August 03, 2024, 07:24:02 AM
Forum Rules: Read This Before Posting

### Topic: Gibbs Free Energy and G=Go+RTlnK derivation  (Read 59643 times)

0 Members and 1 Guest are viewing this topic.

#### Technicalhuman

• Regular Member
• Posts: 61
• Mole Snacks: +1/-14
##### Re: Gibbs Free Energy and G=Go+RTlnK derivation
« Reply #15 on: July 30, 2013, 08:32:07 AM »
ΔG°=ΔG/2

I was thinking ΔG° is when 1 mole of products is formed while ΔG would be for the number of moles of products according to the equation. And in the example given, 2 moles of products were formed. So shouldn't ΔG=2ΔG°?

Thanks for the great help.

#### Corribus

• Chemist
• Sr. Member
• Posts: 3521
• Mole Snacks: +540/-23
• Gender:
• A lover of spectroscopy and chocolate.
##### Re: Gibbs Free Energy and G=Go+RTlnK derivation
« Reply #16 on: July 30, 2013, 09:59:45 AM »
Ok, let's try this one more time.  The standard Gibbs energy of formation for a substance, ΔG°f, at a given temperature, is essentially the free energy change for the formation of 1 mole of that substance from its component elements at their standard states.  For a conversion from one substance to another (a chemical reaction), the standard Gibbs energy change, ΔG°, is the difference in standard Gibbs energies of formation of the products and reactants (ΣΔG°f {products} - ΣΔG°f {reactants}).  So, essentially, ΔG° is the change in free energy that would occur from taking a molar equivalent of the reactant, breaking it down into its component elements (at their standard states) and then reassembling to the products.  That is, for a given temperature, it is the total change in enthalpy and entropy that would occur to break down 1 mole of the reactants (and stoichoimetric equivalents) and then reassemble to products.  Obviously, the reaction doesn't actually happen this way, but since ΔG° is a state function, you can choose whatever path you want and the value is the same.

My general chemistry book gives this example:

For the reaction N2(g) + 3H2(g) 2NH3(g)  ΔG° = -33.3 kJ

The ΔG° value represents the change in free energy (enthalpy and entropy at a given temperature) that occurs when 1 mole of nitrogen at 1 atm reacts with 3 moles of hydrogen gas at 1 atm to produce 2 moles of gaseous ammonia at 1 atm.

The equilibrium constant for this reaction is 6.8 x 105.

The value of ΔG° does not change if the amount of stuff in the reaction changes - only ΔG does.  Also note that ΔG depends on the ratio of products to reactants.  Which means if you double the numerator and denominator in the ratio of reactants to products, ΔG does not change.  That is to say, ΔG depends on Q and ΔG° depends on K.

Let's look at three examples using the above reaction:

Example 1: if the pressure of ammonia = 1.00 atm, the pressure of nitrogen = 1.47 atm and the pressure of hydrogen = 1 x 10-2 atm, then Q = 6.8 x 105, and therefore ΔG = 0 because Q = K.  The reaction is already at equilibrium.

Example 2: If the pressure of ammonia and pressure of nitrogen are both 1 atm and the pressure of hydrogen is 1 atm, then Q = 1 and ΔG = ΔG°.  This is because the system is in its standard state (all reactants at a pressure of 1 atm).

Example 3: If the pressures of all the gasses are doubled compared to example 1 (pressure of ammonia is 2 atm, pressure of nitrogen is 2.94 atm and pressure of hydrogen is 2 x 10-2 atm), now Q = 1.7 x 105, and therefore ΔG = -3.458 kJ (using ΔG = ΔG° + RT ln Q and the value of ΔG° = -33.3 kJ).  This tells us that for this specific amount of reactants and products, the reaction is not at equilibrium and will spontaneously form more products in order to go toward equilibrium.  This should make sense: if you double the number of moles of reactants AND products, you are adding more reactants that products becauase of the stoichiometry.  Therefore to get back to equilibrium, you need to form a higher proportion of products.  Hence ΔG is negative.  Note that when I change the number of reactants and products and calculate a new ΔG, I do NOT change the value of ΔG°.  It stays the same because it is a reference value determined with products at standard state.

No matter what you do with the proportions of the reactants and products in the reaction vessel, ΔG° does not change.  The only thing you need to worry about that will change ΔG°is changing the temperature or the nature/identity of the reactants or products.
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

• Regular Member
• Posts: 20
• Mole Snacks: +1/-2
##### Re: Gibbs Free Energy and G=Go+RTlnK derivation
« Reply #17 on: July 30, 2013, 11:59:53 AM »
Actually it is not a very complicated derivation at all. This is it in great detail: http://en.wikipedia.org/wiki/Chemical_equilibrium#Thermodynamics. You will need a little background thermodynamics, could probably cover that as and when you need.

#### Technicalhuman

• Regular Member
• Posts: 61
• Mole Snacks: +1/-14
##### Re: Gibbs Free Energy and G=Go+RTlnK derivation
« Reply #18 on: July 31, 2013, 08:49:20 PM »
Hi Corribus thank you so much for the amazing post. I have a better understanding now but I still feel lacking in certain areas sorry about the draggy post.

For the chemical reaction, ΔG depends on the relative concentrations Q while ΔG° is fixed as the conditions of ΔG° are fixed (concentration is fixed, temperature and pressure). So if now the concentrations, temperature and pressure are the same, ΔG=ΔG° so RTlnQ=0 so Q=1. But this doesn't mean that the reaction is at equilibrium right? The reaction in this case can either be negative and positive. But now the ΔG would tend to go to 0 right?

I'm still not too sure this process takes place from changing from ΔG=/=0 to ΔG=0.

Thanks so much I have gotten better understanding through that post.
« Last Edit: May 29, 2017, 10:51:27 AM by Arkcon »

#### Corribus

• Chemist
• Sr. Member
• Posts: 3521
• Mole Snacks: +540/-23
• Gender:
• A lover of spectroscopy and chocolate.
##### Re: Gibbs Free Energy and G=Go+RTlnK derivation
« Reply #19 on: July 31, 2013, 10:07:41 PM »
If all the reactants and products are such that Q = 1, then ΔG = ΔG°.  However I wouldn't get caught up too much with this specific scenario.  What matters as far as reaction dynamics go is what Q is in relationship to K, or, alternatively, whether ΔG is positive or negative.  If ΔG is nonzero, the mixture is not at equilibrium and the relative concentrations (or partial pressures for gases) will change to bring Q toward K, or ΔG will move toward zero.  Nothing else really matters.

So the general procedure is: from enthalpies/entropies of formation, determine ΔG°.  From ΔG° determine K (if necessary).  This gives you the reference values for the reaction.  Now for the mixture in any state, use the given concentrations of reactants and products to calculate Q.   From this and ΔG°, determine ΔG.  This tells you how the system in its current state will behave.  If ΔG < 0, more products will form.  If ΔG > 0, more reactants will form.

This is all you really need to know.
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

#### curiouscat

• Chemist
• Sr. Member
• Posts: 3006
• Mole Snacks: +121/-35
##### Re: Gibbs Free Energy and G=Go+RTlnK derivation
« Reply #20 on: August 01, 2013, 12:43:49 AM »

So if now the concentrations, temperature and pressure are the same, ΔG=ΔG° so RTlnQ=0 so Q=1. But this doesn't mean that the reaction is at equilibrium right?

Absolutely not. Equilibrium needs ΔG=0

I don't think ΔG=ΔG° has much significance at all.

#### Technicalhuman

• Regular Member
• Posts: 61
• Mole Snacks: +1/-14
##### Re: Gibbs Free Energy and G=Go+RTlnK derivation
« Reply #21 on: August 02, 2013, 08:24:48 AM »
If all the reactants and products are such that Q = 1, then ΔG = ΔG°.  However I wouldn't get caught up too much with this specific scenario.  What matters as far as reaction dynamics go is what Q is in relationship to K, or, alternatively, whether ΔG is positive or negative.  If ΔG is nonzero, the mixture is not at equilibrium and the relative concentrations (or partial pressures for gases) will change to bring Q toward K, or ΔG will move toward zero.  Nothing else really matters.

So the general procedure is: from enthalpies/entropies of formation, determine ΔG°.  From ΔG° determine K (if necessary).  This gives you the reference values for the reaction.  Now for the mixture in any state, use the given concentrations of reactants and products to calculate Q.   From this and ΔG°, determine ΔG.  This tells you how the system in its current state will behave.  If ΔG < 0, more products will form.  If ΔG > 0, more reactants will form.

This is all you really need to know.

Hi Corribus thanks again for the wonderful post.

Would this indicate that every reversible reaction would have a certain equilibirum constant at different temperatures? But if a reaction is non reversible than this equation would be valid for it?

Also, I have a question regarding melting. In my notes we use ΔG=0 so TΔS=ΔH and we solved for T to get the melting point of the ice which gives us 273K.

However, I'm not sure how to apply this to the formula ΔG=ΔG°+RTlnQ. The equation for this reaction would be H2O(l) H2O(s) so it would mean 1 mole of liquid water turns into 1 mole of gaseous ice and all at 1 atm pressure. But now when my temperature is 298K, the ΔG would eventually reach 0 but how can this be possible? And even if it does reach 0 now my temperature of melting is 298K and not 273K? so I'm quite confused about why the "melting points" are different now.

Hope you can clear up this misconception. Thanks so much for all the help.

#### curiouscat

• Chemist
• Sr. Member
• Posts: 3006
• Mole Snacks: +121/-35
##### Re: Gibbs Free Energy and G=Go+RTlnK derivation
« Reply #22 on: August 02, 2013, 08:41:36 AM »

Would this indicate that every reversible reaction would have a certain equilibirum constant at different temperatures?

Yes. It varies with T though.

Quote
But if a reaction is non reversible than this equation would be valid for it?

Technically all reactions are reversible. Only to small or large extents.

Irreversibility is an idealized abstraction.

• Regular Member
• Posts: 20
• Mole Snacks: +1/-2
##### Re: Gibbs Free Energy and G=Go+RTlnK derivation
« Reply #23 on: August 02, 2013, 10:18:08 AM »
However, I'm not sure how to apply this to the formula ΔG=ΔG°+RTlnQ. The equation for this reaction would be H2O(l) H2O(s) so it would mean 1 mole of liquid water turns into 1 mole of gaseous ice and all at 1 atm pressure. But now when my temperature is 298K, the ΔG would eventually reach 0 but how can this be possible? And even if it does reach 0 now my temperature of melting is 298K and not 273K? so I'm quite confused about why the "melting points" are different now.

Hope you can clear up this misconception. Thanks so much for all the help.

don't want to hijack corribus' excellent explanations but basically ΔG will not reach 0 for this reaction if held at 298.15 K, RTP assumed. equilibrium constant still exists but you'll not see Q=K till T comes back to 273K. and at 273K, equilibrium constant is the one for melting you talked about K273. K298 is purely theoretical never reached.

#### Technicalhuman

• Regular Member
• Posts: 61
• Mole Snacks: +1/-14
##### Re: Gibbs Free Energy and G=Go+RTlnK derivation
« Reply #24 on: August 03, 2013, 05:20:30 AM »

Would this indicate that every reversible reaction would have a certain equilibirum constant at different temperatures?

Yes. It varies with T though.

Quote
But if a reaction is non reversible than this equation would be valid for it?

Technically all reactions are reversible. Only to small or large extents.

Irreversibility is an idealized abstraction.

Hello Curiouscat thanks for the reply.

But what about melting? We would say at 0°C equilibrium is reacted so ΔG=0. But even at a temperature above 0 or below 0 shouldn't equilibrium also be reached? But by using ΔH=TΔS we would get a fixed temperature of 273K only so I'm still not sure about why this happens..

Thanks for the help

#### curiouscat

• Chemist
• Sr. Member
• Posts: 3006
• Mole Snacks: +121/-35
##### Re: Gibbs Free Energy and G=Go+RTlnK derivation
« Reply #25 on: August 03, 2013, 08:42:46 AM »
But what about melting? We would say at 0°C equilibrium is reacted so ΔG=0. But even at a temperature above 0 or below 0 shouldn't equilibrium also be reached?

Why? At 1 atm Pressure water can only be equilibrium with ice at  0°C. So, no, at T above or below 0 no equilibrium will result.

• Regular Member
• Posts: 20
• Mole Snacks: +1/-2
##### Re: Gibbs Free Energy and G=Go+RTlnK derivation
« Reply #26 on: August 03, 2013, 09:31:54 AM »
But what about melting? We would say at 0°C equilibrium is reacted so ΔG=0. But even at a temperature above 0 or below 0 shouldn't equilibrium also be reached?

Why? At 1 atm Pressure water can only be equilibrium with ice at  0°C. So, no, at T above or below 0 no equilibrium will result.

In other words T will end up coming to 0 (at 1 atm) at same time as equilibrium is reached.

#### Technicalhuman

• Regular Member
• Posts: 61
• Mole Snacks: +1/-14
##### Re: Gibbs Free Energy and G=Go+RTlnK derivation
« Reply #27 on: August 04, 2013, 05:56:09 AM »
But what about melting? We would say at 0°C equilibrium is reacted so ΔG=0. But even at a temperature above 0 or below 0 shouldn't equilibrium also be reached?

Why? At 1 atm Pressure water can only be equilibrium with ice at  0°C. So, no, at T above or below 0 no equilibrium will result.

I was thinking since for another temperature the reaction might be spontaneous at first but after a while the Q would turn into K then ΔG=0 so shouldn't the reaction go into equilibrium even at a temperature above 0?

Also, while thinking about this I thought of the basic ΔG=ΔH-TΔS, I was wondering if all reactions were to go into zero, how can we plug in values of ΔH, T and ΔS to get the ΔG to determine if its spontaneous? What does that mean since the reaction would go to ΔG=0? Does that mean the initial ΔG? Or something else? So I'm a bit confused on these 2.

Thanks so much for all the help.

#### era1236

• Very New Member
• Posts: 1
• Mole Snacks: +0/-0
##### Re: Gibbs Free Energy and G=Go+RTlnK derivation
« Reply #28 on: May 29, 2017, 08:40:09 AM »
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.

Derivation of this equation is as follows:
dG = Vdp - SdT
If temperature is constant then:
dT = 0 and dG = Vdp
From Clapeyron equation:
V = nRT/p
After inserting V and integrating:
G = RT ln(p^n) + C
C which is the integration constant is equal to standard Gibbs Free Enthalpy G0, so:
G = RT ln(p^n) + G0
If you assume that you sum this free enthalpy all over the reactants (with possitive sign for products and negative for substrates) then you obtain following version of equation:
G= RT ln (K) + G0
because K= (activities of products)/(activities of substrats) raised to appropriate powers which are number of moles of substrates and products present in reaction. Activities or more precisely volatilities of gasses in equilibrium are equal to partial pressures p of gasses, and  for all steady state reactants all activities are equal to 1.
Example:
Reaction of decompositon of metal oxide on 2nd degree of oxidation is as follows:
2MeO(s) = O2(g) + 2Me(s)
K=(a(Me)^2*p(O2)/a(MeO)^2)= p(O2)
because activities of steady state reactants a(MeO) and a(Me) are equal to 1
Don't forget that G0 is (standard free enthalpy of products) / (standard free enthalpy of substrats) and it is not equal to 1 for steady state reactants.
I hope that it helps you.
« Last Edit: May 29, 2017, 09:21:28 AM by era1236 »