Chemical Forums
Chemistry Forums for Students => Physical Chemistry Forum => Topic started by: AdiDex on December 08, 2016, 08:19:29 AM
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Tomorrow is my exam , And I got a silly doubt.
In this equation ,
Both of the Enthalpy are negative (Since exothermic). So If I want a High value of C , so The Enthalpy of Liquefaction should be higher than the heat of adsorption ??
Or I just don't have to consider the signs so the just reverse should be answer.
I got this confusion because in my book there is a topic Langmuir Isotherm from BET theory .
In that they take an assumption C>>1 and p°>>p .
Basically in Langmuir we considered the mono layer adsorption , that means the enthalpy of liquefaction should be negligible in front of enthalpy of adsorption .
And from this , for getting an High Value of C , I have don't have to consider the sign of enthalpy.
What is the problem with my understanding .
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Here are the BET and Langmuir Isotherm Result.
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I have one more doubt ,
There are some kind of graphs in my book .
If I see the Langmuir isotherm , Its something like y/x=b+ax , a parabola .
So for monolayer adsorption it should be like 3rd . But this is absurd , As it must get constant at higher pressure , because of saturation .
Please reply within 14 Hrs.
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Sign Confusion , please fast
How will not eating help you with your sign confusion?
Seriously though - we are busy people, we will help you when we can. You should know that by now. Don't ask for a fast response, it won't help get you one. It's your responsibility to do your revision in good time.
Now, it should be clear to you that if you want C to be large, and the magnitude of EL is much smaller than that of E1, the numbers must be positive. The true thermodynamic enthalpy of adsorption ΔHads is negative; E here is a positive quantity equal in magnitude to the enthalpy of adsorption. (I always distrust the symbol E and the waffly "heat" in these contexts; they often have the "wrong" signs from a strict thermodynamic viewpoint.)
If you review the derivation of the BET isotherm, you will see that C = exp(-(ΔHads - ΔHL)/RT).
The Langmuir isotherm is not a parabola. It is like y = x/(1+x), not y/x = b+ax.