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JEE Main 2026 Preparation: Question Papers, Solutions, Mock Tests & Strategy Unacademy » JEE Study Material » Chemistry » Gas Law Formula Table
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Gas Law Formula Table

Boyle's Law, Charles' Law, and Gay-Lussac's merged in the gas law. Learn about them in-depth.

Table of Content
  •  

Hydrogen peroxide is a chemical compound with the formula H ₂O ₂. In its pure form, it is a very pale blue liquid, It is used as a good oxidising as well as reducing agent . Hydrogen peroxide is used in pollution contol treatment of industrial effulents. Hydrogen peroxide is a non-planar molecule made up of two oxygen atoms bonded together by a single covalent bond called the peroxide bond.

Structure of Hydrogen Peroxide 

  1. Peroxide, a chemical molecule, is formed by a connection between two oxygen atoms. It’s a very strong oxidant.
  2. Hydrogen peroxide is a non-planar molecule made up of two oxygen atoms bonded together by a single covalent bond called the peroxide bond.
  3. Each oxygen atom has a single connection connecting it to a hydrogen atom. The two bonds do not lie in the same plane because lone pairs of electrons repel each other over oxygen atoms. The interplanar angle between the two planes is 111.5o when hydrogen peroxide is gaseous, but the angle is 90.2o when it is crystalline. This is caused by intramolecular hydrogen bonding.

Preparation of Hydrogen Peroxide

Laboratory preparation

  1. From sodium peroxide (Merck’s Process):

 In this laboratory method of preparation, sodium peroxide is added in minute amounts to a weak sulphuric acid (20%) solution covered by ice and regularly stirred—crystals of Na2SO4. 10H2O occurs as the solution is cooled further, and they can be filtered out. 

                                              Na2O2+H2SO4→Na2SO4+H2O2

  1. From barium peroxide: 

A 20 per cent ice-cold sulphuric acid solution is used to cure a hydrated barium peroxide (BaO2.8H2O) paste produced in ice-cold water. The BaSO4 precipitate, which is white in appearance, is removed using filtration. Only around 5% of the H2O2 is left in the solution.

                                      BaO2⋅8H2O + H2SO4→BaSO4(White ppt) + H2O2 + 8H2O

This approach is ineffective because barium sulphate forms a protective coating around H2O2 that prevents it from continuing the chemical reaction. The Ba2+ ions progressively decompose hydrogen peroxide in the solution. As a result, the solution is not suitable for long-term storage. Instead of sulphuric acid, phosphoric acid is used to test this. In the absence of Ba2+ ions, the barium phosphate that develops is completely precipitated, and there is no risk of hydrogen peroxide breakdown.

                                      3BaO2⋅8H2O + 2H3PO4 → Ba3(PO4)2(ppt) + 24H2O + 3H2O2

Industrial Preparation

Here are the commercial methods of preparation to know.

  1. By the electrolysis of a sulphuric acid solution

A 50 percent sulphuric acid solution is electrolyzed in a cell. The anode produces peroxodisulfuric acid, which causes hydrogen to be released at the cathode.

                                              H2SO4 → HSO4–  + H+

At anode: HSO4–  → H2S2O8 (peroxide sulphuric acid) + 2e–

At cathode: 2H+ + 2e– → H2

The sulfuric acid peroxide in the cell is extracted and destroyed with water to yield hydrogen peroxide.

                                   H2S2O8 + 2H2O → 2H2SO4 + H2O2

High-boiling-point sulphuric acid does not distil, but hydrogen peroxide does. The generation of hydrogen peroxide can be boosted by electrolyzing a mixture of ammonium sulphate and sulphuric acid in equal quantities.

Hydrogen peroxide is obtained by condensing the ammonium peroxide sulphate produced at the anode with water.

               (NH4)2S2O8 + 2H2O → 2NH4HSO4(Ammonium hydrogen sulphate) + H2O2

  1. With the help of 2-Ethyl anthraquinone

In the presence of a palladium catalyst, hydrogen gas is transported through 2-ethyl anthraquinone dissolved in benzene. It is decomposed into 2-ethyl anthraquinone. The air is then pumped with a mixture of 2-ethyl anthraquinone, benzene, and cyclohexanol. When 2-ethyl anthraquinone is oxidised back to hydrogen peroxide, hydrogen peroxide is produced.

  1. Isopropyl alcohol is oxidised: 

Under this commercial method of preparation, it acts as an initiator when a small amount of hydrogen peroxide is mixed with isopropyl alcohol. Oxygen is transmitted through the solution at roughly 340K and with a small amount of pressure. As a result of the oxidation reaction, acetone and hydrogen peroxide are generated. 

CH3CHOHCH3(Isopropyl alcohol) + O2 → CH3COCH3(Acetone) + H2O2

Properties of Hydrogen Peroxide

The properties of Hydrogen Peroxide are as follows:

  • In its purest form, hydrogen peroxide is almost colourless (very pale blue).
  • Its boiling point has been calculated at 150.2 degrees Celsius, which is over 50 degrees Celsius higher than the boiling point of water.
  • Hydrogen Peroxide has a melting point of -0.43 degrees Celsius.
  • In water, it hydrates and produces a homogenous mixture in all amounts.
  • The molar mass of hydrogen peroxide is 34.0147 g/mol.
  • It has a slightly acrimonious odour.
  • In aqueous solution, it has a density of 1.11 g/cc, and in pure form, it has a density of 1.450 g/cc.

Conclusion

That’s a wrap to the commercial methods of preparation. Hydrogen peroxide is the most basic kind of peroxide. It’s a colourless liquid that’s utilised in aqueous solutions to keep things safe. It is used as a disinfectant as well as a bleaching agent. Gaseous hydrogen peroxide is produced by photochemical processes in the earth’s atmosphere. It serves as both a disinfectant and a bleaching agent. At 298 degrees Fahrenheit, it’s acidic, with a pH of 6 to 7.

faq

Frequently Asked Questions

Get answers to the most common queries related to the JEE Examination Preparation.

If 4L of H2 gas at 1.43 atm is at standard temperature, and the pressure was to increase by 2/3, what is the final volume of the H2 gas?

To solve this question, you need to use Boyle’s Law: P...Read full

I will assign a value of 1 to V1 and allow it to double. I will assign a value of 1 to T1 and allow its value to quadruple.

[(P1)(1)] / 1 = [(P...Read full

To solve this question, you need to use Boyle’s Law:

P1V1=P2V2

Keeping the key variables in mind, temperature and the amount of gas is constant and therefore can be put aside, the only ones necessary are:

Initial Pressure: 1.43 atm

Initial Volume: 4 L

Final Pressure: 1.43×1.67 = 2.39

Final Volume: V2

Plugging these values into the equation, you get:

V2=(1.43atm x 4 L)/(2.39atm) = 2.38 L

  • Solve this: If the volume of an ideal gas is doubled while its temperature is quadrupled, does the pressure (a) remain the same, (b) decrease by a factor of 2, (c) decrease by a factor of 4, (d) increase by a factor of 2, or (e) increase by a factor of 4?

Solution:

1) Write the combined gas law:

P1V1 / T1 = P2V2 / T2

[(P1)(1)] / 1 = [(P2)(2)] / 4

P1 = P2 / 2

2P1 = P2

the answer is (d) increase by a factor of 2

Any volume unit is fine for V1, but the temperature unit must be understood to be Kelvin. In other words, do not select 1 °C, allow it to change to 4 °C and then convert those values to K.

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Attempt 2023’s and previous year’s JEE Main
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Attempt Free Test Series for JEE Main 2023

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