# Sodium Bicarbonate Thermal Decomposition

1. Jan 31, 2016

### Robert Flynn

I have a question about the thermal decomposition of sodium bicarbonate (baking soda). I need to use baking soda for an experiment but will not be using a burner to heat it. We are heating it to about 90-100C and I was wondering if we put vessel the baking soda is in under vacuum -0.5bar or pressure about 2 bar would this speed up the reaction? If so which one? Sorry this is a bit of an odd question and the background to this question is long so I left it out but I was wondering if anyone had any insight

2. Jan 31, 2016

### Wee-Lamm

Vacuum will permit water to boil at room temperature, pressure would actually have a cooling effect.

3. Jan 31, 2016

### Staff: Mentor

I don't see water in the original setup.

Generally speaking the lower the pressure the faster the decomposition should be, especially if you can maintain that lower pressure during the reaction.

Not that I am convinced it will decompose fast enough at these temperatures, they can be too low.

4. Jan 31, 2016

### Robert Flynn

Thank you for the help I thought that vacuum would be preferable and will see more tomorrow when I can continue the experiment just wanted to get an idea of the direction that would work best. Maintaining a vacuum will not be an issue -0.5 bar or more we can easily achieve as well as the vessel is steam heated so we should not have an issue.

5. Jan 31, 2016

### Wee-Lamm

Heat can be used to break the bond between H and the O it adheres to?
I would expect this to be the same reaction that causes water to boil.

It is my understanding that decomposition occurs above 50C, where decomposition will be fast at 200C. 90 -100C seems fast enough to be noticeable yet slow enough to be measurable.

6. Jan 31, 2016

### Staff: Mentor

Can't say it makes much sense to me. We are pretty far from the thermal decomposition of water (which is what breaking the H-O bond is). Plus, I wouldn't consider boiling a reaction.

7. Jan 31, 2016

### Robert Flynn

Borek,

So if you were to take a guess vacuum would be the best option?

8. Jan 31, 2016

### Wee-Lamm

https://pubchem.ncbi.nlm.nih.gov/compound/bicarbonate#section=3D-Conformer
Bicarbonate consists of a latice of 3 oxygen surrounding 1 carbon, with 1 hydrogen attached to 1 of the oxygen.

https://pubchem.ncbi.nlm.nih.gov/compound/bicarbonate#section=Melting-Point
Shows Decomposition begins at 50C

https://pubchem.ncbi.nlm.nih.gov/compound/bicarbonate#section=Decomposition
Shows that when heated, it separates CO2 from HO.

Either way, if we heat HO or H2O, H is unbound from O, and releases H. If we happen to have at least 2 of these complex molecules, and I suspect there will be several hundred in the test sample, the result will certainly be liquid. In fact, HO is known to form strong bonds with water, because they are so nearly identical, save for the excess Hydrogen . Really, we are much closer to the thermal decomposition of water than we first imagined. :-)

9. Jan 31, 2016

### Wee-Lamm

10. Jan 31, 2016

### Robert Flynn

11. Jan 31, 2016

### Wee-Lamm

My reference to the boiling temperature of water.

We can presume that the internal pressure of the water molecules against other water molecules, should be the same as the surrounding air pressure. When the air pressure is lower than the water, some water on the surface will evaporate and rise into the atmosphere, in reaction to the difference in pressure. If we heat the water, Hydrogen separates from it's bond to Oxygen which means they now occupy more space, which will increase the pressure within the liquid, which results in accelerated evaporation.

Water inside a sealed container will evaporate at the same rate as that in open air, but the gaseous nature of the evaporate gives those molecules a larger surface area which allows them to cool faster, where they will return to a liquid state as other molecules from the surface evaporate. If we heat water in a sealed container, the same process occurs but our added heat impairs the dissipation of heat from the gaseous matter and to cannot condense quick enough and will exert pressure against the container, in it's effort to reach equilibrium with the surrounding atmosphere.

Now, if we instead draw a vacuum on the sealed container, we greatly reduce the atmospheric pressure against the surface of the water, which causes it to evaporate as though we had heated it. Slightly off topic, if we greatly increased the atmospheric pressure inside the sealed container sufficiently, the water will freeze though depending on the container, it may rupture before increasing the pressure sufficiently. :-)

The change of state of the bicarbonate will begin at 50C, anything hotter than 200C will have the change happen so quickly you will barely notice it, in the lab environment.
Instead of heating the container to 100C, which would have the added impact of greatly increasing the atmospheric pressure inside the sealed container and increased potential of outward rupturing of the container, we can instead draw a vacuum on it and greatly reduce it's internal atmospheric pressure to reach the internal temperature we desire, as witnessed in the evaporation of water.

To answer your last question then, we are using the effect of the change of state of the atmospheric pressure inside the sealed container, to create the environment required to allow the Bicarbonate to decompose at a rate fast enough to witness yet slow enough to measure.

12. Feb 1, 2016

### Staff: Mentor

OK, I see what you mean - but I am still confident you are wrong. I have a feeling you are mixing up personal ideas and made up facts, supported here and there but some true fact taken out of context.

No, I don't expect to see the liquid water, at least not in quantities that would matter. While we are well below the boiling point, even at 50°C pressure of water vapor is high enough for the produced water to become gaseous instantly. Yes, in increased pressures it can look differently, but we are not talking about increased pressures, we are talking about heating in vacuum.

Thermal decomposition of water requires high temperatures. You need above 2000°C for an observable fraction of H2O to decompose (compare http://www.eolss.net/sample-chapters/c08/e3-13-03-01.pdf). O-H bond in the hydrogencarbonate and O-H bond in water molecule are not easily comparable.

Nobody ever talked about a sealed container, please don't make up facts, it doesn't help the thread and only confuses those involved.

And I still strongly doubt the decomposition rate of bicarbonate at 50°C will be large enough for practical purposes. Do you have a thermogravimetric curve? Thats the only sure way of seeing how it behaves on heating, simple statements like the one from pubchem carry no practical information here.

13. Feb 1, 2016

### Wee-Lamm

...
First off, if you read OP's opening question, they stated they were not using a burner to heat the Bicarbonate. They then asked, "if they put it in a vessel should they use a vacuum or positive pressure to speed up the process of decomposition." In order to maintain vacuum or positive pressure, the vessel would have to be sealed. I did not "Make anything up" on this point, it is the conditions of the OP's question.

On the matter of positive pressure, it would actually cause more than noticeable decrease in the temperature of the water. fwiw, I have more than a mere passing interest in refrigeration, and would certainly exchange opinions on cooling and superheat properties in another forum, we seem to be hijacking OP's thread.[/QUOTE]

Respectfully, I suggest you boil some water on the stove and see decomposition for yourself. The water will not get any hotter that 100C, but the steam certainly can. I am confident that the water will decompose completely, long before it reaches 2000. In fact, the Decomposition of the glass container is likely to occur long before 2000C, even Borosilicate glass has a melting temperature of 1648C

Beyond the fact that I was using water as a comparative item to explain the relationship of Negative pressure and Temperature, it has no relevance to the OP's question. But, It may be worth reading the links I posted above, https://pubchem.ncbi.nlm.nih.gov/compound/bicarbonate#section=Decomposition . It speaks to the process of decomposition of Bicarbonate and the residual products of doing so, which indicates that Water is one of the expected byproducts.

Again, OP's opening question said, "they are heating it to 90-100C." Perhaps it is coincidental that 100C is the boiling point of water and it may be equally coincidental that 100C falls in the middle of the temperature range suitable for observing the Decomposition of Bicarbonate, but the design of the exercise seems a little less arbitrary than that.

Last edited: Feb 1, 2016
14. Feb 1, 2016

### Staff: Mentor

No. To maintain vacuum you need to connect the vessel to a vacuum pump, and in a typical lab setup you don't remove the gases and seal the vessel. Quite the opposite, you keep the vessel connected to the vacuum pump all the time. That in turn means gases produced in decomposition are sucked out, leaving only solid in the vessel.

You don't see decomposition in such conditions, all you you see is evaporation. No idea where your claims come from, but they are seriously wrong.

Source for this claim please, one that is based on research and thermodynamics. Wikipedia claims at 2200°C only about 3% of water is decomposed, I don't conssder wiki a sure source but this is more or less in line with results presented in the pdf I linked to earlier, so sounds good to me.

Water being a product and water being present as a liquid are two different things.

15. Feb 1, 2016

### Wee-Lamm

A vacuum condition in a vessel exists, only when there is more atmospheric pressure being removed from the vessel than being permitted to enter the vessel. I won't pretend to claim that a mere rubber stopper is sufficient to create a true "sealed system", but certainly the rubber stopper would be more efficient than the open top of the vessel. As proof of concept, see especially definition noun 2, http://dictionary.reference.com/browse/vacuum?s=t.

In an attempt to reduce the amount of 'Bit Twiddling", perhaps "Evacuated System" would be more accurate than "Sealed System". Still, this dilutes my answer to OP's initial question.

OP is studying the decomposition of "Sodium Bicarbonate". OP further asked if using a 'vacuum or pressure' inside the vessel, would speed up the reaction of raising the temperature inside the vessel, in the absence of an external heat source. I think we at least agree that a vacuum will speed up the reaction whereas an increase in pressure would slow down the reaction.

Initially, my comparison to the boiling temperature of water at different atmospheric pressures, was to allow OP to see the change in reaction by first applying the process to a substance they would be more familiar with. I think that most of us have boiled water at some point in our lives, which gives a common reference on the expected outcome of process. However, for the purpose of answering OP's question; water does not have to decompose, it is sufficient to separate it from NaHCO3 through evaporation.

I will concede to your point on decomposition versus evaporation of Water, I had wrongly presumed evaporation was equal to decomposition, but it is sufficient for this exercise to remove it from the mixture, regardless of it's composition outside of the mixture.

http://www.tatachemicals.com/europe...rbonate/technical_thermogravimetricBicarb.pdf
http://antoine.frostburg.edu/chem/senese/101/inorganic/faq/carbonate-decomposition.shtml
http://repository.ias.ac.in/37727/1/37727.pdf
http://www.chemikinternational.com/pdf/2012/11_2012/11_12_CHEMIK _03.pdf
Even Wikipedia seems to agree, https://en.wikipedia.org/wiki/Sodium_bicarbonate#Thermal_decomposition

Whether water is a merely a product or is present as a liquid, it can still be removed through dehydration.

16. Feb 1, 2016

### Robert Flynn

Ok I was able today to try to conduct part of the experiment. I was not able to use a vacuum as our vacuum system was not working. I started out with 4g of sodium bicarb for both the pressure and "control" no pressure no vacuum step. I kept everything consistent

Control- placed in vessel for 15 mins at 98C. I had a 0.01g change in weight

Pressure- placed in vessel for 15 mins. The first 5 mins (no pressure) at 98c then 10 mins under pressure with max pressure being 1.5 bar. The pressure step does involve heated air with a temp of 90C if it helps we injected the air at about 1.25 Nm3/h. I had a change of about 0.15g (I can't recall the exact number but it was a lot lot more than the control type)

Vacuum- Could not conduct will do later as well as redo pressure and control trials

My only question is realistically I expected this to be much closer to the "control" than it was. The physical appearance afterwards was much more clumpy like was some water vapor did not escape the container that the baking soda was in. It does not appear that I lost any of the sample. Before I even posted this question I thought I recalled reading about pressure speeding up this reaction but I can not seem to find it or 100% recall which way speed up the reaction. The only this I could find is this link does this have any merit on the possible findings? I really was not expecting this at all

http://www.chemguide.co.uk/physical/basicrates/pressure.html

17. Feb 2, 2016

### Staff: Mentor

Strange.

Total pressure should be less important than partial pressures of CO2 and H2O.

Under vacuum you are removing decomposition products, thus shifting the equilibrium to the right, and speeding up the decomposition. This is just LeChatelier's principle at work.

When you introduce additional gas to increase the pressure you are most likely changing the humidity (ie partial pressure of water) and amount of CO2 present, you are also changing the speed with which the heat is distributed in the vessel. Thus, having a different result is not strange - but the difference is much larger than what I would expect and is counterintuitive.

That all being said, there are more things in chemistry, than are dreamt of in our philosophy. Exclusions and specific cases are abundant. Do you have access to the primary literature? Google search is not the best tool to access it. Librarians can be able to help locate papers on the subject. As I wrote earlier, first thing to check is to find thermogravimetric curve, it should tell you what to expect in general. Actually, NaHCO3 is so common substance it should be easy to find even using just google, without searching paid databases. Sorry, no time to try.

18. Feb 2, 2016

### Wee-Lamm

When you redo this test and include the vacuum scenario, you will find that the result of the vacuum stage will have a much more profound effect on the evaporation rate of the water in the mixture, both because the water vapor will be evacuated from the container and also because there will be less atmosphere in the container to absorb heat.

Consider first, an open top container sitting on your lab table. The atmospheric pressure on the inside of the container will be the same as the pressure on the outside of the container. Effectively, the temperature of the air inside should also be equal to the outside, save for things that may interact with the material of the container such as sunlight, moving air in the environment, but these differences will be negligible if they are present at all.

The link you provided speaks to the interaction of gases and shows that increasing the pressure on gases encourages them to clump together which in turn concentrates their combined reactions within a more confined space. Increasing the atmospheric pressure inside your vessel is comparable to shrinking the internal size of your container. This link does have a small impact on what you experienced in that there is some water present in Bicarbonate that is being evaporated, which is the conversion of a liquid to a gas.

Consider also the difference between liquid and gaseous states. In a liquid form, like molecules are more likely to clump together such that portions of their surface area will be touching neighboring like chemicals and not the surrounding atmosphere. In gas state there is more surface area exposed to atmosphere, which provides a larger surface area to allow the transfer of heat.

In this regard, applying heated pressure as you did in your experiment would have an increase on the reaction inside the container largely because of the heat, but this would be largely focused to areas where water had already evaporated. This in turn would accelerate the evaporation rate of the newly exposed surface areas of water but this would have an upward limit of acceleration related to the rate of condensation as the gaseous water would also cool more rapidly.

Consider now what you can expect to see when drawing a vacuum on your vessel, where you are removing the atmosphere. When you then apply a heat source, the heat will be absorbed by what is left in the vessel, in the same way it was in the other 2 examples, except there is simply less material there to absorb the heat. In this then, each surface will absorb more heat per area where the most of the area is the sodium bicarbonate. This will largely focus on areas where water is already evaporated, which is a larger percentage of what is in the vessel, as compared to the other examples. Drawing the vacuum has the added benefit of removing the evaporated water, which will also remove some of the heat but this will be negligible in comparison to the heat source.

Thank you for following up with your results so far, I look forward to seeing the results of your vacuum test as well. :-)

19. Feb 4, 2016

### Robert Flynn

Ok so it looks like pressure was actually more beneficial than vacuum for our process. Vacuum would be beneficial if the vessel was not so large for our process and the lack of intimate contact with the heating jacket seems to slow the process. Thank you for your help!

20. Feb 4, 2016

### Wee-Lamm

Not exactly.

The heat would be absorbed by all matter in the vessel, including atmosphere.

By increasing the pressure, all matter is crammed together more tightly which would concentrate the pockets of heat in more confined regions. This would somewhat increase the apparent effect of the additional heat you added, which you stated was at 90C, so that it occurred in less time but at the same temperature as you experienced in the first test.

A vacuum, on the other hand, would greatly reduce the pockets of atmosphere and the heat would then be absorbed by any matter that was still in the vessel. This results in the evaporation of water in about the same length of time, but at a much lower temperature. Heating to 90C while under a vacuum would cause speed up evaporation even more dramatically, and would be much faster than both previous tests.

Merely to define the comparison and not intended as actual time, if heated with no pressure took 15 minutes to decompose, and heated with positive pressure took 14 minutes to decompose, then half heated under vacuum could take 11 minutes to decompose and fully heated under vacuum could take 8 minutes to decompose but could also include unneeded heat which may serve to impact the integrity of the vessel.

Do answer back with your results after performing the test under vacuum, I suspect heating temperature will be closer to 50C than 90C. :-)