--%>

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 : Problem on volumetric flow rate Methane

    Methane containing 4 mol% N2 is flowing through a pipeline at 105.1 kpa and 22 °C. To check this flow rate, N2 at the same temperature and pressure are introduced to the pipeline at the rate of 2.83 m3/min. At the end of the pipe (

  • Q : Gibberella fusarium in bioremediation

    in bioremediation gibberella fusarium is used to break down____?

  • Q : Avogadros hypothesis how avogadros

    how avogadros hypothesis used to deduce the atomicity of elementry gases ?

  • Q : Strength of dilute acid of Sulfuric acid

    Select the right answer of the question.10ml of conc.H2SO4 (18 molar) is diluted to 1 litre. The approximate strength of dilute acid could be: (a)0.18 N (b)0.09 N (c) 0.36 N (d)1800 N

  • Q : Molarity of Nacl solution When 5.85 g

    When 5.85 g of NaCl (having molecular weight 58.5) is dissolved in water and the solution is prepared to 0.5 litres, the molarity of the solution is: (i) 0.2 (ii) 0.4 (iii) 1.0 (iv) 0.1

  • Q : Problem based on molarity Select the

    Select the right answer of the question. If 18 gm of glucose (C6H12O6) is present in 1000 gm of an aqueous solution of glucose, it is said to be: (a)1 molal (b)1.1 molal (c)0.5 molal (d)0.1 molal

  • Q : Number of moles of potassium chloride

    Choose the right answer from following. The number of moles of KCL in 1000ml of 3 molar solution is: (a)1 (b)2 (c)3 (d)1.5

  • Q : Problem on molality Select the right

    Select the right answer of the question. Calculate the molality of 1 litre solution of 93% H2SO4 (weight/volume). The density of the solution is 1.84 g /ml : (a) 10.43 (b) 20.36 (c) 12.05 (d) 14.05

  • Q : Procedure to judge that organic

    Describe briefly the procedure to judge that the given organic compound is pure or not?

  • Q : Calculating weight of acid Give me

    Give me answer of this question. The formula weight of H2SO4 is 98. The weight of the acid in 400mi of solution is: (a)2.45g (b) 3.92g (c) 4.90g (d) 9.8g