In Gibbs Free Energy, why multiply -T?

[bubblewrap]

When you get the equation for the Gibbs free energy, why do you multiply -T? Can't you multiply just T? I thought about this and came up with an answer of my own, which goes like 'Because its Gibbs 'Free' Energy, when you have that free energy ( \Delta G >0) you have the energy to do 'unfavorable' reactions, but when you don't ( \Delta G <0), you can only do 'favorable' reactions" I'm not so sure about this interpretation so I'd like to get some comments for this. Thank you in advance.

 

[DrDu]

I would say the sign of G is fixed by the demand to be an energy like quantity, as it's name suggests. Specifically, if S=0, G=E.

 

[bubblewrap]

I would say the sign of G is fixed by the demand to be an energy like quantity, as it's name suggests. Specifically, if S=0, G=E.

Would the interpretation(?) I wrote above be suitable?

 

[DrDu]

$\Delta G$ tells you whether an isothermal, isobaric reaction to occur is possible or not. I don't know what you mean with "favourable" vs. "unvavourable" reactions.

 

[bubblewrap]

$\Delta G$ tells you whether an isothermal, isobaric reaction to occur is possible or not. I don't know what you mean with "favourable" vs. "unvavourable" reactions.

I haven't heard that term used with Gibbs free energy as I really don't know much about this. But I've learned that when ΔG is positive, a favorable (spontaneous) reaction will occur and non-spontaneous reactions when its negative.

 

[DrDu]

Only if $\Delta G$ is negative, any reaction will occur at all, hence it makes little sense to say that an "unvavourable reaction occurs".

 

[bubblewrap]

Sorry I meant favorable reactions at negative Gibbs Energy. Also are you saying that reactions only occur at negative Gibbs energy?

 

[DrDu]

Yes.

 

[bubblewrap]

But here it says that unfavorable and non-spontaneous reactions occur at positive Gibbs Energy, and my textbook also says it.http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch21/gibbs.php

 

[DrDu]

Where do they say that unvavourable reactions actually "occur"?

PS: There are some point in that presentation I don't like. E.g. a negative value of Delta G does not really mean that a reaction occurs spontaneously. E.g. a mixture of hydrogen and oxigen gas won't react spontaneously to form water, although it is strongly exergonic. Another example is the decomposition of an explosive. These reactions need a trigger and one has to differentiate between thermodynamical possibility and the kinetics of the reaction.

 

[bubblewrap]

You lost me, so they don't occur? But... why?

 

[DrDu]

Maybe you are referring to the following passage: "The change in the free energy of the system that occurs during a reaction measures the balance between the two driving forces that determine whether a reaction is spontaneous."
This is at best misleading. You will not be able to observe a reaction to occur where Delta G increases in the same way as you won't observe a stone jumping up spontaneously gaining it's kinetic energy from the heat of the flor.

 

[bubblewrap]

So what's actually happening? I think my textbook said that reactions occur at negative Gibbs energy also, are they missing something?

 

[bubblewrap]

I mean positive gibbs energy

 

[DrDu]

Get a better textbook. :-)

 

[bubblewrap]

Yeah I too thought the textbook wasn't that good, only covered the basics and didn't explain much. So I'd like to read a more scientific explanation, do you have to provide energy when Gibbs energy is positive? Please bear with me sir, I really want to understand this. :s

 

[bubblewrap]

Get a better textbook. :)

If you give energy to something with negative Gibbs energy, then you can make reactions happen right? But not when you don't give energy.

Like the reaction here; 6CO2+6H2O−→−−−−−−LightEnergyC6H12O6+CO2ΔG∘=+2870kJ/mol

 

[Ygggdrasil]

Ok, so we've established that thermodynamically favorable reactions have a ΔG < 0, and that the two components that determine the free energy change are the change in enthalpy of the system (ΔH) and change in entropy of the system (ΔS). In general, will reactions that absorb heat from the surroundings (ΔH > 0) or reactions that release heat to the surroundings (ΔH < 0) be more thermodynamically favorable? Similarly, will reactions that increase the entropy of the system (ΔS > 0) or decrease the entropy of the system (ΔS < 0) be more thermodynamically favorable? What do these answers tell you about the signs of ΔH and ΔS as they relate to ΔG?

 

[bubblewrap]

What do these answers tell you about the signs of ΔH and ΔS as they relate to ΔG?

I don't know where you're getting at, but I know the relations between ΔH,ΔS and ΔG. When H is negative and S is positive, the reaction is favorable and when H is positive and S is negative, the reaction is unfavorable.

 

[Ygggdrasil]

I don't know where you're getting at, but I know the relations between ΔH,ΔS and ΔG. When H is negative and S is positive, the reaction is favorable and when H is positive and S is negative, the reaction is unfavorable.

Right. So, which equation: ΔG = ΔH +TΔS or ΔG = ΔH -TΔS is consistent with these ideas.

 

[Chestermiller]

A positive standard Gibbs free energy change does not mean that the reaction will not occur at all. It simply means that the conversion at which the reactants will be at equilibrium with the products will be small.

Chet

 

[DrDu]

A positive standard Gibbs free energy change does not mean that the reaction will not occur at all. It simply means that the conversion at which the reactants will be at equilibrium with the products will be small.

Chet

I have difficulties to figure out the meaning of your last sentence.

 

[Bystander]

Chet's just pointing out that Gibb's free energy has nothing to do with whether a reaction "goes" or does "not go." It limits the degree of "completion" of a reaction, allowing quantitative calculation of quantities/concentrations of products.

 

[DrDu]

Chet's just pointing out that Gibb's free energy has nothing to do with whether a reaction "goes" or does "not go." It limits the degree of "completion" of a reaction, allowing quantitative calculation of quantities/concentrations of products.

I don't think so. The Delta G value always refers to specific concentrations of the reactants, which may even be held constant, e.g. in a continuously stirred reactor. Now if Delta G>0, you will never observe an increase of the amount of moles of the products. So you really have the information that the reaction "does not go" in this direction.
You are right in that from Delta G <0 you can't infer that the reaction really will "go", as, there may be high barriers of activation or more favourable alternative reactions.
I also want to mention that the negative of Delta G is known as chemical affinity and one can show that the reaction speed is a linear function of the affinity, sufficiently close to equilibrium.

 

[Bystander]

you will never observe an increase of the amount of moles of the products.

Correct. You will also never observe zero concentrations of products, no matter how positive the free energy for the reaction is once you have mixed the reactants.

 

[DrDu]

Correct. You will also never observe zero concentrations of products, no matter how positive the free energy for the reaction is once you have mixed the reactants.

...mainly because the fee energy can never be positive if one of the concentrations of the products is 0.

 

[Chestermiller]

I have difficulties to figure out the meaning of your last sentence.

All I was saying was that the standard free energy change of a reaction determines the equilibrium constant for the reaction, and even if the standard free energy change is positive, the equilibrium constant is not equal to zero.

Chet

 

[gracy]

. You will also never observe zero concentrations of products, once you have mixed the reactants.

So is it impossible ?

 

[Bystander]

So is it impossible ?

Is "what" impossible?

 

[gracy]

I have quoted your words.That's what i am asking is it impossible that we mixed reactants but product concentration is zero?

 

[Bystander]

Yes. The concentration of "product(s)" might be so small that it can't be measured, but it is non-zero.

 

[gracy]

Ok.Then why equilibrium constant is sometimes zero?

 

[Chestermiller]

Ok.Then why equilibrium constant is sometimes zero?

Who says the equilibrium constant is sometimes zero?

Chet

 

[Bystander]

equilibrium constant is sometimes zero?

It's never zero. You might have seen "lnK_eq = 0" which means the equilibrium constant = 1, but K itself can never equal zero; that condition gives you an infinite negative free energy for whatever reaction is represented by "K" and that reaction proceeds some infinitesimal amount to reduce the magnitude of that negative free energy.

 

[gracy]

Who says the equilibrium constant is sometimes zero?

I have gone through this site
http://www.answers.com/Q/What_happens_when_the_equilibrium_constant_is_zero_for_a_reaction

 

[Chestermiller]

I have gone through this site
http://www.answers.com/Q/What_happens_when_the_equilibrium_constant_is_zero_for_a_reaction

And you actually accepted this?

Chet

 

[gracy]

No.I was confused that's why asked here.

 

[Chestermiller]

No.I was confused that's why asked here.

As Bystander and I have both said, the equilibrium constant can't be zero.

Chet

 

[gracy]

As Bystander and I have both said, the equilibrium constant can't be zero.

Yes.I can surely trust you.Now i will no longer be confused about this.

 

[gracy]

It's never zero. You might have seen "lnK_eq = 0" which means the equilibrium constant = 1, but K itself can never equal zero; that condition gives you an infinite negative free energy for whatever reaction is represented by "K" and that reaction proceeds some infinitesimal amount to reduce the magnitude of that negative free energy.

Can you please explain why gibb's free energy becomes zero when k i.e equilibrium constant is 1.I know this comes from the formula G=-RT ln K but i want to understand the concept behind this.Why system has no available energy to do work if it's reactants and products have same concentration?

 

[Chestermiller]

Can you please explain why gibb's free energy becomes zero when k i.e equilibrium constant is 1.I know this comes from the formula G=-RT ln K but i want to understand the concept behind this.Why system has no available energy to do work if it's reactants and products have same concentration?

You seem to be ascribing some special significance to the situation where the equilibrium constant is unity. This value of K is no more special than any other value of K. Remember also that the standard free energy change refers to the change in free energy starting out with the pure reactants in stoichiometric proportions at 1 atm., and ending up with the pure products in corresponding proportions at 1 atm. In the equilibrium constant K, you are talking about the reactants and products not necessarily in stoichiometric proportions nor at 1 atm pressure, and the partial pressures of the reactants and products even don't have to be equal; in fact, even, if there are only single moles involved, the only requirement is that the product of the reactant partial pressures must match the product of the partial pressures of the products. If the final number of moles is different from the initial number of moles, not even this is required.

Chet

 

[Ygggdrasil]

There seems to be some confusion in this thread between the actual change in free energy of a reaction (I'll denote this ΔG) and the standard change in free energy associated with a reaction (I'll denote this ΔG^o). The ΔG of a reaction (not the ΔG^o) tells you how much useful work can be extracted from a particular process. The ΔG^o tells you the amount of work that can be extracted when the reactants and products are at a particular concentration (generally taken to be 1M). The two are related by the equation:

ΔG = ΔG^o + RT ln(Q) (eq 1)

Where Q is the reaction quotient. When a reaction is at equilibrium, transforming product into reactant is equally thermodynamically favorable as transforming reactant into product. Thus, the change in free energy for the forward reaction (ΔG_f) is equal to the change in free energy for the reverse reaction (ΔG_r). Since ΔG_f = –ΔG_r, these two equations tell us that ΔG_f = ΔG_r = 0. Thus, at equlibrium ΔG = 0 and Q = K_eq, so rearrangement of equation 1 gives the relationship everyone learns in gen chem: ΔG^o = -RT ln(K)

The way I like to think about this is that all reactions have an equilibrium set by the thermodynamic properties of the reactant and products. Moving a system towards equilibrium gives a decrease in free energy that one can couple to another process in order to perform work. Moving a system away from its equilibrium, however, requires a net input of free energy to the system.

 

[gracy]

ΔG = 0 and Q = Keq, so rearrangement of equation 1 gives the relationship everyone learns in gen chem: ΔGo = -RT ln(K)

So ,you mean at equilibrium ΔG =0 but ΔGo is not zero instead ΔGo = -RT ln(K).Right?
So in this video
At time 1:13 is he wrong?

 

[gracy]

What does it really mean when we say if delta G is positive ,i.e product has larger free energy than reactant.It means product has greater tendency to do work than product,right?Does it mean greater amount of energy is released when product is formed than when reactants are formed from product?
I think it is other way around,i.e greater amount of energy is stored in the bonds of product than reactants so less amount of energy is released when product is formed than when reactant is formed from product.

 

[Bystander]

delta G is positive ,i.e product has larger free energy than reactant.It means product has greater tendency to do work than product,right?

Backwards. If the product has a larger free energy (in the arithmetic sense, more positive) it means it's necessary to put energy/work into the reaction to form the products.

 

[gracy]

If the product has a larger free energy (in the arithmetic sense, more positive) it means it's necessary to put energy/work into the reaction to form the products.

How does it imply that product will have larger tendency to do work?Is it like this
As energy or work has been given to form the product ,product will store that energy in it's bonds so it will have larger tendency to do work or release energy.

 

[Bystander]

Is it like this

Yes. You have "pushed" the "chemistry" up the energy hill, and stored energy that can be released.

 

[gracy]

Can you please answer my post 43?

 

[Bystander]

at equilibrium ΔG =0 but ΔGo is not zero instead ΔGo = -RT ln(K).Right?

Correct.

So in this video
At time 1:13 is he wrong?

I left this for Yggg to handle since it was in response to something he posted, but it's been long enough ----.
When he says "ΔG⁰ < 0" means the reaction is spontaneous, he's talking about the special case where reactants and products are in their standard states. An example would be to consider the "reaction" water vapor condensing to liquid water at 298 K. ΔG⁰_298K(H₂O,liq.) = -56.69 kcal/mol, and ΔG⁰_298K(H₂O,vap.) = -54.64 kcal/mol. The difference in standard state free energies of formation is - 2.05 kcal/mol, so the reaction is spontaneous, and the vapor condenses. If we talk about ΔG = 0 for the same reaction, ΔG(H₂O,liq.) = ΔG⁰_298K(H₂O,liq.) + RTln(a_H2O) = -56.69 kcal/mol since a_H2O = 1, and ΔG(H₂O,vap.) = ΔG⁰_298K(H₂O,vap.) + RTln(P_σ/P_std.) = -56.69 kcal/mol since the saturation pressure of water vapor in equilibrium with liquid water at 298 K is 0.031 atm.

Help any?

 

[gracy]

Where should I use ΔG and where should I use ΔG0?I mean the formula
ΔG = ΔH - TΔS Is this right?or it is only applicable for standard gibbs free energy ..
similarly ΔG = - nFE or ΔG0= - nFE?