scores in the subject. The overall weightage of Chemistry practical mentioned in the CBSE Class 11 Chemistry syllabus is 30 marks. So, students must try their best to score well in practicals along with theory. It will help in increasing their overall academic score.

CBSE Class 11 Chemistry Practical Syllabus

The experiments will be conducted under the supervision of subject teacher. CBSE Chemistry Practicals is for 30 marks. This contribute to the overall practical marks for the subject.

The table below consists of evaluation scheme of practical exams.

CBSE Syllabus for Class 11 Chemistry Practical

Micro-chemical methods are available for several of the practical experiments. Wherever possible such techniques should be used.

A. Basic Laboratory Techniques
1. Cutting glass tube and glass rod
2. Bending a glass tube
3. Drawing out a glass jet
4. Boring a cork

B. Characterization and Purification of Chemical Substances
1. Determination of melting point of an organic compound.
2. Determination of boiling point of an organic compound.
3. Crystallization of impure sample of any one of the following: Alum, Copper Sulphate, Benzoic Acid.

C. Experiments based on pH

1. Any one of the following experiments:

• Determination of pH of some solutions obtained from fruit juices, solution of known and varied concentrations of acids, bases and salts using pH paper or universal indicator.
• Comparing the pH of solutions of strong and weak acids of same concentration.
• Study the pH change in the titration of a strong base using universal indicator.

2. Study the pH change by common-ion in case of weak acids and weak bases.

D. Chemical Equilibrium
One of the following experiments:

1. Study the shift in equilibrium between ferric ions and thiocyanate ions by increasing/decreasing the concentration of either of the ions.
2. Study the shift in equilibrium between [Co(H₂O)₆] ²⁺ and chloride ions by changing the concentration of either of the ions.

E. Quantitative Estimation
i. Using a mechanical balance/electronic balance.
ii. Preparation of standard solution of Oxalic acid.
iii. Determination of strength of a given solution of Sodium hydroxide by titrating it against standard solution of Oxalic acid.
iv. Preparation of standard solution of Sodium carbonate.
v. Determination of strength of a given solution of hydrochloric acid by titrating it against standard Sodium Carbonatesolution.

F. Qualitative Analysis
1) Determination of one anion and one cation in a given salt
Cations‐ Pb²⁺, Cu²⁺, As³⁺, Al³⁺, Fe³⁺, Mn²⁺, Ni²⁺, Zn²⁺, Co²⁺, Ca²⁺, Sr²⁺, Ba²⁺, Mg²⁺, NH₄ ⁺
Anions – (CO₃)^2‐ , S^2‐, NO₂^‐ , SO₃^2‐, SO^2‐ , NO ^‐ , Cl^‐ , Br^‐, I‐, PO₄^3‐ , C₂O²‐ ,CH₃COO^‐
(Note: Insoluble salts excluded)

2) Detection of ‐ Nitrogen, Sulphur, Chlorine in organic compounds.

G) PROJECTS
Scientific investigations involving laboratory testing and collecting information from other sources.

A few suggested projects are as follows:

• Checking the bacterial contamination in drinking water by testing sulphide ion
• Study of the methods of purification of water.
• Testing the hardness, presence of Iron, Fluoride, Chloride, etc., depending upon the regional
  variation in drinking water and study of causes of presence of these ions above permissible
  limit (if any).
• Investigation of the foaming capacity of different washing soaps and the effect of addition of
  Sodium carbonate on it.
• Study the acidity of different samples of tea leaves.
• Determination of the rate of evaporation of different liquids Study the effect of acids and
  bases on the tensile strength of fibres.
• Study of acidity of fruit and vegetable juices.

Note: Any other investigatory project, which involves about 10 periods of work, can be chosen with the approval of the teacher.

Practical Examination for Visually Impaired Students of Class 11

Below is a list of practicals for the visually impaired students.

A. List of apparatus for identification for assessment in practicals (All experiments)
Beaker, tripod stand, wire gauze, glass rod, funnel, filter paper, Bunsen burner, test tube, test tube stand,
dropper, test tube holder, ignition tube, china dish, tongs, standard flask, pipette, burette, conical flask, clamp
stand, dropper, wash bottle
• Odour detection in qualitative analysis
• Procedure/Setup of the apparatus

B. List of Experiments A. Characterization and Purification of Chemical Substances
1. Crystallization of an impure sample of any one of the following: copper sulphate, benzoic acid
B. Experiments based on pH
1. Determination of pH of some solutions obtained from fruit juices, solutions of known and varied
concentrations of acids, bases and salts using pH paper
2. Comparing the pH of solutions of strong and weak acids of same concentration.

C. Chemical Equilibrium
1. Study the shift in equilibrium between ferric ions and thiocyanate ions by increasing/decreasing
the concentration of eitherions.
2. Study the shift in equilibrium between [Co(H2O)6]2+ and chloride ions by changing the
concentration of either of the ions.

D. Quantitative estimation
1. Preparation of standard solution of oxalic acid.
2. Determination of molarity of a given solution of sodium hydroxide by titrating it against standard
solution of oxalic acid.

E. Qualitative Analysis
1. Determination of one anion and one cation in a given salt
2. Cations – NH+4
Anions – (CO3)2-, S2-, (SO3)2-, Cl-, CH3COO-
(Note: insoluble salts excluded)
3. Detection of Nitrogen in the given organic compound.
4. Detection of Halogen in the given organic compound.

Note: The above practicals may be carried out in an experiential manner rather than recording observations.

We hope students must have found this information on CBSE Syllabus useful for their studying Chemistry. Learn Maths & Science in interactive and fun loving ways with ANAND CLASSES (A School Of Competitions) App/Tablet.

Frequently Asked Questions on CBSE Class 11 Chemistry Syllabus

Q1

How many units are in the CBSE Class 11 Chemistry Syllabus?

There are 9 units in the CBSE Class 11 Chemistry Syllabus. Students can access various study materials for the chapters mentioned in this article for free at ANAND CLASSES (A School Of Competitions).

Q2

What is the total marks for practicals examination as per the CBSE Class 11 Chemistry Syllabus?

The total marks for the practicals as per the CBSE Class 11 Chemistry Syllabus is 30. It includes volumetric analysis, content-based experiment, salt analysis, class record, project work and viva.

Q3

Which chapter carries more weightage as per the CBSE Syllabus for Class 11 Chemistry?

The organic chemistry chapter carries more weightage as per the CBSE Syllabus for Class 11 Chemistry.