TBR: Enthalpy v. Stability Question

[justadream]

TBR Gen Chem page 173 #41



I’m confused about enthalpy.



First, TBR says that less stable molecules have “greater enthalpy of reaction” By this, does TBR mean



“less stable molecules have a more POSITIVE enthalpy of reaction”





For example, if TBR also gives a table with these two values for enthalpy:



C2H2 = +226.8

C2H4 = +52.4



Then C2H2 is less stable because it has a greater enthalpy. This would make sense because if you look at the hydrogenation reaction (C2H2 + H2 => C2H4), the deltaH is negative, meaning that C2H2 releases energy to become C2H4 (implying that C2H2 was less stable originally).





Likewise, if the enthalpies of Substance X = -1000 and Substance Y = -500, then is Substance Y is less stable? After all, Substance Y has a “more positive” enthalpy.



This second example is contradictory to me because if Substance X releases MORE energy, then it should be less stable.

---

Members don't see this ad.

 

[justadream]

bump!

 

[Czarcasm]

bump!

The overall enthalpy of reaction is determined by the sum of the energy input (to break bonds of reactants) and the energy released (by bonds formed in products). The sum of these energies ultimately determine whether a reaction is exothermic or endothermic. Recall that products that are very stable release a ton of energy. The reason being is they formed considerably stronger bonds than the reactants and therefore give off more energy when produced. Therefore, the more stable a bond is for a given product, the more exothermic the reaction. In contrast, the less stable the bond, the more endothermic. In this later scenario, it might help to understand that unstable bonds aren't favorable and therefore, it would require a great deal of energy to break the bonds of the reactants to form semi-unstable bonds. The net energy yield would be very small, or even positive depending on how unstable the products are.

Rather than focusing so much on positive or negative numbers, think about what the universe wants to attain spontaneity. The free energy of a reaction = deltaH - T(deltaS). Recall that the more negative deltaG is, the more spontaneous the reaction. Therefore, the more negative deltaH is, the more spontaneous or favorable the reaction. Forming stable bonds is something the universe wants, so you might expect this to have a very negative enthalpy. Substance X, releasing 1000J of energy vs. Substance Y, releasing only 500J would result in a larger energy difference between products and reactants. Ultimately, it's this energy difference that indicates Substance X is more stable than Y.

 

[Czarcasm]

This second example is contradictory to me because if Substance X releases MORE energy, then it should be less stable.

Energy released is a good thing. Again, energy input is required to break bonds. Energy is released to form bonds. If the overall enthalpy is negative, that indicates the energy released (products) "beats out" energy input (reactants). Energy input is very costly and so the universe tries to minimize that as much as possible. In contrast, energy released is very favorable.

I suppose a real world example is, say you invest 100 dollars on a stock only to get a return of 1,000 dollars a week later (you benefited from this). Another possibility is the stock could have dropped to -1,000 (very costly to you). Well, for chemical reactions, it's a bit like this (but backwards when considering the sign conventions). Sorry, my explanations are sorta weak, but hope this helps.

 

[Teleologist]

the more negative deltaH is, the more spontaneous or favorable the reaction. Forming stable bonds is something the universe wants, so you might expect this to have a very negative enthalpy. Substance X, releasing 1000J of energy vs. Substance Y, releasing only 500J would result in a larger energy difference between products and reactants. Ultimately, it's this energy difference that indicates Substance X is more stable than Y.

You forgot entropy. There are endothermic yet spontaneous reactions too, bruh. Unless you at that level where entropy = enthalpy.

 

[Teleologist]

I suppose a real world example is, say you invest 100 dollars on a stock only to get a return of 1,000 dollars a week later (you benefited from this). Another possibility is the stock could have dropped to -1,000 (very costly to you). Well, for chemical reactions, it's a bit like this (but backwards when considering the sign conventions). Sorry, my explanations are sorta weak, but hope this helps.

Bruh LOL just think of relationships; forming a relationship = bond forming in chemistry. Forming a relationship also lowers your energy; i.e. you aint bouncing around in clubs every night anymore; you gotta settle down and stuff. Same in chemistry - bond forming is exothermic/energy lowering.

 

[Teleologist]

Ultimately, it's this energy difference that indicates Substance X is more stable than Y.

Meh, not really, you have to first define stability. If you define stability as a lack of a tendency to react, that's not necessarily true. Substance X may still have a tendency to react. Stability is a tricky subject to simplify without going into Eact and transition state or intermediate stability, etc.

 

[justadream]

Substance X, releasing 1000J of energy vs. Substance Y, releasing only 500J would result in a larger energy difference between products and reactants. Ultimately, it's this energy difference that indicates Substance X is more stable than Y.

@Teleologist

I found the statements below in another SDN thread a while ago.

If the heat of hydrogenation of X is -100, and the heat of hydrogenation of Y is -500, X is MORE stable than Y. Y gives off more heat because it has more energy (is less stable) to begin with.

If you're talking about a reaction like X-->Y having an enthalpy of -100 and X-->Z enthalpy of -500, then Z is more stable than Y because there was a greater energy change to Z, which means Z is at a lower energy level (is more stable) than Y.

Doesn't point #1 contradict what you (Czarcasm) said? Since your substance X releases more energy, doesn't that imply that substance X was less stable to begin with?

Also, if I'm taking these statements to be true, then it appears I had it wrong in my OP. I think point #1 implies that more stable things have HIGHER (more positive, less negative) enthalpies.

 

[Teleologist]

@Teleologist

I found the statements below in another SDN thread a while ago.

If the heat of hydrogenation of X is -100, and the heat of hydrogenation of Y is -500, X is MORE stable than Y. Y gives off more heat because it has more energy (is less stable) to begin with.

If you're talking about a reaction like X-->Y having an enthalpy of -100 and X-->Z enthalpy of -500, then Z is more stable than Y because there was a greater energy change to Z, which means Z is at a lower energy level (is more stable) than Y.

Doesn't point #1 contradict what you (Czarcasm) said? Since your substance X releases more energy, doesn't that imply that substance X was less stable to begin with?

Also, if I'm taking these statements to be true, then it appears I had it wrong in my OP. I think point #1 implies that more stable things have HIGHER (more positive, less negative) enthalpies.

Broski, you completely right. Czarcasm was just being sarcastic (I hope). The higher the enthalpy of combustion of something, the less stable that something was (to start with).

 

[Teleologist]

Also, if I'm taking these statements to be true, then it appears I had it wrong in my OP. I think point #1 implies that more stable things have [B]HIGHER (more positive, less negative) enthalpies. [/B]

Just to be clear, things don't have enthalpies. You must be talking about dH or change in enthalpy.

 

[justadream]

@Teleologist

I'm talking about enthalpy of formation (deltaHf).

So if TBR gives a table with these two values for enthalpy of formation:

C2H2 = +226.8
C2H4 = +52.4


Do I conclude that C2H2 is less stable than C2H4 (since to go from C2H2 to C2H4, it is exothermic)?

If so, this would mean that higher deltaHf means lower stability.

 

[Teleologist]

@Teleologist

I'm talking about enthalpy of formation (deltaHf).

So if TBR gives a table with these two values for enthalpy of formation:

C2H2 = +226.8
C2H4 = +52.4


Do I conclude that C2H2 is less stable than C2H4 (since to go from C2H2 to C2H4, it is exothermic)?

If so, this would mean that higher deltaHf means lower stability.

I don't see how you can go from C2H2 to C2H4 with the data given. Unless you're doing some wicked math to find the BDE of each C-H bond and C-C bond.

 

[justadream]

@Teleologist

The equation would be: C2H2 + H2 => C2H4

deltaHf of products - deltaHf of reactants

52.4 - (226.8 + 0.0) = some number less than 0

The deltaHf for H2 is zero by definition (since it is hydrogen in its "natural" state).

 

[Czarcasm]

Consider the actual bonds being formed. C2H2, has a double bond between two carbons (1 pi bond) while C2H4 has 4 sigma bonds instead. Collectively, these sigma bonds are not only stronger than pi bonds but more stable too. By forming more of these sigma bonds, the molecule as a whole is more stable. When considering the enthalpy of reaction values for both of these molecules, than ethane, having a less positive (unfavorable) enthalpy, is more stable overall. This is a relative comparison we are making between the reactants and products we are forming.

Heat of hydrogenation is a little bit different than enthalpy of reaction or heat of formation (what my explanation referred to early on). The reference point is entirely different here, which only adds the the confusion. Here, the final product yielded is the same (ethyne to ethane or ethene to ethane). In this scenario, the energy differences has to do with the replacement of pi bonds with sigma bonds. The more pi bonds replaced, the more energy given off. A reactant that gives off less energy must indicate it was more stable to begin with.

The pi bond is reacting with H2. For each pi bond that reacts, two sigma bonds to hydrogen are formed. The main thing to realize is that regardless of the initial reactant (ethene or ethyne), they both reach the same final place (ethane). The difference here in terms of energy yield has to do with the amount of pi bonds we had initially. (Ethene has 1 pi bond vs. Ethyne which has 2 pi bonds). In this scenario, the pi bond is the only reactive portion we are considering. The more pi bonds (the more unstable the molecule), the more energy that will be released upon forming the more stable ethane molecule. This is why ethyne, which occupies a higher energy level initially, has a bigger release of energy upon hydrogenation to the same final point than does ethene.

In some scenarios, the difference may have to do with other factors such as aromaticity or relative orientation (cis vs. trans double bond). A geminal alkene having more strain due to steric hindrance is more unstable than a trans alkene which is less effected by this. Therefore, a geminal alkene would give off more energy to produce the same final alkane. (Again, notice this is entirely different than heat of reaction values). Take a look at this diagram:

 

[Teleologist]

Consider the actual bonds being formed. C2H2, has a double bond between two carbons

Woah woah there calm down browski before you come tumbling down like the Dow jones. Acetylene has how many bonds between its carbons???!!11!!111oneoneoneone!!

while C2H4 has 4 sigma bonds instead.

Bruh, are you being sarcastic? C2H4 has five sigma bonds.

----

----

A reactant that gives off less energy must indicate it was more stable to begin with.

Okay, we're getting warmer Sherlock. You corrected the error you made in your post above.

The more pi bonds (the more unstable the molecule)

So N triple bond N is less stable than N single bond N?

 

[Czarcasm]

Woah woah there calm down browski before you come tumbling down like the Dow jones. Acetylene has how many bonds between its carbons???!!11!!111oneoneoneone!!



Bruh, are you being sarcastic? C2H4 has five sigma bonds.



----


----



Okay, we're getting warmer Sherlock. You corrected the error you made in your post above.



So N triple bond N is less stable than N single bond N?

Nice catch, but why go through all that effort to point out a careless error I made when you could have simply addressed it with a sentence. Either you fail at trolling or have a lot of time on your hands, maybe both. Find a better hobby dude.

As I explained before, pi bonds are weaker than sigma bonds due to less overlap. Obviously, this assumes we're obeying the octet rule. The example you gave with nitrogen is unstable only because it fails the octet rule. When considering the relative stabilities of hydrocarbons it's due to something entirely differently. In each instance (ethyne, ethene, ethane) all have a full octet. The difference in energies is attributed to actual bond energies. Although triple bonds are collectively stronger than single bonds, pi bonds are relatively unstable which explains the reactivity of alkynes and alkenes triple and double bonds. Once again, it's these pi bonds that are reacting with hydrogens (for heat of hydrogenation) to form more stable sigma bonds (giving off energy in the process).

 

[Teleologist]

Nice catch, but why go through all that effort to point out a careless error I made when you could have simply addressed it with a sentence. Either you fail at trolling or have a lot of time on your hands, maybe both. Find a better hobby dude.
.

also Bruh were here to learn. I could post my own answer but let's first fix what's messed up rather than post more (conflicting) info for the OP to digest. We're not here just to stroke your ego, this is a place of learning.

 

[Teleologist]

The example you gave with nitrogen is unstable only because it fails the octet rule.

What's "it"? If you mean N2 then it obeys the octet rule.

Also now that I thought of it, nitrogen always obeys the octet rule. It isn't hypervalent. It can't be. I don't see what your grasping for. How does the octet rule factor into pi bond strength? I have a feeling: it doesn't.

 

[justadream]

@Czarcasm

Aren't all the "types" of enthalpy the same in the end (they're just different ways to measure it - but they should have roughly the same magnitude)?

So I think you are distinguishing between deltaH and heat of hydrogenation? I was only considering deltaHformation because the TBR passage used it (in the context of hydrogenation)

What I took from your post is that since C2H2 has more pi bonds, it can release more energy. This is consistent with C2H2 having a higher deltaHformation since if it originally took MORE energy to form C2H2, C2H2 should be able to release more energy. However, you are saying I shouldn't make this conclusions since I should not link heat of hydrogenation with deltaHformation?

So overall, with regard to all the deltaH types then, is there no "simple" rule such as:

Higher deltaEnthalpy (more positive, less negative) = less stable?

lol all I want is a nice guideline like that to remember.

 

[Teleologist]

lol all I want is a nice guideline like that to remember.

This is chemistry. There aren't any simple rules. Source: me.

Aren't all the "types" of enthalpy the same in the end (they're just different ways to measure it - but they should have roughly the same magnitude)?

Enthalpy is a state function and if you're talking about a reversible process then the magnitude of the process's enthalpy change is the exact same forwards and backwards.

So I think you are distinguishing between deltaH and heat of hydrogenation?

What's delta H?

What I took from your post is that since C2H2 has more pi bonds, it can release more energy. This is consistent with C2H2 having a higher deltaHformation since if it originally took MORE energy to form C2H2, C2H2 should be able to release more energy. However, you are saying I shouldn't make this conclusions since I should not link heat of hydrogenation with deltaHformation?

You see, I was trying to warn him about people making erroneous conclusions from his post, but he wouldn't budge. Having more pi bonds =! you "can release more energy," whatever that nebulous statement means. In the context of hydrogenation, then okay, that seems to be true.

Also don't compare dH of formation with dH of hydrogenation. Two completely different processes.

 

[Teleologist]

The deltaHf for H2 is zero by definition (since it is hydrogen in its "natural" state).

You mean H2 is in Rousseau's "state of nature"?