--%>

Explain the process of adsorption of solids on gases.

The extent of adsorption of a gas on a solid adsorbent is affected by the following factors:

1. Nature of the gas

Since physical adsorption is non-specific in nature, every gas will get adsorbed on the surface of any solid to a lesser or greater extent. However, under given conditions of temperature and pressure, the easily liquefiable gases like NH3, HCl, CO2 etc. are adsorbed in a greater extent than the permanent gases such as H2, O2, N2 etc. it is because the van der Waal forces or molecular forces are more predominant in the former than in later category.

The ease with which a gas can be liquefied is mainly determined by its critical temperature Tc, Critical temperature of a gas is the temperature above which the gas cannot be liquefied irrespective of the pressure applied. A gas having higher critical temperature can be liquefied more easily and hence is adsorbed on the solid to greater and extent of adsorption for some gases.

We know that chemisorption is specific in nature. Therefore, in case of chemisorption a gas gets adsorbed on the solid only if it forms chemical bond with it.

2. Effect of nature of the adsorbent

The extent of adsorption of a gas also depends on the nature of adsorbent. Activated charcoal more easily adsorbs toxic gases like CH4, CO, etc. allows its frequent use in gas masks. Finely divided transition metals like Ni, CO, etc. adsorb permanent gases like H2N2O2, etc.

3. Specific area of the solid

Specific area of an adsorbing solid is the surface area available for adsorption per gram of the adsorbent. Greater the specific area of the solid, greater would be its adsorbent power. That is why porous or finely divided forms of adsorbents adsorb more extensively. However, the pores should be large enough to allow the gas molecules to allow the gas molecules to enter them.

4. Effect of pressure of the gas

In order to understood the effect of pressure on the adsorption of a gas on some solid, we must keep in mind that adsorption is a reversible process and is accompanied by decrease in pressure. Therefore, it is expected that a given temperature, the extent of the adsorption increases with the increase in pressure. The extent of the adsorption is generally expressed as x/m where m is the mass of the adsorbent and x is that of the adsorbate when equilibrium has been attained. A graph drawn between extent of adsorption (x/m) and the pressure p of the gas at constant temperature is called adsorption isotherm. Adsorption isotherms of different shapes have been observed experimentally. Two most common types of adsorption isotherms are Freundlich adsorption isotherms and Langmuir adsorption isotherm.

5. Effect of temperature

As already discussed the adsorption at a surface initially increases till a saturation point is achieved. At this juncture an equilibrium is established as represented below.
                                         
Adsorption  727_Physical adsorption.png  Desorption; Δ H = +ve

As adsorption is accompanied by evolution of heat, so in accordance with Le-Chatelier's principle, the magnitude of adsorption should decrease with rise in temperature and this is actually so.

A graph drawn between extent of adsorption (x/m) and temperature (t) at constant pressure is called adsorption isobar. 

6. Activation of adsorbent

Activation of an adsorbent means increasing the adsorbing power of the adsorbent. This can be done in various different ways. One possible way of doing it is to increase the specific area of the adsorbent. This can be done either by making the surface of adsorbent or by breaking it into small pieces. However, if the particles are made very small, then the interparticle spaces will be too small to allow the penetration of gas molecules and hence, the extent of adsorption may increase by certain specific treatments. For example, wood charcoal can be activated by heating it between 650 K and 1330 K in vacuum, air or super-heated steam.

   Related Questions in Chemistry

  • Q : Ionization Potential Second ionization

    Second ionization potential of Li, Be and B is in the order (a)Li>Be>B (b)Li>B>Be (c)Be>Li>B (d)B>Be>Li

  • Q : Advantages of doing your own chemistry

    What are the advantages of doing your own chemistry assignments? State your comment?

  • Q : Determining Mole fraction of water Can

    Can someone please help me in getting through this problem. The mole fraction of water in 20% aqueous solution of H2O2 is: (a) 77/68 (b) 68/77 (c) 20/80  (d) 80/20

  • Q : Calculating total number of moles

    Choose the right answer from following. While 90 gm of water is mixed with 300 gm of acetic acid. The total number of moles will be: (a)5 (b)10 (c)15 (d)20

  • Q : Question based on maximum vapour

    Provide solution of this question. Which has maximum vapour pressure: (a) HI (b) HBr (c) HCl (d) HF

  • Q : Question based on vapour pressure and

    Benzene and toluene form nearly ideal solutions. At 20°C, the vapour pressure of benzene is 75 torr and that of toluene is 22 torr. The parial vapour pressure of benzene at 20°C for a solution containing 78g of benzene and 46g of toluene in torr is: (a) 50 (b)

  • Q : What is Henry law constant and its

    1. The units of Henry Law constant are same as those of pressure, i.e. torr or h bar. 2. Different gases have dissimilar values of Henry law constant. The values of KH for some gases in water are given in tabl

  • Q : Vapour pressure over mercury Choose the

    Choose the right answer from following. At 300 K, when a solute is added to a solvent its vapour pressure over the mercury reduces from 50 mm to 45 mm. The value of mole fraction of solute will be: (a)0.005 (b)0.010 (c)0.100 (d)0.900

  • Q : Problem on reversible and irreversible

    The second law states that  dS ≥ (dQ/T), where dS = dQ/T for a reversible process and dS > dQ/T for an irreversible process.   a. Show that since dW12 = -dW21 (dWreverse = -dWforward) for a r

  • Q : Vapour pressure of the pure hydrocarbons

    Give me answer of this question. A solution has a 1 : 4 mole ratio of pentane to hexane. The vapour pressure of the pure hydrocarbons at 20°C are 440 mmHg for pentane and 120 mmHg for hexane. The mole fraction of pentane in the vapour phase would be: (a) 0.549 (b)