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JEE Chapter: Principles Related to Practical Chemistry

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Introduction to Principles Related to Practical Chemistry

Introduction to 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

Important Terms

Explanation

Titrant

Titrant is the name given to a solution of known strength that is used in the burette.

Titrate

The titrant is the drug whose concentration is to be determined by titration.

Equivalence Point

It is the moment in a titration where the quantity of titrant is precisely sufficient for the completion of the stoichiometric reaction to be completed with titrate. A physical property, such as indicator colour, pH, or conductivity, changes abruptly at this point.

Indicator

An indicator is a substance with a physical property (typically colour) that changes dramatically around the equivalence point of a chemical process. It denotes that the endpoint has been reached.

Standard solution

A standard solution is a solution with a known concentration.

Standardisation

It is the procedure for determining the concentration of a reagent by reacting it with a known quantity of a known concentration of a second reagent.

Primary standard substance

The primary standard is a reagent that is pure enough that its standard solution may be made directly by dissolving a specific weight of it in a certain volume of solvents, such as crystalline oxalic acid or Mohr's salt.

Secondary standard substance

Secondary standards are substances or reagents whose standard solution cannot be made directly, such as KMnO4, NaOH, and KOH.

Redox titrations

The passage of electrons between reacting ions in aqueous solutions is the basis for these titrations.

Acid-base titrations

The strength of an acid or base is evaluated by titrating it against a reference solution of base or acid, respectively, in acid-base titration. It entails a reaction of neutralisation.

 

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.

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26th June 2022 - exam-strategy
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July
24th July 2022 - Maths, Physics and Chemistry
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24th July 2022 - Maths, Physics and Chemistry
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24th July 2022 - Maths, Physics and Chemistry
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25th July 2022 - Maths, Physics and Chemistry
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25th July 2022 - Maths, Physics and Chemistry
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26th July 2022 - Maths, Physics and Chemistry
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27th July 2022 - Maths, Physics and Chemistry
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28th July 2022 - Maths, Physics and Chemistry
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June, July
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JEE Main 2022 Book Solutions and PDF Download

JEE Main 2022 Book Solutions and PDF Download

View all JEE Main Important Books
In order to prepare for JEE Main 2022, candidates should know the list of important books i.e. RD Sharma Solutions, NCERT Solutions, RS Aggarwal Solutions, HC Verma books and RS Aggarwal Solutions. They will find the high quality readymade solutions of these books on Vedantu. These books will help them in order to prepare well for the JEE Main 2022 exam so that they can grab the top rank in the all India entrance exam.
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books
Maths
NCERT Book for Class 12 Maths
books
Physics
NCERT Book for Class 12 Physics
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JEE Main Mock Tests

JEE Main Mock Tests

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JEE Main 2022 free online mock test series for exam preparation are available on the Vedantu website for free download. Practising these mock test papers of Physics, Chemistry and Maths prepared by expert teachers at Vedantu will help you to boost your confidence to face the JEE Main 2022 examination without any worries. The JEE Main test series for Physics, Chemistry and Maths that is based on the latest syllabus of JEE Main and also the Previous Year Question Papers.
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JEE MAIN MOCK TEST - 1
3 hr  • 75 questions • OBJECTIVE
JEE MAIN MOCK TEST - 3
3 hr  • 75 questions • OBJECTIVE
JEE MAIN MOCK TEST - 2
3 hr  • 75 questions • OBJECTIVE
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JEE Main 2022 Cut-Off

JEE Main 2022 Cut-Off

JEE Main Cut Off
NTA is responsible for the release of the JEE Main 2022 June and July Session cut off score. The qualifying percentile score might remain the same for different categories. According to the latest trends, the expected cut off mark for JEE Main 2022 June and July Session is 50% for general category candidates, 45% for physically challenged candidates, and 40% for candidates from reserved categories. For the general category, JEE Main qualifying marks for 2021 ranged from 87.8992241 for general-category, while for OBC/SC/ST categories, they ranged from 68.0234447 for OBC, 46.8825338 for SC and 34.6728999 for ST category.
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JEE Main 2022 Results

JEE Main 2022 Results

JEE Main 2022 June and July Session Result - NTA has announced JEE Main result on their website. To download the Scorecard for JEE Main 2022 June and July Session, visit the official website of JEE Main NTA.
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Rank List
Counselling
Cutoff
JEE Main 2022 state rank lists will be released by the state counselling committees for admissions to the 85% state quota and to all seats in NITs and CFTIs colleges. JEE Main 2022 state rank lists are based on the marks obtained in entrance exams. Candidates can check the JEE Main 2022 state rank list on the official website or on our site.
The NTA will conduct JEE Main 2022 counselling at https://josaa.nic.in/. There will be two rounds of counselling for admission under All India Quota (AIQ), deemed and central universities, NITs and CFTIs. A mop-up round of JEE Main counselling will be conducted excluding 15% AIQ seats, while the dates of JEE Main 2022 June and July session counselling for 85% state quota seats will be announced by the respective state authorities.
NTA is responsible for the release of the JEE Main 2022 June and July Session cut off score. The qualifying percentile score might remain the same for different categories. According to the latest trends, the expected cut off mark for JEE Main 2022 June and July Session is 50% for general category candidates, 45% for physically challenged candidates, and 40% for candidates from reserved categories. For the general category, JEE Main qualifying marks for 2021 ranged from 87.8992241 for general category, while for OBC/SC/ST categories, they ranged from 68.0234447 for OBC, 46.8825338 for SC and 34.6728999 for ST category.
Want to know which Engineering colleges in India accept the JEE Main 2022 scores for admission to Engineering? Find the list of Engineering colleges accepting JEE Main scores in India, compiled by Vedantu. There are 1622 Colleges that are accepting JEE Main. Also find more details on Fees, Ranking, Admission, and Placement.
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FAQs on JEE Chapter: Principles Related to Practical Chemistry

FAQ

1. What is the relationship between principles and practical chemistry?

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.

2. What are Organic Chemistry Basic Principles?

Organic chemistry is the study of carbon molecules that are always carbon-based and have no additional components. Organic compounds came from plants and animals, indicating that their ultimate source was a living organism.

3. Is it necessary to study practical chemistry for JEE?

Practical chemistry is not difficult to memorise; nonetheless, you will remember it best if you are interested in it. It is preferable to practise all NCERT questions. Understand instrumental Physics with the most significant portion called mistakes, and the same goes for Physics.