[Needaask]

I've read on many websites that uses enthalpy change to explain the different natures of the reactants. For example, in thermal decomposition of carbonates the enthalpy change increases as we go down the group 2 cation. And for the oxidising ability of the halogens, the overall enthalpy change increases down the group.

But I don't really see the link between the nature of the compound and the enthalpy change. For the first reaction, I understand that more energy is absorbed as we go down the group but how does that explain that the reactants are more thermally unstable as we go down the group?

And for the second, I understand that in solution reactions, going down the group less energy is released in accepting the electron. So why would this mean that the oxidizing ability decreases down the group? I don't understand the link between the amount of energy given out during the process and the ease of its formation. I'm thinking F2 gives out more energy than Cl2 when it forms hydrated ions. But I don't see how it giving out more energy equates to it a better oxidising agent.

Thanks in advance for explaining them to me I just don't see the direct link between their nature and their enthalpy change usually..

[Corribus]

Maybe this thread will help, where recently the carbonate stability was discussed.

http://www.chemicalforums.com/index.php?topic=66412.0

[Needaask]

Maybe this thread will help, where recently the carbonate stability was discussed.

http://www.chemicalforums.com/index.php?topic=66412.0

Hi thanks for the reply

I read the chemguide link and i understood their explanation about the polarizing power or the cation. But I don't quite understand how the enthalpy change tells us these properties (also for the oxidizing ability or the halogens). Doesn't enthalpy change tell us the net change in energy level or how much energy is gained or lost? I don't see how it could explain the stability or oxidizing ability.

Thanks so much for the help

[Corribus]

Doesn't enthalpy change tell us the net change in energy level or how much energy is gained or lost? I don't see how it could explain the stability or oxidizing ability.

When you create a molecule, you essentially do two things: you form chemical bonds and other intermolecular interactions (and break them), and you also disturb the environment by creating a change in volume or pressure.  Both of these processes require (or consume) energy.  As a static quantity, enthalpy of a (chemical) system is the amount of actual energy stored in chemical bonds and other interatomic interactions, plus the amount of energy that was used (or expended) in changing the environment when a given molecule was formed.  We call the former part "internal energy", U. 

This is why H = U + pV.

For a process, the enthalpy change is essentially the change in the amount of energy stored in all bonds and interactions and the net change in energy required to disturb the environment.  Enthalpy is a type of potential energy.  It repesents the ability of the system to do work.  Systems tend to move from a state of high potential energy to a state of low potential energy.  (Think a ball rolling down a hill, in the case of gravitational potential energy.)  A chemical change that results in molecule with less stored energy (or which require less energy in form of environmental change) is thermodynamically favorable because it is a change that goes from a state of high potential energy to a state of low potential energy.  This kind of process releases that potential energy in the form of heat.  It is why exothermic processes are usually favorable.  The only confounding factor is entropy, of course.

But let's not confuse the issue with entropy.

Enthalpy impacts many chemical properties because it governs how much energy is required to form bonds, or how much energy is released when bonds break.  The more energy it takes to form the bonds of a molecule, the less likely it is to happen.  The more energy released if a bond breaks, the more likely it is to happen.  Provided, of course, the sum total of the resulting bonds and interactions have less total enthalpy than the starting material.  As with anything in chemical thermodynamics, you must consider effects of a process on the entire system, not just in an isolated molecule. 

This all feeds into the Gibbs energy, which determines a process's "favorability" and itself feeds into kinetics concepts.

Just as an aside, stability of a molecule encompasses both thermodynamic and kinetic counterparts. The latter in particular depends on more than just the net change in thermodynamical parameters like enthalpy and entropy.  Even ignoring nonthermodynamical considerations, the activation energy of a process can be large, even for a favorable process.  The transition state, for example, may involve a very unfavorable geometry.  In other words, if a reaction requires a large initial input of energy, it can be so slow as to be negigible, even if the ultimate energetic payout is large.  This is why I keep saying that it's important to keep the two concepts of thermodynamics and kinetics separate in your mind.  In the case of the carbonates - while the stability certainly is related to the internal energy of the carbonate bonds, it also relates to how close each carbonate is to the transition state, which in turn greatly impacts the kinetics of decomposition.

[Needaask]

Hi thanks again for the great reply So sorry for not understanding the kinetic and thermodynamic stability here, didn't learned much about them

Oh so we are ignoring activation energy here. So we are just saying that as we go down the group, we need more energy to decompose the carbonate. Hence, this shows that the higher up the group, the more thermally stable their carbonates are. So we are just using the fact that each compound absorbs more energy down the group?

What about for the oxidising power of the halogens? I still can't understand that Cl2(aq)+2e->2Cl-(aq) gives out more energy than I2(s)+2e->2I-(aq) so it should be more oxidising than I2. The first one makes sense now because it requires more energy to be absorbed to react so its more stable in that sense that i have to heat the lower down carbonate more for it to react. But i can't wrap my mind over exothermic changes and how they explain properties. Could you help me out with this?

Thanks so much for all the help

[Corribus]

Oh so we are ignoring activation energy here. So we are just saying that as we go down the group, we need more energy to decompose the carbonate. Hence, this shows that the higher up the group, the more thermally stable their carbonates are. So we are just using the fact that each compound absorbs more energy down the group?

No, no! Activation energy is everything!!  Well, not everything, but a big thing.  Smaller counterions are more polarizing.  When carbonate is bonded to a smaller ion, it is more polarized - the carbonate starts out looking more like a CO₂ + O^2-.  One of the C-O bond orders is smaller than the other two.  Therefore it takes less energy to break that bond, therefore the activation energy is lower, therefore the rate of the reaction is higher.  Therefore stability is less for smaller counterions than larger counterions.

What about for the oxidising power of the halogens? I still can't understand that Cl2(aq)+2e->2Cl-(aq) gives out more energy than I2(s)+2e->2I-(aq) so it should be more oxidising than I2. The first one makes sense now because it requires more energy to be absorbed to react so its more stable in that sense that i have to heat the lower down carbonate more for it to react. But i can't wrap my mind over exothermic changes and how they explain properties. Could you help me out with this?

Enthalpy can get you far, but you really can't discount entropic changes when considering reactivity down a group, especially with halogens, and especially in aqueous solutions.  If you go by enthalpies alone, you reach a lot of incorrect conclusions.  Without entropy, for example, HF would be a strong acid.  But it is not, because fluorine is a very small, highly charged ion that causes a lot of highly organized reorganization of water.  Dissociation of HF is entropically unfavorable, which is why the Ka (rate of ionization) of HF is about 5-8 orders of magnitude less than other HX acids. 

Because the size of the X-X bond increases as you go down the group, you'd expect the bond enthalpy of X2 to decrease as you go down the group.  This is what you observe - EXCEPT in the case of fluorine, which actually has a bond enthaply close to that of iodine!  Why?  Because fluorine is so small that electron-electron repulsion destabilizes the bond

Getting to your original question, when trying to understand the redox reactivity of the halogens, you have to consider the enthalpy of X-X bond breaking, but you also have to consider the enthalpy of hydration of the ions, the enthalpy of ionization, and so on, before you even get to entropy at all.  All these relationships become very complicated very quickly and so trends are not always so clear as you think they should be.

So while yes in principle you should be able to rationalize anything in terms of the enthalpies involved, in practice it's not always so easy to see a path toward a rational explanation.

[Needaask]

Oh so we are ignoring activation energy here. So we are just saying that as we go down the group, we need more energy to decompose the carbonate. Hence, this shows that the higher up the group, the more thermally stable their carbonates are. So we are just using the fact that each compound absorbs more energy down the group?

No, no! Activation energy is everything!!  Well, not everything, but a big thing.  Smaller counterions are more polarizing.  When carbonate is bonded to a smaller ion, it is more polarized - the carbonate starts out looking more like a CO₂ + O^2-.  One of the C-O bond orders is smaller than the other two.  Therefore it takes less energy to break that bond, therefore the activation energy is lower, therefore the rate of the reaction is higher.  Therefore stability is less for smaller counterions than larger counterions.

What about for the oxidising power of the halogens? I still can't understand that Cl2(aq)+2e->2Cl-(aq) gives out more energy than I2(s)+2e->2I-(aq) so it should be more oxidising than I2. The first one makes sense now because it requires more energy to be absorbed to react so its more stable in that sense that i have to heat the lower down carbonate more for it to react. But i can't wrap my mind over exothermic changes and how they explain properties. Could you help me out with this?

Enthalpy can get you far, but you really can't discount entropic changes when considering reactivity down a group, especially with halogens, and especially in aqueous solutions.  If you go by enthalpies alone, you reach a lot of incorrect conclusions.  Without entropy, for example, HF would be a strong acid.  But it is not, because fluorine is a very small, highly charged ion that causes a lot of highly organized reorganization of water.  Dissociation of HF is entropically unfavorable, which is why the Ka (rate of ionization) of HF is about 5-8 orders of magnitude less than other HX acids. 

Because the size of the X-X bond increases as you go down the group, you'd expect the bond enthalpy of X2 to decrease as you go down the group.  This is what you observe - EXCEPT in the case of fluorine, which actually has a bond enthaply close to that of iodine!  Why?  Because fluorine is so small that electron-electron repulsion destabilizes the bond

Getting to your original question, when trying to understand the redox reactivity of the halogens, you have to consider the enthalpy of X-X bond breaking, but you also have to consider the enthalpy of hydration of the ions, the enthalpy of ionization, and so on, before you even get to entropy at all.  All these relationships become very complicated very quickly and so trends are not always so clear as you think they should be.

So while yes in principle you should be able to rationalize anything in terms of the enthalpies involved, in practice it's not always so easy to see a path toward a rational explanation.

Oh! Sorry I'm quite blur. Correct me if I'm wrong here but there are 3 ways to look at stability 1. energy level of the compound 2. enthalpy change of the same type of reaction and 3. activation energy of the same type of reaction so we have to look at all 3 to see if something is more stable than another?

Regarding the previous post, I actually didn't really get this '"The more energy it takes to form the bonds of a molecule, the less likely it is to happen.  The more energy released if a bond breaks, the more likely it is to happen. " Because I thought that bond formation gives out energy and bond breaking absorbs energy. Do you mean that the more energy it takes to break the bonds of a molecule the less likely it is to happen and the more energy released when a bond forms the more likely it is to happen? But in these statements do you just mean the individual bond breaking/forming process of a reaction or the reaction as a whole?

Also, what do you mean by "Provided, of course, the sum total of the resulting bonds and interactions have less total enthalpy than the starting material." This would mean that the reaction had to be exothermic right? But prior to that you explained both exothermic reactions and endothermic reactions. Could you explain this part again for me? Thanks so much

Thanks so much for taking the time to type out all these explanations

[Corribus]

The take home lesson here is to be careful the way you write things and to re-read everything (twice!) carefully before you hit “post”.  Because: I see that in my haste to write out a detailed response, I accidentally wrote out almost everything in that one paragraph completely backward!  Wow, I should really be ashamed of myself.  My only defense is that, even after doing this stuff for well over a decade now, it’s still easy to get thermodynamics screwed up if you aren’t really careful. 

So, now that my tail is tucked firmly between my legs, please give me the opportunity to correct myself:

Forming any individual bond releases energy.  You can see this easily in a Morse potential, which can be used to describe formation of a bond between two atoms as they approach.  A bond represents a chemical potential well.  High energy atoms meet, form a bond, and release energy.  To break that bond requires energy, like lifting a heavy object out of a pit takes gravitational work.  Standard heats of formation for most molecules are negative, which is akin to saying that forming molecules usually releases energy.  It takes no great detective work to see this – if chemical bonds weren’t lower energy than individual atoms, there’d be no molecules.   And reactions (ignoring entropy again) tend to proceed toward molecules with more stable bonds, because they have deeper potential wells than less stable bonds.

When I said, “enthalpy impacts many chemical properties because it governs how much energy is required to form bonds, or how much energy is released when bonds break,” what I should have said was… well, switch “required” and “released”.  Do note that you have to consider all bonds broken and formed during a reaction to get a full enthalpic picture. When you break some bonds, you form others.  The overall enthalpy of the process is determined by the net energy gain (or loss) from the energy produced form each individual bond formed and from the energy consumed for each bond broken. (All intermolecular interactions formed and broken need to go into it as well, and things like ionization and so forth also have associated enthalpies.)  If the net is positive, endothermic.  If negative, exothermic. (At constant pressure!)  And finally – the net heat gained or lost from a reaction is different from the amount actually required to make a reaction go.  Even forming strong, stable bonds (in molecules) usually requires an initial energy investment (activation energy) for an ultimate energy payout because typically you’ve got to break bonds to form new ones.  If that investment isn’t available because the temperature is too low, then the reaction doesn’t go*, even if it is a favorable exothermic reaction.  Kinetics and thermodynamics, remember!

I hope that clears up that bit of confusion.  And I hope I didn’t make any new silly errors for which I will have to redeem myself.

*Well, it will go, the rate will just be slow.  Even the most unfavorable reactions will EVENTUALLY go to "completion" (equilibrium) at nonzero temperatures if they are thermodynamically favorable, because probability is never strictly zero.  Eventually the rate becomes so slow that it is "practically zero".

[Needaask]

Hi thanks so much this cleared up a lot now.

But I don't really see how the net enthalpy change explained here would tell us about their chemical properties. For instance in this link http://www.chemguide.co.uk/inorganic/group7/halogensasoas.html they explained that going down the group, the total process' enthalpy change decreases. And they used that to explain that the oxidizing strength decreases down the group.

This is the part I usually don't understand. For an exothermic reactions I don't see why if the enthalpy change is more negative as we go down the group it would mean that A. The stability of the reactants decreases down the group and B. The ease of reaction increases and vice versa if an exothermic reaction becomes more positive.

And similar for endothermic reactions going down a group, if it becomes even more positive down the group it would mean that A. The reactants become more stable down the group and B. it becomes harder for the reaction to occur. And vice versa for an endothermic reaction that becomes more negative down the group.

Thanks so much for the help let me know if I'm unclear here

[Corribus]

I think the reason for your confusion in this case is that you are only considering half of the reaction.  In the case of the halogens, the page you linked to lists the overall enthalpy change for each of the halogens undergoing a reduction (being an oxidizing agent).  Just so you don't have to go back and look, these values are:

F: -755, Cl: -592, Br: -547, I: -481 kJ/mol

These values include the sum total of atomization energy, electron affinity, and hydration enthalpy.

So your question is - why do these values relate to the favorability of the oxidation strength?

Remember that when something is doing the oxidizing, something else is being oxidized.  Sticking just to the halogens, from these values we can predict why chlorine will oxidize bromine, but not vice-versa.

The -592 kJ/mol value for chlorine represents the total energy released when chlorine is atomized (taken from aqueous to gas phase), reduced (adds an electron) and then the ion re-solvated.  This is an estimate of the half reaction.  But for this to happen, the chlorine must steal those electrons from the bromide ions.  So bromine must be de-solvated, oxidized and then de-atomized (resolvated).  The energy required to make this happen is going to be +547 kJ/mol.  Therefore the net enthalpy for the reaction would be predicted to be -45 kJ/mol. Exothermic.  The reverse reaction (bromine taking electrons from chlorine) would have a net enthalpy of +45 kJ/mol.  It would be endothermic, and therefore unfavorable.

Thus, when you mix chlorine and bromide together, the chlorine will tend to form chloride ions and bromine because exothermic reactions are more favorable.  But when you mix bromine and chloride ions together, you will tend to keep the starting materials.  (All of this ignoring entropic changes, of course, and keeping in mind there will always be an equilibrium reached.  It's not that bromine won't oxidize SOME chlorides.  The equilibrium will heavily favor chloride ions and di-bromine no matter what materials you start with.)

[Needaask]

I think the reason for your confusion in this case is that you are only considering half of the reaction.  In the case of the halogens, the page you linked to lists the overall enthalpy change for each of the halogens undergoing a reduction (being an oxidizing agent).  Just so you don't have to go back and look, these values are:

F: -755, Cl: -592, Br: -547, I: -481 kJ/mol

These values include the sum total of atomization energy, electron affinity, and hydration enthalpy.

So your question is - why do these values relate to the favorability of the oxidation strength?

Remember that when something is doing the oxidizing, something else is being oxidized.  Sticking just to the halogens, from these values we can predict why chlorine will oxidize bromine, but not vice-versa.

The -592 kJ/mol value for chlorine represents the total energy released when chlorine is atomized (taken from aqueous to gas phase), reduced (adds an electron) and then the ion re-solvated.  This is an estimate of the half reaction.  But for this to happen, the chlorine must steal those electrons from the bromide ions.  So bromine must be de-solvated, oxidized and then de-atomized (resolvated).  The energy required to make this happen is going to be +547 kJ/mol.  Therefore the net enthalpy for the reaction would be predicted to be -45 kJ/mol. Exothermic.  The reverse reaction (bromine taking electrons from chlorine) would have a net enthalpy of +45 kJ/mol.  It would be endothermic, and therefore unfavorable.

Thus, when you mix chlorine and bromide together, the chlorine will tend to form chloride ions and bromine because exothermic reactions are more favorable.  But when you mix bromine and chloride ions together, you will tend to keep the starting materials.  (All of this ignoring entropic changes, of course, and keeping in mind there will always be an equilibrium reached.  It's not that bromine won't oxidize SOME chlorides.  The equilibrium will heavily favor chloride ions and di-bromine no matter what materials you start with.)

Oh! That's right I have to look at the counterpart as well! I completely didn't think of that. So because of having gave out so much more energy down the group it allows for a more exothermic reaction if we keep the reducing agent a constant making it more favorable! This cleared up a lot for me thanks so much for this

One last question here, actually if i had a set of exothermic reactions with different reactants and the trend showed that the enthalpy change decreases. So why would it mean that the reaction becomes more favorable? Because I thought that the change only tells us the net amount of energy released? So I can't seem to link that and favorbility of the reaction..

Thanks again for all the help The explanation on the halogen was awesome

[Corribus]

Well, you can't explictly link the enthalpy change to absolute thermodynamic favorability because you need the entropy contribution.  This will allow you to determine equilibrium concentrations of the reaction based on the Gibbs energy.  Nevertheless, as the enthalpy change gets more negative, the Gibbs energy decreases - so, relatively speaking, more exothermic reactions tend to be more favorable.