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Topic: Gibbs Free Energy and G=Go+RTlnK derivation  (Read 60365 times)

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kev90123

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Gibbs Free Energy and G=Go+RTlnK derivation
« on: July 20, 2013, 02:27:27 PM »
So I know G = H-TS but haven't been able to figure out how this is equation is used to derive G=Go+RTlnK. In classes I've taken the equation is just given to us without derivation and I haven't been able to find any clear answers/derivations. I've just been told its "out of the scope of this class" and what not. Could someone take me through a step by step process of this derivation? Thanks.

Yggdrasil

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Re: Gibbs Free Energy and G=Go+RTlnK derivation
« Reply #1 on: July 20, 2013, 03:39:36 PM »
The derivation from first principles is usually covered in a statistical thermodynamics or statistical mechanics course.  The gist of the derivation relies on an understanding of the patition function, and how this function relates the quantum mechanical properties of the molecules to the thermodynamics of the system.  Essentially, the relationship comes from considering the most favorable way of distributing the energy of the system among the various degrees of freedom of the reactants and products.

Most statistical mechanics text books should cover the derivation if you are interested in reading more.  For example, see chapter 9 of Statistical Mechanics by McQuarrie.

Technicalhuman

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Re: Gibbs Free Energy and G=Go+RTlnK derivation
« Reply #2 on: July 22, 2013, 09:17:23 PM »
The derivation from first principles is usually covered in a statistical thermodynamics or statistical mechanics course.  The gist of the derivation relies on an understanding of the patition function, and how this function relates the quantum mechanical properties of the molecules to the thermodynamics of the system.  Essentially, the relationship comes from considering the most favorable way of distributing the energy of the system among the various degrees of freedom of the reactants and products.

Most statistical mechanics text books should cover the derivation if you are interested in reading more.  For example, see chapter 9 of Statistical Mechanics by McQuarrie.

Hello, may I ask a question about this formula here?

When ΔG=0, ΔG°=-RTlnK so for the T variable, what would the temperature be? It wouldn't be at 298K right? Rather it should be at the temperature where ΔG=0? So unless ΔG=ΔG°=0 then the T can be 298K?

If so, I have a revision exercise question that puzzles me here it is.
3. The overall reaction for the corrosion (rusting) of iron by oxygen is

4 Fe(s)   +   3 O2(g)   ⇌   2 Fe2O3(s)

Using the following data, calculate the equilibrium constant for this reaction at 25°C given R = 8.314 J mol-1K-1.

Substance
DHfo / kJmol-1
So / Jmol-1K-1
Fe2O3 (s)
-826
90
Fe(s)
0
27
O2(g)
0
205

So now they are asking for the K at equilibrium and it is at 298K too. So technically, if what I said was true then ΔG=ΔG°=0. However, when I solved for ΔG° it gave a value that was non zero. Is there something wrong with this question? Or is my concept faulty?

Thank you in advance for the help.

Yggdrasil

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Re: Gibbs Free Energy and G=Go+RTlnK derivation
« Reply #3 on: July 23, 2013, 06:55:18 PM »
The equation here is  ΔG = ΔG° + RT ln Q, which describes how to calculate the free energy change of a reaction.  In order to calculate this ΔG value, you need to know the temperature at which you're running the reaction, the ΔG° of your reaction at the specific temperature (in other words ΔG° changes with temperature), and the concentrations of products and reactants (which you use to calculate Q, the reaction quotient.  Q is simply the concentration of products divided by the concentration of reactants.  For example, for the simple reaction W + X --> Y + Z, the reaction quotient Q = [Y][Z] / [W][X]).

Why do we need these values to know the free energy change of the reaction?  Well, in general there are two major factors that affect the free energy of a chemical reaction.  The first factor is obviously the free energy change associated with breaking bonds in the reactants and reforming the bonds in the products.  This free energy change is largely accounted for by ΔG°.

However, the relative concentrations of products and reactants also affect the free energy change of a reaction.  For example (ignoring any differences in free energy between the products and reactants), if all the particles in the system are reactants, the entropy of the system is much lower than if the system contained a mix of products and reactants.  Therefore, the forward reaction is much more favorable at high concentrations of products and low concentrations of reactants.  The RT ln Q term reflects this entropic contribution to the overall free energy change of the reaction.

Now, at equilibrium we know that the forward and reverse reactions are equally favorable, so ΔG = 0.  Plugging this into the original equation, we can find the relationship:

ΔG° = -RT ln K

Where K, the equilibrium constant, is simply the reaction quotient for a system at equilibrium.

« Last Edit: July 23, 2013, 09:54:33 PM by Yggdrasil »

Technicalhuman

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Re: Gibbs Free Energy and G=Go+RTlnK derivation
« Reply #4 on: July 24, 2013, 08:39:31 AM »
The equation here is  ΔG = ΔG° + RT ln Q, which describes how to calculate the free energy change of a reaction.  In order to calculate this ΔG value, you need to know the temperature at which you're running the reaction, the ΔG° of your reaction at the specific temperature (in other words ΔG° changes with temperature), and the concentrations of products and reactants (which you use to calculate Q, the reaction quotient.  Q is simply the concentration of products divided by the concentration of reactants.  For example, for the simple reaction W + X --> Y + Z, the reaction quotient Q = [Y][Z] / [W][X]).

Why do we need these values to know the free energy change of the reaction?  Well, in general there are two major factors that affect the free energy of a chemical reaction.  The first factor is obviously the free energy change associated with breaking bonds in the reactants and reforming the bonds in the products.  This free energy change is largely accounted for by ΔG°.

However, the relative concentrations of products and reactants also affect the free energy change of a reaction.  For example (ignoring any differences in free energy between the products and reactants), if all the particles in the system are reactants, the entropy of the system is much lower than if the system contained a mix of products and reactants.  Therefore, the forward reaction is much more favorable at high concentrations of products and low concentrations of reactants.  The RT ln Q term reflects this entropic contribution to the overall free energy change of the reaction.

Now, at equilibrium we know that the forward and reverse reactions are equally favorable, so ΔG = 0.  Plugging this into the original equation, we can find the relationship:

ΔG° = -RT ln K

Where K, the equilibrium constant, is simply the reaction quotient for a system at equilibrium.

Hi thank you for the informative post. Sorry for the late reply.

From what I learned, ΔG° is at standard state where 298K so say equilibrium is reached at 298K. So now wouldn't my ΔG=ΔG°=0? So using the formula, ΔG=ΔG°+RTlnK, 0=0+R(298K)lnK so in this case shouldn't R(298K)lnK be 0 too?

But in the question given, even though they stated that equilibrium is reached at 298K, the ΔG=/=0. So I'm wondering if there's a problem with the question. Hope you can help out. Thanks.

Thanks for the great response.

Corribus

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Re: Gibbs Free Energy and G=Go+RTlnK derivation
« Reply #5 on: July 24, 2013, 10:46:21 AM »
Technicalhuman -

Practically speaking, ΔG° is basically a reference value which specifies what the relative concentrations of products and reactants are at equilibrium for a particular temperature.  As such, ΔG° can be related to the equilibrium constant by the equation ΔG° = -RT ln K.  Recall that the equilibrium constant is essentially determined by the relative thermodynamic activities/potentials of the reactants and products, which is why it can be related to the Gibbs energy (also an expression of chemical potential).  As a reference value for potential energy, ΔG° can also be determined by heats of formation and entropies of formation (themselves determined relative to reference values), which is what your problem wants you to do.  In other words, you can predict the value of the equilibrium constant for a process just by knowing the amount of heat energy released/absorbed through breaking/forming of bonds as well as the change in entropy, which itself is sort of a measure of the difference in the number of ways the available heat energy can be distributed.

So, to sum up here, you calculate ΔG° from ΔG° = ΔH° - TΔS°, and you calculate ΔH° and ΔS° in the usual way, using tabulated values for heats of formation and entropies of formation using a Hess law formalism.

Once you have ΔG°, this serves as a reference point for any other condition that the system may find itself in.  Let's say for a simple reaction A B, the equilibrium constant (at temperature 298 K) is 10.  This means that at equilibrium, at 298 K, the concentration of B is ten times what the concentration of A is (assuming we can approximate the activites as concentrations, which is generally the case in solution).  Using the equation ΔG° = -RT ln K, we can also determine that ΔG° for this reaction at this temperature is -5708 J/mol.  This tells us that the conversion of A to B at room temperature will tend to be exergonic, that is, A has a higher chemical potential than B.  Most of the time we do things in the opposite direction, though: if you start with ΔG°, calculated from heats and entropies of formation, you can learn what the relative values of A and B are once equilibrium is reached, at 298 K.  You can usually easily get an estimate of ΔG° from tabulated thermodynamic reference values, and hence estimate what the equilibrium constant is likely to be.

What information we really want to know as chemists, however, is this: if I put a certain amount of A and a certain amount of B in a reaction pot, what is likely to happen?  This is where ΔG comes in.  From ΔG°, we can determine what the relative concentrations are going to be for A and B at equilibrium, and we already know, from above, that [B.] will be ten times that of [A].  We can guess from intuition that if we put a higher relative concentration of B in the pot than 10:1, we're going to generate more A (reaction will go backward), because we know the system is always going to approach equilibrium and form a 10:1 ratio, and the only way to do that is to reduce the amount of B and increase the amount of A.  If we put a lower relative concentration of B in than 10:1, we are going to form more B - reaction will go forward.  We express the real relative concentrations of B to A as the reaction quotient, Q.  And from Q, using the reference value of ΔG°, we can determine a ΔG, which is the potential energy of the system when the reactants and products have any particular relative activities.  ΔG quantifies the expected reaction direction.

So if we start out with a ratio of B to A of 5, Q = 5 and thus (using the ΔG° value above and the equation ΔG = ΔG° + RT ln Q) ΔG = -1718.3 J/mol.  A negative ΔG means the specified process (conversion of A to B under the conditions specified) is exergonic and will happen spontaneously.  Which makes sense: when the ratio of B to A is 5, this means we have less B (relative to A) than we know we should have at equilibrium.  To go toward equilibrium, we need to form more B, so there should be a spontaneous formation of A to B.  Hence ΔG we expect to be negative for A B.

On the other hand, if we start with a ratio of B to A of 20, Q = 20 and ΔG = 1717.49 J/mol.  A positive ΔG means the specified process (conversion of A to B under the conditions specified) is endergonic and will not happen spontaneously. This also makes sense: when the ratio of B to A is 20, this means we have MORE B (relative to A) than we know we should have at equilibrium.  To go toward equilibrium, then, we need to form more A, so there should not be spontaneous formation of A to B (in fact, there should be spontaneous formation of B to A).  Hence ΔG we expect to be positive for A B.

Of course, if we start with a ratio of B to A of 10, Q = 10 and ΔG = 0.  This shouldn't be surprising because we are already at equilibrium!  There's no driving force to convert A to B or B to A.

So you see, ΔG gives us a real measure of the thermodynamic driving force (difference in potantial energy between reactants and products) for any particular set of starting conditions.  ΔG° is a reference value that basically specifies what the equilibrium point is.  By calculating ΔG, it will be possible to predict how the system will behave - is more A likely to be formed or is more B likely to be formed, relative to the starting point, once equilibrium is reached?  Note also that the system is dynamic: Q will tend to change until it reaches the equilibrium constant K, at which point ΔG is zero and no more gross change will be observed... unless the system is disturbed (temperature change, more A added, etc).  Note also that ΔG doesn't say anything specific about the rate at which equilibrium will be reached, because kinetics is not determined exclusively by thermodynamic driving force.

I hope this clears up the difference between ΔG and ΔG°.  ΔG is equal to zero when the system is at equilibrium, but ΔG° is a reference value that doesn't change for a specific set of conditions (temperature, e.g.).
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

Technicalhuman

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Re: Gibbs Free Energy and G=Go+RTlnK derivation
« Reply #6 on: July 25, 2013, 04:57:19 AM »
Hello Corribus, thanks for the great post. I have some queries regarding this.

If I were to say that at 1000K, equilibirum is reached does it mean at another temperature, the reaction wouldn't be a reversal one anymore?

And also for the example you gave where equilibirum. Is reached at 298K, my ΔG=0 but since the ° in ΔG° represents the Gibbs free energy at 298K too, so shouldn't the ΔG°=0 as well? So won't I have to use 0=-R(298K)lnK to solve for K?

Thank you for the great help.
« Last Edit: May 29, 2017, 10:49:35 AM by Arkcon »

Yggdrasil

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Re: Gibbs Free Energy and G=Go+RTlnK derivation
« Reply #7 on: July 25, 2013, 09:07:51 AM »
ΔG° represents the free energy change associated with a reaction when the reactants and products are at their standard concentrations (usually taken to be 1M).  Unlike in other cases, the o refers to standard concentrations rather than standard temp and pressure.  The reaction will have a different change in free energy if the concentrations of reactants and products are different.

Corribus

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Re: Gibbs Free Energy and G=Go+RTlnK derivation
« Reply #8 on: July 25, 2013, 10:08:35 AM »
Technicalhuman,

Perhaps an analogy will help.  Suppose you have a spring with one end attached to a wall or other solid surface.  If you know anything about springs, you know that there is a certain rest length where the spring is most happy.  If you stretch the spring out longer than this rest position, it pulls back.  If you compress the spring so it is shorter than this rest position, it pushes out.  In other words, there's a restoring force which acts opposite to the direction that you displace the spring from it's rest position.  This restoring force, captured by Hooke's Law, can also be expressed as a potential energy.

Anyway, now let us say you know nothing about this spring, and I asked you what will happen to the spring if I make it twelve inches long.  And what will happen if I make it six inches long?  eight inches?  20 inches?  2 inches?  In each of those cases, will the spring tend to shorten or lengthen?

It is impossible to answer this without knowing what the rest position is.  If I say that the rest (equilibrium) position is eight 7 inches, now you can easily say that when the spring is eight, twelve or twenty inches long, it will shorten, and when it is six or two inches long, it will lengthen.  The magnitude of the restoring force, as you can guess, is also somehow inversely related to the difference between the actual position and the rest position.  And if I give you a second spring, it will likely have a different rest position - say, 10 inches - which will specify how this second spring behaves.  If both springs are set to eight inches long, the first will shorten but the second will lengthen, because the relationship between the starting position and the rest position is different in the two cases.

One more point of elaboration, but it's important to make the analogy strictly correct.  Implicitly when we're comparing actual displacements to the reference displacement, what we're really doing is comparing forces and/or potential energies.  If the actual length is greater than the rest length, we know the force has a negative sign. In Hooke's Law, F = -k(x-xo); where x = the actual displacement and x0 is the reference displacement. And k is a constant for the spring. Usually we set the xo to zero for convenience so we can speak of relative displacements, but for the analogy to work better, I'm not doing that here.  Force, in other words, is a vector related to real, positional space.  Notice if we expand Hooke's law, we can say that F = -kx + kx0.  In other words, the force the spring experiences for any displacement is determined by an actual force (-kx), which depends on the actual displacement, and a "reference" force (kx0), which is always the same no matter what the actual displacement is.  This "reference force" basically puts the actual force in context, and depends only on the rest position.  Since the rest position doesn't change, neither does this reference force.  Without the reference force, the actual force has no useful meaning.

I think you will agree this is all pretty much common sense.

In this analogy, a chemical system is the spring.  K is the rest position of the spring.  Q is the actual position of the spring.  ΔG° relates to the reference force of the spring.  ΔG relates to the actual force of the spring.  By knowing ΔG°, we can easily predict how the system will be have given any test condition, embodied as ΔG.  By combining ΔG° - which doesn't change for a given set of external conditions - with ΔG, using K and Q as our "rest position" and "actual position" of the system, the actual force on the system is determined.  Chemists often speak of a "driving force" for a reaction.  This is why.  In a way, ΔG is a vector, but the space we're in isn't positional; it's a reaction coordinate.  The directions are "toward products" and "toward reactants".

One more thing.  Going back to the spring for a moment, what happens if we change the temperature of the spring?  As you can imagine, what happens most likely is the spring constant changes.  When the temperature goes up, maybe the material the spring is made of weakens.  This relaxes the spring and increases the magnitude of the rest position.  This basically "recalibrates" the system.  If the rest position is ten inches for the higher temperature, this will change the behavior of the restoring force as a function of displacement.  Take for example a displacement of 8 inches.  At the lower temperature, eight inches is longer than the reference point, so the restoring force will be negative.  At the higher temperature, eight inches is shorter than the new reference point (ten inches), so now the restoring force will be positive.  The reference point, thus, can change as a function of temperature, and so when you go to a new temperature, the equilibrium of system is altered.

Back to a chemical reaction: when you change the temperature, the equilibrium point also changes.  Instead of changing the spring constant, we change the equilibrium constant, which manifests as a change in the reference force, ΔG°.  For a certain Q - at a lower temperature the driving force might be forward because Q<K.  However at a higher temperature, there is a new K, and thus a new ΔG°, and in this case the same Q might be greater than K, which would cause a backward driving force.  This is why temperature can impact chemical equilibrium.

Now I do feel compelled to add that, as with all analogies, this one certainly isn't perfect.  But maybe being able to visualize how it works in something a little less abstract than a generic chemical reaction might help you understand what's going on.
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

Technicalhuman

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Re: Gibbs Free Energy and G=Go+RTlnK derivation
« Reply #9 on: July 27, 2013, 02:29:54 AM »
ΔG° represents the free energy change associated with a reaction when the reactants and products are at their standard concentrations (usually taken to be 1M).  Unlike in other cases, the o refers to standard concentrations rather than standard temp and pressure.  The reaction will have a different change in free energy if the concentrations of reactants and products are different.

Hi Yggdrasil. Thank you for the great response.

Oh so the ΔG° is not the same as ΔG if the reaction is carried out at standard conditions? So what temperature would ΔG° be carried out then?

I found a yahoo answers regarding this and it seems to contradict http://answers.yahoo.com/question/index?qid=20101105201638AAJXwis so I'm not too sure how to go about this. Initially, I had the same thoughts as the yahoo answers person whereby ΔG° is at standard conditions. So I thought if ΔG is 0 at 298K then ΔG° should also be 0. But it seems that that's not the case in my questions.

Thanks so much.

Hi Corribus, sorry but I'm not too good with visualizing but I think I got the meaning. But I was wondering if ΔG=0 at standard conditions won't the reference point (ΔG°) be the same as the actual position (ΔG)?

Thanks for the insightful post it helped me a lot.

« Last Edit: July 27, 2013, 02:52:59 AM by Technicalhuman »

Yggdrasil

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Re: Gibbs Free Energy and G=Go+RTlnK derivation
« Reply #10 on: July 27, 2013, 11:45:08 AM »
The ΔG of a reaction depends not only on the temperature and pressure at which the reaction takes place, but also the concentrations of reactants and products.  ΔGo represents the ΔG of the reaction at standard temperature and pressure AND with the concentrations of all reactants and products at 1M.

Under these conditions, you can show using the relationship ΔG = ΔG° + RT ln Q, that ΔG = ΔG° when the reactants and products are all at 1M concentration.  For example, for the reaction A --> B, the reaction quotient Q = [ B ]/[A].  If both are at 1M concentration then Q = 1 and RT ln Q = 0, so ΔG = ΔG°.

Below is an illustration of the relationship between the concentration of product in a reaction and the ΔG of the reaction.  It assumes the reaction A --> B and we have somehow fixed [A] at 1M while varying the concentration of B.  When [ B ] = 1M, the ΔG of the reaction is equal to ΔGo.  The concentration of B where ΔG = 0 tells you the concentration of B at equilibrium.
« Last Edit: July 27, 2013, 01:31:35 PM by Yggdrasil »

Technicalhuman

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Re: Gibbs Free Energy and G=Go+RTlnK derivation
« Reply #11 on: July 28, 2013, 02:10:57 AM »
The ΔG of a reaction depends not only on the temperature and pressure at which the reaction takes place, but also the concentrations of reactants and products.  ΔGo represents the ΔG of the reaction at standard temperature and pressure AND with the concentrations of all reactants and products at 1M.

Under these conditions, you can show using the relationship ΔG = ΔG° + RT ln Q, that ΔG = ΔG° when the reactants and products are all at 1M concentration.  For example, for the reaction A --> B, the reaction quotient Q = [ B ]/[A].  If both are at 1M concentration then Q = 1 and RT ln Q = 0, so ΔG = ΔG°.

Below is an illustration of the relationship between the concentration of product in a reaction and the ΔG of the reaction.  It assumes the reaction A --> B and we have somehow fixed [A] at 1M while varying the concentration of B.  When [ B ] = 1M, the ΔG of the reaction is equal to ΔGo.  The concentration of B where ΔG = 0 tells you the concentration of B at equilibrium.

Hi Yggdrasil thanks for the informative post.

I think I better understand this now. Would it be right to say that in the question even though ΔG is at 298K and presumably 1 bar, it is not stated that it is at 1M so we cannot assume ΔG=ΔG°?

But actually, if it isn't at 1M why wouldn't the ΔG=ΔG°? I thought the concentration of the reactants doesn't affect the Keq of a reversible reaction?

Thanks for the help.

Yggdrasil

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Re: Gibbs Free Energy and G=Go+RTlnK derivation
« Reply #12 on: July 28, 2013, 04:20:25 PM »
When a system is at equilibrium, ΔG = 0.

From the relation ΔG = ΔG° + RT ln Q, we can see that at equilibrium, ΔG° = –RT ln Qeq.  The reaction quotient of a reaction at equilibrium Qeq is by definition the equilibrium constant Keq.  Therefore, to calculate the equilibrium constant for a particular reaction, you only need to know the temperature of the reaction, and the ΔG° for the reaction at that temperature:

Keq = exp(-ΔG°/RT)

ΔG depends on the concentration of reactants and products, but because ΔG does not factor directly into the above equation for Keq, Keq does not depend on the concentration of products and reactants.

Technicalhuman

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Re: Gibbs Free Energy and G=Go+RTlnK derivation
« Reply #13 on: July 29, 2013, 07:54:44 AM »
When a system is at equilibrium, ΔG = 0.

From the relation ΔG = ΔG° + RT ln Q, we can see that at equilibrium, ΔG° = –RT ln Qeq.  The reaction quotient of a reaction at equilibrium Qeq is by definition the equilibrium constant Keq.  Therefore, to calculate the equilibrium constant for a particular reaction, you only need to know the temperature of the reaction, and the ΔG° for the reaction at that temperature:

Keq = exp(-ΔG°/RT)

ΔG depends on the concentration of reactants and products, but because ΔG does not factor directly into the above equation for Keq, Keq does not depend on the concentration of products and reactants.

Hi thank you for the great post.

I think I get why ΔG depends on the number of moles as if I had two times the number of moles the ΔH and ΔS would be doubled so the ΔG would also be doubled.

But in the reaction the actual ΔG would be for 2 moles of Fe2O3 produced while for ΔG° it is the formation of 1 mole of Fe2O3 instead. So if the ΔG° and ΔG are both at 298K and 1 bar, since ΔG=0 and ΔG°=ΔG/2 shouldn't ΔG°=0 too? I don't quite get why the two change in Gibbs free energy should be different actually.

Thanks so much for the help.

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Re: Gibbs Free Energy and G=Go+RTlnK derivation
« Reply #14 on: July 29, 2013, 08:12:50 AM »