Integrated Rate Laws: Definition, Equation, Reactions & Half Life, Solved Problems, FAQs

This article helps us learn about Integrated Rate Law and its derivation for various reactions. We will also learn how to solve numerical problems based on the Rate Law.

What is Rate Law?

Rate Law is defined as the molar concentration of the reactants with each term raised to some powers which may or may not be the same as the stoichiometric coefficient of the reactant in a balanced chemical equation.

For a Chemical Balanced reaction i.e., A + B 

Rate = k[A]α[B]β

Where,

  • α and β are the concentration of the A and B respectively
  • k is Rate Constant

Rate Constant

The rate constant is defined as the rate of reaction when the molar concentration of all reactants is unity. For a balanced chemical equation of second order, i.e., A + B ⇢ C + D, if [A]=[B]= 1 mol/liter.

Therefore, Rate = K.

Note: Rate Constant depends on the molar concentration of the reactant, so it means the greater the value of the rate constant, the more quickly the reaction takes place.

Unit of Rate Constant: The unit of the rate constant depends on the order of the reaction (time, concentration).

Order of Reaction

The sum of power to which the molar concentrations in the rate law equation are raised to express the observed rate of the reaction is known as the order of the reaction.

Example: What is the Order of Reaction given as follows:

2N2O5 ⇢ 4NO2  + O2

Solution:

For the given Reaction,

Rate of Reaction, R = [N2O5]1

As, rate of reaction only depends on the concentration of N2O5

Thus, Order of Reaction is 1

Note: Order of the reaction is basically integers numbers like 0,1,2 but for some complex species, it can be in fractions as well.

What is Integrated Rate Law?

Integrated Rate Law provides a mathematical representation of the rate of the reaction using an initial and actual concentration of one or more reactants at any given time (t). Using this law we can calculate the rate constant as well as the mechanics of the reactions as well. For each order of reaction, the rate raw is given differently.  Let’s discuss the Integrated Rate Law for zeroth, first and second order reaction in detail.

Integrated Rate Law Zero Order

Integrated Rate Law for zero order reaction states that the rate of reaction does not depend on the concentration of the reactants. For a reaction, A ⇢ Product zero order Integrated Rate Law is represented as:

Rate ∝ [A]0

Where,

[A] is the concentration of the Reactant

After integrating and simplifying the above condition, we get the integrated rate law for a zero-order reaction, i.e.,

[A] = [A0] – kt

Where,

  • [A] is the concentration of the reactant A at time t
  • [A0] is the initial concentration of reactant A
  • k is the rate constant of the reaction
  • t is the reaction time

Integrated Rate Law First Order

This law states that the rate of the first-order reaction depends upon the first power of the reactant. For a reaction, A ⇢ Product first order Integrated Rate Law is represented as:

Rate ∝ [A]1

After integrating and simplifying the above condition, we get the integrated rate law for a first-order reaction, i.e.,

[A] = [A₀]e-kt

Where,

  • [A] is the concentration of the reactant A at time t
  • [A0] is the initial concentration of reactant A
  • k is the rate constant of the reaction
  • t is the reaction time
  • e is the mathematical constant i.e., Euler number (e = 2.71828)

Integrated Rate Law Second Order

Integrated Rate Law Second Order states that the rate of the second order reaction depends upon either the second power of the concentration. For a reaction, 2A ⇢ Product first order Integrated Rate Law is represented as:

Rate ∝ [A]2

After integrating and simplifying the above condition, we get the integrated rate law for a second-order reaction, i.e.,

Integrated Rate Law is one of the fundamental concepts in the field of chemical kinetics, which is the branch of chemistry that deals with the speed or rate of reactions and various other factors affecting them. Integrated Rate Law tells us about the rate of the reaction for various different reactions such as zeroth order, first order, and second order, etc. Rate Law helps us from measuring rates to predicting the concentration of the reactants, which further helps scientists and scholars to unfold the mysteries of chemical transformations. 

Where,

  • [A] is the concentration of the reactant A at time t
  • [A0] is the initial concentration of reactant A
  • k is the rate constant of the reaction
  • t is the reaction time

Integrated Rate Equations

For the general reaction aA + bB ⇢ cC + dD, the Rate of this reaction is given as:

Rate = -dR/dt = k[A]a[B]b

Where negative sign signify the decrease in concentration.

This form of the equation is called the differential rate equation. This form is not convenient to determine the rate law and hence the order of the reaction. This is because the instantaneous rate has to be determined to form the slope of the tangent at time t in the plot of concentration versus time. 

To overcome the above difficulty, we integrate the differential equation for the reaction of any order. This gives us an equation related directly to the experimental data, i.e., time, concentrations at different times, and rate constant.

Integrated Rate Equation for Zero-Order Reaction

Any reaction for which the rate doesn’t depend on the concentration of the reactant is called Zero Order Reaction. In other words, form A ⇢ Product; is Zero Order Reaction if its rate is independent of the concentration of A.

Consider the general reaction: A ⇢ products

If it is a reaction of zero-order,

Rate = – d[A]/dt = k[A]0 = k 

⇒ d[A] = – k dt

Integrating both sides, we get: 

[A] = – kt + C                . . . (i) 

Where C is the constant of Integration.

at t = 0, the Concentration of the reaction is [A0], thus

[A0] = 0 + C ⇒ C = [A0]

Substituting this value of C in Eqn. (i), we get 

[A] = – kt + [A0               . . . (ii)

⇒ kt = [A0] – [A] 

⇒ k = {[A0] – [A]}/t                . . . (iii)

Characteristics of Reactions of Zero Order

The characteristics of the zero-order reaction are,

  • Any reaction of zero-order must obey equation, (ii) As it is an equation of a straight line (y = mx + c), the plot of [A] versus t will be a straight line with slope = – k and intercept on the concentration axis = [A0],
  • Half-Life Period: The half-life period (t1/2) is the time in which half of the substance has reacted.

This implies that when [A] = [A0]/2, t = t1/2.

Substituting these values in Eqn. (iii), we get

t1/2 = 1/k  { [A0] – [A0]/2 } = [A0]/2k  

⇒ t1/2 =[A0]/2k                . . . (iv) 

Thus, the half-life period of a zero-order reaction is directly proportional to the initial concentration, i.e., t1/2 ∝ [A]0.

  • Units of Rate Constant (k) for Zeroth Order Reaction = molar conc./time = mol L-1 /time = mol L-1  time-1 .

Integrated Rate Equation for First Order Reactions

A reaction is said to be of the first order if the rate of the reaction depends upon one concentration term only. Thus, we may have

  • For the reaction: A ⇢ products, rate the reaction ∝ [A]. 
  • For the reaction: 2A ⇢  products, rate the reaction ∝ [A] only.
  • For the reaction: A + B ⇢  products, rate the reaction ∝ [A] or [B] only.

Let us consider the simplest case,

A  ⇢  Products

Suppose we start with moles per liter of the reactant A. After time t, suppose x moles per liter of it have decomposed. Therefore, the concentration of A after time t = (a – x) moles per liter. Then according to the law of mass action, 

Rate of reaction ∝ (a – x), 

i.e., 

dx/dt  ∝ (a – x)  

⇒ dx/dt = k (a – x)                . . . (i)

where k is called the rate constant or the specific reaction rate for the reaction of the first order. 

The expression for the rate constant k may be derived as follows:

Equation (i) may be rewritten in the form  

dx/a-x =k dt                . . . (ii)

Integrating equation (ii), we get

-In (a – x) = kt + I                 . . . (iii)

where I is a constant of integration.

In the beginning, when t=0, x=0 

Putting these values in equation (iii), we get 

– In (a – 0) = k x 0 + I 

⇒ – In a = I                . . . (iv)

Substituting this value of I in equation (iii), we get 

– In (a – x) = kt + (- In a) 

⇒ kt = a – In (a -x) 

⇒ kt = In a/a-x                . . . (v)

⇒ k = 1/ t In a/a-x 

⇒ k=2.303/t log a/a-x                . . . (vi)

If the initial concentration is [A]0 and the concentration after time t is [A], then putting a = [A]0 and (a – x ) = [A],

Equation (vi) becomes:

k = 2.303/t log [A0]/[A]                . . . (vii)

Putting a = [A0] and (a – x) = [A] in eqn. (v), 

We get, 

kt = In [A0]/[A]                . . . (viii)

which can be written in the exponential form as:

[A0]/[A] = ekt  

⇒ [A]/[A0] =e-kt  

[A] = [A0] e-kt                  . . . (ix)

Characteristics of First-Order Reactions

The characteristics of the zero-order reaction are,

  • Any reaction of the first order must obey equations (vi), (vii), and (ix).
  • Half-life Period:The time in which half of the substance is reacted with each other and is converted into the product is called the half-life of the reaction.

As we know, the time taken for substance from concentration a to a-x is given by:

t = 2.303/k log [a/(a – x)]

When half of the reaction is completed, x = a/2. Representing the time taken for half of the reaction to be completed by t1/2

t1/2 = 2.303/k log a/(a-a/2) 

⇒ t1/2 = 2.303/k log 2 

t1/2 = 0.693/k

Where, 

  • t1/2is the half life of reaction, and 
  • k is the rate constant.

Integrated Rate Equation for First-Order Gas-Phase Reaction

Consider the general first-order gas-phase reaction:

A (g) ⇢ B (g) + C (g)

Suppose the initial pressure of A = P0 atm. After time t, suppose the pressure of A decreases by p atm. 

Now, as 1 mole of A decomposes to give 1 mole of B and 1 mole of C, the pressure of B and C will increase by p each. Hence, we have 

  P0 atm   00
  P0 – p  p atmp atm

The total pressure of the reaction mixture after time t, 

Pt = (P0 – p) + p + p = P0 + p atm

⇒ p = Pt – P0

So, pressure of A after time t (PA) = P0 – p = P0 –  (Pt – P0) = 2 P0 – Pt

But initial pressure of A (P0) ∝ initial conc. of A, i.e., [A]0

and pressure of A after time t(PA) ∝ conc. of A at time t, i.e., [A]

Substituting these values in the first-order rate equation,

k = 2.303/t log [A]0/[A], We get 

k = 2.303/t log [P0/(2P0 – Pt)]

Where, 

  • P0 is the Intial Pressure of Reactant,
  • Pt is the total pressure after time t,
  • t is the time in which pressure changes from P0 to Pt, and 
  • k is the rate constant.

Summary of Integrated Rate Law

This is the most common method for studying the kinetics of a chemical reaction. For example, consider the reaction: 

nA ⇢ products 

If we start with a moles/liter of A and in time t, x moles/liter have reacted so that the concentration after time t is (a – x) moles/liter, then 

if the reaction is of the first order, dx/dt = k(a – x), and if the reaction is of the second order, dx/dt = k (a – x)2, and so on.

These differential equations can be integrated to get expressions for the rate constants. These are given below for zero, first, and second-order reactions:

  • For zero-order reaction, k = 1/t {[A0] – [A]} 
  • For first-order reaction, k = 2.303/t log  [A0]/[A]
  • For second-order reaction, k = 1/t {1/[A] – 1/[A0]}

The advantage of the integrated method is that these integrated forms of equations contain the concentration of a reactant at different times and hence can be solved to find the value of k from the data of the run of one experiment only and need not start with different initial concentrations. Moreover, they can be used to find the time for any fraction of the reaction to complete.  

Differential vs Integrated Rate Laws

There are some key differences between both differential and integrated rate laws, which are listed in the following table:

Describes the rate of change of concentration with respect to time.Describes the relationship between concentration and time.
Represents the rate equation in terms of initial concentrations and rate constants.Represents the concentration of reactants or products as a function of time.
Provides information about the instantaneous rate of a reaction at a specific moment in time.Provides information about the overall change in concentration over a given time interval.
Can vary with time and is dependent on the concentration of reactants.Remains constant for a specific reaction under constant conditions.
Typically expressed as a differential equation, involving derivatives.Usually expressed as a mathematical equation, relating concentrations and time.
Helpful in determining the order of a reaction and the reaction rate constant.Useful for determining reaction orders and obtaining information about reaction mechanisms.
Represents the rate of change at a specific point on the reaction progress curve.Represents the cumulative effect of the reaction at different time points.

Sample Problems on Integrated Rate Law

Problem 1: At 373 K, the half-life period for the thermal decomposition of N2O5 is 4.6 sec and is independent of the initial pressure of N2O5. Calculate the specific rate constant at this temperature.

Solution:

Since the half-life period is independent of the initial pressure, this shows that the reaction is of the first order.

For a reaction of the first order, we know that t1/2 = 0.693/k 

or 

k = 0.693/t1/2 = 0.693/4.6s 

   = 0.1507 s-1

Problem 2: A first-order reaction is found to have a rate constant, k = 5.5 x 10-14s-1. Find the half-life of the reaction.

Solution:

For a first order reaction, t1/2 = 0.693/k 

t1/2 = 0.693/5.5×10-14s-1

      = 1.26 x 1013 s-1.

Problem 3: Show that in the case of a first-order reaction, the time required for 99.9% of the reaction to take place is about ten times that required for half the reaction.

Solution:

For reaction of first order, 

t1/2 = 2.303/k log a/a – a/2 

      = 2.303/k log 2

     = 2.303/k (0.3010)

t99.9% = 2,303/k log 

         = a/a-0.999a

t99.9% = 2.303/k log 10-3 

         = 2.303/k x 3  

Therefore, t99.9%/t1/2 = 3/0.3010 ≅ 10

Problem 4: The initial concentration of N2O5 in the first-order reaction, N2O5(g) ⇢ 2 NO2(g)+ 1/2 O2(g), was 1.24 x 10-2 mol L-1 at 318 K. The concentration of N2O5 after 60 minutes was 0.20 x 10-2 mol L-1. Calculate the rate constant of the reaction at 318 K.

Solution:

k = 2.303/t log [A]0/[A] = 2.303/t log [N2O5]0/[N2O5]t 

 2.303/60min log 1.24 x 10-2 mol L-1/0.2 x 10-2 mol L-1

= 2.303/60 log 6.2 min-1 = 2.303/60 x 0.7924min-1 

= 0.0304 min-1.

Problem 5: A first-order reaction is found to have a rate constant k = 7.39 x 10-5 sec-1. Find the half-life of the reaction (log 2 = 0.3010).

Solution:

For a first order reaction, k = 2.303/t log a/a-x

For t = t1/2, x = a/2

t1/2 = 2.303/k log a/ a-a/2 

= 2.303/k log 2 

= 2.303/7.39×10-5s-1 x 0.3010 

= 9.38 x 103 s-1.

FAQs on Integrated Rate Law

1. What is Integrated Rate Law?

An integrated rate law is an equation that describes the relationship between the concentration of a reactant and time during a chemical reaction. It integrates the rate equation, which is typically derived from the rate of change of reactant concentrations with respect to time.

2. What are Integrated Rate Laws for Different Order Reactions?

For different order reactions, rate law is given as follows:

  • For Zero-order Reaction: Rate ∝ [A]0
  • For First-order Reaction: Rate ∝ [A]
  • For Second-order Reaction: Rate ∝ [A]2

Where [A] is the concentration of the reactant.

3. What are Common Forms of Integrated Rate Equations?

The common form for  integrated rate Equation for various differnt order reactions as follows:

  • For Zero-order reaction: [A] = [A₀] – kt
  • For First-order reaction: ln[A] = ln[A₀] – kt
  • For Second-order reaction: 1/[A] = 1/[A₀] + kt

Where,

  • k is the rate constant,
  • [A₀] is the inital concentration of the reactant,
  • [A] is the required concentration after time t, and 
  • ln represents the natural logarithm. 

4.  Can Integrated Rate Laws be Used for All Types of Reactions?

Integrated rate laws are typically derived for elementary reactions (reactions that occur in a single step) and reactions with simple rate expressions. They may not be directly applicable to complex reaction mechanisms involving multiple steps or reactions with non-elementary rate laws. In such cases, alternative mathematical approaches or computational methods may be required to analyze the reaction kinetics.

5. What are Units for Rate Constant in Integrated Rate Laws?

The units of the rate constant (k) in integrated rate laws depend on the order of the reaction, and given as follows:

  • For zero-order reactions, the units of k are concentration/time (e.g., M/s). 
  • For first-order reactions, the units of k are 1/time (e.g., 1/s). 
  • For second-order reactions, the units of k are 1/(concentration·time) (e.g., 1/(M·s)).

6. Can Integrated Rate Laws be Applied to Reversible Reactions?

Integrated rate laws can be used for reversible reactions if the reaction is in a steady state or if the forward and backward reactions have reached equilibrium. In such cases, the concentrations of reactants or products can be treated as constant, allowing for the application of integrated rate laws. However, for reversible reactions that are not at equilibrium, integrated rate laws may not accurately describe the kinetics, and more complex mathematical models are necessary.

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.

CBSE Class 12 Chemistry Syllabus Download PDF

Below is the CBSE Class 12 Syllabus along with the marking scheme and time duration of the Chemistry exam.

S.NoTitleNo. of PeriodsMarks
1Solutions107
2Electrochemistry129
3Chemical Kinetics107
4d -and f -Block Elements127
5Coordination Compounds127
6Haloalkanes and Haloarenes106
7Alcohols, Phenols and Ethers106
8Aldehydes, Ketones and Carboxylic Acids108
9Amines106
10Biomolecules127
Total70

CBSE Class 12 Chemistry Practical Syllabus along with Marking Scheme

The following is a breakdown of the marks for practical, project work, class records, and viva. The total number of marks for all parts is 15. The marks for both terms are provided in the table below.

Evaluation Scheme for ExaminationMarks
Volumetric Analysis08
Salt Analysis08
Content-Based Experiment06
Project Work and Viva04
Class record and Viva04
Total30

CBSE Class 12 Chemistry Syllabus (Chapter-wise)

Unit -1: Solutions

  • Raoult's law.
  • Colligative properties - relative lowering of vapour pressure, elevation of boiling point, depression of freezing point, osmotic pressure, determination of molecular masses using colligative properties, abnormal molecular mass.
  • Solutions, Types of solutions, expression of concentration of solutions of solids in liquids, solubility of gases in liquids, solid solutions.
  • Van't Hoff factor.

Unit -2: Electrochemistry

  • Redox reactions, EMF of a cell, standard electrode potential
  • Nernst equation and its application to chemical cells
  • Relation between Gibbs energy change and EMF of a cell
  • Kohlrausch's Law
  • Electrolysis and law of electrolysis (elementary idea)
  • Dry cell-electrolytic cells and Galvanic cells
  • Conductance in electrolytic solutions, specific and molar conductivity, variations of conductivity with concentration.
  • Lead accumulator
  • Fuel cells

Unit -3: Chemical Kinetics

  • Rate of a reaction (Average and instantaneous)
  • Rate law and specific rate constant
  • Integrated rate equations and half-life (only for zerfirst-order order reactions)
  • Concept of collision theory (elementary idea, no mathematical treatment)
  • Factors affecting rate of reaction: concentration, temperature, catalyst;
  • Order and molecularity of a reaction
  • Activation energy
  • Arrhenius equation

Unit -4: d and f Block Elements  

  • Lanthanoids- Electronic configuration, oxidation states, chemical reactivity and lanthanoid contraction and its consequences.
  • Actinoids- Electronic configuration, oxidation states and comparison with lanthanoids.
  • General introduction, electronic configuration, occurrence and characteristics of transition metals, general trends in properties of the first-row transition metals – metallic character, ionization enthalpy, oxidation states, ionic radii, color, catalytic property, magnetic properties, interstitial compounds, alloy formation, preparation and properties of K2Cr2O7 and KMnO4.

Unit -5: Coordination Compounds  

  • Coordination compounds - Introduction, ligands, coordination number, color, magnetic properties and shapes
  • The importance of coordination compounds (in qualitative analysis, extraction of metals and biological system).
  • IUPAC nomenclature of mononuclear coordination compounds.
  • Bonding
  • Werner's theory, VBT, and CFT; structure and stereoisomerism

Unit -6: Haloalkanes and Haloarenes  

  • Haloarenes: Nature of C–X bond, substitution reactions (Directive influence of halogen in monosubstituted compounds only). Uses and environmental effects of - dichloromethane, trichloro methane, tetrachloromethane, iodoform, freons, DDT.
  • Haloalkanes: Nomenclature, nature of C–X bond, physical and chemical properties, optical rotation mechanism of substitution reactions.

Unit -7: Alcohols, Phenols and Ethers   

  • Phenols: Nomenclature, methods of preparation, physical and chemical properties, acidic nature of phenol, electrophilic substitution reactions, uses of phenols.
  • Ethers: Nomenclature, methods of preparation, physical and chemical properties, uses.
  • Alcohols: Nomenclature, methods of preparation, physical and chemical properties (of primary alcohols only), identification of primary, secondary and tertiary alcohols, mechanism of dehydration, and uses with special reference to methanol and ethanol.

Unit -8: Aldehydes, Ketones and Carboxylic Acids   

  • Carboxylic Acids: Nomenclature, acidic nature, methods of preparation, physical and chemical properties; uses.
  • Aldehydes and Ketones: Nomenclature, nature of carbonyl group, methods of preparation, physical and chemical properties, mechanism of nucleophilic addition, the reactivity of alpha hydrogen in aldehydes, uses.

Unit -9: Amines    

  • Diazonium salts: Preparation, chemical reactions and importance in synthetic organic chemistry.
  • Amines: Nomenclature, classification, structure, methods of preparation, physical and chemical properties, uses, and identification of primary, secondary and tertiary amines.

Unit -10: Biomolecules     

  • Proteins -Elementary idea of - amino acids, peptide bond, polypeptides, proteins, structure of proteins - primary, secondary, tertiary structure and quaternary structures (qualitative idea only), denaturation of proteins; enzymes. Hormones - Elementary idea excluding structure.
  • Vitamins - Classification and functions.
  • Carbohydrates - Classification (aldoses and ketoses), monosaccharides (glucose and fructose), D-L configuration oligosaccharides (sucrose, lactose, maltose), polysaccharides (starch, cellulose, glycogen); Importance of carbohydrates.
  • Nucleic Acids: DNA and RNA.

The syllabus is divided into three parts: Part A, Part B, and Part C. Part A consist of Basic Concepts of Chemistry, which covers topics such as atomic structure, chemical bonding, states of matter, and thermochemistry. Part B consists of Topics in Physical Chemistry, which includes topics such as chemical kinetics, equilibrium, and electrochemistry. Part C consists of Topics in Organic Chemistry, which covers topics such as alkanes, alkenes, alkynes, and aromatic compounds.

Basic Concepts of Chemistry:

  • Atomic structure: This section covers the fundamental concepts of atomic structure, including the electronic configuration of atoms, the Bohr model of the atom, and the wave nature of matter.
  • Chemical bonding: This section covers the different types of chemical bonds, including ionic, covalent, and metallic bonds, as well as the concept of hybridization.
  • States of the matter: This section covers the three states of matter - solid, liquid, and gas - and the factors that influence their properties.
  • Thermochemistry: This section covers the principles of thermochemistry, including the laws of thermodynamics and the concept of enthalpy.

Chapters in Physical Chemistry:

  • Chemical kinetics: This section covers the study of the rate of chemical reactions and the factors that influence it, including the concentration of reactants, temperature, and the presence of catalysts.
  • Equilibrium: This section covers the principles of chemical equilibrium, including the concept of Le Chatelier's principle and the equilibrium constant.
  • Electrochemistry: This section covers the principles of electrochemistry, including the concept of half-cell reactions, galvanic cells, and electrolysis.

Chapters in Organic Chemistry:

  • Alkanes: This section covers the properties and reactions of alkanes, including their structure, isomerism, and combustion.
  • Alkenes: This section covers the properties and reactions of alkenes, including their structure, isomerism, and addition reactions.
  • Alkynes: This section covers the properties and reactions of alkynes, including their structure, isomerism, and addition reactions.
  • Aromatic compounds: This section covers the properties and reactions of aromatic compounds, including their structure, isomerism, and electrophilic substitution reactions.

In addition to the topics covered in the syllabus, the CBSE Class 12 Chemistry exam also tests students on their analytical and problem-solving skills, as well as their ability to apply the concepts learned in the classroom to real-world situations.

Students can also check out the Tips for the Class 12 Chemistry Exam. They can easily access the Class 12 study material in one place by visiting the CBSE Class 12 page at ANAND CLASSES (A School Of Competitions). Moreover, to get interactive lessons and study videos, download the ANAND CLASSES (A School Of Competitions) App.

Frequently Asked Questions on CBSE Class 12 Chemistry Syllabus

Q1

How many chapters are there in the CBSE Class 12 Chemistry as per the syllabus?

There are 10 chapters in the CBSE Class 12 Chemistry as per Syllabus. Students can learn all these chapters efficiently using the study materials provided at ANAND CLASSES (A School Of Competitions).

Q2

What is the marking scheme for CBSE Class 12 Chemistry practical exam according to the syllabus?

The marking scheme for CBSE Class 12 Chemistry practical exam, according to the syllabus, is 8 marks for volumetric analysis, 8 marks for salt analysis, 6 marks for the content-based experiment, 4 marks for the project and viva and 4 marks for class record and viva.

Q3

Which is the scoring chapter in Chemistry as per CBSE Class 12 syllabus?

The chapter Electrochemistry in Chemistry is the scoring chapter as per CBSE Class 12 syllabus.