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« on: May 31, 2016, 02:46:58 PM »
Hi !!

I have a doubt about the phisical-chemistry  meaning above the  mathematical  expression of "Langmuir isotherm"...
The expression  is this:

θ = K[A] /  1+K[A]

where θ is the degree of adsorption (how much the solid surface is filled up). 0≤θ≤1
n.b: if I have a gas I can put [A]= p (pression)

Now at lesson we have done a graphical plot of θ (y) against [A]( the indipendent variable x)  and the graph is similar at:
y=√x  where [A]=x

Now my doubt is the phisical meaning of this graph and the meaning of [A]  in this esxpression ( N.B. I 'm considering a GAS adsorbed on the surface...so I could substitute this with pression p)

1) what is [A] in langmuir isotherm?   The  different INITIAL  concenentration that I use in my expreiments? Or the different  values of concentration of A in the gas phase that i can get at EQUILIBRIUM ??

2) phisical-chemistry meaning of this isotherm:
WHY at [A] --> 0 is  θ≈0   while at [A] --> ∞  is θ≈1 ??

IF (and this is my first doubt)  [A] represents  the equilibrium concentration of A in the gas  phase why I should have low values of degree of adsorption ,θ, on the surface  when I have a small concentration of [A] and big values of θ  when [A] at equilibrium is high ??

I don't manage to find a correlation about the concentration of A at equilibrium and the degree of adsorbtion...
Someone can halp me??

THANKS

#### mjc123

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##### Re: langmuir isotherm: adsorption equation
« Reply #1 on: June 01, 2016, 04:24:18 AM »
Quote
IF (and this is my first doubt)  [A] represents  the equilibrium concentration of A in the gas  phase why I should have low values of degree of adsorption ,θ, on the surface  when I have a small concentration of [A] and big values of θ  when [A] at equilibrium is high ??
Why do you think it should be any different?