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Topic: Calcium Oxide Decomposition  (Read 33687 times)

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Offline Corribus

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Re: Calcium Oxide Decomposition
« Reply #30 on: March 20, 2013, 10:02:49 AM »
I guess I'm having a little difficulty understanding the questions being asked by both of you.

In response to this:

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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.
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 #31 on: March 20, 2013, 12:07:14 PM »
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 :)
« Last Edit: March 10, 2016, 08:36:30 AM by Arkcon »

Offline Corribus

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Re: Calcium Oxide Decomposition
« Reply #32 on: March 20, 2013, 10:15:10 PM »
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.     

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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. 

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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.
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 #33 on: March 21, 2013, 08:37:41 AM »
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
« Last Edit: March 10, 2016, 08:36:59 AM by Arkcon »

Offline Corribus

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Re: Calcium Oxide Decomposition
« Reply #34 on: March 21, 2013, 10:54:47 AM »
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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

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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.

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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.

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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.
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 #35 on: March 22, 2013, 04:26:38 AM »
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 :)

« Last Edit: March 10, 2016, 08:41:24 AM by Arkcon »

Offline Corribus

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Re: Calcium Oxide Decomposition
« Reply #36 on: March 25, 2013, 10:09:31 AM »
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.

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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!

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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. :)
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 #37 on: March 25, 2013, 12:05:30 PM »
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 :)
« Last Edit: March 10, 2016, 08:39:08 AM by Arkcon »

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