Uniqueness of Isotopic Properties

[Amaton]

Hello Might be here with another dumb question.

So I'm looking at a list of the elements which is organized as a table of properties according to each element. These properties include...

 

• mass density

• phase at STP

• standard atomic weight (or rather, mass)

• number of naturally occurring isotopes

• categorization (alkali, transition, etc.)

• melting and boiling points

• specific heat capacity

• electronegativity

• natural occurrence (as in primordial, in trace, or synthetic)

 

Now, we know that each element exists as various isotopes, artificially or not. I'm wondering if some of the properties above differ among the given isotopes of a particular element (of course excluding those derived from or concerning the isotopes in general). Most lists make it seem as though these properties are tacitly identified with the element as a whole, rather than any particular isotope.

 

For example, let's say we have the nuclides with 1 proton. Shouldn't any of the measurements of density, electronegativity, heat capacity, etc. differ among protium, deuterium, and tritium? If so, are these properties just identified with the most common or stable of the natural nuclides?

 

I'm only taking high school chemistry as of now, so sorry if I'm making a trivial misconception. Just curious.

[Amaton]

P.S. - I don't know if Chemistry is the appropriate subforum for this thread. I'm not sure if it's physics, chemistry or, maybe best said, nuclear chemistry. Sorry if it seems out of place.

[swansont]

AFAIK, the bulk properties are calculated with the naturally occurring mix of the isotopes. If you had isotopically pure samples, or a sample with a different mix, then properties could change. e.g. the freezing and boiling points of heavy water are different from "normal" water

[John Cuthber]

"Shouldn't any of the measurements of density, electronegativity, heat capacity, etc. differ among protium, deuterium, and tritium?"

Yes, they do.

https://en.wikipedia.org/wiki/Deuterium#Data_for_elemental_deuterium

can be compared to

https://en.wikipedia.org/wiki/Hydrogen#Properties

 

Generally, the quoted figures for elements like chlorine or copper which are a mixture of isotopes refer to the value for the "normal" mixture.

For some elements like selenium and lead the ratios of the isotopes varies with the mineral in which they are found and the "atomic weight " for these elements isn't very accurately defined.

[Amaton]

Thanks for the responses.

 

Generally, the quoted figures for elements like chlorine or copper which are a mixture of isotopes refer to the value for the "normal" mixture.

 

Okay. So a "normal mixture" of boron, I'm guessing, would be approx. (80%) boron-11 and (20%) boron-10, as to reflect to their natural variation. Right?

 

(swansont) AFAIK, the bulk properties are calculated with the naturally occurring mix of the isotopes.

 

I see. So how is the natural composition of an element calculated?

[swansont]

I see. So how is the natural composition of an element calculated?

 

I imagine one would obtain a number of samples and do measurements with a mass spectrometer. There's probably a journal or two where such measurements are published, which would give details.

[John Cuthber]

Originally, they measured the atomic masses by doing chemistry and weighing things so, for example if you take some known mass of hydrogen and pass it over heated silver chloride you get hydrogen chloride and silver.

If you weigh the silver and the leftover mixture of silver and silver chloride then you can find out how much chlorine was used reacting with the known weight of hydrogen.

It turns out to be about 35.5 grams of chlorine per gram of hydrogen.

Atomic masses were originally measured using the mass of hydrogen as the unit so the atimic weight of chlorine was measured to be 35.5

That's consistent with the fact that chlorine has two isotopes of mass 35 and 37 in the ratio of about 3 to 1

 

(It gets a bit more complicated when you have valencies other than 1)

[Enthalpy]

Properties that result from electrons (electronegativity, boiling point...) don't depend significantly on the number of neutrons (hydrogen being the one exception), so you can take the mixture's properties and use them for individual isotopes.

 

Properties that result from the atom's mass (density...) can be scaled for each isotope from its mass compared with the mean mass of the natural mix.

 

So element tables give properties only for the natural mix of the isotopes, to keep the table's size reasonable, and because the rest can be deduced if needed. A nice table there:

http://www.webelements.com/

including isotopes when, at the chosen element, you ask for Nuclear properties > Isotopes, example

http://www.webelements.com/boron/isotopes.html

 

Most elements have a nearly constant proportion of isotopes on Earth, but because this proportion can be counted down to individual atoms, special uses exploit the tiny difference: ¹³C datation, origin of uranium ore...

 

The proportions vary a bit with altitude or with depth, and so on. A few natural elements are less constant - generaly when their isotopes result from the radioactive decay of various elements that differ chemically. Artificial elements like the transuraniums vary a lot. The proportion varies a lot among our Sun's planets but is remarkably the same for our Moon.

[swansont]

Properties that result from electrons (electronegativity, boiling point...) don't depend significantly on the number of neutrons (hydrogen being the one exception), so you can take the mixture's properties and use them for individual isotopes.

 

I would expect boiling point to depend on the mass, since KE depends on mass. It's true for Helium's BP http://en.wikipedia.org/wiki/Helium-3 though the effect should become proportionally less as you have heavier elements. (It's also true for molecules, as I mentioned with heavy water)

[Enthalpy]

All right, helium is the other abnormal element... Just cool it below 2K and the isotopes separate spontaneously. ³He goes in one phase, ⁴He in the other. Plus many properties that make helium really special.

 

As for the boiling point, the number of neutrons has very little effect, because heat gives a uniform (mean) energy to the molecules rather than a uniform speed, and this kinetic energy compares directly to the sticking energy that keeps the molecules in the liquid. So the isotopic effect if far smaller than the mass ratio - very small in fact.

 

Hydrogen is special in that its chemical bonds depend on the nucleus' mass. Because the proton is not so heavy, it is significantly delocalized (call it ground state vibration energy if you wish), so that the electrons make less strong bonds, because the proton can't be centered precisely at the best possible place that minimizes electrons' energy. Deuterium is heavier, so electron bonds can localize its nucleus better to minimize their energy, and the bonds are stronger.

 

This makes an interesting difference at the Van der Waals' forces which aren't very strong and define the melting and boiling points of many substances. For water, it's a hydrogen bond, which is oriented by the molecular H-O bond, hence sensitive to the diffuse position of the proton.

 

The ease of ionization in acidic conditions is also sensitive to hydrogen's nuclear mass. This is used to separate heavy water using a simple acidic cycle, without first separating hydrogen from oxygen of all water just to keep the deuterium. Thanks to that, heavy water is readily available at moderate price, while other pure isotopes are expensive.

[swansont]

As for the boiling point, the number of neutrons has very little effect, because heat gives a uniform (mean) energy to the molecules rather than a uniform speed, and this kinetic energy compares directly to the sticking energy that keeps the molecules in the liquid. So the isotopic effect if far smaller than the mass ratio - very small in fact.

 

No, that's not quite right. There is an energy distribution, but collisions are involved and momentum is conserved. A lighter atom will have a higher speed, on average, such that mv is equal for the two particles. So lighter atoms will preferentially have more than half the energy in those interactions.

 

This isotope effect is discussed in the literature. One should google for "thermodynamic isotope effect"

[John Cuthber]

Properties that result from electrons (electronegativity, boiling point...) don't depend significantly on the number of neutrons (hydrogen being the one exception), so you can take the mixture's properties and use them for individual isotopes.

It is perfectly common to be able to separate isotopically substituted chemicals by chromatography.

For example The deuterium labelled versions of chemicals (things like C6D6 or C10D8 the "heavy" versions of benzene and naphthalene) are used as standards in analytical chemistry.

You might imagine that the "heavy" compounds would stick to the column better and come off after the normal versions, but it's often the other way round.

The effect arises from the way the electrons behave- the nuclei are not involved.

 

The differences are small, but they are measurable.

[Enthalpy]

I believe collisions tend to equalize the energy (as a mean result, sure) rather than the momentum. By the way, "momentum conserved" means the vector sum of the colliding molecules, not the way they share the momentum relative to their common center of mass.

 

The equipartition theorem (wiki article) tells the same as I do, unless I missed something. The mean energy is equally shared among all degrees of freedom, which includes the translation energy of all species.

 

About the thermodynamic isotope effect, this paper seems to understand it the same way as I do:

http://old.iupac.org/goldbook/T06319.pdf

"zero-point energy difference" on vibration of protium, deuterium and tritium.

Looks like this effect has a long name and theory to apply essentially to hydrogen, a tiny bit to carbon, by vibration rather than translation.

 

If I try to put figures, imagine temporarily that thermal equilibrium gives a uniform mean momentum rather than energy.

- CH₃D (17g) would have 3% less translation energy than CH₄ (16g) so it would need 115K to boil instead of 112K @1atm - way overvalued.

- H₂¹⁸O would be segregated by several per-cent at each evaporation, but its at most 10^-4.

- Heavy water separation would also enrich in H₂¹⁸O, but this does not happen.

[swansont]

I believe collisions tend to equalize the energy (as a mean result, sure) rather than the momentum. By the way, "momentum conserved" means the vector sum of the colliding molecules, not the way they share the momentum relative to their common center of mass.

 

It means both. Momentum's going to be conserved in any frame you pick.

 

The equipartition theorem (wiki article) tells the same as I do, unless I missed something. The mean energy is equally shared among all degrees of freedom, which includes the translation energy of all species.

 

Degrees of freedom of motion, not degrees of freedom of different isotopes.

 

About the thermodynamic isotope effect, this paper seems to understand it the same way as I do:

http://old.iupac.org/goldbook/T06319.pdf

"zero-point energy difference" on vibration of protium, deuterium and tritium.

 

I wonder why the several examples I found all talked about the change in boiling point.

 

 

Looks like this effect has a long name and theory to apply essentially to hydrogen, a tiny bit to carbon, by vibration rather than translation.

 

If I try to put figures, imagine temporarily that thermal equilibrium gives a uniform mean momentum rather than energy.

- CH₃D (17g) would have 3% less translation energy than CH₄ (16g) so it would need 115K to boil instead of 112K @1atm - way overvalued.

- H₂¹⁸O would be segregated by several per-cent at each evaporation, but its at most 10^-4.

- Heavy water separation would also enrich in H₂¹⁸O, but this does not happen.

 

Again, we're not talking about heterogenous molecules. I'd like to see the calculations, though.