Litmus Test for Reversibility of Reactions?

[The Head]

I'm a little confused by what exactly determines if a reaction is reversible or irreversible. I thought I have heard that if the delta G of a reaction is negative (which means it's spontaneous), that it's then irreversible, but that doesn't make a whole lot of sense when I think about it because the vast majority of reactions don't have a delta of zero and there are a lot of reversible reactions. Obviously there are reactions like combustion reactions that are essentially irreversible, but outside of these few reactions I'm not sure how to judge what factors are important.

I get that changing the temperature can change whether something is endergonic or exergonic, but when asking if a reaction is reversible, I'm assuming they don't mean that you can change the temperature. Is there a litmus test that one can when looking at a reaction to determine reversibility?

 

[Borek]

Actually reversible/irreversible is not a well defined concept. Way too often it is a bit handwavy.

 

[The Head]

Yeah, perhaps that's why I've been a bit confused. Could you provide some of the most important considerations though so that I'd be better equipped at determining if a reaction is reversible, at least in the classical sense? Basically, if I'm looking at the chemical equation and know whether something is exo/endothermic, what else should I consider? It can't be so simple to say exothermic or irreversible, because there are reactions in equilibrium (or where all the reactants don't react) that don't have a delta G of zero.

 

[Ygggdrasil]

The thermodynamic definitions of reversible/irreversible probably don't mesh well with the popular definitions of the words. Most of the general populace would say that ice melting is reversible. In thermodynamic terms, ice melting at 10^oC is an example of an irreversible process, but ice melting at 0^oC is reversible. In thermodynamics, reversibility basically has to do with whether the system is at equilibrium at all times during the transition or whether the system is traveling along a non-equilibrium path to the final state.

So, if you're asking about thermodynamics, you've already given the answer: reversible processes are ones where the ΔG = 0. If you're asking about the popular definition of the word, there is no set of definitive guidelines because the popular definition of the word reversible is not scientifically defined.

 

[Chestermiller]

In my judgment, every reaction is reversible, and so there is no valid criterion for reversibility. The criterion of $\Delta G^0$ being negative is only a crude rule of thumb. Even for combustion reactions, if there is a huge enough overabundance of products relative to reactants, the reaction will proceed spontaneously in the reverse direction to a small extent before equilibrium is achieved. If you use the $\Delta G^0$ for a combustion reaction to calculate the equilibrium constant, you will find that the value for the equilibrium constant is very, very, large, but not infinite. So there is no such thing as a truly irreversible reaction (again, in my judgment).

Chet

 

[Tiiba]

Well, my definition would be "removing the cause of the change brings the system back to its original state (at least in terms of composition)." So if ice melted because it was heated, cooling it will make it ice again. If salt was dissolved in water, evaporating the water will precipitate the salt.

So, it would seem that the answer is, the reaction is reversible if the initial state of the system had the least possible Gibbs free energy of any arrangement of particles that the reaction could create. Thus, if the salt was NaCl, you get back NaCl. But if it was NaCl + KBr, you'll get NaCl + KBr + NaBr + KCl, because that allows more entropy.

Is all this correct?

 

[Chestermiller]

Well, my definition would be "removing the cause of the change brings the system back to its original state (at least in terms of composition)." So if ice melted because it was heated, cooling it will make it ice again. If salt was dissolved in water, evaporating the water will precipitate the salt.

So, it would seem that the answer is, the reaction is reversible if the initial state of the system had the least possible Gibbs free energy of any arrangement of particles that the reaction could create. Thus, if the salt was NaCl, you get back NaCl. But if it was NaCl + KBr, you'll get NaCl + KBr + NaBr + KCl, because that allows more entropy.

Is all this correct?

Is there a valid reference on all this, or is this just your own personal speculation? This does not seem to relate to what the OP was asking.

 

[Tiiba]

Is there a valid reference on all this, or is this just your own personal speculation? This does not seem to relate to what the OP was asking.

It's more of a request for confirmation of my understanding than an answer, really, so no references.

OP asked for a test for reversible reactions. Yggdrasil said that reversibility is defined differently in thermodynamics than in colloquial speech, and that the colloquial definition is too hazy to give an answer. So I tried to formalize the colloquial usage.

 

[lavoisier]

Very interesting.
I tend to agree with @Chestermiller . I seem to remember that 'irreversible' was basically the limit for ΔG → -∞, therefore not something that actually exists, but a limiting case.
In practice though, I am not sure it's really possible to reverse 'any' reaction.

There was an example in one of my school textbooks, a nitrogen oxide (NO₂ maybe?) that was in equilibrium with its dimeric form. The experiment consisted in filling a flask with this gas and then changing the temperature by dipping the flask in hot or cold water. Apparently this caused the equilibrium to shift left or right, and you could see that because the colours of the two gases were different. This is what I would call a clearly reversible reaction.
In other cases, e.g. acid-catalysed esterification of a carboxylic acid, you need to distill away the water, i.e. remove one of the products of the reaction, otherwise you will reach an equilibrium with only some ester formed and still some carboxylic acid present. This for me would be another reversible reaction.

Consider instead throwing a piece of sodium metal in a bucket of water. Apart from the fireworks bit, what you get is a solution of sodium hydroxide in water, and H₂ will just fly off. Now, how would you turn that back into water and sodium? Is that even theoretically possible?

And to be frank, in all my career as an organic chemist I think I've mostly been using 'irreversible' reactions, or at least reactions 'with very negative ΔG', because indeed our aim is to get a certain product.
Suppose for instance that you make an ether by reacting sodium alkoxide with an alkyl halide. You get the ether and sodium halide. If this were really reversible, it would be sufficient to provide energy to a mixture of ether and sodium halide to get back the reactants. Why doesn't it happen? I know you could cleave the ether with hydrogen halide, but that for me is a different reaction (and in fact there is an acid-catalysed etherification of alcohols).

And what about entropy? Hasn't reversibility more to do with ΔS than with ΔG? Aren't processes always going to flow in the direction of larger entropy, in a closed system? So if you wanted to reverse a process that has a very large ΔS, you would need to do a lot of work and overall end up with a positive ΔS in the whole system.
Is there any reaction with overall negative ΔG but also negative ΔS, i.e. sufficiently negative ΔH to overcome the entropic cost?

Not sure if I'm making any sense at all, thermodynamics has never been my favourite subject.

 

[Chestermiller]

Very interesting.
I tend to agree with @Chestermiller . I seem to remember that 'irreversible' was basically the limit for ΔG → -∞, therefore not something that actually exists, but a limiting case.
In practice though, I am not sure it's really possible to reverse 'any' reaction.

There was an example in one of my school textbooks, a nitrogen oxide (NO₂ maybe?) that was in equilibrium with its dimeric form. The experiment consisted in filling a flask with this gas and then changing the temperature by dipping the flask in hot or cold water. Apparently this caused the equilibrium to shift left or right, and you could see that because the colours of the two gases were different. This is what I would call a clearly reversible reaction.
In other cases, e.g. acid-catalysed esterification of a carboxylic acid, you need to distill away the water, i.e. remove one of the products of the reaction, otherwise you will reach an equilibrium with only some ester formed and still some carboxylic acid present. This for me would be another reversible reaction.

Consider instead throwing a piece of sodium metal in a bucket of water. Apart from the fireworks bit, what you get is a solution of sodium hydroxide in water, and H₂ will just fly off. Now, how would you turn that back into water and sodium? Is that even theoretically possible?

And to be frank, in all my career as an organic chemist I think I've mostly been using 'irreversible' reactions, or at least reactions 'with very negative ΔG', because indeed our aim is to get a certain product.
Suppose for instance that you make an ether by reacting sodium alkoxide with an alkyl halide. You get the ether and sodium halide. If this were really reversible, it would be sufficient to provide energy to a mixture of ether and sodium halide to get back the reactants. Why doesn't it happen? I know you could cleave the ether with hydrogen halide, but that for me is a different reaction (and in fact there is an acid-catalysed etherification of alcohols).

And what about entropy? Hasn't reversibility more to do with ΔS than with ΔG? Aren't processes always going to flow in the direction of larger entropy, in a closed system? So if you wanted to reverse a process that has a very large ΔS, you would need to do a lot of work and overall end up with a positive ΔS in the whole system.
Is there any reaction with overall negative ΔG but also negative ΔS, i.e. sufficiently negative ΔH to overcome the entropic cost?

Not sure if I'm making any sense at all, thermodynamics has never been my favourite subject.

You have most of it right, but your memory failed you a little. If a reaction is at equilibrium, the best way to get it to go in the reverse direction is to add products to the mixture. This will shift the equilibrium back toward reactants.

I also wanted to mention that the terms reversible/irreversible have two different contexts in thermodynamics. When referring to a process, reversible means that the system can be returned to its original state without causing a significant change in the surroundings. In the case of a chemical reaction, it means that the reaction does not go to completion.

With regard to chemical reactions, it is well established that reversibility (at constant temperature and pressure) is determined by the $\Delta G^0$ for the reaction, and, as you said, the more negative $\Delta G^0$ is, the closer the reaction will get to being irreversible (i.e., closer to going to completion).