Help with free energy (ΔG)

[spoudaios]

Hi guys,

 

I am confused with the concept of ΔG in regards to equilibrium. Which equations should i memorize?

 

ΔG = -nFE

ΔG(o)= -nFE(o) o --> standard state

ΔG = ΔG(o) + RT log (Q)

 

In equilibrium: ΔG = 0 --> ΔG(o) = -RT ln Keq

 

1>. When ΔG(o) = 0 then Keq = 1

Question: I know mathematically that when ΔG(o) = 0, Keq will be equal to 1.

 

a>. Does this mean [product] = [reactant]?

b>. Is this always the case, ie. when a reaction is in equilibrium, the Keq = 1??? I don't think so... what do you guys think?

 

2>. When ΔG(o) < 0 then Keq is > 1. This means reaction will shift to the right.

Question: Why? I thought when Keq is > 1, this means there are more products than the reaction... which means rxn should shift to the left ??

K = [product] / [reactant]

 

3>. When ΔG (o) > 0 then Keq is < 1. This means reaction will shift to the left.

Same question as above. Why?

 

4>. Also, as for no. 2 and 3, it says that rxn will shift towards equilibrum. However, in the beginning, it's stated that it is already in equilibrium which means

ΔG=0. So, all these 3 scenarios are already in equilibrium, correct? If so, then why they say the reaction will shift towards equilibrium???

 

 

 

 

Thanks all

[The Law]

1) When delta g = 0, Keq=1 it means that there is no net change in the energy of the reactants. They're flopping back and forth equally, so you're not getting overall more products or overall more reactnts.

 

2) When delta G is < 0, this means that the reaction is SPONTANEOUS. That means it will go (remember that does not mean it will go fast, just that it will go). Now think about this ... if the reaction is going to go, that means you're going to end up with more products. This is why Keq is >1 (this is what you expect when you expect a lot of product... to remember > means favouring products, think about strong acids, they have a HUUUGE Ka)

 

3) Again, when Delta G > 0 that means the reaction is NON-SPONTANEOUS. It will not go naturally but needs an input of energy. For this reason, Keq < 1 and reactants are favoured... the reaction isn't going sooooo you exepct there to be more reactants.

 

4) You are misunderstanding the meaning of equilibrium. Equilibrium does not have to be a state where you have equal number of products and reactants... in some equilibria the reactants are favoured, in some the products are favoured, in some it's a balance between the 2. Equilibrium revolves around a set point, where that point lies depends on the energy of the system.

 

I hope that helps...

[avenir001]

Hi guys,

 

I am confused with the concept of ΔG in regards to equilibrium. Which equations should i memorize?

 

ΔG = -nFE

ΔG(o)= -nFE(o) o --> standard state

ΔG = ΔG(o) + RT log (Q)

 

In equilibrium: ΔG = 0 --> ΔG(o) = -RT ln Keq

 

1>. When ΔG(o) = 0 then Keq = 1

Question: I know mathematically that when ΔG(o) = 0, Keq will be equal to 1.

 

a>. Does this mean [product] = [reactant]?

b>. Is this always the case, ie. when a reaction is in equilibrium, the Keq = 1??? I don't think so... what do you guys think?

 

no, this is not always true...it's only true under standard conditions...when the reaction is carried out under standard conditions, delta Go = 0 at equilibrium, which means Keq=1.

[spoudaios]

I am still confused...

 

1>. To "the Law": Do you mean ΔG or ΔGo?

 

When a rxn is in equilibrium, which one is zero? Is it ΔG or ΔGo? or does it have to be both???

 

2>. Same question to avenir: do you mean ΔG or ΔGo ???

 

Thanks guys

[avenir001]

for a reaction to be at equilibrium under any kind of condition, delta G must equal zero.

when delta Go = 0, the reaction is in equilibrium under standard conditions (eg 25 degrees, 1 atm, etc)

[Alastriss]

Hey OP, if you get confused, make sure you designate Q as the reaction quotient at any time and Keq as the reaction time.

 

Note: in AG, A is delta, and * is the STP symbol

 

 

AG = AG* + RT ln Q

 

AG* = - RT ln Keq

 

therefore

 

AG = RT ln Q - RT ln Keq

AG = RT ln (Q/Keq)

 

when Q< Keq, so the reaction recently started and is going towards the equilibrium, AG is a negative. As Q --> Keq, then ln 1 = o and AG becomes 0 in value. This is where the maximum entropy of the system is reached.

Notice that Q>Keq yeilds a AG value that is positive. In fact when Q > Keq the reaction will shift so that Q = Keq again.

 

 

hope that helped

 

and btw, at equilibrium, it is not [product] = [reactant] its the rate of product production = rate of reactant consumption, thus the forward rate = the reverse rate.

[spoudaios]

1>. I've never seen this equation (although it's just a manipulation of the previous 2 formulas):

 

ΔG = RT ln Q - RT ln Keq

ΔG = RT ln Q/Keq

 

Is this always useful for any type of problem whether to predict where rxn will shift or whether we have numbers to play with and whether it's standard state or non standard state?

 

2>. So, if rxn is at standard state, this means i have to use ΔGo = -RT ln Keq

If it's in equilibrium, then ΔGo must be ZERO. Correct?

This has nothing to do with ΔG because it's clearly stated that it's standard state.

If it's not zero, then we just have to look at the ΔGo (not ΔG) to predict where rxn will shift. Correct?

 

3>. The same with non standard state, the equation to be used will be:

ΔG = ΔGo + RT ln Q

If it is in equilibrium, then ΔG is ZERO. Does not matter the value of ΔGo. To predict where rxn will shift, we just have to look at ΔG (if it's not zero).

Correct?

 

4>. How do you guys memorize these equations where sometimes you use "log" instead of "ln" ? Any useful mnemonics?

 

 

Thanks again everyone

 

PS:To Alastriss, what do you mean by: Keq as the reaction time ????

[avenir001]

1) yes, that's a useful equation..but for the mcat in general, u have to understand what the equations and formulas mean...rarely can u just memorize one equation that works for all kinds of problems...u have to be able to do simple manipulations

2) yes...but u're making it sound more complicated than it is...delta Go is simply the delta G of the standard state...so when a reaction is carried out under standard state, u look at delta Go...when a reaction is carried out under other conditions, u look at delta G which uses standard state as a reference point to give the change in free energy under any conditions.

3) yes...see the explanation above.

4) memorize the ln form...but ln X = 2.3 log X...

 

I think Alastriss meant Keq at the equilibrium time

[spoudaios]

OK dokey...

 

Thanks all... onto the next topic...

[The Law]

Yeah, I wouldn't try to learn all those equilibrium equations... I doubt they'll test you on that... it's more important to learn the concepts behind entropy, equilibrium, and gibbs free energy. I'm not going to lie, but I honestly did not know any of the equations for Gibbs heading into the exam... but I did not find any problem with those topics since the concepts are what they seemed to be more concerned with.

[Alastriss]

Hey that equation is not an official equation to know, its just to show you how free energy changes relative to the position of the reaction Q at any time.

And one poster said that you need to understand the concept behind the equation, as this is pretty much what many of the tough questions are based on.

[greenlight]

If you stumbled across this page (like I did) looking for review, answers, or to clear your confusion, I wanted to comment on the above answers so you're not mislead.

 

The Law and Alastriss have it right (aside from writing a word or so, which wasn't meant), but avenir001's answers will mislead your thinking and will further create confusion. But it is okay to not be fully correct on something, avenir001, we're all human :-) What would have been not okay if no one addressed this, and had others learning wrong ideas.

 

The Law is all correct in his/her response, except, yes spoudaios, I believe he meant deltaGo (or, AG*, thanks Alastriss) in number one. Otherwise, listen to his good observations.

 

Alastriss, is all correct, and by "Keq as the reaction time," we can safely assume it means "at the time reaction is visibly complete or stationary (at a macroscopic level)." His explanation is really great.

 

Avenir001 is partially correct, but if you don't feel comfortable with this concept, skip over avenir001's answers to avoid further confusion or incorrect thinking. (Sorry if you already learned from it). Yes, AG* is at standard conditions. Because AG depends on reactant and product concentration proportions, smart people figured out that in order to make a list comparing rxns, they needed some consistent system, so they chose 1M [reactant] and 1M [product] and the other standard conditions avenir001 listed. Notice that the starting concentrations are in EQUAL proportions. Does that mean that there are no reactions that have other proportions, once a reaction went to equilibrium? No. Equilibrium does not always mean that the [reactant] and [product] are equal. Here are three reactions where the AG* equilibria is not zero, from a table found at Chem Purdue:

Reaction | deltaGo (kJ) Note: no zero value | K

2 SO3(g) <---> 2 SO2(g) + O2(g) | 141.7 | 1.4 x 10-25

H2O(l) <---> H+(aq) + OH-(aq) | 79.9 | 1.0 x 10-14

AgCl(s) + H2O <---> Ag+(aq) + Cl-(aq) | 55.6 | 1.8 x 10-10

 

So what does this mean? The most quick answer/concept is to understand and remember:

a) If reaction is at equilibrium ==> AG = 0 (and vice-versa) Spoudaios question/statement numbered 2>. is not correct based on the answer he/she received: If it's in equilibrium then AG, not necessarily AG*, is zero. AG* may too be zero, but you can't state that. If it's not zero, look at Keq to predict the proportions.

If AG* = 0 ==> [reactant] = [product] (and vice-versa) It does NOT have to be zero to be in equilibrium, but when it IS zero, the rxn is in a special type of equilibrium, i.e. reactant and product concentrations are equal at equilibrium under these standard conditions (Keq dependent on T).

c) Look at Keq value (like spoudaios originally stated) to see what you get more of, reactants or products. A "shift in equilibrium" is a misleading statement in this context, but some textbooks do use it. A better statement is "proportions of reactants products are greater than the other" at the established equilibrium.

 

Otherwise, thank you, Avenir, for the rest of the answers and contributions Your willingness to help is appreciated.

 

***Sorry to bring up an old thread, but it would be unethical to let it go and confuse others.***