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Offline galpinj

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Gibbs Free Energy and Spontaneity
« on: July 31, 2016, 06:09:32 PM »
I'm having a problem with Gibbs energy on a conceptual level.

Based on what I've read, if Gibbs is negative, the reaction is spontaneous. Does this mean that, over time, all the reactants will be used up to form the products? I know that, in a reaction, if the [products] is too high, the reactants will be favored; however, I don't see how this type of change impacts Gibbs Free Energy (ΔG = ΔH - TΔS). If Gibbs is negative, then the reaction should continue regardless of concentration, and one should ultimately be left with no reactant, right?

On a related note, how is equilibrium reached in a zero order reaction? If a reaction is occurring independently of the [reactant], then how can it attain any sort of equilibrium? Won't it always end up as all products?
« Last Edit: July 31, 2016, 08:39:19 PM by galpinj »

Offline orthoformate

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Re: Gibbs Free Energy and Spontaneity
« Reply #1 on: August 01, 2016, 12:44:28 AM »
I'm having a problem with Gibbs energy on a conceptual level.

Based on what I've read, if Gibbs is negative, the reaction is spontaneous. Does this mean that, over time, all the reactants will be used up to form the products? I know that, in a reaction, if the [products] is too high, the reactants will be favored; however, I don't see how this type of change impacts Gibbs Free Energy (ΔG = ΔH - TΔS). If Gibbs is negative, then the reaction should continue regardless of concentration, and one should ultimately be left with no reactant, right?

On a related note, how is equilibrium reached in a zero order reaction? If a reaction is occurring independently of the [reactant], then how can it attain any sort of equilibrium? Won't it always end up as all products?

If a reaction is at equilibrium, the ΔG=0. The reason equilibrium reactions have a ΔG value you can reference is because that ΔG quantity is calculated with respect to a [1M] value for all components.

ΔG=-RTlnK+RTlnQ is the full equation, but since if you start with [1M] of all components then Q=1 and the ln(1)=0 so you left with the equation most commonly known ΔG=-RTlnK.

So if a reaction is allowed to reach equilibrium from different starting points we would get different values for ΔG. does this make sense?

To address your post script, I think you are mixing thermodynamic and kinetics. Am I misunderstanding this second question?

Offline mjc123

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Re: Gibbs Free Energy and Spontaneity
« Reply #2 on: August 01, 2016, 05:21:01 AM »
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if Gibbs is negative, the reaction is spontaneous.
Not necessarily. You must distinguish between thermodynamic and kinetic stability. A reaction with negative ΔG is thermodynamically favourable, but might not occur spontaneously because of unfavourable kinetics - a very high Ea, for example.
You and the atmosphere are thermodynamically unstable relative to CO2 and water, but spontaneous human combustion happens either never or extremely rarely (depending whether you believe the accounts of it), so you are OK to make plans for tomorrow.
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Does this mean that, over time, all the reactants will be used up to form the products?
(Assuming from here on that the reaction is kinetically feasible) No. There will be a position of equilibrium defined by ΔG° = -RTlnK; the reaction will stop there. But if the value of ΔG° is large (positive or negative) the equilibrium will be very much towards reactants or products. Theoretically, there will always be some reactants at equilibrium, however negative ΔG° is, short of infinity.
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I know that, in a reaction, if the [products] is too high, the reactants will be favored; however, I don't see how this type of change impacts Gibbs Free Energy (ΔG = ΔH - TΔS). If Gibbs is negative, then the reaction should continue regardless of concentration, and one should ultimately be left with no reactant, right?
This is where it is important to distinguish between ΔG (the actual change in G) and ΔG° (the value for standard states, i.e. all concentrations = 1M. Changing the concentrations does not change ΔG°, but it does affect ΔG. Equilibrium occurs when G for the system is a minimum, i.e. dG/dα = 0 (where α is the extent of reaction). See here for a discussion of this topic: http://www.chemicalforums.com/index.php?topic=79826.0.
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So if a reaction is allowed to reach equilibrium from different starting points we would get different values for ΔG. does this make sense?
This is where you must be careful. The difference in G between equilibrium and starting point will obviously differ for different starting points, but this value is not ΔG°, nor is it the quantity that is equal to 0 at equilibrium. See the thread referred to.

Offline Corribus

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Re: Gibbs Free Energy and Spontaneity
« Reply #3 on: August 01, 2016, 09:16:07 AM »
Not necessarily. You must distinguish between thermodynamic and kinetic stability. A reaction with negative ΔG is thermodynamically favourable, but might not occur spontaneously because of unfavourable kinetics - a very high Ea, for example.
You and the atmosphere are thermodynamically unstable relative to CO2 and water, but spontaneous human combustion happens either never or extremely rarely (depending whether you believe the accounts of it), so you are OK to make plans for tomorrow.
Personally, I would argue that such a reaction still qualifies as spontaneous. If we define a spontaneous reaction as one in which the free energy of the products is lower than the free energy of the reactants, then it is spontaneous no matter how long it takes for the system to reach equilibrium - even if the amount of time is practically infinite under "normal" conditions. In other words, I would argue that, as a purely thermodynamic concept related only to the Gibbs energy change, kinetics has nothing to do with the classification of spontaneity (or nonspontaneity). Just so, as I'm sure you'd agree, concluding that a reaction has a negative Gibbs energy change provides no conclusive information about the reaction speed.
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

Offline galpinj

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Re: Gibbs Free Energy and Spontaneity
« Reply #4 on: August 02, 2016, 01:35:25 AM »
I'm having a problem with Gibbs energy on a conceptual level.

Based on what I've read, if Gibbs is negative, the reaction is spontaneous. Does this mean that, over time, all the reactants will be used up to form the products? I know that, in a reaction, if the [products] is too high, the reactants will be favored; however, I don't see how this type of change impacts Gibbs Free Energy (ΔG = ΔH - TΔS). If Gibbs is negative, then the reaction should continue regardless of concentration, and one should ultimately be left with no reactant, right?

On a related note, how is equilibrium reached in a zero order reaction? If a reaction is occurring independently of the [reactant], then how can it attain any sort of equilibrium? Won't it always end up as all products?

If a reaction is at equilibrium, the ΔG=0. The reason equilibrium reactions have a ΔG value you can reference is because that ΔG quantity is calculated with respect to a [1M] value for all components.

ΔG=-RTlnK+RTlnQ is the full equation, but since if you start with [1M] of all components then Q=1 and the ln(1)=0 so you left with the equation most commonly known ΔG=-RTlnK.

So if a reaction is allowed to reach equilibrium from different starting points we would get different values for ΔG. does this make sense?

To address your post script, I think you are mixing thermodynamic and kinetics. Am I misunderstanding this second question?

Thank you everyone for the responses! I feel like I'm getting a much better grasp on the topic, though still a little shaky in some places. I've spent a lot of time looking at the equation posted by orthoformate (ΔG=-RTlnK+RTlnQ). Can I ask if you know how this equation was derived? I understand that at standard conditions, RTlnQ will equal zero, and therefore ΔG=-RTlnK; however, I don't know how this connects with with the equation ΔG = ΔG°+RTlnQ. I understand that, if ΔG° = -RTlnKeq, then I can just input this to get the equation ΔG=-RTlnK+RTlnQ, but I'm still having conceptual problems with the initial part. Based off the equation ΔG = ΔG°+RTlnQ, ΔG° could also equal ΔG - RTlnQ, which makes absolutely no sense to me!

Offline Corribus

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Re: Gibbs Free Energy and Spontaneity
« Reply #5 on: August 02, 2016, 09:23:01 AM »
You can find a derivation here, under the section heading of Thermodynamics.

https://en.wikipedia.org/wiki/Chemical_equilibrium

You'll have to start with the fact that Gibbs energies are basically related to chemical potentials and activities.

https://en.wikipedia.org/wiki/Thermodynamic_activity

Based on your posts so far, this may be a hard derivation for you to follow at your current level. My recommendation would be to focus your energies on what the equation means to chemical systems rather than where it comes from.
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

Offline galpinj

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Re: Gibbs Free Energy and Spontaneity
« Reply #6 on: August 02, 2016, 12:48:12 PM »
Yes the derivation does seem above my level. Could you explain, on a conceptual level, why we include ΔG° In the equation ΔG = ΔG° + RTln(Q)?  Why does the change in Gibbs energy at standard conditions have to be included whenever we want to determine ΔG for a system?

Offline Corribus

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Re: Gibbs Free Energy and Spontaneity
« Reply #7 on: August 02, 2016, 01:45:52 PM »
Could you explain, on a conceptual level, why we include ΔG° In the equation ΔG = ΔG° + RTln(Q)?  Why does the change in Gibbs energy at standard conditions have to be included whenever we want to determine ΔG for a system?

You may check out this thread:

http://www.chemicalforums.com/index.php?topic=69815.0
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

Offline galpinj

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Re: Gibbs Free Energy and Spontaneity
« Reply #8 on: August 02, 2016, 03:03:32 PM »
Could you explain, on a conceptual level, why we include ΔG° In the equation ΔG = ΔG° + RTln(Q)?  Why does the change in Gibbs energy at standard conditions have to be included whenever we want to determine ΔG for a system?

You may check out this thread:

http://www.chemicalforums.com/index.php?topic=69815.0

Your response in the other forum was outstanding! Really helped me understand the difference between ΔG° and ΔG! I am still a little unclear on one point, and I'm really hoping you can help clarify it for me.

Given that ΔG = ΔG° + RTln(Q), and ΔG° = -RTln(K), ΔG = -RTln(K) + RTln(Q)
This makes sense, and if we were in standard conditions (everything held at 1M), then ΔG = -RTln(K) + RTln(1/1), which would mean that ΔG = -RTln(K) (in this instance, ΔG is essentially ΔG° (i.e. ΔG° = ΔG° + 0)

I also understand that when different concentrations are used the ΔG will deviate from ΔG° (because the  RTln(Q) value will change) , and we can use this different number for ΔG to predict what will happen in the reaction. (All of this follows from your post in the other forum, Corribus).

However, the other forum also asks how the equation ΔG = ΔG° + RTln(Q) is derived/related to ΔG = ΔH - TΔS. I have read a lot about ΔG = ΔH - TΔS, and I understand how it measures thermal energy transfer between the system and the environment; however, I still can't understand how  ΔG = ΔG° + RTln(Q) was derived, particularly why we are relying on ΔG°. I completely agree that the equation works (as just shown), but I still don't understand why we would include ΔG°. Why does the free energy at standard conditions matter/impact the results for other free energy calculations? If I want to calculate the free energy for a reaction at a certain time with a certain concentration/temperature, why should I care about/include the free energy of the equation at standard conditions? It seems somewhat random to throw that into the equation.

Edit: Does it somehow fit into this derived form of the Gibbs Equation? ΔS(universe) = - ΔH/T(system) + ΔS(system) ; Is ΔG° perhaps a representation of the change in entropy of the system (i.e. the reversible process)?

Offline Corribus

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Re: Gibbs Free Energy and Spontaneity
« Reply #9 on: August 02, 2016, 03:18:42 PM »
If you have a value for ΔG for a system, this is sufficient to tell you how the system will evolve. But how do you know what ΔG is?

The value of the equation ΔG = ΔG° + RT ln(Q) is that it provides an estimate for ΔG (and thus information on how a system will likely evolve) based only on the reaction quotient (relative amounts of the reactants and products at the starting point) and a reference value ΔG°, which can generally be determined by consulting tables of thermodynamic data.

Students often have a difficult time distinguishing between standard thermodynamic values (those with the °) and non-standard ones, and understanding what their respective use and value is. The non-standard ones are useful toward predicting the behavior of "real" systems. Of course these values can be determined experimentally, but this undermines the purpose of prediction. By using thermodynamic quantities determined at standardized conditions, which can be looked up, we can project what non-standard thermodynamic values are likely to be - and therefore predict the behavior of the chemical system at any possible starting point.

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

Offline galpinj

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Re: Gibbs Free Energy and Spontaneity
« Reply #10 on: August 02, 2016, 07:57:07 PM »
If you have a value for ΔG for a system, this is sufficient to tell you how the system will evolve. But how do you know what ΔG is?

The value of the equation ΔG = ΔG° + RT ln(Q) is that it provides an estimate for ΔG (and thus information on how a system will likely evolve) based only on the reaction quotient (relative amounts of the reactants and products at the starting point) and a reference value ΔG°, which can generally be determined by consulting tables of thermodynamic data.

Students often have a difficult time distinguishing between standard thermodynamic values (those with the °) and non-standard ones, and understanding what their respective use and value is. The non-standard ones are useful toward predicting the behavior of "real" systems. Of course these values can be determined experimentally, but this undermines the purpose of prediction. By using thermodynamic quantities determined at standardized conditions, which can be looked up, we can project what non-standard thermodynamic values are likely to be - and therefore predict the behavior of the chemical system at any possible starting point.

Thank you once again Corribus for all the help.
So, is the ΔG in ΔG = ΔG° + RT ln(Q) different from the ΔG  in ΔG = ΔH - TΔS? Could I find a ΔG value using an equation like ΔG =RT ln(Q) (ignoring  ΔG°)? I can definitely see how adding ΔG° is important when we want to compare what a reaction will do with its equilibrium condition, but is it actually necessary to include that value (ΔG°) when determining  ΔG?

Edit:
I've spent a couple hours researching this topic, and still feel lost; however I have some more info to add. ΔG° = ΔH° - TΔS°, therefore  ΔG° = -RT ln(Keq) = ΔH° - TΔS°
This means that ΔG° can be replaced in ΔG = ΔG° + RT ln(Q) to give ΔG =ΔH° - TΔS° + RT ln(Q). This really makes me think that ΔG should itself equal RT ln(Q) without influence by ΔG° . If I found the value of ΔG to be -90KJ through ΔG = ΔH - TΔS, then would it be the same if I used the equation ΔG = ΔG° + RT ln(Q)? If so, why?

I'm really grateful for any help
« Last Edit: August 02, 2016, 09:42:12 PM by galpinj »

Offline orthoformate

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Re: Gibbs Free Energy and Spontaneity
« Reply #11 on: August 03, 2016, 01:22:19 AM »
You can say: ΔH-TΔS=-RTlnK+RTlnQ

Consider this: you are conducting a simple [A] :rarrow: [ B] reaction with [1M] starting concentrations for both [A] and [ B] at S.T.P.

The equilibrium constant for this reaction is 3, meaning that 0.5M of [A] converted to [ B] giving final concentrations of [A]=0.5M and [ B]=1.5 M, hence [ B]/[A]=3=K

This means that 0.5 mols of [A] were converted to [ B] and you must consider this in your calculations for ΔH-TΔS. ΔH as well as ΔS are on a per/mol basis, ΔH and ΔS are extensive properties that increase with the amount of mols you convert.


So, if you wanted to convert more mols that just [0.5] you would have to add the term RTlnQ to ΔG=-RTlnK+RTlnQ. Since at standard conditions you can never convert more than 0.5 mols given the equilibrium constant.

Does this help?

Offline mjc123

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Re: Gibbs Free Energy and Spontaneity
« Reply #12 on: August 03, 2016, 10:27:42 AM »
You are confusing concentrations and amounts. If you had 1000 L of 1M slution, would you convert 0.5 mol? Or if you had 1 mL? But is K different for the 3 scenarios?
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ΔH as well as ΔS are on a per/mol basis, ΔH and ΔS are extensive properties that increase with the amount of mols you convert
These two statements are contradictory. The enthalpy you would measure for a particular system, e.g. with a calorimeter, is a quantity in J which will vary with the size of the sample. The molar heat of reaction, which is what you will find in tables, is an intensive quantity in J/mol, which doesn't vary with size of sample. ΔH = nΔHm
Really we should put subscript m on molar quantities, but it is tedious and generally assumed.
Which one do we use in ΔG° = -RTlnK? (Hint: what are the units of R?)

Offline orthoformate

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Re: Gibbs Free Energy and Spontaneity
« Reply #13 on: August 03, 2016, 12:09:58 PM »
No, you are correct, K is constant no matter what concentration or volume you begin with. If we had only 1 L though in my example, wouldn't it be a correct way of demonstrating the link between the two equations?

What do you mean when you say which one? which one of what?: "Which one do we use in ΔG° = -RTlnK? (Hint: what are the units of R?)"


Offline mjc123

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Re: Gibbs Free Energy and Spontaneity
« Reply #14 on: August 03, 2016, 12:36:07 PM »
Extensive (J) or intensive (J/mol)?

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