15
$\begingroup$

Is it meaningful to characterize the relative strength of desiccants? For example, is there a measure of hygroscopy, or an ordering of desiccants, such that higher ones will always dry lower ones?

Let us restrict the question to "practical" desiccants, which are hygroscopic substances that can be "recharged" (i.e., dried out) by heating. I'm not certain I know all possible mechanisms of water sorption within this class of chemicals. But is it safe to say that for all such chemicals, at standard temperature and pressure, there is some water vapor pressure around a desiccant that has sorbed any water?

I am trying to imagine a series of tests in which two different desiccants are put in a humid chamber in adequate quantities that either could completely sorb all the water. At hydrous equilibrium will the "stronger" desiccant contain virtually all of the water? Or is the equilibrium distribution of water a function of the desiccants' relative "hygroscopy," rates of sorption, or some other factor(s)?


This is ultimately a practical question. One example: I have a hygroscopic, hydrated salt X. Can I dry it out by sticking it in a sealed box at STP with a "more powerful" desiccant? If not what determines the limit of dehydration that would be achieved by exposure to other desiccants?

$\endgroup$
  • 2
    $\begingroup$ Probably the easiest way would be to compare equilibrium water vapor pressure over various desiccants. I have a feeling that I have seen the relevant chart, but I can't remember where. $\endgroup$ – permeakra Aug 24 '15 at 5:44
  • $\begingroup$ You also have to consider particle sized. Finely divided desiccants work much better than coarse grains of exactly the same material. $\endgroup$ – Gimelist Aug 24 '15 at 13:35
  • $\begingroup$ @Michael: Certainly. That would be part of the practical characterization of the desiccant. My guess is that the "form" of the desiccant is more indicative of its capacity than of its "strength." ("Form" could also include such specifications as a desiccant needing to be stirred to maintain its strength.) $\endgroup$ – feetwet Aug 24 '15 at 13:39
3
$\begingroup$

Yes, it is meaningful, but often ignored for practical reasons. In the lab, e.g. in a desiccator, you would just use a large excess of the desiccant of choice which would always work. It’s often more meaningful to classify them as to whether they are acidic, basic or neutral.

But let’s assume we wanted to create that scale. For simplicity reasons, let’s check a single desiccant first. We need to consider the following equilibrium:

$$\ce{H2O_{(g)} <=> H2O_{(x)}}$$

With (x) again being the water bound to the desiccant in whichever way. (Do not think of this as chemical bonding; it usually is not). It is important to remember that this is an equilibrium with an equilibrium constant k.

$$k_{\mathrm{x}} = \frac{[\ce{H2O_{(x)}}]}{[\ce{H2O_{(g)}}]}$$

One could rearrange this in numerous ways resulting in a representation that essentially says the partial pressure of water in the surrounding atmosphere is more or less proportional to the concentration of water in the desiccant.

$$p_{(\ce{H2O})} \approx c \cdot [\ce{H2O_{(x)}}]$$

Now let’s add a second desiccant into our consideration. The first equation now rearranges itself to the following:

$$\ce{H2O_{(y)} <=> H2O_{(g)} <=> H2O_{(x)}}$$

For each of the two subsystems we again get an equilibrium constant $k$ which we can rearrange to proportionalise the partial pressure to the concentration of water in x or y. $k_{\mathrm{x}} \neq k_{\mathrm{y}}$. There will be one partial pressure of water where both the left side and the right side of the double equation will be at equilibrium; that partial pressure is the final equilibrium that will be reached. This gives us the following conclusions:

  • Never will the air be completely dry in the presence of (only) a (or two, or a hundred) desiccant(s).

  • Neither of the two desiccants will fall completely dry again; rather unlike when heating (in an open atmosphere).

  • Say you start the experiment with the weaker desiccant and then add a dry sample of the stronger desiccant:

    • The stronger desiccant will further reduce the vapour pressure;
    • The reduced vapour pressure will remove water from the weaker desiccant;
    • That water will partly be absorbed by the stronger desiccant
    • However neither will the weaker desiccant be completely dried nor will the partial pressure of water in the atmosphere be completely zero.
  • How dry (or wet) either desiccant will end up/how low the partial pressure of water will end up, depends on the strengths of the desiccants. If you use a very strong and a very weak one, the strong one will absorb much more water than the weak one. If the two are similar in strength, they will absorb similar amounts of water.

$\endgroup$
6
$\begingroup$

I'd say no, because there is also a reactivity dimension to desiccants. Some drying agents react with some solvents and therefore can't be used as a desiccant. This happens for many organic solvent and desiccant pairs. Some desiccants also react as desiccants and this changes their drying power. This is the case for phosphorous pentoxide, for example, which gradually forms a viscous layer of phosphoric acid which stops any drying. So any linear ordering you are attempting is likely to be overwhelmed by the reactivity side effects of the many desiccants.

On the other hand, if you are only interested in some theoretical point in time estimate of desiccant efficiency, this might be possible but would have virtually no practical import. A desiccant has to react over time and the side effects cannot be ignored.

$\endgroup$
  • $\begingroup$ Interesting. To keep the question tractable let's assume that we're only talking about water in a standard atmosphere; no other volatile compounds present. The question is still widely applicable with that assumption. $\endgroup$ – feetwet Aug 24 '15 at 2:14
  • $\begingroup$ No that's still not true. Phosphorous pentoxide is used as a general very strong desiccant, but still has the viscous phosphorous acid problem. What you are proposing is a very controlled environment which would likely never exist in real conditions (i.e. we always expose new phosphorous pentoxide). This is technically possible, but has almost no practical application. Are you only interested in theoretical but perhaps non-achievable results? $\endgroup$ – user467 Aug 24 '15 at 2:20
  • $\begingroup$ No, I'm interested in the practical application. All desiccants reach sorption limits. I found this page which shows capacity and rates for a few common desiccants. In my proposed test we would put enough phosphorous pentoxide in the test chamber that it could absorb all of the water before its sorption rate slope goes below 1, or hits zero, or something like that. (Presumably, as $\cf{P4O10}$ is a popular practical desiccant, it is used in quantities or methods in which the $\cf{H3PO4}$ doesn't degrade its desiccation strength.) $\endgroup$ – feetwet Aug 24 '15 at 3:13
6
+50
$\begingroup$

At hydrous equilibrium will the "stronger" desiccant contain virtually all of the water? Or is the equilibrium distribution of water a function of the desiccants' relative "hygroscopy," rates of sorption, or some other factor(s)?

"Rates" aren't relevant for questions about equilibrium.

I am trying to imagine a series of tests in which two different desiccants are put in a humid chamber in adequate quantities that either could completely sorb all the water.

For that particular test, consider two dessicants $\ce{X}$ and $\ce{Y}$. The winner of the test will be determined by the thermodynamics of the two reactions:

$$\ce{H2O(g) <=> H2O_{(X)}}$$ $$\ce{H2O(g) <=> H2O_{(Y)}}$$

which represents the the absorption of water vapor into phase $\ce{X}$ (or $\ce{Y}$) respectively. The two reactions can be combined to yield

$$\ce{H2O_{(X)} <=> H2O_{(Y)}}$$

and the free energy change $\Delta G_r$ of that reaction will determine which dessicant is "stronger" in the sense of that particular test.

This test doesn't take into account many practical considerations that other answers have noted, such as reactivity.

I'd bet that liquid sulfur trioxide would win most desicating contests.$%edit$

$\endgroup$
  • 1
    $\begingroup$ It almost sounds like you have the answer to the question! If I understand what you're saying, there is a binding energy associated with each desiccant (which may not be constant over hydration level, but which nonetheless can be characterized by a monotonic curve?), and in the limit water will be bound to minimize free energy? E.g., if X has higher binding strength than Y over all hydration levels attainable in the test then all water will ultimately be bound to X? If so then what is this "binding strength" desiccant characteristic called in the art? $\endgroup$ – feetwet Aug 24 '15 at 13:00
  • 1
    $\begingroup$ Note also: Chemicals like $\ce{SO3}$ would not be included in this question as "practical" desiccants because they are not "rechargeable" by simple baking in a standard atmosphere (AFAIK). $\endgroup$ – feetwet Aug 24 '15 at 13:58
0
$\begingroup$

It depends on temperature as well. Sure metaphosphoric acid is extremely difficult to dehydrate to phosphorus pentoxide, but with heat and liquid $\ce{SO3}$, it should be rather easy if it's under vacuum, and everything is in separate containers inside a larger one. This way the equilibrium is broken since the $\ce{SO3}$ absorbs the water and can't get back to the resulting pentoxide. So despite $\ce{SO3}$ being a weaker dessicant, it can dehydrate hot metaphosphoric acid under certain conditions.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.