--%>

What is chemisorption or chemical adsorption?

When the forces of attraction existing between adsorbate particles and adsorbent almost of the same strength as chemical bonds, the adsorption is called chemical adsorption. This type of adsorption is also known as chemisorptions. Since forces of attraction existing between adsorbent and adsorbate are relatively strong, therefore, this type of adsorption cannot be easily reversed. 

Characteristics of chemisorptions

Some important characteristics of chemisorptions are as follows:

(i) High specificity: chemisorptions is highly specific in nature. It occurs only if there is a possibility of bond formation between adsorbent and adsorbate molecules. For example O2 is adsorbed on metals by virtue of oxide formation and H2 is absorbed by transition metals due to hydride formation.

(ii) Irreversibility: as chemisorptions involve the compound formation between adsorbent and adsorbate, it is generally irreversible.

(iii) Enthalpy of adsorption: attractive forces between adsorbent and adsorbate molecules are strong chemical bonds and therefore, molar heat of adsorption is high and is of the order of 200-400 kJ mol-1.

(iv) High activation energy: although chemisorptions is exothermic, yet the process is slow at low temperature. It is because of high activation energy required for chemical process to occur. Like most of the chemical changes the extent of chemisorption increases initially with rise in temperature. High pressure is also supporting for chemisorption.

(v) Surface area: like physisorption, chemisorption also increases with increase in surface area of adsorbent.

(vi) State of adsorbate: since chemical reaction takes place in this type of adsorption, therefore, the molecular state of adsorbate molecules may be altered. For example, oxygen exists as O2, but on the surface where it is chemisorbed, it may exist as O2-, O22-, O-, O, O3-, etc.

(vii) Activation energy: chemical adsorption involves a chemical reaction between adsorbent and adsorbate; therefore, it requires high activation energy.

The adsorption of N2 on iron under two different conditions provides distinction between physisorption and chemisorption. At 83 K nitrogen gas undergoes physical adsorption on iron surface. N2 molecules are amount of N2 adsorbed decreases with further rise in temperature. At room temperature there is almost again shows adsorption as N atoms on the iron surface. This mode of adsorption is chemical adsorption as atoms form chemical bonds with iron atoms.

   Related Questions in Chemistry

  • Q : Number of moles present in water

    Provide solution of this question. How many moles of water are present in 180 of water: (a)1 mole (b)18 mole (c)10 mole (d)100 mole

  • Q : Relative lowering of vapour pressure

    Which of the following solutions will have a lower vapour pressure and why? a) A 5% aqueous solution of cane sugar. b) A 5% aqueous solution of urea.

  • Q : Molarity of acid solution If 20ml of

    If 20ml of 0.4N, NaoH solution completely neutralises 40ml of a dibasic acid. The molarity of the acid solution is: (a) 0.1M (b) 0.2M  (c) 0.3M (d) 0.4M Choose the right answer fron above.

  • Q : Molality of a glucose solution What

    What will be the molality of a solution containing 18g of glucose (having mol. wt. = 180) dissolved in 500g of water: (i) 1m  (ii) 0.5m  (iii) 0.2m  (iv) 2m

  • Q : Relative lowering in vapour pressure of

    Give me answer of this question. "Relative lowering in vapour pressure of solution containing non-volatile solute is directly proportional to mole fraction of solute". Above statement is: (a) Henry law (b) Dulong and Petit law (c) Raoult's law (d) Le-Chatelier's pri

  • Q : Problem on reversible process a. For a

    a. For a reversible process involving ideal gases in a closed system, Illustrate thatΔS = Cv ln(T2/T1) for a constant volume process ΔS = Cp ln(T2/T1) for a constant pressu

  • Q : Strength of any solution Give me answer

    Give me answer of this question. A solution contains 1.2046 x 1024 hydrochloric acid molecules in one dm3 of the solution. The strength of the solution is: (a) 6 N (b) 2 N (c) 4 N (d) 8 N

  • Q : Problem on mole fraction of glucose

    Provide solution of this question. While 1.80gm glucose dissolve in 90 of H2O , the mole fraction of glucose is: (a) 0.00399 (b) 0.00199 (c) 0.0199 (d) 0.998

  • Q : Question associated to vapour pressure

    Choose the right answer from following. The vapour pressure lowering caused by the addition of 100 g of sucrose(molecular mass = 342) to 1000 g of water if the vapour pressure of pure water at 25degree C is 23.8 mm Hg: (a)1.25 mm Hg (b) 0.125 mm Hg (c) 1.15 mm H

  • Q : Molar mass Select the right answer of

    Select the right answer of the question. Which is heaviest: (a)25 gm of mercury (b)2 moles of water (c)2 moles of carbon dioxide (d)4 gm atoms of oxygen