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Gibbs free energy in standard state vs. equilibrium

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Gibbs free energy in standard state vs. equilibrium

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1

$begingroup$

I have a problem with the definition of the standard Gibbs energy and its connection to the equilibrium constants.

I think, that I've basically understood what the different equation mean but there is one thing, I'm unable to understand:

On the one hand:

One may describe a chemical reaction with $Delta G=Delta G^circ + RTlnQ$. In equilibrium $Delta G = 0$ and the equation reads $Delta G^circ = -RT lnK$.

On the other hand:

The definition of standard state is very clear: pressure = 1 bar and all reactants and products must have activity = 1.

If I consider these two aspects separately, everything seems to be fine. But these two concepts have to be valid at the same time, what leads to $Delta G^circ = 0$ (always), since $K=1$ (all activities are per definition = 1).

Therefore, $Delta G^circ$ would be always zero. I know that this isn't true, but I don't understand why.

Can anyone explain this to me?

Thanks!

physical-chemistry reaction-mechanism equilibrium free-energy

share|improve this question

asked 5 hours ago

user76122user76122

61

New contributor

user76122 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  "since $K=1$" not necessarily true; it is $Q = 1$. Nobody said that at standard state the system must be in equilibrium.
  $endgroup$
  – orthocresol♦
  5 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  ok, but the formulae say: At standard state $Delta G^circ = -RTlnK$ and K is the ratio of the activities of reactants and products in equilibrium, due to the standard state definition K = 1, because it says: each $a=1$. Doesn't this definition disagree with your comment? I find this all really confusing.
  $endgroup$
  – user76122
  4 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I find that terribly confusing and wrong if it claims $K = 1$.
  $endgroup$
  – orthocresol♦
  4 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I know it is wrong, but I don't get why :( Did you understand my problem, that is the pure formula seems to contradict the general understanding...
  $endgroup$
  – user76122
  3 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @user76122 Orthocresol is right. Your definition of standard state in the context of $Delta G^0$ is wrong. The standard state of a pure material entails that its activity is 1. But in the context of the equation for $Delta G$ you deal conceptually with mixtures and thus not pure materials. If you follow my derivation of the formula (see here) you can see which assumptions go into $Delta G^0$. It is true that it is defined for standard pressure/concentration, but activity being equal to 1 is not presumed.
  $endgroup$
  – Philipp
  2 hours ago
  
  

  
  
  
  

add a comment | 

1

$begingroup$

I have a problem with the definition of the standard Gibbs energy and its connection to the equilibrium constants.

I think, that I've basically understood what the different equation mean but there is one thing, I'm unable to understand:

On the one hand:

One may describe a chemical reaction with $Delta G=Delta G^circ + RTlnQ$. In equilibrium $Delta G = 0$ and the equation reads $Delta G^circ = -RT lnK$.

On the other hand:

The definition of standard state is very clear: pressure = 1 bar and all reactants and products must have activity = 1.

If I consider these two aspects separately, everything seems to be fine. But these two concepts have to be valid at the same time, what leads to $Delta G^circ = 0$ (always), since $K=1$ (all activities are per definition = 1).

Therefore, $Delta G^circ$ would be always zero. I know that this isn't true, but I don't understand why.

Can anyone explain this to me?

Thanks!

physical-chemistry reaction-mechanism equilibrium free-energy

share|improve this question

asked 5 hours ago

user76122user76122

61

New contributor

user76122 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  "since $K=1$" not necessarily true; it is $Q = 1$. Nobody said that at standard state the system must be in equilibrium.
  $endgroup$
  – orthocresol♦
  5 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  ok, but the formulae say: At standard state $Delta G^circ = -RTlnK$ and K is the ratio of the activities of reactants and products in equilibrium, due to the standard state definition K = 1, because it says: each $a=1$. Doesn't this definition disagree with your comment? I find this all really confusing.
  $endgroup$
  – user76122
  4 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I find that terribly confusing and wrong if it claims $K = 1$.
  $endgroup$
  – orthocresol♦
  4 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I know it is wrong, but I don't get why :( Did you understand my problem, that is the pure formula seems to contradict the general understanding...
  $endgroup$
  – user76122
  3 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @user76122 Orthocresol is right. Your definition of standard state in the context of $Delta G^0$ is wrong. The standard state of a pure material entails that its activity is 1. But in the context of the equation for $Delta G$ you deal conceptually with mixtures and thus not pure materials. If you follow my derivation of the formula (see here) you can see which assumptions go into $Delta G^0$. It is true that it is defined for standard pressure/concentration, but activity being equal to 1 is not presumed.
  $endgroup$
  – Philipp
  2 hours ago
  
  

  
  
  
  

add a comment | 

$begingroup$

I have a problem with the definition of the standard Gibbs energy and its connection to the equilibrium constants.

I think, that I've basically understood what the different equation mean but there is one thing, I'm unable to understand:

On the one hand:

One may describe a chemical reaction with $Delta G=Delta G^circ + RTlnQ$. In equilibrium $Delta G = 0$ and the equation reads $Delta G^circ = -RT lnK$.

On the other hand:

The definition of standard state is very clear: pressure = 1 bar and all reactants and products must have activity = 1.

If I consider these two aspects separately, everything seems to be fine. But these two concepts have to be valid at the same time, what leads to $Delta G^circ = 0$ (always), since $K=1$ (all activities are per definition = 1).

Therefore, $Delta G^circ$ would be always zero. I know that this isn't true, but I don't understand why.

Can anyone explain this to me?

Thanks!

physical-chemistry reaction-mechanism equilibrium free-energy

share|improve this question

asked 5 hours ago

user76122user76122

61

New contributor

user76122 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|improve this question

asked 5 hours ago

user76122user76122

61

New contributor

user76122 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|improve this question

asked 5 hours ago

user76122user76122

61

New contributor

user76122 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked 5 hours ago

user76122user76122

61

New contributor

user76122 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked 5 hours ago

user76122user76122

61

2 Answers
2

active

oldest

votes

2

$begingroup$

As explained in the comments, the standard state conditions lead to $Q=1$ and therefore $$Delta G=Delta G^circ+ RTln1=Delta G^circ$$ On the other hand at equilibrium $Q=K$ and so $$Delta G=Delta G^circ + RTlnK$$ This of course leads to $Delta G^circ = -RTlnK$ since at equilibrium $Delta G=0$.

So you might want to think of it as three statements:

1. For the conversion of reactants to products in their standard states $Q=1$


2. At equilibrium $Delta G=0$
   


3. At equilibrium $Q=K$
   

The first statement is consistent with the definition of standard states.
The second statement follows from combination of the first and second laws of thermodynamics.
The third statement is a definition of $K$.

share|improve this answer

answered 1 hour ago

Night WriterNight Writer

2,433223

$endgroup$

add a comment | 

1

$begingroup$

What you enter into $K$ are not the activities of the pure reactants and pure products at standard state (if you did then, yes, $K$ would be 1). Rather, it is their activities at equilibrium (raised, of course, to the power of their respective stochiometric coefficients). And, at equilibrium, these activities are generally not equal to one.

share|improve this answer

edited 26 mins ago

answered 35 mins ago

theoristtheorist

2288

$endgroup$

add a comment | 

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2

$begingroup$

As explained in the comments, the standard state conditions lead to $Q=1$ and therefore $$Delta G=Delta G^circ+ RTln1=Delta G^circ$$ On the other hand at equilibrium $Q=K$ and so $$Delta G=Delta G^circ + RTlnK$$ This of course leads to $Delta G^circ = -RTlnK$ since at equilibrium $Delta G=0$.

So you might want to think of it as three statements:

1. For the conversion of reactants to products in their standard states $Q=1$


2. At equilibrium $Delta G=0$
   


3. At equilibrium $Q=K$
   

The first statement is consistent with the definition of standard states.
The second statement follows from combination of the first and second laws of thermodynamics.
The third statement is a definition of $K$.

share|improve this answer

answered 1 hour ago

Night WriterNight Writer

2,433223

$endgroup$

add a comment | 

2

$begingroup$

As explained in the comments, the standard state conditions lead to $Q=1$ and therefore $$Delta G=Delta G^circ+ RTln1=Delta G^circ$$ On the other hand at equilibrium $Q=K$ and so $$Delta G=Delta G^circ + RTlnK$$ This of course leads to $Delta G^circ = -RTlnK$ since at equilibrium $Delta G=0$.

So you might want to think of it as three statements:

1. For the conversion of reactants to products in their standard states $Q=1$


2. At equilibrium $Delta G=0$
   


3. At equilibrium $Q=K$
   

The first statement is consistent with the definition of standard states.
The second statement follows from combination of the first and second laws of thermodynamics.
The third statement is a definition of $K$.

share|improve this answer

answered 1 hour ago

Night WriterNight Writer

2,433223

$endgroup$

add a comment | 

$begingroup$

As explained in the comments, the standard state conditions lead to $Q=1$ and therefore $$Delta G=Delta G^circ+ RTln1=Delta G^circ$$ On the other hand at equilibrium $Q=K$ and so $$Delta G=Delta G^circ + RTlnK$$ This of course leads to $Delta G^circ = -RTlnK$ since at equilibrium $Delta G=0$.

So you might want to think of it as three statements:

1. For the conversion of reactants to products in their standard states $Q=1$


2. At equilibrium $Delta G=0$
   


3. At equilibrium $Q=K$
   

The first statement is consistent with the definition of standard states.
The second statement follows from combination of the first and second laws of thermodynamics.
The third statement is a definition of $K$.

share|improve this answer

answered 1 hour ago

Night WriterNight Writer

2,433223

$endgroup$

share|improve this answer

answered 1 hour ago

Night WriterNight Writer

2,433223

answered 1 hour ago

Night WriterNight Writer

2,433223

answered 1 hour ago

Night WriterNight Writer

2,433223

1

$begingroup$

What you enter into $K$ are not the activities of the pure reactants and pure products at standard state (if you did then, yes, $K$ would be 1). Rather, it is their activities at equilibrium (raised, of course, to the power of their respective stochiometric coefficients). And, at equilibrium, these activities are generally not equal to one.

share|improve this answer

edited 26 mins ago

answered 35 mins ago

theoristtheorist

2288

$endgroup$

add a comment | 

1

$begingroup$

What you enter into $K$ are not the activities of the pure reactants and pure products at standard state (if you did then, yes, $K$ would be 1). Rather, it is their activities at equilibrium (raised, of course, to the power of their respective stochiometric coefficients). And, at equilibrium, these activities are generally not equal to one.

share|improve this answer

edited 26 mins ago

answered 35 mins ago

theoristtheorist

2288

$endgroup$

add a comment | 

$begingroup$

What you enter into $K$ are not the activities of the pure reactants and pure products at standard state (if you did then, yes, $K$ would be 1). Rather, it is their activities at equilibrium (raised, of course, to the power of their respective stochiometric coefficients). And, at equilibrium, these activities are generally not equal to one.

share|improve this answer

edited 26 mins ago

answered 35 mins ago

theoristtheorist

2288

$endgroup$

share|improve this answer

edited 26 mins ago

answered 35 mins ago

theoristtheorist

2288

answered 35 mins ago

theoristtheorist

2288

answered 35 mins ago

theoristtheorist

2288