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Hydrazine
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[*] posted on 22-10-2008 at 06:57
Delta Heat of Formation question


Hello everyone,

I have a question about units of conversion.

I have been trying to find the heat of formation of H2O2 in the 30%-50% range. Preferably in units of "kcal/gram" but all I could find is a table in "kcal/gram mol".

For example:
The heat of formation of a 30% solution of H2O2 is 45.61 kcal/gram mol.

The part I don't get is "gram mol"... I need the heat of formation in just "Grams"

I suppose I could just divide (45.61 kcal)/mol.... But what is the value of "mol"???

water mol wt = 18
H2O2 mol wt = 34

I know for a fact that the heat of formation of 30% peroxide does not equal (45.16kcal/18) nor does it equal (45.16kcal/34).

So either way, "mol" does not equal the mol wt of water or peroxide... So what is "mol"?

How can I convert this table into kcal/gram?

Thank you,
Hydrazine
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sparkgap
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[*] posted on 22-10-2008 at 07:11


The conversion is obviously to divide the "kcal/mol" value by the molar mass in units of "g/mol". Cancel the units and it is now "kcal/g".

:)

"gram mol" is an archaic way of referring to the plain old mole we know. There's a history behind the terminology that I'm now hazy as to the details, but I think the chemical engineers might remember.

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watson.fawkes
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[*] posted on 22-10-2008 at 09:12


Quote:
Originally posted by sparkgap
"gram mol" is an archaic way of referring to the plain old mole we know. There's a history behind the terminology that I'm now hazy as to the details, but I think the chemical engineers might remember.
Other mass units for a mole are the number of particles such that the numerical value for a certain particle mass, measured in that alternate unit, is the same. For example, the mass of 1 gram-mole of C<sub>12</sub> is 12 grams, the mass of 1 kilogram-mole is 12 kilograms, and the mass of 1 pound-mole is 12 pounds. The kilogram-mol is the same as the kmol, an SI unit with "k" as a multiplier. The SI standard mole is the gram-mole. Thankfully, it's about the only one in use.
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Hydrazine
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[*] posted on 22-10-2008 at 09:33


Hmmmm. I'm still not clear on this.

"The conversion is obviously to divide the "kcal/mol" " I wish it was simply kcal/mol, but its kcal/gmol. So I can't just divide it by g/mol.

Since I'm slow on this, lets use 50% (w/w) peroxide as an example.

In the Solvey chart the delta Hf = -45.49 (kcal/gmol) H2O2

How is this converted to kcal/g ?

I know the real answer is around -2.55 kcal/g. And this suggests mol=18 = water
But I don't trust this answer. It seems more coincidental.
See below.

---------

In the Solvey chart:
100% peroxide = -44.85 (kcal/gmol)

And I know 100% peroxide is -1.32 kcal/g.

-44.85/-1.32 = 33.97 = mol wt of H2O2 ???

-------

In the Solvey chart:
100% water = 45.67 (kcal/gmol)

And I know 100% water = -3.79 kcal/g

-45.67/-3.79 = 12.05?.... What's this?... Nothing in water has a mol wt less than 18.

Something isn't adding up correctly.

If this doesn't add up correctly, how can any of it be trusted?

Something is wrong. What am I missing?

[Edited on 22-10-2008 by Hydrazine]
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sparkgap
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[*] posted on 22-10-2008 at 09:49


Like I said, "gram mole" and "mole" are two different names for one unit.

Quote:

In the Solvey chart:
100% peroxide = -44.85 (kcal/gmol)

And I know 100% peroxide is -1.32 kcal/g.

-44.85/-1.32 = 33.97 = mol wt of H<subb>2</sub>O<sub>2</sub>


...which is correct.

Quote:

In the Solvey chart:
100% water = 45.67 (kcal/gmol)

And I know 100% water = -3.79 kcal/g

-44.85/-3.79 = 11.83?.... What's this?


Oddly, I see from my tables that (liquid) water's enthalpy of formation is -285.8 kJ/mol or -68.3 kcal/mol or -3.79 kcal/g, as you have said. :o Something's up, can't put my finger exactly on it, though.

sparky (~_~)

[Edited on 23-10-2008 by sparkgap]




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Hydrazine
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[*] posted on 22-10-2008 at 09:51


I had to edit the water section of my prior post. There was a small error in it, but still.

Something isn't adding up.
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Hydrazine
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[*] posted on 22-10-2008 at 09:55


Quote:
Originally posted by sparkgap
Like I said, "gram mole" and "mole" are two different names for one unit.



OK so gmol = mol ?... That makes me feel better. At least I know what that is.;)
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[*] posted on 22-10-2008 at 10:02


"Something isn't adding up."

Looks like it. The listed enthalpy for pure water on your table is off. Have you no other Solvey charts?

sparky (~_~)




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Hydrazine
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[*] posted on 22-10-2008 at 10:20


Somehow I doubt that.
I believe the listed enthalpy is correct.

I have another chart from a Boeing Handbook of Propellant chemisty. Its chart is essentially identical to the chart from Solvey.

With both charts in agreement we can take it as a mutual validation for accuracy.

But still.... how are these values reliably converted to kcal/g?
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Hydrazine
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[*] posted on 22-10-2008 at 10:46


Here is the Solvey chart.

Peroxide heat of formation of solutions
% btu/lb cal/gmol
0 -2413 -45600
10 -2413 -45590
20 -2412 -45570
30 -2410 -45540
40 -2407 -45490
50 -2404 -45420
60 -2399 -45330
70 -2393 -45220
80 -2386 -45090
90 -2378 -44940
100 -2369 -44780
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[*] posted on 22-10-2008 at 14:32


Hydrazine I can't comment on your discrepancies in your tables and computed values. What I want to comment on is the concept of a mole, or mol.

When I took chemistry courses it was always understood that a mole (or mol) was a gram-mole, ie, a mass of a given chemical equal to its molecular weight in grams.

In my chemical engineering classes, being that I live in the US, grams were not an especially useful quantity because the scale is too small. We seldom worked in kgs either because kgs are not the usual engineering unit of mass. So we used lbs and tons. Likewise we used lb-moles and ton-moles. It was the same idea, simply a mass of a chemical equal to the molecular weight expressed in the particular unit of interest. Ounce-moles or grain-moles would work just as well if you wanted to work at a smaller scale.

In the chemistry literature I think that it is assumed that when a mole (or mol) is specified it is understood that the writer is referring to gram-moles, unless otherwise indicated.
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Hydrazine
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[*] posted on 22-10-2008 at 15:34


I don't have a problem with moles. 1 mol = 6.0225 × 10^23 number of molecules. That's the easy part.


I just can't get the units in the tables to make sense. So even if gmol = mol, what value goes in the denominator?
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[*] posted on 22-10-2008 at 17:26


I don't have a lot of experience using heats of formation,etc, but it does not make sense to me to put anything other than the molecular weight of H2O2 in the denominator. If it was an aqueous solution of NaCl I would place the MW of NaCl in the denominator. Anything else seems incredibly clumsy and non-sensical.
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Hydrazine
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[*] posted on 22-10-2008 at 19:57


If the mol wt of H2O2 (34) is plugged into the denominator of anything but 100% peroxide it doesn't work.

So what is the correct answer?
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[*] posted on 22-10-2008 at 20:37


Quote:

Here is the Solvey chart.

Peroxide heat of formation of solutions
% btu/lb cal/gmol
0 -2413 -45600
10 -2413 -45590
20 -2412 -45570
30 -2410 -45540
40 -2407 -45490
50 -2404 -45420
60 -2399 -45330
70 -2393 -45220
80 -2386 -45090
90 -2378 -44940
100 -2369 -44780


From Chemical Process Principles, Part I, 2nd ed, by Hougen et al, Table 29:

Heat of formation for H2O2 = -44.84 kcal/g-mole
Heat of solution at infinite dilution = -0.84 kcal/g-mole

So the "net heat of formation" at infinite dilution = -44.84 - 0.84 = -45.68 kcal/g-mole

These two figures, for undiluted H2O2 and the infinitely diluted H2O2, pretty well match your Solvey table. Then the intermediate values are similarly determined to include the appropriate heat of solution at the lesser dilutions. This is my guess.
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Hydrazine
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[*] posted on 23-10-2008 at 06:54


That sounds plausible.

I'll plug a table of "heat of dilution" into excel, add it to the Solvey table and report back.
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[*] posted on 23-10-2008 at 18:54


Shoot... It doesn't work.

The heat of infinite dilution of 0% peroxide should equal the heat of formation of pure water. But it doesn't add up either.

The same with any value less than 100% peroxide doesn't work either. So it suggests a coincidence that it only works at 100% peroxide.

Still stuck in a dead end.

I don't what else to try.

Any suggestions???

There has to be a whizbang chemist somewhere in this forum that could figure this one out.
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[*] posted on 23-10-2008 at 19:19


Quote:

The heat of infinite dilution of 0% peroxide should equal the heat of formation of pure water.


That is not true.

For heats of solution what this means is that you have one mole of H2O2 dissolved in enough water that the heat of solution is no longer changing. The heat of formation at "0% H2O2" is for one mole of H2O2 at that dilution. It is not really at 0% concentration but is virtually so. This is the scientific vernacular that is used.

So, if you want the heat of formation of one gram of H2O2 when it is dissolved at 30% dilution, take the value from your Solvey table and divide it by the MW of H2O2. Done.
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Hydrazine
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[*] posted on 24-10-2008 at 13:39


Hmmm.... Your saying that 0% is actually some infinitely small value greater than 0?

In any case, this is the table for "Heat of Solution".
It may be more relevant than the "Heat of infinite dilution"

Peroxide heat of solution
% cal/gmol
0 -817
10 -809
20 -709
30 -757
40 -706
50 -640
60 -549
70 -441
80 -311
90 -162
100 0

With this data, can show me an example?



[Edited on 24-10-2008 by Hydrazine]
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[*] posted on 24-10-2008 at 14:55


Quote:

Hmmm.... Your saying that 0% is actually some infinitely small value greater than 0?

In any case, this is the table for "Heat of Solution".
It may be more relevant than the "Heat of infinite dilution"

Peroxide heat of solution
% cal/gmol
0 -817
10 -809
20 -709
30 -757
40 -706
50 -640
60 -549
70 -441
80 -311
90 -162
100 0

With this data, can show me an example?



It doesn't have to even be infinitely small. The vernacular is: "at infinite dilution." But it is actually just the heat of solution at the dilution where it stops changing. I think the "0%" nomenclature is poor as it leads one to think that it is for pure solvent. This faked me out at first also unitl I dusted off an old textbook and could see that it was the same value as at "infinite dilution," which I feel is better nomenclature.

The book I cited before, on p. 319, defines the heat of formation of solutes thus: "The total standard heat of formation is the sum of the standard heat of formation of the solute delta Hf and it's standard integral heat of solution delta Hs2, at the specified concentration."

So, from the table above, the heat of solution is -0.817 kcal/mole at infinite dilution. How is that different from the value I gave you earlier of -0.84 kcal/mole? No diference, just different investigators at different times.

So, from my book the delta heat of formation for H2O2 is -44.84 kcal/mole. Add to that the heat of solution at 30% dilution, ie, -0.757kcal/mole and you get -45.6kcal/mole for the heat of formation. Your Solvey table gives -45.54kcal/mole.

This would become clearer to you if you looked at some graphs of integral heats of solution vs moles of water per mole of solute. I suggest the book I cited. But any book on physical chemistry or thermochemistry should have them.
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Hydrazine
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[*] posted on 24-10-2008 at 16:21


Yes, I agree adding the heat of formation of H2O2 and the heat of dilution will add up to the Solvey tables.

So lets use the 30% peroxide example you cited and try to convert it to kcal/g

"So, from my book the delta heat of formation for H2O2 is -44.84 kcal/mole. Add to that the heat of solution at 30% dilution, ie, -0.757kcal/mole and you get -45.6kcal/mole for the heat of formation. Your Solvey table gives -45.54kcal/mole."

My question still comes back to the value in the denominator. I still need the solution value expressed in kcal/g.

And I know the heat of formation of a 30w/70w peroxide/water solution does not equal (-45.54kcal/gmol)/34 or -1.339kcal/g
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[*] posted on 24-10-2008 at 16:29


Quote:

And I know the heat of formation of a 30w/70w peroxide/water solution does not equal (-45.54kcal/gmol)/34 or -1.339kcal/g


How do you know that? I think it does. What we are talking about here is just the H2O2. The heat of formation of water is not involved.
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[*] posted on 24-10-2008 at 18:40


I know this because when that value is plugged into a rocket propellant chemistry program (Propep), it gives a totally unrealistic heat of decomposition value Tch, gamma and Isp.

Adiabatic decomposition temperature (Tch) for a 30% peroxide solution should not be anywhere near 2000'F. We also know there isn't enough energy in a 30% solution to vaporize all the water. Therefore we know DeltaHf cannot be correct.

I know this much. If the propellant computational program is showing a Tch anywhere in the neighborhood of 2000'F, the DeltaHf input value is obviously wrong.

I'm looking for the heat of formation of a 30% solution (peroxide and water as a whole).
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[*] posted on 24-10-2008 at 20:42


You might just be forgetting the water here. Try mwt/mass fraction, then use the kcal/mol value.

Basically, 34.01/0.3 = 113.37, -45.54 kcal/mole/113.37 = -0.402.

What do you get using this and other fractions with this program? E.g. reported of 75% H2O2 is 117 Isp and 630 K combustion temperature.

[Edited on 24-10-2008 by Formatik]
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[*] posted on 24-10-2008 at 21:01


Quote:

I'm looking for the heat of formation of a 30% solution (peroxide and water as a whole).


That has to be where the disconnect is. This is not the way the general literature presents the delta Hf for 30% H2O2, so I didn't understand what you really wanted.

Here's what I would calculate: We know that for the H2O2 component of the 30% solution the deltaHf = -1.339kcal/g, and for water the deltaHf = -3.792 kcal/g. So, for a 30 wt% H2O2 solution the deltaHf = (0.3)(-1.339) + (0.70)(-3.792) = -0.402 + (-2.654) = -3.056 kcal/g.

Does that give a reasonable answer from your program?
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