Necr0sis713

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I'm a little confused here about ATP hydrolysis and it's use for energy. I had a TPR review practice exam that asked:

"The phosphoanhydride bond of ATP most likely has a bond dissociation energy that is:"

I chose "negative, and the breaking of bond releases energy", but the answer was "positive, and the breaking of the bond requires energy".

Here's the thing, I know that breaking bonds requires energy, but I thought for ATP it's an exception because of the strong repulsion between phosphate groups. A lot of sources also seem to say that it releases energy.

The only thing I can think of is maybe it's EXOTHERMIC (heat releasing) but not EXERGONIC (energy releasing)?
 

Astra

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Yup.

G = H - TS

So in the case of bond breaking, it is always endergonic ( requires input of energy and + G)

so then,

+ = H - TS

Lets think about enthalpy. There is heat being released by the breakage of the bonds of ATP ( so -H)

+ = (-) - TS

So that means that the entropy of the system decreases which allows the positive G value.

There might be cases where

S is positive and the system has to have a +H in order to still break the bonds ( +G).

ATP doesn't and that is what allows it to be so useful in organisms.
I'm a little confused here about ATP hydrolysis and it's use for energy. I had a TPR review practice exam that asked:

"The phosphoanhydride bond of ATP most likely has a bond dissociation energy that is:"

I chose "negative, and the breaking of bond releases energy", but the answer was "positive, and the breaking of the bond requires energy".

Here's the thing, I know that breaking bonds requires energy, but I thought for ATP it's an exception because of the strong repulsion between phosphate groups. A lot of sources also seem to say that it releases energy.

The only thing I can think of is maybe it's EXOTHERMIC (heat releasing) but not EXERGONIC (energy releasing)?
 
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Astra

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For ATP yea. I am not sure if that holds true for all chemical bonds.
 

theonlytycrane

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deltaG = deltaH - T * deltaS

Wouldn't deltaH be positive because it takes energy to break the bond? And deltaS be positive because entropy increases. And finally deltaG is negative as free energy is released for reaction coupling.
 

spicynuggetfiend

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Bond dissociation energies are tabulated and are relevant in calculations when they are in the gas phase. To my understanding, the ATP hydrolysis aspect of that question is a red herring. Any molecule will have a bond dissociation energy that is positive, requiring energy to break the bond. I can't think of a counterexample.
 

Necr0sis713

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Ok so it turns out that breaking the bond requires energy, but the formation of the end products of the reaction is more exergonic, so the net reaction has a negative delta G. I just don't get how this would make coupling with ATP favorable, because it's contribution would be a positive delta G.

If you look here, it shows hydrolysis of ATP to be negative

http://2.bp.blogspot.com/-759W2oWdpRc/TrA6tlrBH0I/AAAAAAAALqE/sP5hzpOLmA4/s1600/tmp.bmp.

But hold on, I think I figured it out: The question asked what the bond dissociation energy is for the bonds in ATP. Is this different from hydrolysis?
 

theonlytycrane

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Breaking the bond requires energy (endothermic), but free energy is released exergonic for reaction coupling.
 
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I think what this question is trying to get at is the activation energy. Even if something is -H or -G overall, you need to initially input energy to overcome Ea.
 

aldol16

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Okay, so to frame everything here, all of you must first realize that BDE's are completely ground state effects. They are not kinetic effects. I repeat: BDE is a ground state phenomenon.

Okay, with that said, this is a case of overthinking. BDEs must necessarily be positive because if energy is released from a bond being broken, that bond would not exist in the first place. It would just break. Delta G is irrelevant here because breaking a bond necessarily creates two items from one and thus increases entropy. Even if the bond-breaking was intramolecular, bond-breaking would increase the degrees of freedom of the molecule. There's also such a thing called the BDFE, which does take into account entropy as well.

Wouldn't deltaH be positive because it takes energy to break the bond? And deltaS be positive because entropy increases. And finally deltaG is negative as free energy is released for reaction coupling.

You're confusing the BDE of a single bond with the delta H of an entire reaction, which includes all bonds broken and formed.

Thanks. So it's ENDERGONIC and EXOTHERMIC, correct?

For ATP yea. I am not sure if that holds true for all chemical bonds.

No, endergonic and exothermic are terms that refer to a reaction, or process. The BDE of a single bond is not a process unless you're talking about a reaction that breaks or forms only a single bond. BDE is a simple tabulated value that can be used to calculate overall delta H and delta G for a reaction. Do not confuse these concepts.

Bond dissociation energies are tabulated and are relevant in calculations when they are in the gas phase. To my understanding, the ATP hydrolysis aspect of that question is a red herring. Any molecule will have a bond dissociation energy that is positive, requiring energy to break the bond. I can't think of a counterexample.

The reason you can't think of a counterexample is that a counterexample simply cannot exist physically. A bond with a negative BDE would not exist because it would just roll downhill and exist as two separate items.

Breaking the bond requires energy (endothermic), but free energy is released exergonic for reaction coupling.

The free energy released has to do with the sum of all bonds broken and formed and not just one. Whether the delta H of the overall reaction is positive or negative depends on all the bond breaking and forming. So if you replace two C-H bonds which has a BDE of 100 kcal/mol (good value to remember if you're a chemist) with a C-C bond which has a BDE of about 80 kcal/mol and two N-H bonds with a BDE of about 90 kcal/mol, you could calculate the delta H of such a reaction and find it to be energetically downhill (exothermic) because the sum of the bonds formed are stronger than the sum of the bonds broken.
 
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Necr0sis713

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I'm just concerned because of the many ways such questions can be worded. So in terms of pure BDE, it's positive since a negative BDE would mean it wouldn't form in the first place (loved learning this today). The way I see it, is that coupling ATP hydrolysis with reactions makes it favorable because the electrostatic repulsion of the phosphates is a very unstable situation, and ADP + Pi is lower in energy, so overall it's favorable (negative delta G)
 

Astra

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Okay, so to frame everything here, all of you must first realize that BDE's are completely ground state effects. They are not kinetic effects. I repeat: BDE is a ground state phenomenon.

Okay, with that said, this is a case of overthinking. BDEs must necessarily be positive because if energy is released from a bond being broken, that bond would not exist in the first place. It would just break. Delta G is irrelevant here because breaking a bond necessarily creates two items from one and thus increases entropy. Even if the bond-breaking was intramolecular, bond-breaking would increase the degrees of freedom of the molecule. There's also such a thing called the BDFE, which does take into account entropy as well.



You're confusing the BDE of a single bond with the delta H of an entire reaction, which includes all bonds broken and formed.





No, endergonic and exothermic are terms that refer to a reaction, or process. The BDE of a single bond is not a process unless you're talking about a reaction that breaks or forms only a single bond. BDE is a simple tabulated value that can be used to calculate overall delta H and delta G for a reaction. Do not confuse these concepts.



The reason you can't think of a counterexample is that a counterexample simply cannot exist physically. A bond with a negative BDE would not exist because it would just roll downhill and exist as two separate items.



The free energy released has to do with the sum of all bonds broken and formed and not just one. Whether the delta H of the overall reaction is positive or negative depends on all the bond breaking and forming. So if you replace two C-H bonds which has a BDE of 100 kcal/mol (good value to remember if you're a chemist) with a C-C bond which has a BDE of about 80 kcal/mol and two N-H bonds with a BDE of about 90 kcal/mol, you could calculate the delta H of such a reaction and find it to be energetically downhill (exothermic) because the sum of the bonds formed are stronger than the sum of the bonds broken.


I knew you would come and clear up our confusions! ;)

So for the reaction ATP + H20 --> ADP + pi + H+

The reaction involves the breaking of a bond in ATP.

The bond dissociation energy is the energy required to break this bond. This number will be positive.

How does this value come into play for G = H -TS ?
 

aldol16

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I knew you would come and clear up our confusions! ;)

So for the reaction ATP + H20 --> ADP + pi + H+

The reaction involves the breaking of a bond in ATP.

The bond dissociation energy is the energy required to break this bond. This number will be positive.

How does this value come into play for G = H -TS ?

:) You're not only breaking a bond in ATP but also in water. Let's walk through the mechanism in more detail. Deprotonate the water molecule first, breaking a O-H bond (it doesn't matter which step this occurs in because of Hess's law). That generates the proton. Then attack the phosphodiester bond in ATP with the hydroxide, forming a P-O bond and breaking a P-O bond (these will have similar BDEs). If you assume that the two P-O bonds are similar in energy (which they are), then the overall delta H becomes simply the BDE of the O-H bond. That's worth about 110 kcal/mol. So the delta H for the overall reaction is not the BDE of the P-O bond but rather the O-H bond in water.

Now, delta G is a function of not only enthalpy but also entropy. Therefore, delta S will come into play. So how do we understand this? All of this occurs in the aqueous phase so we can rule out any phase change effects. So ATP goes to ADP and Pi. This is a huge increase in entropy. This is an increase in entropy because you go from one molecule to two and the two molecules can move about a lot more (draw out the structures - you gain several degrees of freedom in the atoms that were previously bound in ATP). What about water? Well, you can basically imagine water going from H20 to H+ (roughly). You still have one molecule so entropy doesn't change much. This is a very rough estimate, of course, and I can't find any tabulated values for the delta H and delta S of ATP hydrolysis, but you would expect the delta S times T term to be large and positive and greater in magnitude than the delta H term because ATP hydrolysis does in fact release so much energy.

I'm just concerned because of the many ways such questions can be worded. So in terms of pure BDE, it's positive since a negative BDE would mean it wouldn't form in the first place (loved learning this today). The way I see it, is that coupling ATP hydrolysis with reactions makes it favorable because the electrostatic repulsion of the phosphates is a very unstable situation, and ADP + Pi is lower in energy, so overall it's favorable (negative delta G)

That's a great way of thinking about it. It will give you the same result. BDEs by themselves don't tell you whether a reaction is spontaneous or not, exergonic or not. They only tell you how strong bonds are.
 
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Astra

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:) You're not only breaking a bond in ATP but also in water. Let's walk through the mechanism in more detail. Deprotonate the water molecule first, breaking a O-H bond (it doesn't matter which step this occurs in because of Hess's law). That generates the proton. Then attack the phosphodiester bond in ATP with the hydroxide, forming a P-O bond and breaking a P-O bond (these will have similar BDEs). If you assume that the two P-O bonds are similar in energy (which they are), then the overall delta H becomes simply the BDE of the O-H bond. That's worth about 110 kcal/mol. So the delta H for the overall reaction is not the BDE of the P-O bond but rather the O-H bond in water.

Now, delta G is a function of not only enthalpy but also entropy. Therefore, delta S will come into play. So how do we understand this? All of this occurs in the aqueous phase so we can rule out any phase change effects. So ATP goes to ADP and Pi. This is a huge increase in entropy. This is an increase in entropy because you go from one molecule to two and the two molecules can move about a lot more (draw out the structures - you gain several degrees of freedom in the atoms that were previously bound in ATP). What about water? Well, you can basically imagine water going from H20 to H+ (roughly). You still have one molecule so entropy doesn't change much. This is a very rough estimate, of course, and I can't find any tabulated values for the delta H and delta S of ATP hydrolysis, but you would expect the delta S times T term to be large and positive and greater in magnitude than the delta H term because ATP hydrolysis does in fact release so much energy.



That's a great way of thinking about it. It will give you the same result. BDEs by themselves don't tell you whether a reaction is spontaneous or not, exergonic or not. They only tell you how strong bonds are.

Made perfect sense. I have the full picture now!

Where were you doing my undergrad career lol

I swear you could write a prep book by yourself and it would be amazing.

I for one would buy it
 
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