Why is Equilibrium constant K > 1 when DeltaG is - and the Ecell is +????

[HannibalLecter]

So if we have a galvanic cell obviously our Ecell is positive meaning its a spontaneous reaction thus deltaG would be negative.

But I am getting hung on the concept surrounding Equilibrium constant and why it has to be greater than 1. Yes I know you can plug in a negative deltaG into the equation: DeltaG = -RT ln Keq and obtain a K greater than 1

but concept wise wouldn't the Equilibrium constant be less than one because its a spontaneous reaction and we would want it to go forward so the reactants concentration would be higher than the products concentration making the K less than 1 instead of greater than 1.

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[FROGGBUSTER]

So if we have a galvanic cell obviously our Ecell is positive meaning its a spontaneous reaction thus deltaG would be negative.

But I am getting hung on the concept surrounding Equilibrium constant and why it has to be greater than 1. Yes I know you can plug in a negative deltaG into the equation: DeltaG = -RT ln Keq and obtain a K greater than 1

but concept wise wouldn't the Equilibrium constant be less than one because its a spontaneous reaction and we would want it to go forward so the reactants concentration would be higher than the products concentration making the K less than 1 instead of greater than 1.

delta G = -RT ln Keq at standard conditions, when all the concentrations equal to 1. This means Q = 1. At standard conditions, Q will always be 1.

If K > 1, this means K > Q. This means you want to go forward in the reaction to achieve equilibrium.

If K < 1, this means K < Q, and the system will spontaneously convert products to reactants.

 

[rmm30]

So if we have a galvanic cell obviously our Ecell is positive meaning its a spontaneous reaction thus deltaG would be negative.

But I am getting hung on the concept surrounding Equilibrium constant and why it has to be greater than 1. Yes I know you can plug in a negative deltaG into the equation: DeltaG = -RT ln Keq and obtain a K greater than 1

but concept wise wouldn't the Equilibrium constant be less than one because its a spontaneous reaction and we would want it to go forward so the reactants concentration would be higher than the products concentration making the K less than 1 instead of greater than 1.

Lets back up a couple steps because this can get tricky:

The determine delta G for any rxn that isn't at equalibrium use the equation: delG = del Gstand + RT lnQ.

So above we have del G or rxn. del G standard (which comes from a book, never changes and is delG when standard conditions are present ie 1atm 298K and 1M concentration of all products and reactants), R is const, T is temp, and Q is rxn quotient (concentration of products and reactants at a givin instant.)

Now you can see why this equation is useful. Because for a given rxn, the stand del G is fixed! If you add RT ln Q to that (rem Q is intantaneous rxn concentrations) you will get the del G under those condidtions.

Lets take note of a few things here bf continuing.
( At standard conditions all concentrations of products and reacants = 1M)
Consider rxn A-----> B. If at this moment you have way more reactants than product, the result will be a Q that is very small (smaller than K more than likely). So intuitively you can see that with lots of reactant, the rxn will be proceeding toward the right (LeChatlie's principle). If you were to plug in the numbers for Q ( a very small number seeing as how there are many more reactants than products), you would indeed get a negative numerical value for del G indicating a spontaneous process for A---->B occuring in the direction written.


But your question is about K. Well if we are discussing K we must be talking about equilibrium right? So here are a few things you NEED to know about Equilibrium, del G, and Ecell.
@Equilibrium:
1)Q=K
2)del G = 0
3)Ecell = 0

You have to know that @ equilibrium, ALL of these statements are true. Furthermore if you know any of these statements are true, then they ALL are true and you are @ equilbrium. It also important to note that the del G and Ecell = 0 does NOT refer to standard values; remember those come from a book and do not change. So knowing this lets take a look at the original equation and manipulate it to derive an equation that represents equilibrium condititions.

We started with delG = delGstand + RT ln Q. At equilibrium, del G = 0
and Q=K. Sub in del G = 0 and Q=K and you have new equation that relates del G standard to equilibrium constant and is as follows:

delG stand = -RTlnK.

Here is where I think you may be a little mixed up:
"but concept wise wouldn't the Equilibrium constant be less than one because its a spontaneous reaction and we would want it to go forward so the reactants concentration would be higher than the products concentration making the K less than 1 instead of greater than 1"

Remember that we are at EQUILIBRIUM. That means there is no more going forward and backwards to approach equilibrium since we are already there. Think about that for a second of two. That's why del G and Ecell = 0 at this point; bc there is no more "free energy" or "electromotive force" left to move rxn.

So if we are at equilibrium (where conc of products and reactants is not changing any more) and delGstand is < 0; indicating that the reaction progressed in the forward direction forming lots of product. And when you have lots of product at equilibrium, K>>1.

I think that's your hangup. You are thinking as if the rxn is still moving but it isn't. Consider rxn A--->B at equilibrium. Lets look at some possible values for K keeping in mind K = {Prod}^x/(React)^y.
K>>1 means lots of product at equilib
K<<1 means lots of reactant at equilib
K = 1 means prod and reactant concentrations are equal at equilib.

 

[HannibalLecter]

delta G = -RT ln Keq at standard conditions, when all the concentrations equal to 1. This means Q = 1. At standard conditions, Q will always be 1.

If K > 1, this means K > Q. This means you want to go forward in the reaction to achieve equilibrium.

If K < 1, this means K < Q, and the system will spontaneously convert products to reactants.

Frogger I know what you are trying to say but in your terms with the concentrations of reactants and products being at 1, a K > 1 means a K > Q, but a K = 2 for example means there is more products than reactants in the given equation so how can the reaction go forward when we already have enough products.

Lets back up a couple steps because this can get tricky:

The determine delta G for any rxn that isn't at equalibrium use the equation: delG = del Gstand + RT lnQ.

Ohh so you are saying the moment we are trying to calculate out Keq is the moment after equilibrium right because why would we be talking about Keq anyways if it weren't after equilibrium. I think I get the way the problem is stated. The problem clearly doesn't state that you have to look at two points in the reaction:

1) Before equilibrium: Where the Q < 1, Ecell = positive, &#916;G is negative, and the galvanic cell is going to run

2) Once equilibrium has been reached and it is after it: &#916;G = 0, Ecell = 0, Since we are at equilibrium and we have reached it we can start talking about Keq which would have a value greater than one because the spontaneous reaction has gone through and produced a lot of products.

I think I get it how these problems are trying to describe the situation. Thank you guys.

 

[rmm30]

Frogger I know what you are trying to say but in your terms with the concentrations of reactants and products being at 1, a K > 1 means a K > Q, but a K = 2 for example means there is more products than reactants in the given equation so how can the reaction go forward when we already have enough products.



Ohh so you are saying the moment we are trying to calculate out Keq is the moment after equilibrium right because why would we be talking about Keq anyways if it weren't after equilibrium. I think I get the way the problem is stated. The problem clearly doesn't state that you have to look at two points in the reaction:

1) Before equilibrium: Where the Q < 1, Ecell = positive, &#916;G is negative, and the galvanic cell is going to run

2) Once equilibrium has been reached and it is after it: &#916;G = 0, Ecell = 0, Since we are at equilibrium and we have reached it we can start talking about Keq which would have a value greater than one because the spontaneous reaction has gone through and produced a lot of products.

I think I get it how these problems are trying to describe the situation. Thank you guys.

Yeah you almost have it. Use Del G = Del G stand + RTlnQ when you are trying to figure out if the reaction (voltaic/galvanic rxn in this case) is spontaneous or not under NONSTANDARD states (ie when concentrations of all prod and reactants do not = 1).

Remember how Q and K relate to one another. You don't compare Q in terms of 1. You compare it in terms of K:

If Q>K; rxn will procede left toward reactants
If Q<K; rxn will proced right toward products
If Q = K; rxn is at equilibrium



Btw didn't you already put up a good DAT score? Are you headed for a retake?

 

[Dusk]

Lets back up a couple steps because this can get tricky:

The determine delta G for any rxn that isn't at equalibrium use the equation: delG = del Gstand + RT lnQ.

So above we have del G or rxn. del G standard (which comes from a book, never changes and is delG when standard conditions are present ie 1atm 298K and 1M concentration of all products and reactants), R is const, T is temp, and Q is rxn quotient (concentration of products and reactants at a givin instant.)

Now you can see why this equation is useful. Because for a given rxn, the stand del G is fixed! If you add RT ln Q to that (rem Q is intantaneous rxn concentrations) you will get the del G under those condidtions.

Lets take note of a few things here bf continuing.
( At standard conditions all concentrations of products and reacants = 1M)
Consider rxn A-----> B. If at this moment you have way more reactants than product, the result will be a Q that is very small (smaller than K more than likely). So intuitively you can see that with lots of reactant, the rxn will be proceeding toward the right (LeChatlie's principle). If you were to plug in the numbers for Q ( a very small number seeing as how there are many more reactants than products), you would indeed get a negative numerical value for del G indicating a spontaneous process for A---->B occuring in the direction written.


But your question is about K. Well if we are discussing K we must be talking about equilibrium right? So here are a few things you NEED to know about Equilibrium, del G, and Ecell.
@Equilibrium:
1)Q=K
2)del G = 0
3)Ecell = 0

You have to know that @ equilibrium, ALL of these statements are true. Furthermore if you know any of these statements are true, then they ALL are true and you are @ equilbrium. It also important to note that the del G and Ecell = 0 does NOT refer to standard values; remember those come from a book and do not change. So knowing this lets take a look at the original equation and manipulate it to derive an equation that represents equilibrium condititions.

We started with delG = delGstand + RT ln Q. At equilibrium, del G = 0
and Q=K. Sub in del G = 0 and Q=K and you have new equation that relates del G standard to equilibrium constant and is as follows:

delG stand = -RTlnK.

Here is where I think you may be a little mixed up:
"but concept wise wouldn't the Equilibrium constant be less than one because its a spontaneous reaction and we would want it to go forward so the reactants concentration would be higher than the products concentration making the K less than 1 instead of greater than 1"

Remember that we are at EQUILIBRIUM. That means there is no more going forward and backwards to approach equilibrium since we are already there. Think about that for a second of two. That's why del G and Ecell = 0 at this point; bc there is no more "free energy" or "electromotive force" left to move rxn.

So if we are at equilibrium (where conc of products and reactants is not changing any more) and delGstand is < 0; indicating that the reaction progressed in the forward direction forming lots of product. And when you have lots of product at equilibrium, K>>1.

I think that's your hangup. You are thinking as if the rxn is still moving but it isn't. Consider rxn A--->B at equilibrium. Lets look at some possible values for K keeping in mind K = {Prod}^x/(React)^y.
K>>1 means lots of product at equilib
K<<1 means lots of reactant at equilib
K = 1 means prod and reactant concentrations are equal at equilib.

Awesome you would take so much time to help others, wish all of sdn were like you.

 

[HannibalLecter]

Yeah you almost have it. Use Del G = Del G stand + RTlnQ when you are trying to figure out if the reaction (voltaic/galvanic rxn in this case) is spontaneous or not under NONSTANDARD states (ie when concentrations of all prod and reactants do not = 1).

Remember how Q and K relate to one another. You don't compare Q in terms of 1. You compare it in terms of K:

If Q>K; rxn will procede left toward reactants
If Q<K; rxn will proced right toward products
If Q = K; rxn is at equilibrium

Btw didn't you already put up a good DAT score? Are you headed for a retake?

No my back story is that I already did get a 21 once on the DAT but my GPA was so sh*tty I didn't get any interviews, so I am taking my DAT again because it expired and get a better score than a 21 so it can compensate for my post-postbac GPA of 3.1.

Ohh yes it was wrong of me to compare the value of Q to 1. So you say never to compare Q values to 1 instead just compare the Q values to Keq to determine which way the reaction will go if left all by itself. Good to know. Thank you very much rmm30.