Adsorption Isotherm
The relation between the amount of gas adsorbed and the gas pressure is known as an adsorption isotherm.
A model for the adsorption process, and particularly for the chemisorption process, was presented by Langmuir in 1916 and led him to a simple but important theoretical derivation of an adsorption isotherm. The chemisorption process is pictured as leading ultimately to a monomolecular film over the surface of the adsorbent, and the derived adsorption isotherm results from an investigation of the equilibrium that is set up between the gas phase and the primarily formed monolayer. When the gas is at a pressure P, the fraction of the surface that is covered is represented by θ. The equilibrium state can be interpreted in terms of the dynamic equilibrium that results from an equal rate of evaporation of the adsorbed material and rate of condensation of the gas phase molecules.
The Langmuir theory suggests that the rate of evaporation can be taken to be proportional to the fraction of the surface covered and can therefore be written as k_{1}θ, where k_{1} is some proportionality constant. This simple proportionality assumption ignores the fact that the enthalpy of adsorption generally depends on the extent of coverage . (Then the simple assumption of an evaporation rate is taken to be proportional to k_{1}θ might not hold). The rate of the condensation, furthermore, is taken to be proportional both to the gas pressure P, which according to the kinetic-molecular theory determines the number of molecular collisions per unit area per unit time, and to the fraction of the surface not already covered by adsorbed molecules, i.e. to 1 – θ. It is assumed that only collisions with this exposed surface can lead to the sticking of a molecule to the surface. The relation between equilibrium surface coverage and gas pressure is then obtained by equating the expressions deduced for the rate of evaporatio
n and the rate of condensation, i.e.
k_{1}θ = k_{2} P(1 – θ), where k_{2} is another proportionality constant. Rearrangement gives
Introduction of a = k_{1}/k_{2} allows this result to be written as
θ = P/(a + P)
inspection of equation shows that a chemisorption type isotherm is obtained from this theory. At small values of P, where P in the denominator can be neglected compared with a, reduces to a simple proportionality between θ and P, and this behavior is that corresponding to the initial steep rise of the isotherm curve. At higher pressures the value of P in the proportionality of the isotherm curve. At higher pressures the value of P in the denominator contributes appreciably, and the increasing denominator leads to values of θ that do not increase proportionality to the increase in P. for sufficiently large values of P, then θ approaches the constant value of unity.
Experimental isotherm data consist of the amount of gas absorbed by a given amount of absorbent as a function of the gas pressure. For adsorption, up to a monolayer, the amount of gas y adsorbed at some pressure P and the amount of gas ym needed to form a monolayer are related to θ according to
y/y_{m} = θ
Equation becomes
y = (y_{m} P)/(a + P)
Experimental results can be compared with the Langmuir theory most easily equation is rearranged to
Example: Estimate the surface area per unit mass of the absorbent for which the isotherm was obtained.
Solution: The Langmuir plots of the isotherm data gives the following slope and intercept values
Slope = 1/y_{m} = 745 L^{-1}
Intercept = a/y_{m} = 5.3 bar L^{-1}
The values of ym and a are calculated as
y_{m} = 0.00134 L a = 0.0071 bar
And the adsorption isotherm is represented by the equation
The result is obtained, therefore, that the surface of 1 g of adsorbent would be covered by an amount of H_{2} which occupies a volume of about 0.00134 L at STP i.e. by 0.
00134/24.8) × 6.03 × 10^{23} = 3.3 × 10^{19} molecules.
The surface area is obtained if the area covered by this much H_{2}_{ }can be estimated. The easiest, if rather crude, method to is to use the bulk volume of liquid H_{2 }and to calculate the effective area per molecule as (V_{liq}/ N)2/3, where V_{liq} is the volume of 1 mol of liquid H_{2}. The density of 0.070 g mL^{-1} for liquid hydrogen gives a molar volume of 2.0 g/(0.070 g mL^{-1}) = 29 mL = 29 × 10^{-6} m^{3}. The area covered by one molecule is estimated as
The area of 1 g of this copper powder adsorbent is therefore
(3.3 × 10^{19}) (13 × 10^{-20}) = 4.3 m^{2}
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