JEE Chapter: Principles Related to Practical Chemistry

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.

Important Topics of Chemical Thermodynamics

• Inorganic Chemistry

• Qualitative Chemical Analysis

• Chemical Analysis

• Flame Test

• Qualitative Analysis of Organic Compounds

• Physical Chemistry

Important Definitions of Chemical Thermodynamics

Inorganic Chemistry

Qualitative Analysis

• The qualitative analysis is concerned with identifying the numerous ingredients that make up a specific substance. 

• Its main goal is to detect ions in aqueous solutions. 

• It's a type of analytical chemistry that determines the elemental composition of different substances. 

• Preliminary tests, flame tests, and wet testing for anions and cations, among other things, are all part of this analysis.

Preliminary Test

• Inorganic salts are subjected to preliminary testing that includes colour, odour, and heating, as described below:
  - Colour: Co2+ (pink), Cu2+ (blue), Mg2+ (white), Fe3+ (brown), etc.
  - Heating: When inorganic salts are heated, they undergo several modifications.
  When some salts are heated, they change colour.
  CuSO4.5H2O, for example, changes colour from blue to white.
  Some inorganic salts make sound, such as when NaCl is heated and produces a crackling sound.

- Odour: NH4+ (ammoniacal smell), CH3COO- (vinegar smell), S2- (smell of rotten eggs).

Flame Tests

• Flame tests are used to determine the presence of a metal atom or ion in a chemical. In this test, a sample is placed on the burner's flame, and the resulting colour or sound aids in the identification of the element. 

• When a salt with a few drops of HCl is placed on a flame with the help of a metal wire, the metal ion is determined by the colour of the flame. 

• The presence of Ba+ in this experiment is shown by the colour of the flame being apple green. Similarly, if the flame is golden yellow in colour, Na+ is to blame.

Wet Tests

• Wet tests are those in which the salts are dissolved in water, acid, or base. Wet tests are used to determine if the radicals are acidic or basic.

• Group I acidic radicals: When a salt is dissolved in dilute HCl or dilute H2SO4, this test is performed.
  - The presence of group I acidic radicals is thus indicated by the development of gases.
  - Here are a few examples:
  (a) The existence of carbonate ions is indicated by the evolved CO2 when Na2CO3 is dissolved with dilute H2SO4.
  (b) The existence of sulphite ion is indicated by the developed SO2 when Na2SO4 is dissolved with dilute H2SO4.

• Group II acidic radicals: When a salt is dissolved in concentrated HCl or concentrated H2SO4, this test is performed.
  - The presence of group II acidic radicals is thus indicated by the development of gases.
  - Here are a few examples:
  (a) The existence of chloride ions is shown by the development of HCl gas when NaCl is dissolved with concentrated H2SO4.
  (b) The existence of bromide ions is shown by the development of Br2 gas when NaBr is dissolved with concentrated H2SO4.

• Group III acidic radicals: Acid, whether dilute or concentrated, has no effect on these radicals.
  - Identification of these radicals necessitates the use of specialised tests.
  - Here are a few examples:
  (a) In this scenario, Na2SO4 reacts with BaCl2 to generate a white BaSO4 precipitate.
  (b) In this scenario, Na3BO3 reacts with H2SO4 and, following a series of reactions, ethyl borate is generated, which burns with a green flame.

• Basic radicals: Salts are dissolved in a suitable solvent to produce transparent solutions.
  - The salts are usually dissolved in distilled water.
  - When this solution is made, it reveals the presence of basic radicals, as seen below:
  (a) The presence of transition metal ions causes the solution to be coloured.
  (b) It is the presence of Pb2+ if a solution is made in HCl and white precipitate forms.
  (c) The presence of Co2+ is indicated when a solution is made in water and the colour changes from pink to blue.

Organic Chemistry

Qualitative analysis

• Organic compounds can contain oxygen, nitrogen, sulphur, halogens, and phosphorus in addition to carbon and hydrogen. 

• The qualitative examination of organic molecules entails using appropriate chemical assays to detect all of the primary constituents present. 

• Extra elements are elements that are not C, H, or O. 

• In most cases, the elements are analysed as ions. Organic substances do not ionise since they are covalent in nature. 

• To transform elements included in organic molecules into ions, the organic is distilled with sodium metal, which is used as a catalyst. 

• Lassaigne's extract (L.E.) or sodium extract is the name given to the filtrate.

Nitrogen Detection

• Iron sulphate reacts with sodium fusion extract. 

• The solution is then acidified with strong sulphuric acid, resulting in the creation of Prussian blue, which indicates the presence of nitrogen.

Sulphur Determination

• In this scenario, sodium extract is combined with acetic acid before being added to lead acetate. 

• As a result, the presence of sulphur is evidenced by the production of a black precipitate of lead sulphide.

Halogens Detection

• Sodium fusion extract is acidified with nitric acid and then treated with silver nitrate in this procedure. 

• If a white precipitate occurs, the presence of chlorine is present; if a yellowish precipitate forms, the presence of bromine or iodine is present.

Physical Chemistry

Quantitative Analysis

• A quantitative analysis determines the amount or concentration of a specific species in a sample with accuracy and precision. 

• Essentially, it is the process of determining the percentage composition of constituents in a compound.

Volumetric Analysis

• Volumetric analysis is a technique for determining the concentration or strength of an unknown chemical solution by measuring the volume of the solution involved in a chemical reaction. 

• Titration is the most important part of this analysis.

Titration

• Titration is the process of determining the strength of one solution by comparing it to another solution of known strength under volumetric conditions.

Solved Examples from the Chapter

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.

Conclusion

In this chapter, we look at the numerous methods and processes for determining the quality and quantity of various constituents. All of these approaches are classified according to the nature of the substances, which can be inorganic, organic, or physical.