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Topic: Calcium Oxide Decomposition  (Read 33641 times)

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Offline Needaask

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Re: Calcium Oxide Decomposition
« Reply #15 on: March 17, 2013, 01:20:35 PM »
Have you ever seen heated sugar caramelizing?

Yup :) just watched a youtube video on it. I think its the process of the melting then boiling and decomposition? I'm not very sure about this though. So is it something like initially the intermolecular forces are broken (melting/boiling) then intramolecular forces? So during the the breaking of the intermolecular forces, all the energy is put into breaking those bonds only?

But I think there are covalent molecules that can decompose without melting or boiling first?

And thanks so much for the help :)

Offline Borek

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Re: Calcium Oxide Decomposition
« Reply #16 on: March 17, 2013, 03:57:24 PM »
Actually sucrose starts to decompose BEFORE it melts.
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Offline Needaask

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Re: Calcium Oxide Decomposition
« Reply #17 on: March 17, 2013, 10:08:40 PM »
So in this case the intramolecular bonds start to break before the intermolecular bonds breaks? Are there also cases where me,ting or boiling occurs before?

Offline Borek

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Re: Calcium Oxide Decomposition
« Reply #18 on: March 18, 2013, 04:57:22 AM »
So in this case the intramolecular bonds start to break before the intermolecular bonds breaks?

Yes.

Quote
Are there also cases where me,ting or boiling occurs before?

Isn't it the standard situation? Melting first, then boiling, decomposition much later.
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Offline Needaask

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Re: Calcium Oxide Decomposition
« Reply #19 on: March 18, 2013, 08:28:21 AM »
So in this case the intramolecular bonds start to break before the intermolecular bonds breaks?

Yes.

Quote
Are there also cases where me,ting or boiling occurs before?

Isn't it the standard situation? Melting first, then boiling, decomposition much later.

Oh but sucrose decomposes before melting or boiling?

Offline Borek

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Re: Calcium Oxide Decomposition
« Reply #20 on: March 18, 2013, 08:35:43 AM »
Oh but sucrose decomposes before melting or boiling?

Yes.

Now get back to this question:

What may cause the compound to decompose before melting or boiling?
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Offline Needaask

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Re: Calcium Oxide Decomposition
« Reply #21 on: March 18, 2013, 01:00:33 PM »
Oh but sucrose decomposes before melting or boiling?

Yes.

Now get back to this question:

What may cause the compound to decompose before melting or boiling?

I'm thinking when i heat something up even the bonds in the molecule itself starts to weaken just that in most cases its not enough to break any of those bonds so the intermolecular bonds break before those intramolecular bonds break? So I guess that if those intramolecular forces are extremely weak then decomposition occurs first?

I don't think my answer is right though.. it makes no sense for a covalent bond to be weaker than the intermolecular forces..

Thanks so much for the replies :)

Offline Borek

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Re: Calcium Oxide Decomposition
« Reply #22 on: March 18, 2013, 03:15:08 PM »
I'm thinking when i heat something up even the bonds in the molecule itself starts to weaken just that in most cases its not enough to break any of those bonds so the intermolecular bonds break before those intramolecular bonds break? So I guess that if those intramolecular forces are extremely weak then decomposition occurs first?

I don't think my answer is right though.. it makes no sense for a covalent bond to be weaker than the intermolecular forces.

But that's the way it happens.

In the case of small molecules it is rarely a problem, but in the case of large molecules it is not uncommon. In the case of heated sugars decomposition quite often happens before melting. I suppose Dan would be able to give much more precise answer, but the way I see it you have plenty of strongly interacting hydroxyl groups, and you are pretty close from them leaving the carbon skeleton and creating simple water molecules - which are a great, stable product, plus they nicely increase the entropy - so the decomposition is pretty likely.
« Last Edit: March 18, 2013, 03:28:30 PM by Borek »
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Offline Needaask

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Re: Calcium Oxide Decomposition
« Reply #23 on: March 19, 2013, 02:24:09 AM »
But that's the way it happens.

In the case of small molecules it is rarely a problem, but in the case of large molecules it is not uncommon. In the case of heated sugars decomposition quite often happens before melting. I suppose Dan would be able to give much more precise answer, but the way I see it you have plenty of strongly interacting hydroxyl groups, and you are pretty close from them leaving the carbon skeleton and creating simple water molecules - which are a great, stable product, plus they nicely increase the entropy - so the decomposition is pretty likely.
[/quote]

Oh wow that's really cool. Is also due to the large number of electrons so the induced dipole-induced dipole interactions are stronger than those bonds? Also, was I right to say that when I heat an object both of its bonds inter and intra are being overcomed just that usually it's not enough to overcome the intra bonds?

Thanks so much for the help here :) if possible could you go through ionic substances as well?

Also, I have a general question for endothermic reactions such as decomposition. Why would the compounds prefer to form new substances after those bonds are broken, rather than to return back to their initial states? Since their initial states have a lower energy level, shouldn't it be better for those bonds to reform rather than to form new bonds?
« Last Edit: March 10, 2016, 08:33:16 AM by Arkcon »

Offline Corribus

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Re: Calcium Oxide Decomposition
« Reply #24 on: March 19, 2013, 10:09:01 AM »
Oh but sucrose decomposes before melting or boiling?

This depends a lot on the sugar.  Glucose has a fairly low melting point and can be melted easily before caramelization.  Sucrose is harder because it has a higher melting point which is close to the caramelization point.  Some people will tell you that the process of melting and caramelization of sucrose are inseparable, that the temperatures are so close that you can't melt without beginning decomposition.  Others will tell you that the melting point of sucrose depends a lot on the rate that you heat it and that if you heat it carefully, you can melt it before it caramelizes because the melting point is still technically lower than the decomposition point.  It doesn't take too long with a search engine to see this confusion first hand - look up the melting point of sucrose in 10 places, you'll get 10 different answers.  One I found lists the melting point of granulated table sugar as about 320 F and the caramelization point at about 338 F (link below).  Wikipedia lists the caramelization point at 160 C and the melting point at 186 C.  Which is right?

One thing is for certain: no matter what the relative melting points and caramelization point, over a kitchen burner there's not a whole lot of difference.  In a lab environment it may be easier to first melt, then caramelize.  Even so, you can still try this at home if you want by putting table sugar in a pan and heating gently.  You'll see the sugar turn into a thick clear syrup first and then gradually turn brown (then black).  Many people will call the former melting and the latter caramelization ("decomposition").  But regardless of terminiology, if you remove the heat from "melted" sucrose before any browning is observed, it has been shown (see reference below) by DSC that the cooled substance is no longer sucrose.  Which means it didn't just melt - it also decomposed.

In any case, in common parlance you'll see people (chefs) refer to "melting" sugar first and then caramelizing.  In reality, these processes seem to be impossible to separate, even when using highly sensitive heating instrumentation like DSC.  Small chance a chef can do it in a pan over a hot flame!  Most people when making caramel do add a little bit of water to the sugar before caramelizing it - maybe a tablespoon or two - which helps even out the heat distribution, preventing the reaction from happening unevenly.  I.e., preventing burning the sugar and making bad caramel.  But no matter what, when the sugar liquifies, it's more than just melting, so remember that the next time a chef tells you to melt sugar and then caramelize it.

Here's a nice website about the impact of heat on carbohydrates from a culinary point of view, which does refer to melting and caramelization as separate processes:

http://chefsblade.monster.com/training/articles/215-food-science-basics-effects-of-heat-on-starches-and-sugars

Here's the reference I mentioned about the irreversible "melting" of sucrose:

Joo Won Lee, Leonard C. Thomas, Shelly J. Schmidt. "Can the Thermodynamic Melting Temperature of Sucrose, Glucose, and Fructose Be Measured Using Rapid-Scanning Differential Scanning Calorimetry (DSC)". Journal of Agricultural and Food Chemistry, 2011; 59 (7): 3306.

Here's a popular science article about that reference, in case you find it easier to read:

http://voices.yahoo.com/scientists-discover-why-table-sugar-always-melts-at-10495765.html

I think the take home message here is that caramelization reactions (like other reactions in cooking like Maillard) aren't particularly well understood, which is why cooking is still more an art than a science. ;)
What men are poets who can speak of Jupiter if he were like a man, but if he is an immense spinning sphere of methane and ammonia must be silent?  - Richard P. Feynman

Offline Needaask

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Re: Calcium Oxide Decomposition
« Reply #25 on: March 19, 2013, 10:58:55 AM »
Oh that was very helpful (by any chance are you also a chef haha). But now I feel a little lost over what the heat does. Does it break all types of bonds (or try to) so for example in that sucrose case, 100J of heat is required to melt it. So that 100J is used both to break those intermolecular bonds as well as those intramolecular bonds? So if 130J of heat is required to actually decompose it, we'd expect it to be melted since heat is 'impartial' and 'chooses' to break everything?

is that the right concept here? Also, going back to the previous question, why would the sucrose start to form new bonds at 338F/186°C? Why won't it continue to boil and once we stop the heating process, go back and condense and solidify to its own initial energy level? What would make the sucrose/any endothermic reaction choose to form new bonds if we continue heating it and go to a higher energy level making it more unstable? Or would the reaction proceed if we actually continuously heat the substance?

Thanks for the help :)
« Last Edit: March 10, 2016, 08:34:59 AM by Arkcon »

Offline Corribus

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Re: Calcium Oxide Decomposition
« Reply #26 on: March 19, 2013, 12:30:21 PM »
Heat, unlike light, is an indiscriminate agent of chemical and physical change.  It feeds into all available molecular states with no preference because it is basically just a collective manifestation of a body of vibrating, rotating molecules moving with a distribution of speeds and bumping into each other with high frequency.  If two reaction pathways have exactly the same energetic barrier to get to their respective products, then they will occur with equal rates.  (Well, not really - there are other things that impact kinetics of a reaction.  But if we pretend the rate only depends in enthalpy, and we assume both reactions are mostly irreversible...) 

I mean, consider some hypothetical molecule A-B-C, which can either decompose to form AB + C or A + BC, and the "strength" of the bonds A-B and B-C are identical.  Heat - transfer of kinetic or vibratory motions - is just as likely to feed energy into the A-B bond as it is the B-C bond, and because these bonds take exactly the same amount of energy to break, the decomposition of A-B-C would tend to yield equivalent amounts (all things being equal) of AB + C and A + BC products.  If one bond is more stable than the other, the reaction which requires breaking the less stable bond will be favored.  That doesn't mean you won't see ANY of the less favored product.  All these concepts are subject to statistical distributions of energies, and all molecular processes are, at their heart, a matter of probability. 

(The A-B-C reactions described above are actually very far from reality - in any molecule like A-B-C, the vibration of A-B and B-C are not independent of each other, so this kind of dichotomous reaction mechanism would be virtually impossible.  In other words, any heat that goes into A-B actually goes into B-C as well, and vice-versa.  Heat goes into vibrational modes, which involve all atoms/bonds in a molecule, not into individual vibrating bonds.  In A-B-C, you'd have some kind of symmetric vibrational mode, an asymmetric mode and a bending mode, for example.  But the point stands - in real reaction dynamics, every chemical reaction proceeds in some way through one (or more) unique vibrational modes.  Feeding energy into other vibrational modes cannot yield the reaction (although they could yield other reactions), but heat is fed into all vibrational modes, so it's a pretty inefficient way to make reactions go.  Light, on the other hand, can be used to excite one vibrational mode without exciting others, so you can in principle drive a chemical reaction with light a lot more efficiently than you can with heat, because it provides a way to selectively put energy into reactive vibrational modes.  )

Intermolecular bonds are significantly weaker than covalent bonds, but they operate under the same kinds of rules.  A molecule like water has a high specific heat capacity (takes a lot of energy to heat it up) because there are lots of intermolecular bonds that can absorb that energy.  This is also the primary reason why water has a pretty high boiling point despite its low mass.  (Compare to methane - almost the same mass, far lower melting and boiling point.)  In physical chemistry lingo we call this factor - the amount of bonds (actually vibrational modes) and such that can absorb energy - "degrees of freedom".

Having said all that, you have to realize that the chemical decomposition of sucrose and the melting of sucrose are two nominally independent processes that happen to require around the same amount of heat to make happen.  The bonds between atoms in sucrose are strong, but so is the crystal lattice energy.  The reason you can't just melt sucrose and then cool it down and have sucrose again (unlike, say, water) is because while that much heat is sufficient to break down the crystal lattice of sucrose to form a liquid, it's also enough heat to excite internal sucrose vibrational modes to the extent that some of them break as well.  However I'll point out that as I kind of did before that the favorability of a reaction is not only dependent on having enough energy to break bonds - it's also on the stability of the products formed.  You may very well have enough energy to break bonds, but if the products aren't particularly stable, that same heat can get them to react back the other way almost instantly, which is basically just a way of saying that the equilibrium favors the reactants.  In the case of sucrose, though, the products of caramelization are probably various polymers and so forth which are pretty stable, so it makes sense that the reaction is favorable.  And thank the heavens for that, else we'd not have yummy caramel to put on top of our ice cream. :)

Anyway, that was kind of a mess of a response which threw around a lot of concepts in no particular order.  I hope I answered your question without confusing you too much.

ps.  Not a chef, but I like to cook.  There's an incredible amount of chemistry that goes on in cooking, and virtually everyone loves food, so I always view cooking examples as a great way to engage students and teach them what would otherwise be boring chemical concepts.
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Offline Big-Daddy

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Re: Calcium Oxide Decomposition
« Reply #27 on: March 19, 2013, 03:18:27 PM »
Does the caramel have any reformed covalent bonds, or have covalent bonds only been broken by the heat? And are the same intermolecular forces in place, or do different ones operate? (For the specific case of sugar caramelizing, because sucrose is I think in a covalent lattice and so it's interesting to note whether we'll get different lattices forming or not if we let the caramel solidify - if we do, wouldn't that suggest we'd broken the "intermolecular" bonds as well, i.e. the bonds between the molecules, of sucrose, to form the caramel with new and different bonds in the lattice?)

Offline Needaask

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Re: Calcium Oxide Decomposition
« Reply #28 on: March 20, 2013, 04:53:31 AM »
Hi thanks for the great explanation.

So is it when I apply 100J of heat on an object, and the object requires 100J of energy for its covalent bonds to break and 100J for its isn't intermolecular bonds to break, will both bonds break.

However, I have a query about why would the products form in that manner. In both cases (the A-B-C and sucrose one and probably all other decomposition reactions) when the bonds are broken new bonds are formed. However, shouldn't those reactants reform to their original states since that'll mean that there is no gain in energy making them even more unstable than usual?

Thanks so much for the help :) and i agree food draws people together haha
« Last Edit: March 10, 2016, 08:35:32 AM by Arkcon »

Offline Needaask

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Re: Calcium Oxide Decomposition
« Reply #29 on: March 20, 2013, 05:16:21 AM »
Does the caramel have any reformed covalent bonds, or have covalent bonds only been broken by the heat? And are the same intermolecular forces in place, or do different ones operate? (For the specific case of sugar caramelizing, because sucrose is I think in a covalent lattice and so it's interesting to note whether we'll get different lattices forming or not if we let the caramel solidify - if we do, wouldn't that suggest we'd broken the "intermolecular" bonds as well, i.e. the bonds between the molecules, of sucrose, to form the caramel with new and different bonds in the lattice?)

I'd think if you could perfectly heat sucrose, it would first melt (breaking the intermolecular bonds partially) then in that perfect situation if we cooled it it would solidify to the same product. But as explained by Corribus heat is indiscriminate and that same heat is also breaking the intramolecular covalent bonds. So if we increased the temperature of it by a little in that perfect situation those covalent bonds would break as well. But I'm not sure if new bonds would form if i continue to heat the sucrose now. Would it remain as broken apart atoms? Or would new bonds immediately form despite the heat?

But I think your question is about having a substance decompose without melting (right?). So it's weird cos the intermolecular forces aren't broken by the heat (its not enough to destroy that but its enough to destroy the covalent bonds) and the act of breaking those covalent bonds would mean that those intermolecular forces would have to change as well. So I'd guess that even though the heat is not enough to destroy the van der waals forces by destroying what causes those van der waals causes a change in the van der waals? I'm not sure though..

I think that in that case the covalent bonds are broken first and new smaller molecules are formed thereby causing the induced dipole-induced dipole to become weakened. But it might also cause the other intermolecular forces to become stronger as well (beside id-id forces there are still other kinds of intermolecular forces going on). So I'd guess the act of breaking those covalent bonds causes the change in the intermolcular bonds of the system now. But the change could make the bonds stronger or weaker and that would depend on the new substances being formed.

However, if no new bonds are being formed so only atoms lie around. Then the as a result of the covalent bond breaking the intermolecular forces (can we even it that anymore here?) are weakened because having only atoms there can only be induced dipole forces going on.

But I'm not sure about this. I think someone with greater knowledge should be better here..

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