March 28, 2024, 06:21:27 PM
Forum Rules: Read This Before Posting


Topic: What would happen first with diamond?  (Read 2478 times)

0 Members and 1 Guest are viewing this topic.

Offline pnacze199204

  • Regular Member
  • ***
  • Posts: 40
  • Mole Snacks: +0/-0
What would happen first with diamond?
« on: December 08, 2019, 09:13:50 AM »
I've found an interesting question on the Internet, that haven't been answered yet, so let me cite it here:

"It is frequently stated that although graphite is the more stable allotrope of carbon at STP, the activation energy of the diamond-to-graphite transformation is so high that our diamonds will never spontaneously turn into black dust.

Some sources add (correctly) that nothing is ever never, and reactions merely slow down with lower temperature, they never stop entirely. (The Arrhenius equation comes here). But nobody ever quantifies the non-neverness of the death of a diamond. So my question is:

Under normal conditions at room temperature, which of the following will happen first; when, and how fast?

1. Diamond transforming to graphite.
2. Diamond evaporating. Marshall and Norton ,J. Am. Chem. Soc., 1950, 72 (5), pp 2166–2171, seem to say that latent heat of carbon is 170kcal/mole, but I haven't got access to the whole paper to see what they say about the rate of evaporation.
3. Diamond spontaneously combusting to CO2.
4. Carbon decaying to iron via tunnelling. Barrow and Tipler, in The Anthropic Cosmological Principle (Oxford, 1986), p.654, quote 10^1500 years for this".

Online billnotgatez

  • Global Moderator
  • Sr. Member
  • ***
  • Posts: 4399
  • Mole Snacks: +223/-62
  • Gender: Male
Re: What would happen first with diamond?
« Reply #1 on: December 08, 2019, 11:19:50 AM »
Is this a continuation of your past posting
https://www.chemicalforums.com/index.php?topic=95325.0

Offline Enthalpy

  • Chemist
  • Sr. Member
  • *
  • Posts: 4041
  • Mole Snacks: +304/-59
Re: What would happen first with diamond?
« Reply #2 on: December 08, 2019, 07:32:29 PM »
4. is definitely slower than all others. The barrier is like 1MeV.

You can evaluate 2. because you can extrapolate the vapour pressure down to room temperature, and the vapour pressure gives you the evaporation rate (At least in theory. Sometimes my attempts failed, I don't know why).

As experimental verification won't happen, a wrong result is about as good as a correct one.

Offline Enthalpy

  • Chemist
  • Sr. Member
  • *
  • Posts: 4041
  • Mole Snacks: +304/-59
Re: What would happen first with diamond?
« Reply #3 on: December 16, 2019, 02:03:10 PM »
I feel reasonable to say that room-temperature combustion is faster than room-temperature evaporation, because any evaporated carbon (whether C2, C3 or whatever one prefers) would react quickly with the surrounding oxygen.

My intuition, and nothing more, tells that oxidation at room temperature depends fundamentally on trace impurities in air: ozone, nitrogen oxides. These, not molecular oxygen, are able to react without heating. It's observed at tyres for instance. So the oxidation rate would not be a constant for "air" but would be a consequence of pollution.

I also suspect that the four listed causes are so slow that other ones are faster. For instance cosmic rays (and their daughter particles) arrive at Earth's surface at a rate like 1/s/m2. Many of them have enough energy to rip a carbon atom from the crystal. This may well happen more often than 26meV thermal energy concentrating by chance to 7.4eV to evaporate an atom.

Following a personal message by pnacze199204:

Here's a paper dealing with the relationship between vapour pressure and evaporation rate
http://www.nano.lu.se/data/nanometer/users/ftf-jjo/file_element/846f307d82d0db1bdcf8f3d5379611fd/lecture2.pdf
with some formulas. As I said already, these formulas are inaccurate, I got wrong predictions from them. But for your estimates they may suffice.

The theory behind them (I guess only an end number is important to you) is that
  • Pressure results from a number of gas molecules impinging per time and area unit, and from their momentum.
  • At equilibrium, the solid sublimes as much vapour at it receives. Definition of vapour pressure.
  • Now without an external partial pressure, we could deduce at which rate vapour sublimes from the solid (or the liquid).
Looks convincing, perhaps even unavoidable? But it fails. Proofs are for maths. Physics is an experimental science, where we make models. Sometimes they succeed, here they fail.

Offline Enthalpy

  • Chemist
  • Sr. Member
  • *
  • Posts: 4041
  • Mole Snacks: +304/-59
Re: What would happen first with diamond?
« Reply #4 on: December 17, 2019, 01:22:36 PM »
For diamond these questions are of purely theoretical interest. But if some day we have secondary kilogram standards made of silicon (the primary resulting now from Planck's constant), they may become important.

Copies of the kg drifted by few 10-8 in a century, which would mean a complete change in 10 billion years. I trust Pt and Ir better than C and Si for that task.

Offline pnacze199204

  • Regular Member
  • ***
  • Posts: 40
  • Mole Snacks: +0/-0
Re: What would happen first with diamond?
« Reply #5 on: December 17, 2019, 04:56:13 PM »
Thank you for your response! Let me ask you if was it explained why exactly the former prototype of kg lost its weight or what factors could influence that? I read somewhere that it could be also related with maintenance and cleaning it, but I don't know if that's the truth. 

Offline Enthalpy

  • Chemist
  • Sr. Member
  • *
  • Posts: 4041
  • Mole Snacks: +304/-59
Re: What would happen first with diamond?
« Reply #6 on: December 18, 2019, 12:14:19 PM »
According to Wiki, the drift causes are not known. The former Pt+Ir kilogram standard was cleaned from time to time, drift is supposed because the two primary standard differ measurably now and they differ from the secondary standards. We don't even know whether mass was lost or gained.

I suppose any analysis would degrade the standards more than the usual procedure does, so analysis has not been done, and we have only hypotheses. Adsorption, dust, scratches are among them.

I trust Pt+Ir more because the unreactive surface stays clean. Si builds a layer of oxide that changes the mass. The proposal with Si includes probably to create from the beginning a thick layer of oxide that won't evolve any more, but how certain is that? And how reproducible would that layer be: if not, this defeats the idea of extremely reproducible monocrystalline silicon. I also wonder about adsorption and absorption by the oxide layer.

Anyway, the new standard is completely different.

Offline Enthalpy

  • Chemist
  • Sr. Member
  • *
  • Posts: 4041
  • Mole Snacks: +304/-59
Re: What would happen first with diamond?
« Reply #7 on: January 21, 2020, 07:32:08 AM »
Following a personal message from pnacze199204, here's an attempted estimation of carbon's vapour pressure at room temperature.

From Wiki's https://en.wikipedia.org/wiki/Vapor_pressures_of_the_elements_(data_page)
Press  Temp  Activation
100kPa 3908K
             95663K
 10kPa 3572K
             95588K
  1kPa 3289K
             95781K
 100Pa 3048K
             95334K
  10Pa 2839K


Much can go horribly wrong:
  • Data is for graphite. But diamond is only 1.9kJ/mol away, or 0.8×298K. Less than one magnitude error, my least concern here. It could even be computed away.
  • Huge extrapolations use to fail the big way.
  • Wiki doesn't tell if the vapour is C1, C2, C3 or C4. This open question changes completely the activation energy.
What encourages me to go on:
  • The activation energy isn't constant in Wiki's data, which could be experimental hence.
  • The activation energy (793kJ at 10Pa-100Pa) resembles the vaporisation enthalpy Hv (717kJ/mol for C1 at RT according to the CRC Hdbk of Chem&Phys)
So let's imagine that C1 evaporates at RT and that 95334K activation energy applies. For C1 it can't be very different from Hv anyway, but that difference makes magnitudes on the pressure. An improvement would take the experimental heat capacity of graphite or diamond, the theoretical heat capacity of C1, and correct the heat of vaporisation over the temperature range - feel free to do it, I won't.

Then the extrapolated graphite vapour C1 pressure at 298K is, tadaa,
10-124×10Pa
which is sub-nothing.

The rate of evaporation isn't measurable at 298K. Take 103J/mol mean kinetic energy perpendicular to a diamond surface, then 10-2kg C have mean 103m/s and 10-23kg×m/s momentum per vapour molecule, so 10-124Pa suggest 10-101 shocks×m-2×s-1. This model is knowingly wrong by 0 to 2 magnitudes.

Even if some means measures individual gaseous carbon atoms, it won't see any single gas atom in years. But it would be a funny experiment at temperatures less cold, perhaps by ionisation and mass spectroscopy.

And if a diamond is 10-3m=1027 atoms×m-2 thick, evaporation in all directions takes
10127s = 10120yr

I'm confident that a decent cosmic ray rips one carbon atom off. That's hugely faster than 10-101 atom×m-2×s-1 evaporation. Ozone and nitrogen oxides are faster too, as we see at rubber.

Sponsored Links