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While I was learning about Langmuir's adsorption isotherm in my chemistry class, my teacher talked about the situation when the rate of adsorption and desorption will become equal. However, he mentioned that the rate of adsorption is dependent on pressure but the rate of desorption is not and when he gave an explanation I wasn't satisfied with it. I would imagine that greater the external pressure on the adsorbent lesser will be the tendency of the adsorbate particles to leave the surface. But that isn't what my teacher said. Please explain.

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Without knowing exactly what you were told its hard to be exact, but here is a description of the situation.

The Langmuir model makes assumptions (a) adsorption is complete when the surface is filled with one gas molecule per site, (b) all adsorption sites are equivalent (i.e. the same) and (c) adsorption and desorption are separate processes, i.e. one does not depend on the other, i.e if one site is occupied it does not affect adjacent sites an any way. Note that this is a model and this determines how we look at the problem. The model is then $$M(g) + S(surface) = GM(surface) $$ where G is gas and S surface and rate constant for adsorption is $k_a$ and for desorption $k_d$.

We let $\theta$ be the extent of adsorption which varies from 0 to 1, i.e. how full the surface is, then the rate of change of $\theta$ depends on the rate constant for adsorption , pressure and fraction of vacant sites or $N(1-\theta)$ for N vacant sites. Thus for adsorption $$ \frac{d\theta}{dt} = k_aPN(1-\theta) $$

and for desorption $$ \frac{d\theta}{dt} = -k_dN\theta $$

At equilibrium we make the rate of change equal to zero, i.e. rate of adsorption equals rate of desorption and with a bit of algebra find that $$\theta=\frac{KP}{KP+1}$$ where K is the equilibrium constant or $K=k_a/k_d$ This equation has the fancy name as the Langmuir Isotherm.

You can see that the rate going onto the surface must be equal to the pressure (no gas, no adsorption) and the area free to adsorb onto $(1-\theta)$. (We assume in this model than no molecule double adsorbs; this is dealt with in more advanced treatments). The rate leaving has to be proportional to the amount already there , i.e. proportional to N the total number of surface sites multiplied by the fraction covered.

Ask more questions in the comments if this is not clear.

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  • $\begingroup$ I understood Langmuir's isotherm. I didn't understand why rate of adsorption is directly proportional to pressure but rate of desorption is not. $\endgroup$ – EdmDroid Jul 6 '16 at 13:14
  • $\begingroup$ Why is the inverse process concentration-independent? $\endgroup$ – Marcel Jan 3 '17 at 18:27
  • $\begingroup$ @Marcel it has to be proportional to the number on the surface , which is $\theta N$ so is fraction of concentration. No molecules on surface off rate zero, full surface; maximum off rate :) $\endgroup$ – porphyrin Jan 4 '17 at 8:50
  • $\begingroup$ @porphyrin why is desorption not proportional to , say, the decrease of pressure? $\endgroup$ – sajjad islam Jan 12 '18 at 18:33
  • $\begingroup$ @sajja islam because we consider adsorption and desorption as separate processes: if there are no molecules on the surface the rate of desorption must be zero and it has to be at a maximum when the surface is completely covered. $\endgroup$ – porphyrin Jan 12 '18 at 22:58
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For adsorption, its forward reaction is precursor concentration dependent and the precursor concentration is represented by pressure. For desorption, the forward reaction is also precursor concentration dependent and the concentration here is theta (or one - you can treat the precursor as solid surface so you have concentration of one).

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Desorption is inverse of adsorption.since pressure is directly proportional to adsorption so desorption is inversely proportional to pressure(provided surface does not reach to a saturation level)

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  • 1
    $\begingroup$ Adsorption and desorption are not quantities so they can't be proportional to something. $\endgroup$ – bon Jul 6 '16 at 8:18
  • $\begingroup$ In any case, this answer is wrong in so many levels... $\endgroup$ – Nando Oct 18 '18 at 14:41

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