Chemical Forums
Chemistry Forums for Students => High School Chemistry Forum => Topic started by: Diamonds on March 10, 2013, 12:01:07 AM
-
Why does CaO not decompose into Ca and O2? I have been looking all over the web and can't find an appropriate equation that represents or explains this. Thanks in advance!
-
I'm new to chemistry but I think I may know what's going on.
There appears to be an equal number of moles on either side when CaO decomposes into Ca and O, so why would CaO decompose into Ca and O2? Naturally oxygen is found as O2, but this is a spontaneous reaction.
But perhaps you're referring to Ca^2+? That would look like Ca(O)2, though.. But then again, I'm not sure.
-
Sorry for the misunderstanding. This isn't a stoichiometry question.
I would expect the reaction to be
2CaO :rarrow: 2Ca + O2, but it is not so. I am asking a question about the actual reaction, not the coefficients.
-
Why should it decompose in the first place, it is much more stable than the mixture of Ca and O2.
-
Why does CaO not decompose into Ca and O2?
Who says it doesn't? Also, why should it? As Borek: asked. How easy did you need it decomposed?
I have been looking all over the web
You should try harder. http://en.wikipedia.org/wiki/Calcium#History
and can't find an appropriate equation that represents or explains this. Thanks in advance!
Since you wrote a balanced chemical equation for Adam2126:, I'm guessing you didn't mean this. But what did you mean?
-
I am sure that, under enough heating, either CaO will liquify/sublime or will decompose into Ca and O. Probably the former, considering that oxides tend to form lattices with very high dissociation enthalpy.
But that's besides the main point. Most plain oxides do not decompose under Bunsen burner temperatures (around 1000 K).
-
I am sure that, under enough heating, either CaO will liquify/sublime or will decompose into Ca and O. Probably the former, considering that oxides tend to form lattices with very high dissociation enthalpy.
A fair guess, Big-Daddy:, but for this specific case, not likely at any reasonable temperature. http://en.wikipedia.org/wiki/Refractory#Refractory_materials
But that's besides the main point. Most plain oxides do not decompose under Bunsen burner temperatures (around 1000 K).
Plain ones being the ones cited above, then, yeah.
-
I can tell you already the reaction is not 2CaO --> 2Ca + O2
Why may you ask? I actually took a practice Olympiad today and one of the questions asked which one does not form O2 when it decomposes? The answer was CaO.
The specific question was "Oxygen gas can be produced by the decomposition of all of the following substances EXCEPT:
A) Calcium oxide
B) Hydrogen peroxide
C) Mercury(II) Oxide
D) Ozone
The answer was A
But as others have said, Calcium Oxide will not decompose on its own. If anything it'll undergo hydration to form calcium hydroxide. But for a reason why it doesn't decompose into it, I'm not so sure.
-
As it was signaled several times, it doesn't decompose because - for thermodynamical reasons - it is pretty stable.
Every substance will decompose if heated strong enough. But if something requires heating to very high temperature, products don't have to be these we know from the STP - and for example at high temperatures oxygen dissociates to the atomic gas.
-
As it was signaled several times, it doesn't decompose because - for thermodynamical reasons - it is pretty stable.
Every substance will decompose if heated strong enough. But if something requires heating to very high temperature, products don't have to be these we know from the STP - and for example at high temperatures oxygen dissociates to the atomic gas.
Hi Borek, from this question I was thinking about decomposition and melting. Both involves the process of bond breaking but the former requires new bonds to be formed as well. So why do some compounds melt and others decompose? Or do some compounds melt then decompose? Then how does boiling come in since that's similar to lattice energy already?
Thanks so much :)
-
Both involves the process of bond breaking but the former requires new bonds to be formed as well. So why do some compounds melt and others decompose? Or do some compounds melt then decompose? Then how does boiling come in since that's similar to lattice energy already?
Start with covalent compounds.
Which bonds break during melting?
Which bonds break during decomposition?
Which bonds break during boiling?
What may cause the compound to decompose before melting or boiling?
-
Both involves the process of bond breaking but the former requires new bonds to be formed as well. So why do some compounds melt and others decompose? Or do some compounds melt then decompose? Then how does boiling come in since that's similar to lattice energy already?
Start with covalent compounds.
Which bonds break during melting?
Which bonds break during decomposition?
Which bonds break during boiling?
What may cause the compound to decompose before melting or boiling?
You mean for simple covalent molecules right?
1) some the intermolecular forces of attraction breaks
2) both the intermolecular forces (not sure if all breaks) and the actual bonds break?
3) all the intermolecular forces of attraction breaks
4) I would guess that if the compound's bond strength is weaker than the bond strength? But in all cases the bond strength is way stronger than any intermolecular forces of attraction so actually thinking about it now, when I boil water is any energy put in to start breaking those O-H bonds or is all the energy put to destroy those good interaction there?
Greatly appreciate the help :)
-
bond strength is weaker than the bond strength?
Think in terms of INTERmolecular forces (bonds) and INTRAmolecular forces (bonds).
-
bond strength is weaker than the bond strength?
Think in terms of INTERmolecular forces (bonds) and INTRAmolecular forces (bonds).
Oops sorry I meant to say the intramolecular forces (the actual covalent bonds) would have to be weaker than the intermolecular forces (van der waals forces). But then again that's impossible right?
Thanks again :)
-
Have you ever seen heated sugar caramelizing?
-
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 :)
-
Actually sucrose starts to decompose BEFORE it melts.
-
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?
-
So in this case the intramolecular bonds start to break before the intermolecular bonds breaks?
Yes.
Are there also cases where me,ting or boiling occurs before?
Isn't it the standard situation? Melting first, then boiling, decomposition much later.
-
So in this case the intramolecular bonds start to break before the intermolecular bonds breaks?
Yes.
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?
-
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?
-
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 :)
-
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.
-
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?
-
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. ;)
-
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 :)
-
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.
-
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?)
-
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
-
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..
-
I guess I'm having a little difficulty understanding the questions being asked by both of you.
In response to this:
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.
The simple answer is yes, but not completely because there's not enough energy to do both. The true answer however is unfortunately that molecular systems are too complicated to describe so simply. There are statistical considerations related to the fact that not every bond is the same at any instant in time, even if they appear to be on paper, and you also have to keep in mind that while there's almost always an initial energy requirement to make any reaction go, the NET effect on heat can be either positive or negative. So even if you're putting 100 J of energy into a closed system, the amount of NET available energy to do reactions could be greater or less than this, depending on whether the individual pathways are exothermic or endothermic. In addition, most reactions are reversible to some degree (equilibrium) so you also have the "backward" reactions to worry about. In a closed system where you're putting in 100J, that equilibrium will be reached rather quickly, and in the simple system you've described, you'll have a little bit of both processes being activated by that heat. In an open system it becomes more complicated because your heat reservoir is a lot bigger.
-
hi :) I have 3 questions one sparked by Big-Daddy I'll make them clearer below
1) In the A-B-C case and all other decomposition or endothermic reactions, after those chemical bonds are broken (ionic/covalent) why would those intermediate particles choose to reform to more substances that are more unstable than before? Because why not return back to the same energy level in the same way solidification gives out energy, rather than to form newer products that are less stable than the reactants/
2) When i decompose something say sucrose. After all those covalent bonds are broken new products should form. But if i continue heating the substance up, would those bonds reform or would it linger around in its intermediate stage?
3) When i have a substance that decomposes without melting, it would imply that the total strength of the covalent bonds present in the substance is weaker that the intermolecular forces. So to put in numbers to it, A requires 50J to break its intramolecular bonds while 100J to break its intermolecular bonds. So when i apply 50J of heat to it, its forms new substances. So by forming new substances what would happen to those 100J required to break the intermolecular bonds? Or by the act of breaking those intra/covalent bonds that 100J is broken?
Thanks so much for the help :)
-
I could spend hours on these, but I'll keep it brief.
1) In the A-B-C case and all other decomposition or endothermic reactions, after those chemical bonds are broken (ionic/covalent) why would those intermediate particles choose to reform to more substances that are more unstable than before? Because why not return back to the same energy level in the same way solidification gives out energy, rather than to form newer products that are less stable than the reactants/
You have to remember that there is more determining reaction dynamics than just the relative enthalpies of the bonds in the products and reactants. There's entropy for one thing. But even beyond that, thermodynamics and kinetics are two wholly separate issues. A reaction can be very favorable thermodynamically but very slow kinetically. The transition state could be energetically unfavorable, but there are other factors involved as well. Rate is dependent on reactant concentration and it's also dependent on other things like the geometry of collision, spin states of the various species, etc. It's not always easy to predict why a reaction that is thermodynamically favorable may be sluggish.
In the case of a decomposition reaction: even if it's endothermic and "thermodynamically unfavorable", if there's enough heat energy around to surmound the activation energy, it will happen relatively fast because there's only one reactant. No collision may be necessary. However the reverse direction (combination of two decomposition products to reform the reactant) will be required to recollide before they can react. This will be slow, especially in the beginning when there aren't many around. Eventually an equilibrium will be established, assuming a closed system is involved, but decomposition products are often themselves unstable and so there may be other reactions involved. So you see, it's not always easy to reform the reactant even if it would be energetically unfavorable to do so.
2) When i decompose something say sucrose. After all those covalent bonds are broken new products should form. But if i continue heating the substance up, would those bonds reform or would it linger around in its intermediate stage?
They might if it was an isolated reaction and it was thermodynamically favorable. But if you look at any textbook which shows decomposition of a sugar to form caramel, there can dozens of reactions involved. It's not a simple A --> B reaction.
3) When i have a substance that decomposes without melting, it would imply that the total strength of the covalent bonds present in the substance is weaker that the intermolecular forces. So to put in numbers to it, A requires 50J to break its intramolecular bonds while 100J to break its intermolecular bonds. So when i apply 50J of heat to it, its forms new substances. So by forming new substances what would happen to those 100J required to break the intermolecular bonds? Or by the act of breaking those intra/covalent bonds that 100J is broken?
Well everything happens simultaneously - it's not "this then that". It's also kind of strange to be speaking of putting a finite amount of heat energy into a system. Typically we bring a body to a specific temperature - there is a large heat reservoir involved. Reactions that can happen with that amount of latent energy around will happen. Those that don't won't. But ANY temperature (other than absolute zero) will have a heat reservoir, so reactions are ALWAYS happening, even if the probability is low. Raising the temperature doesn't make a single reaction go so much as it shifts equilibria around.
-
Ohh is it possible to explain the 2 different factors for question 1)? I always thought that the energy level explained the tendency for it to form. Like if it has a higher energy level, being more unstable reactants would react and go down another level.
2) Oh what do you mean by isolated and thermodynamically favorable? Also, if this were a perfect situation with infinite amounts of heat shouldn't the energy level keep rising such that they are unable to cool down and give out energy? Or are these 2 process (heating of the the substance and its own reaction) independent of each other. Something like despite the heat being put into the system, it doesn't matter as the product will still form and the additional heat does not factor in that 'giving out of heat/exothermic process'?
I think I'm associating this with the idea of melting whereby if i increase the temperature above its melting point, it would continue to remain in its molten stage and would only reform if i lower the temperature of the molten substance.
3) Oh so its like i heat an object with an unlimited supply of heat energy just at a different temperature. So in the case where the total strength of all the covalent bonds is weaker than the intermolecular forces, by heating it at a constant rate, the covalent bonds would break even before the intermolecular forces are broken. So as a result of that I'm quite confused as this would suggest that the intermolecular forces are 'broken' even when i don't have enough energy to break those intermolecular forces.
Thanks so much Corribus for all the help given :) I really appreciate it :D
-
I always thought that the energy level explained the tendency for it to form.
It does, but the rate is different. It's kind of like if I'm sitting on the couch and would really like a soda, but it make take me awhile to work up the energy to go to the fridge and get one. And what then? Well I'm standing there in the kitchen, which is a high energy state. I'd much rather be back on the couch watching TV. That would be my low energy state. So you might be inclined to say that I ALWAYS go back to the couch after I get up to get my soda from the fridge, because the couch is a lower energy state than the kitchen. And all things being equal that'd probably be the case. But then the wife appears and tells me I gotta take out the trash, and go to the grocery store, and do all the other billions of "high energy" chores I gotta do, and so the end result is that I never make it back to the couch, even though it's the most thermodynamically favorable state for me to be in.
The moral? We'd all be better off living in closed systems. That's why doors with locks were invented. :D
2) Oh what do you mean by isolated and thermodynamically favorable?
I mean, if this was the only reaction we had to worry about, and it was thermodynamically favorable for products to recombine to re-form the original reactants, then it would happen. But often other reactions can happen, and so products don't always recombine to re-form original reactants. There are competiting pathways. And the fastest rate will dominate.
Also, if this were a perfect situation with infinite amounts of heat shouldn't the energy level keep rising such that they are unable to cool down and give out energy?
I don't mean "infinite heat". I mean the temperature is held constant such that the amount of available energy is kept constant. Consider an endothermic reaction. For every molecule that is converted, some available heat is absorbed from the system. This will lower the temperature of the system, which will slow down the reaction. Eventually the temperature will get cool enough that there isn't enough heat left to drive more conversions. This is an equilibrium of sorts. (This basically happens when we sweat.)
In a lab, though, or in a kitchen, we use a heat source (a hot plate or stove) which maintains a constant temperature, even if heat is being consumed by the reaction. This allows us to control the point of equilibrium and drive a reaction to favor productive formation, even if it's endothermic. Then we remove the heat source and now maybe there isn't enough energy to surmount the activation energy for the reverse reaction. So we've completed an "unfavorable" endothermic chemical conversion, and we don't have to worry about the thermodynamically "favorable" reverse conversion because it would be kinetically slow once the heat source has been removed.
3) Oh so its like i heat an object with an unlimited supply of heat energy just at a different temperature. So in the case where the total strength of all the covalent bonds is weaker than the intermolecular forces, by heating it at a constant rate, the covalent bonds would break even before the intermolecular forces are broken. So as a result of that I'm quite confused as this would suggest that the intermolecular forces are 'broken' even when i don't have enough energy to break those intermolecular forces.
Maybe my prior answer will help with this one, too.
-
Hi thanks so much for all the help :) hope I'm not being too draggy here :)
1) That helps a lot haha :) so in this case the couch represents the lower energy while the kitchen represents the higher energy. But what would your wife represent here? If i were to use the decomposition of MgCO3 in this case, MgCO3 would break into Mg2+ CO2 and O2- so in the intermediate stage. So this would prefer to go back to form MgCO3 as it's energetically favorable. However, when i cool down this solution the 'wife' would cause the particles to reform as MgO and CO2 rather than MgCO3. So what is the 'wife' that doing this to the intermediate stage particles?
2) Hmm I'm not very sure what you mean. Because i'm thinking that when i heat up the decomposing substance, bonds would be broken. So now that all the bonds required to be broken are broken, new bonds should form as in all decomposition reactions. However, if i were to leave the heating unit on, would those new bonds be able to form? For example in this reaction, MgCO3 :rarrow: MgO+CO2 the bonds between Mg2+ and CO3 2-, CO2 and O2- are being broken during the heating phase. So now the new bonds MgO has to form. But I'm thinking if i don't let the solution cool down the Mg2+ and the O2- cannot form. So i'd have a gaseous solution of Mg2+, O2- and CO2 (since all bonds are broken technically it has to be a gas rather than a molten liquid)?
Lastly, by 'then we remove the heat source and now maybe there isn't enough energy to surmount the activation energy for the reverse reaction.', do you mean now that we remove all the heat, there is not enough energy energy for the reverse reaction of the decomposition? Like MgO+CO2 :rarrow: MgCO3?
Sorry if I'm not understanding this directly and thanks for all the help :)
-
1) That helps a lot haha :) so in this case the couch represents the lower energy while the kitchen represents the higher energy. But what would your wife represent here? If i were to use the decomposition of MgCO3 in this case, MgCO3 would break into Mg2+ CO2 and O2- so in the intermediate stage. So this would prefer to go back to form MgCO3 as it's energetically favorable. However, when i cool down this solution the 'wife' would cause the particles to reform as MgO and CO2 rather than MgCO3. So what is the 'wife' that doing this to the intermediate stage particles?
My wife represents a second reaction that is taking me farther away from the couch. If my wife wasn't around, I'm not going to stand around in the kitchen. I'm going to go back to the couch becuse that's the lower energy state. But instead my wife tells me I need to run some errands. Having a peaceful home is an even lower energy state than watching TV on the couch, so then instead of going back to the couch I go to the store.
The point here is this:
Given a process A --> B: Even if the process is thermodynamically unfavorable, from simple probability it does happen because there is latent energy in the environment. The thermodynamic favorability and unfavorability only impacts whether the equilibrium tends to favor A or B, not whether the reaction happens. That is, at equilibrium the thermodynamical favorability impacts the average relative concentrations of A and B (or, put another way, if you were to zoom in at a single molecule in the system at a given time, it impacts the relative probability that the molecule would be A or B). This is something students new to chemical thermodynamics don't typically grasp.
In addition, in a real, open system, one must consider the thermodynamical quantities of all possible reactions.
That is, a process like A --> B may be better described as A --> B --> C. Even if B is thermodynamically unfavorable (in which case the rate constant for A --> B is probably slow), C may be most favorable of all (in which B --> C rate contant is enormous). We know that A --> B is unfavorable, but it DOES happen. Even if it is so unfavorable that the ratio of A to B in solution at equilibrium is 1000:1, if the rate constant of the second conversion is very fast, every B that is formed in solution almost automatically gets transformed into C. Going from C back to B might be even MORE unfavorable than going from A to B (say - 10 billion to 1 for equilibrium concentrations). What this means is that over time, all A will be consumed and transformed to C, despite the fact that the first conversion is "thermodynamically unfavorable". The amount of time this takes will depend on the temperature, of course, the level of unfavorability of B wrt A, and other non-thermodynamical factors.
But, given infinite time, it will happen.
In my analogy, my couch is A and the kitchen is B. In an isolated system, my equilibrium will favor A. However in an open system, the world around me is summarized as C. As a representative example I've chosen my wife (a thermodynamical driving force if there ever was one) but it can be anything. While I'm sure we all love sitting on the couch and watching TV, we don't spend our lives there because we have other things that draw us away (jobs, the need to sleep, whatever). So while it's thermodynamically unfavorable to get up and do these other things, we do it because there are other thermodynamically favorable endpoints down the road.
2) Hmm I'm not very sure what you mean. Because i'm thinking that when i heat up the decomposing substance, bonds would be broken. So now that all the bonds required to be broken are broken, new bonds should form as in all decomposition reactions. However, if i were to leave the heating unit on, would those new bonds be able to form? For example in this reaction, MgCO3 :rarrow: MgO+CO2 the bonds between Mg2+ and CO3 2-, CO2 and O2- are being broken during the heating phase. So now the new bonds MgO has to form. But I'm thinking if i don't let the solution cool down the Mg2+ and the O2- cannot form. So i'd have a gaseous solution of Mg2+, O2- and CO2 (since all bonds are broken technically it has to be a gas rather than a molten liquid)?
Well yes, if you kept the heat on indefinitely you would not form stable molecules - at least, in the statistical average. (If you took a snapshot in time, you'd see some molecules anyway because there is a distribution of molecular energies.) Bond formation is a low energy state (a potential well or trough) and you need to remove energy to "trap" molecules there. In a closed fantasy system, you'd probably have some kind of really rapid interconversion between highly excited products and reactants. In reality, if you kept the heat high you'd access all kinds of downstream side products - in common parlance, you'd get charred goop. The trick to any reaction is heating long enough to favor your product state but not so long as to form too many unwanted downstream (and even lower energy) side products. In a kitchen, we call this "overcooking", and it's no good. :) In the scheme above, we want to go from A --> B but cool down before getting too much of C. Of course, you're always going to have some of each because that's the nature of probability and kinetics, but if you use the right combination of temperatures and times, you can heat then cool and trap most of your molecules in a stable B intermediate state. Otherwise known as a burger cooked to medium!
Lastly, by 'then we remove the heat source and now maybe there isn't enough energy to surmount the activation energy for the reverse reaction.', do you mean now that we remove all the heat, there is not enough energy energy for the reverse reaction of the decomposition? Like MgO+CO2 :rarrow: MgCO3?
Yes. Though again there is ALWAYS some kind of equilibrium. In a binary system that's closed, the equilibrium may favor the products so much that the reverse reaction is virtually negligible at the temperature under consideration. This is especially true in real world systems where you might have a chain of fifty chemical conversions before you get to the end product. This is why, for example, cooking a chicken breast is considered "irreversible", because the probability of a single "cooked" molecule going through all the reverse reactions is small - then how low must be the probability of an entire 6 oz hunk of cooked meat (gazillions of molecules!) all going through the reverse process at the same time to yield its raw starting material? Probably so low that it wouldn't happen once in many, many lifetimes of the universe.
It's all probability, friend. That's the universe you live in. :)
-
Thanks Corribus :) this makes a lot more sense now. So it seems that the thermodynamical driving force is a different concept to the energy level? I was too stuck into thinking the energy level is the only variable that determines the equilibrium. But if that were the case all reactions would be exothermic. I'm inclined to ask more about about these other variable and what causes them to happen this way as opposed to just the energy level concept but I think I'll leave this for now haha :D hopefully I'll learn more about this in the near future so I can understand this is greater detail :) but this was very enlightening. Thanks so much for all the detailed explanation :)