Principles Related to Practical Chemistry Chapter - Chemistry JEE Main

This chapter is dedicated to the investigation of various methods and approaches for determining the amount of any given element contained in a compound. Analyses are separated into three groups in this chapter: inorganic, organic, and physical. We can use a variety of assays to discover which element is present in a compound.

There are a variety of real-life applications that we witness in our everyday lives. The following are a few of them: The nitrogen estimation is done using two separate methods: Duma's approach and Kjeldahl's method. The Carius method is used to estimate halogens quantitatively. The following is a visual illustration of the procedure.

JEE Main Chemistry Chapters 2024

Detection of Extra Elements and Functional Groups in Organic Compounds

Identifying the presence of specific elements and functional groups in organic compounds is crucial for both analytical and synthetic chemistry. JEE Main students should have a strong grasp of these detection methods to understand and manipulate organic molecules effectively.

Detection of Extra Elements:

Nitrogen (N): Nitrogen is typically detected through the Dumas method, which involves heating the organic compound in the presence of copper oxide. The liberated nitrogen forms ammonia gas, which is then collected and quantified.

Sulfur (S): Sulfur can be detected using the Carius method. The organic compound is combusted with fuming nitric acid, leading to the formation of sulfuric acid. The sulfur content is determined by precipitating it as barium sulfate.

Halogens (F, Cl, Br, I): Halogens are detected using methods like the Lassaigne's test. The organic compound is heated with sodium, and the resulting sodium halides are then extracted with water. The presence of halogens is confirmed through various chemical tests specific to each halogen.

Detection of Functional Groups:

Hydroxyl (Alcoholic and Phenolic) Groups:

Alcoholic hydroxyl groups are detected using the Lucas test. Alcohols react with Lucas reagent (concentrated HCl and $ZnCl_2$) to form alkyl chlorides.

Phenolic hydroxyl groups are identified through tests like the ferric chloride test, which results in a colored complex with phenols.

Carbonyl (Aldehyde and Ketone) Groups:

Tollens' test is employed to detect aldehyde groups. Aldehydes reduce Tollens' reagent $(Ag(NH_3)^{2+})$ to form a silver mirror on the reaction vessel's walls.

Ketone groups can be identified using the iodoform test, which results in the formation of yellow iodoform precipitate.

Carboxyl Group:

The sodium bicarbonate test is used to detect carboxyl groups. Carboxylic acids react with sodium bicarbonate to produce effervescence ($CO_2$ gas is released).

Amino Groups:

Amino groups can be detected through the Hinsberg test, which involves reacting the amine with benzenesulfonyl chloride (Hinsberg's reagent). Primary amines yield insoluble precipitates, secondary amines yield soluble compounds, and tertiary amines do not react.

Chemistry Involved in the Preparation of Various Inorganic and Organic Compounds

Understanding the chemistry behind the preparation of different inorganic and organic compounds is fundamental for JEE Main students. This knowledge not only contributes to their understanding of chemical reactions but also provides a strong foundation for various applications in chemistry and related fields.

Inorganic Compounds:

1. Mohr's Salt (Ferrous Ammonium Sulfate Hexahydrate):

Chemical Formula: $Fe(NH_4)-2(SO_4)_2\cdot 6H_2O$

Preparation:

Mohr's salt is prepared through a double displacement reaction. Here's a step-by-step explanation of the process:

Step 1: Ferrous Sulfate Solution

A solution of ferrous sulfate $(FeSO_4)$ is prepared.

Step 2: Ammonium Sulfate Solution

A separate solution of ammonium sulfate $((NH_4)_2SO_4)$ is also prepared.

Step 3: Mixing and Precipitation

The ferrous sulfate and ammonium sulfate solutions are mixed.

A double displacement reaction occurs, leading to the formation of ferrous ammonium sulfate (Mohr's salt) and ammonium ions (NH4+).

The ferrous ammonium sulfate precipitates as a green crystalline solid due to its low solubility.

The ammonium ions remain in the solution.

Step 4: Crystallization

The green precipitate is collected and allowed to crystallize. The resulting crystals are Mohr's salt.

2. Potash Alum (Aluminum Potassium Sulfate Dodecahydrate):

Chemical Formula: $KAl(SO_4)_2\cdot 12H_2O$

Preparation:

Potash alum is prepared through a double displacement reaction, similar to Mohr's salt.

Step 1: Aluminum Sulfate Solution

A solution of aluminum sulfate $(Al_2(SO_4)_3)$ is prepared.

Step 2: Potassium Sulfate Solution

A separate solution of potassium sulfate $(K_2SO_4)$ is also prepared.

Step 3: Mixing and Precipitation

The aluminum sulfate and potassium sulfate solutions are mixed.

A double displacement reaction occurs, leading to the formation of potash alum and aluminum ions $(Al^{3+})$.

Potash alum precipitates as colorless, octahedral crystals.

Aluminum ions remain in the solution.

Step 4: Crystallization

The precipitated potash alum is collected and allowed to crystallize, forming the final product.

Organic Compounds:

1. Acetanilide:

Chemical Formula: $C_8H_9NO$

Preparation:

Acetanilide is synthesized by acetylation of aniline, which involves the reaction of aniline with acetic anhydride. The process can be outlined as follows:

Step 1: Formation of Acetic Anhydride

Acetic anhydride is prepared by mixing acetic acid $(CH_3COOH)$ and acetic anhydride $(C_4H_6O_3)$ in the presence of a catalyst.

Step 2: Acetylation of Aniline

Aniline is reacted with the acetic anhydride in the presence of a mild Lewis acid or base catalyst. This reaction leads to the replacement of the hydrogen atom in aniline with an acetyl group $(CH_3CO^-)$.

Step 3: Isolation

The resulting acetanilide is isolated through suitable purification techniques like crystallization.

2. p-Nitroacetanilide:

Chemical Formula: $C_8H_8N_2O_3$

Preparation:

p-Nitroacetanilide is synthesized by nitrating acetanilide. The process involves the following steps:

Step 1: Nitration of Acetanilide

Acetanilide is treated with a mixture of concentrated nitric acid $(HNO_3)$ and concentrated sulfuric acid $(H_2SO_4)$.

This reaction leads to the nitration of the aniline ring in acetanilide, resulting in the introduction of a nitro group $(NO_2)$ at the para position.

Step 2: Isolation

The product, p-nitroacetanilide, is isolated through suitable purification techniques, such as recrystallization.

3. Aniline Yellow (p-Sulfonated Aniline Dye):

Chemical Formula: $C_6H_4N_2Na_2O_7S_2$

Preparation:

Aniline Yellow is a synthetic dye derived from aniline. It is prepared through the sulfonation of aniline followed by diazotization and coupling reactions. The detailed procedure involves multiple steps and chemical transformations.

4. Iodoform (Triiodomethane):

Chemical Formula: $CHI_3$

Preparation:

Iodoform is synthesized through the halogenation of acetone in the presence of iodine and a base (e.g., sodium hydroxide). The reaction can be summarized as follows:

Step 1: Halogenation

Acetone is treated with iodine $(I_2)$ and a basic solution (e.g., NaOH).

The reaction results in the formation of iodoform, which is a yellow solid.

Step 2: Isolation

The iodoform is collected, typically through filtration, and may undergo further purification if required.

Chemistry Involved in Titrimetric Exercises: Acids, Bases, and the Use of Indicators

Titration, a fundamental technique in analytical chemistry, plays a significant role in the determination of various chemical concentrations. For JEE Main students, understanding the chemistry behind titrimetric exercises, particularly in the context of acids, bases, and the use of indicators, is crucial. This knowledge equips them with practical skills and theoretical foundations to perform accurate and precise chemical analyses.

1. Acids and Bases Titration:

a. Use of Indicators:

In acid-base titrations, indicators are essential. Indicators are substances that undergo a color change at or near the equivalence point of the titration. Common indicators include phenolphthalein (pink to colorless in the transition from basic to acidic), methyl orange (red to yellow in the transition from acidic to basic), and bromothymol blue (yellow to blue in the transition from acidic to basic). The choice of indicator depends on the type of titration and the expected pH at the endpoint.

b. Oxalic Acid vs. KMnO4:

Oxalic acid $(H_2C_2O_4)$ is often used as a reducing agent in titrations. When oxalic acid reacts with potassium permanganate $(KMnO_4)$ in an acidic medium, it undergoes oxidation. The balanced chemical equation for the reaction is:

$5H_2C_2O_4 + 2KMnO_4 + 3H_2SO_4 \rightarrow 10CO_2 + 2MnSO_4 + 8H_2O + K_2SO_4$

The oxidation state of carbon in oxalic acid changes from +3 to +4, resulting in the production of carbon dioxide and the reduction of manganese in permanganate from +7 to +2.

2. Mohr's Salt vs. KMnO4:

Mohr's salt, also known as ferrous ammonium sulfate $((NH_4)2Fe(SO_4)_2\cdot 6H_2O)$, is employed in redox titrations. In the presence of potassium permanganate $(KMnO_4)$ and sulfuric acid $(H_2SO_4)$, Mohr's salt undergoes oxidation. The balanced chemical equation for this reaction is:

$5(NH_4)_2Fe(SO_4)_2\cdot 6H_2O + 2KMnO_4 + 8H_2SO_4 \rightarrow 10FeSO_4 + 2MnSO_4 + K_2SO_4 + 5(NH_4)_2SO_4 + 8H_2O$

In this reaction, iron in Mohr's salt is oxidized from +2 to +3, while manganese in KMnO4 is reduced from +7 to +2. The equivalence point is reached when all the iron ions are oxidized, and the purple color of KMnO4 disappears.

Chemical Principles in Qualitative Salt Analysis for JEE Main Students

Qualitative salt analysis is a crucial aspect of analytical chemistry, enabling the identification of various cations and anions in a given salt mixture. This knowledge is invaluable for JEE Main students as it forms the basis for understanding complex chemical reactions and analytical techniques.

Cations:

Lead $(Pb^{2+})$:

Confirmatory Test: Lead ions form a yellow precipitate with potassium chromate $(K_2CrO_4)$ solution.

Copper $(Cu^{2+})$:

Confirmatory Test: $Cu^{2+}$ ions react with potassium ferrocyanide to produce a brown precipitate.

Silver $(Ag^+)$:

Confirmatory Test: $Ag^+$ ions form a white precipitate with chloride ions $(Cl^-)$ in the form of silver chloride (AgCl).

Iron ($Fe^{2+}$ and $Fe^{3+}$):

Confirmatory Test: Iron ions react with thiocyanate ions $(SCN^-)$ to produce a blood-red solution due to the formation of ferric thiocyanate $(Fe(SCN)_3)$.

Zinc $(Zn^{2+})$:

Confirmatory Test: Zinc ions form a white precipitate with sodium hydroxide (NaOH) solution.

Nickel $(Ni^{2+})$:

Confirmatory Test: Nickel ions produce a green precipitate when reacted with dimethylglyoxime (DMG).

Calcium $(Ca^{2+})$, Barium $(Ba^{2+})$, Magnesium $(Mg^{2+})$, and Ammonium $(NH_4^+)$:

These ions do not form distinctive precipitates with common reagents. Their presence is usually confirmed indirectly through other tests or the absence of characteristic reactions with other ions.

Anions:

Carbonate $(CO_3^{2-})$:

Confirmatory Test: Carbonate ions evolve carbon dioxide gas $(CO_2)$ when treated with dilute hydrochloric acid (HCl), resulting in effervescence.

Sulfide $(S^{2-})$:

Confirmatory Test: Sulfide ions form black precipitates with lead(II) nitrate $(Pb(NO_3)_2)$.

Sulfate $(SO_4^{2-})$:

Confirmatory Test: Sulfate ions generate a white precipitate of barium sulfate $(BaSO_4)$ when reacted with barium chloride $(BaCl_2)$ solution.

Nitrate $(NO_3^-)$ and Nitrite $(NO_2^-)$:

Confirmatory Test: Nitrate and nitrite ions do not form precipitates. Their presence is confirmed by the absence of reactions with typical tests.

Chloride $(Cl^-)$, Bromide $(Br^-)$, and Iodide $(I^-)$:

Confirmatory Test: Chloride, bromide, and iodide ions form white, cream, and yellow precipitates, respectively, when reacted with silver nitrate $(AgNO_3)$ solution.

Chemical Principles Involved in Selected Experiments for JEE Main Students

Understanding the chemical principles behind experiments is vital for JEE Main students. It not only enhances their grasp of fundamental chemistry but also equips them with the knowledge to excel in various scientific endeavors. Let's explore the chemical principles behind four essential experiments.

1. Enthalpy of Solution of CuSO4:

This experiment involves dissolving copper sulfate $(CuSO_4)$ in water, and its chemical principles can be explained as follows:

Enthalpy Change: The enthalpy change, often referred to as heat of solution (ΔHsolution), measures the heat absorbed or released when a solute $(CuSO_4)$ dissolves in a solvent (water). It is a thermodynamic parameter that quantifies the energy transfer during dissolution.

Lattice Energy: The lattice energy of the ionic compound $CuSO_4$ plays a crucial role in this experiment. When $CuSO_4$ dissolves, the ionic bonds between copper ions $(Cu^{2+})$ and sulfate ions $(SO_4^{2-})$ are broken, and water molecules surround and solvate these ions. This process releases energy, leading to a temperature rise in the solution.

Hydration: CuSO4 dissolves in water through the process of hydration. Water molecules surround and stabilize the ions, and this interaction is exothermic, contributing to the overall enthalpy change.

2. Enthalpy of Neutralization of Strong Acid and Strong Base:

This experiment involves the reaction between a strong acid (e.g., HCl) and a strong base (e.g., NaOH) to form water and a salt (e.g., NaCl). The chemical principles include:

Neutralization Reaction: The reaction between HCl and NaOH is a classic example of a neutralization reaction. It is an exothermic process in which hydrogen ions (H+) from the acid combine with hydroxide ions (OH-) from the base to form water.

Heat of Reaction: The enthalpy change (ΔH) in this experiment represents the heat evolved during the neutralization reaction. This exothermic process results in a temperature increase in the reaction mixture.

Hess's Law: The enthalpy change in this reaction is consistent with Hess's Law, which states that the enthalpy change of a reaction depends only on the initial and final states and is independent of the reaction pathway.

3. Preparation of Lyophilic and Lyophobic Sols:

Preparation of colloidal solutions involves principles related to colloids:

Lyophilic Sols: These are colloidal solutions where the dispersed phase (colloid) has a strong affinity for the dispersion medium (solvent). The preparation of lyophilic sols typically involves solvating the colloid with the solvent, leading to stable colloidal solutions.

Lyophobic Sols: In contrast, lyophobic sols are colloidal solutions where the dispersed phase has little affinity for the dispersion medium. Preparing lyophobic sols often involves vigorous dispersion methods, such as mechanical agitation or chemical treatments, to disperse the colloidal particles evenly in the medium.

4. Kinetic Study of the Reaction of Iodide Ion with Hydrogen Peroxide:

This experiment explores the reaction kinetics of iodide ions (I-) with hydrogen peroxide $(H_2O_2)$. The chemical principles include:

Rate of Reaction: The rate of this reaction is determined by the rate of collision between iodide ions and hydrogen peroxide molecules. This collision leads to the formation of iodine $(I_2)$ and water $(H_2O)$, and the reaction rate is governed by the concentration of reactants and the activation energy required for the reaction to occur.

Reaction Mechanism: Understanding the reaction mechanism, including any intermediate steps, is crucial. The kinetic study helps identify reaction orders and rate-determining steps.

Catalysis: In some cases, a catalyst (e.g., iodide ion in this reaction) may be involved. The role of the catalyst in lowering the activation energy and increasing the reaction rate can be explored.

JEE Main Principles Related to Practical Chemistry Solved Examples

Example 1: Explain how you'd go about calculating w and q throughout the time it takes for the piston velocity to drop to zero.
Solution: 

• The relationship between V1 and V2 is discovered by equating the piston's effort on the gas.

∴ nRT.ln(V2/V1) = p2(V2-V1); p2(V2-V1) = external work performed on the piston; 

pext = 2p1 = 2nRT/V1.

V2 = (0.2032)V1 

w =  (1.5936)nRT,

q = - w =( -1.5936)nRT.

Key point to remember: The relationship between the number of moles of gas, pressure, and volume: nRT.ln(V2/V1) = p2(V2-V1) i.e. the external work done. Also, the values of R will be the given values for the constant and the values of n and T will be the ones provided in any question of this type. 

Example 2: Liquid nitrogen, with a boiling point of -195.79°C, is utilised as a coolant and a biological tissue preservation. Is nitrogen's entropy higher or lower at -200°C than it is at -190°C? Give an explanation for your response. At -210.00°C, liquid nitrogen freezes to a white solid with a fusion enthalpy of 0.71 kJ/mol. What is fusion entropy? Is it possible to freeze living tissue in liquid nitrogen in a reversible or irreversible manner?

Solution: As the temperature decreases, the entropy increases, as they are in inverse correlation to each other. Hence, when the temperature of liquid nitrogen decreases to -200°C, its entropy is higher than at -190°C. 

Key point to remember: Relation between entropy and temperature: ∆S=∆H/T.

Solved Questions from the Previous Year Question Papers

Question 1:  If an endothermic reaction is nonspontaneous at the freezing point of water and becomes feasible at its boiling point, then

(a) ∆H is –ve, ∆S is +ve

(b) ∆H and ∆S both are +ve

(c) ∆H and ∆S both are –ve

(d) ∆H is +ve, ∆S is –ve

Solution: For an endothermic reaction, ΔH = positive value.

ΔG = ΔH – TΔS 

• For a non-spontaneous reaction, ΔG should be positive.

• At low temperatures, ΔG is positive if ΔH is positive.

• If S is positive, ΔG is negative at high temperatures.

• As a result, option (b) is the correct answer.

Trick: Low Temperatures: ΔG is positive if ΔH is positive, High Temperatures:  ΔG is negative if S is positive.

Question 2: The standard enthalpy of formation of NH3 is - 46 kJ mol–1. If the enthalpy of formation of H2 from its atoms is - 436 kJ mol–1 and that of N2 is -712 kJ mol–1, the average bond enthalpy of N-H bond in NH3 is

(1) - 964 kJ mol–1

(2) + 352 kJ mol–1

(3) + 1056 kJ mol–1

(4) - 1102 kJ mol–1

Solution: 

• The equation of formation of ammonia using nitrogen and hydrogen is given below:
  ½ N2 + ½H2 → NH3

• The standard enthalpy of formation of ammonia, given by ∆Hf of NH3, can thus be determined as follows:
  ∆Hf of NH3 = ((1/2)B.E of N2 + (3/2) B.E of H2 – (3) B.E of N-H)

• Here, B.E stands for Bond energy and ∆Hf  stands for standard enthalpy of formation. 

• In the above equation, the enthalpy of the formation of ammonia is obtained as a summation of the bond energies of the constituent elements and the subtraction of the bond energies of nitrogen and hydrogen bonds.
  - 46 = ((1/2)(712) + (3/2) 436 – (3) B.E of N-H)

- 46 = 356 + 654 – 3 B.E of N-H
3 B.E of N-H = 1056
∴ B.E of N-H = 1056/3 = 352 kJ mol–1

Hence, option (2) is the answer.

Trick: Enthalpy of formation can be calculated by subtracting the sum of bond energies of the bonds that are formed in the chemical reaction from the sum of bond energies of the bonds that are broken in the chemical reaction.

Question 3: During compression of a spring, the work done is 10 kJ and 2 kJ escapes to the surroundings as heat. The change in internal energy ∆U (in kJ) is

(a) – 8

(b) 12

(c) 8

(d) – 12

Solution: The work done is given by, w = 10 kJ.

The heat that escapes into the surrounding is expressed as q = – 2 kJ

The first law of thermodynamics states that ∆U = q + w 

= – 2 + 10 = 8 kJ

As a result, option (c) is the correct answer.

Trick: Work done on the system has a positive magnitude and work done by the system has a negative magnitude. Similarly, the heat released by the system has a negative magnitude and heat supplied by the surroundings to the system has a positive magnitude.

Practice Questions

Question 1: A leak in a Russian spacecraft resulted in a reduction in internal pressure from 1 atm to 0.85 atm. Is this a reversible expansion scenario? Has any work been completed?

Answer: It is irreversible.

Question 2: From the following thermochemical data, calculate the enthalpy of production of OH– ions:

H2O → H+(aq) + OH-(aq); ΔH+298K = +57.32 kJ

H2 + 1/2O2 → H2O(l); ΔH+298K = -285.83 kJ

Answer: - 228.51 kJ.

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Conclusion

The JEE Main Chapter on "Principles Related to Practical Chemistry" provides students with essential insights into the fundamental chemical principles and their practical applications. Through a series of experiments and theoretical knowledge, students learn about thermodynamics, reaction kinetics, and the properties of various chemical compounds. This chapter equips students with the necessary skills to understand and conduct experiments, analyze data, and draw meaningful conclusions. It fosters an appreciation for the underlying principles governing chemical reactions, enabling students to apply this knowledge in real-world scenarios.