Derivation of bet equation

In 1938, Brunauer, Emmett and Teller proposed a model called the BET isotherm for multilayer adsorption. According to them, the first step in adsorption is,

\[\displaystyle {{A}_{{(g)}}}+S\rightleftharpoons AS;{{K}_{1}}=\frac{{{{\theta }_{1}}}}{{{{\theta }_{\nu }}p}}---(1)\]

Here, A = gaseous adsorbate

S = vacant site on the surface

AS = adsorbed molecule of A or an occupied site on the surface

K₁ = equilibrium constant

θ ₁ = fraction of the surface sites covered by single molecules

θ _v = fraction of vacant sites

P = pressure of the gas

They again supposed the additional molecules sit on top of one molecule sit on top of to form a variety of multilayers. This process is a sequence of chemical reactions, each with an appropriate equilibrium constant;

\[\displaystyle {{A}_{{(g)}}}+AS\rightleftharpoons {{A}_{2}}S;{{K}_{2}}=\frac{{{{\theta }_{2}}}}{{{{\theta }_{1}}p}}\]

\[\displaystyle {{A}_{{(g)}}}+{{A}_{2}}S\rightleftharpoons {{A}_{3}}S;{{K}_{3}}=\frac{{{{\theta }_{3}}}}{{{{\theta }_{2}}p}}\]

\[\displaystyle {{A}_{{(g)}}}+{{A}_{{n-1}}}S\rightleftharpoons {{A}_{n}}S;{{K}_{n}}=\frac{{{{\theta }_{n}}}}{{{{\theta }_{{n-1}}}p}}\]

Here, the symbol A_nS = surface site a stack of n A molecules pilled up on it

θ i = fraction of sites on which the stack of A molecules is i layer deep

Interaction between the first A molecules and the surface site is unique, depending on the nature of the particular A molecule and the surface. When the second A molecule sits on the first A molecule, the interaction cannot be very different from the interaction of two A molecules in the liquid.

The same is true when the third molecule sites on the second. All these processes, except the first, can be considered equivalent to liquefaction. The same equilibrium constant, K. the BET treatment is supposed to be,

\[\displaystyle {{K}_{2}}={{K}_{3}}={{K}_{4}}=.....={{K}_{n}}=K---(2)\]

Here, K = equilibrium constant for the reaction A_(g) ↔ A_(Liq)

\[\displaystyle K=\frac{1}{{{{p}^{0}}}}---(3)\]

Here p⁰ = equilibrium vapour pressure of the liquid

We can use the equilibrium conditions to calculate the values of the various θ_i; we get

\[\displaystyle {{\theta }_{2}}={{\theta }_{1}}{{K}_{p}},{{\theta }_{3}}={{\theta }_{2}}{{K}_{p}},{{\theta }_{4}}={{\theta }_{3}}{{K}_{p}}----(4)\]

Combining the first two, we have θ₃ = θ₁(Kp)²

Repeating the operation, we find that,

\[\displaystyle {{\theta }_{i}}={{\theta }_{1}}{{(Kp)}^{{i-1}}}---(5)\]

The sum of all these fractions must be equal to unity;

\[\displaystyle 1={{\theta }_{v}}+\sum\limits_{{i=1}}{{{{\theta }_{i}}=}}{{\theta }_{v}}+\sum\limits_{i}{{{{\theta }_{i}}}}{{(Kp)}^{{i-1}}}----(6)\]

We have equation (5) for θ_i. Let Kp = x, then equation (6) becomes,

\[\displaystyle 1={{\theta }_{v}}+{{\theta }_{1}}(1+x+{{x}^{2}}+{{x}^{3}}+.....)\]

Assume that the process can go on indefinitely, then n → and the series is simply the expansion of 1/(1-x) = 1 + x +x₂ +…… so,

\[\displaystyle 1={{\theta }_{v}}+\frac{{{{\theta }_{1}}}}{{(1-x)}}----(7)\]

Using the equilibrium condition for the first adsorption, we find that θ_v =θ₁/K₁p.

New constant,c =K₁/K, we have θ_v/θ₁/(cx) and equation (7) becomes,

\[\displaystyle 1={{\theta }_{1}}\left( {\frac{1}{{cx}}+\frac{1}{{1-x}}} \right)\]

\[\displaystyle {{\theta }_{1}}=\frac{{cx(1-x)}}{{1+(c-1)x}}----(8)\]

Let N be the number of molecules adsorbed per unit of adsorbent and cs be the total number of surface sites per unit mass. Cs θ₁ is the number of sites carrying one molecule. Cs θ₂ is the number of sites taking two molecules, and so on. Accordingly,

\[\displaystyle N={{c}_{s}}\left( {1{{\theta }_{1}}+2{{\theta }_{2}}+3{{\theta }_{3}}+....} \right)={{c}_{s}}\sum\limits_{i}{{i{{\theta }_{i}}}}----(9)\]

From equation (5), we have i = θ₁x_i-1 so that,

\[\displaystyle N={{c}_{s}}{{\theta }_{1}}\sum\limits_{{i=1}}{{i{{x}^{{i-1}}}}}={{c}_{s}}{{\theta }_{1}}\left( {1+2x+3{{x}^{2}}} \right)---(10)\]

This series is the derivative of the earlier one. Thus,

\[\displaystyle 1+2x+3{{x}^{2}}+....=\frac{d}{{dx}}\left( {1+x+{{x}^{2}}+{{x}^{3}}+...} \right)\]

\[\displaystyle =\frac{d}{{dx}}\left( {\frac{1}{{1-x}}} \right)=\frac{1}{{{{{\left( {1-x} \right)}}^{2}}}}---(11)\]

Using this result, the expression for N becomes,

\[\displaystyle N=\frac{{{{c}_{s}}{{\theta }_{1}}}}{{{{{\left( {1-x} \right)}}^{2}}}}--(12)\]

If the entire surface is covered with a monolayer, then Nm molecules would be adsorbed so that N_mono = cs and,

\[\displaystyle N={{N}_{m}}\frac{{{{\theta }_{1}}}}{{{{{\left( {1-x} \right)}}^{2}}}}---(13)\]

Substituting the value of θ₁ from equation (8), we have,

\[\displaystyle N=\frac{{{{N}_{m}}{{c}_{x}}}}{{\left( {1-x} \right)\left[ {1+(c-1)x} \right]}}---(14)\]

The amount adsorbed is usually reported as the volume of the gas adsorbed measured at STP. The volume is proportional to N so that we have N/N_mono =v/v_mono

\[\displaystyle {{v}_{{total}}}=\frac{{{{v}_{m}}{{c}_{x}}}}{{(1-x)\left[ {1+(c-1)x} \right]}}----(15)\]

Here v_m satnd for v_mono

Recalling that x=Kp and that K=1/P⁰, we obtain the BET isotherm;

\[\displaystyle {{v}_{{total}}}=v=\frac{{{{v}_{m}}cp}}{{({{p}^{0}}-p)\left[ {1+(c-1)\left( {\frac{p}{{{{p}^{0}}}}} \right)} \right]}}---(16)\]

The volume of v_total is measured as a function of p. from the data; we can obtain the value of v_mono and c. when p=p⁰, the equation has a singularity and v →∞ . This accounts for the steep rise of the isotherm, as shown in the figure, as the pressure approaches p0.

To obtain the constant c and v_m, we multiply both sides of equation (16) by (p⁰-p)/p;

Multilayer physisorption

\[\displaystyle \frac{{v({{p}^{0}}-p)}}{p}=\frac{{{{v}_{m}}c}}{{1+(c-1)\left( {\frac{p}{{{{p}^{0}}}}} \right)}}---(17)\]

Taking the reciprocal of both sides, we get,

\[\displaystyle \frac{p}{{v({{p}^{0}}-p)}}=\frac{1}{{{{v}_{m}}c}}+\left( {\frac{{c-1}}{{{{v}_{m}}c}}} \right)\left( {\frac{p}{{{{p}^{0}}}}} \right)---(18)\]

The above equation (18) is the BET equation.

The left-hand side of this equation is experimentally measured. The plot of p/[vp⁰-p] versus p gives a straight line. From the intercept, (1/v_mc) and the slope, (c-1)/(v_mcp⁰), we can calculate the value v_m, that is v_mono.

From the value of v_mono at STP, N_mono can be calculated;

\[\displaystyle {{N}_{{mono}}}=\frac{{{{N}_{A}}{{v}_{{mono}}}}}{{0.022414k{{m}^{3}}mo{{l}^{{-1}}}}}---(19)\]

N_mono = number of molecules required to cover a unit mass of the adsorbent with monolayer, a = area covered by a molecule, we can calculate the area of a unit mass of material;

\[\displaystyle \frac{{Area}}{{unitmass}}={{N}_{{mono}}}a---(20)\]

This way, the surface area of a finely divided solid can be determined.

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