In Gibbs Free Energy, why multiply -T?

[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?

 

[Bystander]

Where should I use ΔG and where should I use ΔG0?

Use "ΔG" when looking at a system for which reactants and products are NOT in their standard states. Use ΔG⁰ and the activity relation when you are trying to find ΔG.

ΔG = ΔH - TΔS Is this right?

This is the general expression for change in Gibb's free energy. It's always valid.

ΔG =nFE or ΔG0=nFE?

ΔG = - nFE gives you the emf of an actual cell, and ΔG⁰ = - nFE⁰ gives the emf of a cell in which reactants and products are in their standard states.

 

[gracy]

Thanks for your all answers.Please bear with me.How can we apply gibbs free energy concept in electrochemical cells?I mean what are reactants and products in electrochemical cells?We will write over all reaction equation (combining oxidation and reduction half cell equation )and can figure out reactants and product of the reaction.Then we can apply all gibbs energy concepts,right?

 

[gracy]

Two different kinds of metal can be placed in a conducting solution to form galvanic cell,right?So one electrode of Na and one of chlorine can not make galvanic cell as chlorine is not a metal,right?

 

[Chestermiller]

Gracy: I think it would better for you to take a step backwards temporarily and first get a good solid understanding of how all this works for ideal gases that are undergoing chemical reaction. After you master that, you can safely move on to more complicated systems with confidence. Are you familiar with the Van't Hoff Equilibrium Box?

Chet

 

[gracy]

Are you familiar with the Van't Hoff Equilibrium Box?

Van't Hoff Equilibrium Box is a kind of vessel which has constant volume and the various substance which take part in any reaction are in equilibrium.The wall of this vessel is permeable to only some substances.This box is only applicable for homogeneous systems.

 

[Chestermiller]

Van't Hoff Equilibrium Box is a kind of vessel which has constant volume and the various substance which take part in any reaction are in equilibrium.The wall of this vessel is permeable to only some substances.This box is only applicable for homogeneous systems.

This is kind of correct, but let's make it more precise. Is there only one semipermeable membrane? What about the temperature?

We've been talking about ΔG and ΔG⁰. If you had a Van't Hoff Equilibrium Box available to you, how would use use it to experimentally measure ΔG⁰ for a reaction involving ideal gases? (Use of supplementary equipment, such as cylinders with pistons, is allowed.)

Chet

 

[gracy]

Is there only one semipermeable membrane?

I think it depends on number of walls of van't hoff box.

What about the temperature?

At constant temperature.I guess.

 

[gracy]

We've been talking about ΔG and ΔG0. If you had a Van't Hoff Equilibrium Box available to you, how would use use it to experimentally measure ΔG0 for a reaction involving ideal gases? (Use of supplementary equipment, such as cylinders with pistons, is allowed.)

I don't know about this.Please help me.

 

[Chestermiller]

I think it depends on number of walls of van't hoff box.

No. The is a separate semipermeable membrane for each reactant and each product of the reaction.

 

[gracy]

No. The is a separate semipermeable membrane for each reactant and each product of the reaction.

So,how many?