July 07, 2022, 04:39:31 AM
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Topic: Understanding Gibbs free energy relating to equilibrium constant  (Read 355 times)

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Roxo

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Understanding Gibbs free energy relating to equilibrium constant
« on: May 06, 2022, 09:44:12 AM »
I am learning chemical thermodynamics and one of my resources is this link: http://rkt.chem.ox.ac.uk/tutorials/equilibrium/entropy_mixGas.html

You will see it eventually arrives at the expression ΔG = RT lnK
I think this is wrong and should be ΔG = - RT lnK
Have I made an error in following their maths or have they omitted the negative sign?

I have a second question as follows:
Assuming the correct relationship is ΔG = - RT lnK , then it seems to me clear that any fractional value for K will result in positive ΔG. My understanding of this is that the reaction is therefore not spontaneous (feasible). However if there is a value for K then surely there must have been a reaction to enable K to have a value. It seems contradictory to me that on the one hand a reaction must have taken place (to reach an equilibrium) but on the other hand ΔG is positive so the reaction cannot take place. I am confused about the interpretation of ΔG in this case. I realise that once equilibrium is reached then ΔG = zero but from the initial conditions to reaching equilibrium some reaction has taken place. I am sure I am missing something here but none of my reference resources nor textbooks address this.
I worked out an example 2HI  H2 + I2 all in gas phase has K= 0.0218 resulting in ΔG = +3.83 (RT)   Under standard conditions ΔH is -26.5 kJ/mol. So we have an exothermic reaction arriving at an equilibrium but with a negative free energy which suggests the reaction is not feasible. I'm very confused.

Corribus

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Re: Understanding Gibbs free energy relating to equilibrium constant
« Reply #1 on: May 06, 2022, 10:06:04 AM »
(1) Yes it appears to be a typo. Since the line before simplifies to

$$\Delta G^o + RT [\ln{\alpha}-\ln{(1-\alpha)}] = \Delta G^o + RT \ln \frac {\alpha}{1-\alpha} = 0$$

(2) You appear to be confusing ΔG with ΔG°. Note, the article you linked to gives an expression for ΔG°, which determines a frame of reference for the reaction (what the concentrations of reactants and products will be at equilibrium). The value of ΔG for any starting point, determined by the reaction quotient Q in terms of the actual concentrations of the reactants and products, will describe - relative to the equilibrium point - whether the system will proceed in a forward or backward direction.

ΔG does not determine "feasibility". It is a way of describing how a system will evolve from a specific point. I.e., where is the starting point relative to the equilibrium point.
« Last Edit: May 06, 2022, 10:16:06 AM by Corribus »
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Babcock_Hall

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Re: Understanding Gibbs free energy relating to equilibrium constant
« Reply #2 on: May 06, 2022, 10:28:45 AM »
Obviously I defer to Corribus regarding the physical chemistry.  I think that different people may use the same word differently.  I try to use the terms "spontaneous" and "non spontaneous" for negative and positive values of ΔG, respectively.  I use the words "favorable" and "unfavorable" for negative and positive values of ΔG°.  I don't know how I would define or use the word "feasibility," but it seems to me that you are using it as a synonym for spontaneous.

It is possible for a reaction to have a negative value of ΔG and a positive value of ΔG°.

Corribus

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Re: Understanding Gibbs free energy relating to equilibrium constant
« Reply #3 on: May 06, 2022, 11:46:10 AM »
Possibly a language/translation issue, not sure. Given the definition of the word "feasible" in common English, I agree that "spontaneous" and "favorable" are probably better descriptors, since feasible could be taken to imply (erroneously) that reaction events do not happen if the ΔG is positive. It's worth bearing in mind that thermodynamic values like ΔG are generally framed for large ensembles of molecules. I.e., they represent the average (expected) behavior.
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

Roxo

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Re: Understanding Gibbs free energy relating to equilibrium constant
« Reply #4 on: May 07, 2022, 05:19:49 AM »
Many thanks for your replies. To be honest I can't say the penny has completely dropped for me yet and I still have some confusion about this but your comments have certainly been helpful and given me more to think about. I'm going to work on this some more and perhaps come back for a bit more help if I need to.

Thanks again.

Corribus

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Re: Understanding Gibbs free energy relating to equilibrium constant
« Reply #5 on: May 07, 2022, 09:40:46 AM »
The difference between ΔG and ΔG° is something many students struggle with and we get a lot of questions here about it. If you find yourself still struggling with this or other related concepts, you can come back here for more help. Alternatively, I advise trying different textbooks. Some are better than others and they all approach thermodynamic topics in different ways. So if reading one isn't helping it click, try another.
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