Gibbs free energy in standard state vs. equilibrium
$begingroup$
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 $Delta G=Delta G^circ + RTln{Q}$. In equilibrium $Delta G = 0$ and the equation reads $Delta G^circ = -RT ln{K}$.
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
asked Mar 24 at 15:47
user76122user76122
334
$endgroup$
|
show 9 more comments
$begingroup$
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 $Delta G=Delta G^circ + RTln{Q}$. In equilibrium $Delta G = 0$ and the equation reads $Delta G^circ = -RT ln{K}$.
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
asked Mar 24 at 15:47
user76122user76122
334
$endgroup$
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♦
Mar 24 at 15:48
$begingroup$
ok, but the formulae say: At standard state $Delta G^circ = -RTln{K}$ 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
Mar 24 at 16:14
$begingroup$
I find that terribly confusing and wrong if it claims $K = 1$.
$endgroup$
– orthocresol♦
Mar 24 at 16:20
$begingroup$
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
Mar 24 at 17:36
$begingroup$
@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
Mar 24 at 18:24
|
show 9 more comments
$begingroup$
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 $Delta G=Delta G^circ + RTln{Q}$. In equilibrium $Delta G = 0$ and the equation reads $Delta G^circ = -RT ln{K}$.
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
asked Mar 24 at 15:47
user76122user76122
334
$endgroup$
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 $Delta G=Delta G^circ + RTln{Q}$. In equilibrium $Delta G = 0$ and the equation reads $Delta G^circ = -RT ln{K}$.
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
physical-chemistry reaction-mechanism equilibrium free-energy
asked Mar 24 at 15:47
user76122user76122
334
asked Mar 24 at 15:47
user76122user76122
334
asked Mar 24 at 15:47
user76122user76122
334
asked Mar 24 at 15:47
user76122user76122
334
asked Mar 24 at 15:47
user76122user76122
334
334
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♦
Mar 24 at 15:48
$begingroup$
ok, but the formulae say: At standard state $Delta G^circ = -RTln{K}$ 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
Mar 24 at 16:14
$begingroup$
I find that terribly confusing and wrong if it claims $K = 1$.
$endgroup$
– orthocresol♦
Mar 24 at 16:20
$begingroup$
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
Mar 24 at 17:36
$begingroup$
@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
Mar 24 at 18:24
|
show 9 more comments
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♦
Mar 24 at 15:48
$begingroup$
ok, but the formulae say: At standard state $Delta G^circ = -RTln{K}$ 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
Mar 24 at 16:14
$begingroup$
I find that terribly confusing and wrong if it claims $K = 1$.
$endgroup$
– orthocresol♦
Mar 24 at 16:20
$begingroup$
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
Mar 24 at 17:36
$begingroup$
@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
Mar 24 at 18:24
2
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♦
Mar 24 at 15:48
$begingroup$
"since $K=1$" not necessarily true; it is $Q = 1$. Nobody said that at standard state the system must be in equilibrium.
$endgroup$
– orthocresol♦
Mar 24 at 15:48
$begingroup$
ok, but the formulae say: At standard state $Delta G^circ = -RTln{K}$ 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
Mar 24 at 16:14
$begingroup$
ok, but the formulae say: At standard state $Delta G^circ = -RTln{K}$ 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
Mar 24 at 16:14
$begingroup$
I find that terribly confusing and wrong if it claims $K = 1$.
$endgroup$
– orthocresol♦
Mar 24 at 16:20
$begingroup$
I find that terribly confusing and wrong if it claims $K = 1$.
$endgroup$
– orthocresol♦
Mar 24 at 16:20
$begingroup$
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
Mar 24 at 17:36
$begingroup$
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
Mar 24 at 17:36
$begingroup$
@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
Mar 24 at 18:24
$begingroup$
@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
Mar 24 at 18:24
|
show 9 more comments
2 Answers
2
active
oldest
votes
$begingroup$
As explained in the comments, the standard state conditions lead to $Q=1$ and therefore $$Delta G=Delta G^circ+ RTln{1}=Delta G^circ$$ On the other hand at equilibrium $Q=K$ and so $$Delta G=Delta G^circ + RTln{K}$$ This of course leads to $Delta G^circ = -RTln{K}$ since at equilibrium $Delta G=0$.
So you might want to think of it as multiple statements:
- For the conversion of reactants to products in their standard states $Q=1$
- At equilibrium $Delta G=0$
- At equilibrium $Q=K$
$Delta G^circ$ is the free energy change for conversion of reactants to products in their standard states.
The first statement is consistent with the definition of standard states.
The second and fourth statements follow from combination of the first and second laws of thermodynamics.
The third statement is a definition of $K$.
$K$ and $Delta G^circ$ are very much connected, but $Delta G^circ$ does not describe the change in free energy from reactants to products at equilibrium. $K$ and $Delta G^circ$ are not points on the reaction progress curve. Rather they are related to the slope of that curve.
answered Mar 24 at 19:51
Night WriterNight Writer
2,721223
$endgroup$
$begingroup$
So, based on your answer and on the comments above, I think that my understanding increases. These are just two different point of views or two different points of the same equation: $Delta G=Delta G^circ + RTln{Q}$. My misunderstanding arises from the equation: $Delta G^circ = - RTln{K}$, since it seems to connect standard ($Delta G^circ$) state with equilibrium ($K$) and that's not the meaning of this formula, correct?
$endgroup$
– user76122
Mar 25 at 10:12
$begingroup$
@user76122 $K$ and $Delta G^circ$ are very much connected. One point that is confusing for a beginner is that $Delta G^circ$ does not describe the change in free energy from reactants to products at equilibrium. I'll add this to my answer.
$endgroup$
– Night Writer
Mar 25 at 11:42
$begingroup$
At equilibrium, there is no change in free energy neither a net conversion from reactants to products (or vice versa). In my understanding, if you consider the "hanging rope"-style curve of $G(xi)$, $Delta G^circ$ and $K$ are just two different "points" on that curve, describing different things. The connection is, that these points lie on the same curve and therefore, can be calculated from each other. This is what is shown by the equations. Did you mean it that way (or at least in a similar way)? I'd like to emphasize that I'm trying to understand and not to correct you :)
$endgroup$
– user76122
Mar 25 at 12:53
$begingroup$
@user76122 I would not interpret $Delta G^circ$ and $K$ as points on the progress curve. Rather they are related to the slope of that curve.
$endgroup$
– Night Writer
Mar 25 at 19:25
1
$begingroup$
Thanks! By digging deeper, I found the following answer: chemistry.stackexchange.com/a/41864/76122 There it is claimed, that $Q=1$ has to lie on the curve, since $Q$ alters from $infty$ to zero.
$endgroup$
– user76122
Mar 28 at 9:42
|
show 3 more comments
$begingroup$
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.
answered Mar 24 at 20:20
theoristtheorist
2488
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "431"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fchemistry.stackexchange.com%2fquestions%2f111475%2fgibbs-free-energy-in-standard-state-vs-equilibrium%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
As explained in the comments, the standard state conditions lead to $Q=1$ and therefore $$Delta G=Delta G^circ+ RTln{1}=Delta G^circ$$ On the other hand at equilibrium $Q=K$ and so $$Delta G=Delta G^circ + RTln{K}$$ This of course leads to $Delta G^circ = -RTln{K}$ since at equilibrium $Delta G=0$.
So you might want to think of it as multiple statements:
- For the conversion of reactants to products in their standard states $Q=1$
- At equilibrium $Delta G=0$
- At equilibrium $Q=K$
$Delta G^circ$ is the free energy change for conversion of reactants to products in their standard states.
The first statement is consistent with the definition of standard states.
The second and fourth statements follow from combination of the first and second laws of thermodynamics.
The third statement is a definition of $K$.
$K$ and $Delta G^circ$ are very much connected, but $Delta G^circ$ does not describe the change in free energy from reactants to products at equilibrium. $K$ and $Delta G^circ$ are not points on the reaction progress curve. Rather they are related to the slope of that curve.
answered Mar 24 at 19:51
Night WriterNight Writer
2,721223
$endgroup$
$begingroup$
So, based on your answer and on the comments above, I think that my understanding increases. These are just two different point of views or two different points of the same equation: $Delta G=Delta G^circ + RTln{Q}$. My misunderstanding arises from the equation: $Delta G^circ = - RTln{K}$, since it seems to connect standard ($Delta G^circ$) state with equilibrium ($K$) and that's not the meaning of this formula, correct?
$endgroup$
– user76122
Mar 25 at 10:12
$begingroup$
@user76122 $K$ and $Delta G^circ$ are very much connected. One point that is confusing for a beginner is that $Delta G^circ$ does not describe the change in free energy from reactants to products at equilibrium. I'll add this to my answer.
$endgroup$
– Night Writer
Mar 25 at 11:42
$begingroup$
At equilibrium, there is no change in free energy neither a net conversion from reactants to products (or vice versa). In my understanding, if you consider the "hanging rope"-style curve of $G(xi)$, $Delta G^circ$ and $K$ are just two different "points" on that curve, describing different things. The connection is, that these points lie on the same curve and therefore, can be calculated from each other. This is what is shown by the equations. Did you mean it that way (or at least in a similar way)? I'd like to emphasize that I'm trying to understand and not to correct you :)
$endgroup$
– user76122
Mar 25 at 12:53
$begingroup$
@user76122 I would not interpret $Delta G^circ$ and $K$ as points on the progress curve. Rather they are related to the slope of that curve.
$endgroup$
– Night Writer
Mar 25 at 19:25
1
$begingroup$
Thanks! By digging deeper, I found the following answer: chemistry.stackexchange.com/a/41864/76122 There it is claimed, that $Q=1$ has to lie on the curve, since $Q$ alters from $infty$ to zero.
$endgroup$
– user76122
Mar 28 at 9:42
|
show 3 more comments
$begingroup$
As explained in the comments, the standard state conditions lead to $Q=1$ and therefore $$Delta G=Delta G^circ+ RTln{1}=Delta G^circ$$ On the other hand at equilibrium $Q=K$ and so $$Delta G=Delta G^circ + RTln{K}$$ This of course leads to $Delta G^circ = -RTln{K}$ since at equilibrium $Delta G=0$.
So you might want to think of it as multiple statements:
- For the conversion of reactants to products in their standard states $Q=1$
- At equilibrium $Delta G=0$
- At equilibrium $Q=K$
$Delta G^circ$ is the free energy change for conversion of reactants to products in their standard states.
The first statement is consistent with the definition of standard states.
The second and fourth statements follow from combination of the first and second laws of thermodynamics.
The third statement is a definition of $K$.
$K$ and $Delta G^circ$ are very much connected, but $Delta G^circ$ does not describe the change in free energy from reactants to products at equilibrium. $K$ and $Delta G^circ$ are not points on the reaction progress curve. Rather they are related to the slope of that curve.
answered Mar 24 at 19:51
Night WriterNight Writer
2,721223
$endgroup$
$begingroup$
So, based on your answer and on the comments above, I think that my understanding increases. These are just two different point of views or two different points of the same equation: $Delta G=Delta G^circ + RTln{Q}$. My misunderstanding arises from the equation: $Delta G^circ = - RTln{K}$, since it seems to connect standard ($Delta G^circ$) state with equilibrium ($K$) and that's not the meaning of this formula, correct?
$endgroup$
– user76122
Mar 25 at 10:12
$begingroup$
@user76122 $K$ and $Delta G^circ$ are very much connected. One point that is confusing for a beginner is that $Delta G^circ$ does not describe the change in free energy from reactants to products at equilibrium. I'll add this to my answer.
$endgroup$
– Night Writer
Mar 25 at 11:42
$begingroup$
At equilibrium, there is no change in free energy neither a net conversion from reactants to products (or vice versa). In my understanding, if you consider the "hanging rope"-style curve of $G(xi)$, $Delta G^circ$ and $K$ are just two different "points" on that curve, describing different things. The connection is, that these points lie on the same curve and therefore, can be calculated from each other. This is what is shown by the equations. Did you mean it that way (or at least in a similar way)? I'd like to emphasize that I'm trying to understand and not to correct you :)
$endgroup$
– user76122
Mar 25 at 12:53
$begingroup$
@user76122 I would not interpret $Delta G^circ$ and $K$ as points on the progress curve. Rather they are related to the slope of that curve.
$endgroup$
– Night Writer
Mar 25 at 19:25
1
$begingroup$
Thanks! By digging deeper, I found the following answer: chemistry.stackexchange.com/a/41864/76122 There it is claimed, that $Q=1$ has to lie on the curve, since $Q$ alters from $infty$ to zero.
$endgroup$
– user76122
Mar 28 at 9:42
|
show 3 more comments
$begingroup$
As explained in the comments, the standard state conditions lead to $Q=1$ and therefore $$Delta G=Delta G^circ+ RTln{1}=Delta G^circ$$ On the other hand at equilibrium $Q=K$ and so $$Delta G=Delta G^circ + RTln{K}$$ This of course leads to $Delta G^circ = -RTln{K}$ since at equilibrium $Delta G=0$.
So you might want to think of it as multiple statements:
- For the conversion of reactants to products in their standard states $Q=1$
- At equilibrium $Delta G=0$
- At equilibrium $Q=K$
$Delta G^circ$ is the free energy change for conversion of reactants to products in their standard states.
The first statement is consistent with the definition of standard states.
The second and fourth statements follow from combination of the first and second laws of thermodynamics.
The third statement is a definition of $K$.
$K$ and $Delta G^circ$ are very much connected, but $Delta G^circ$ does not describe the change in free energy from reactants to products at equilibrium. $K$ and $Delta G^circ$ are not points on the reaction progress curve. Rather they are related to the slope of that curve.
answered Mar 24 at 19:51
Night WriterNight Writer
2,721223
$endgroup$
As explained in the comments, the standard state conditions lead to $Q=1$ and therefore $$Delta G=Delta G^circ+ RTln{1}=Delta G^circ$$ On the other hand at equilibrium $Q=K$ and so $$Delta G=Delta G^circ + RTln{K}$$ This of course leads to $Delta G^circ = -RTln{K}$ since at equilibrium $Delta G=0$.
So you might want to think of it as multiple statements:
- For the conversion of reactants to products in their standard states $Q=1$
- At equilibrium $Delta G=0$
- At equilibrium $Q=K$
$Delta G^circ$ is the free energy change for conversion of reactants to products in their standard states.
The first statement is consistent with the definition of standard states.
The second and fourth statements follow from combination of the first and second laws of thermodynamics.
The third statement is a definition of $K$.
$K$ and $Delta G^circ$ are very much connected, but $Delta G^circ$ does not describe the change in free energy from reactants to products at equilibrium. $K$ and $Delta G^circ$ are not points on the reaction progress curve. Rather they are related to the slope of that curve.
answered Mar 24 at 19:51
Night WriterNight Writer
2,721223
edited Mar 26 at 11:50
answered Mar 24 at 19:51
Night WriterNight Writer
2,721223
answered Mar 24 at 19:51
Night WriterNight Writer
2,721223
answered Mar 24 at 19:51
Night WriterNight Writer
2,721223
2,721223
$begingroup$
So, based on your answer and on the comments above, I think that my understanding increases. These are just two different point of views or two different points of the same equation: $Delta G=Delta G^circ + RTln{Q}$. My misunderstanding arises from the equation: $Delta G^circ = - RTln{K}$, since it seems to connect standard ($Delta G^circ$) state with equilibrium ($K$) and that's not the meaning of this formula, correct?
$endgroup$
– user76122
Mar 25 at 10:12
$begingroup$
@user76122 $K$ and $Delta G^circ$ are very much connected. One point that is confusing for a beginner is that $Delta G^circ$ does not describe the change in free energy from reactants to products at equilibrium. I'll add this to my answer.
$endgroup$
– Night Writer
Mar 25 at 11:42
$begingroup$
At equilibrium, there is no change in free energy neither a net conversion from reactants to products (or vice versa). In my understanding, if you consider the "hanging rope"-style curve of $G(xi)$, $Delta G^circ$ and $K$ are just two different "points" on that curve, describing different things. The connection is, that these points lie on the same curve and therefore, can be calculated from each other. This is what is shown by the equations. Did you mean it that way (or at least in a similar way)? I'd like to emphasize that I'm trying to understand and not to correct you :)
$endgroup$
– user76122
Mar 25 at 12:53
$begingroup$
@user76122 I would not interpret $Delta G^circ$ and $K$ as points on the progress curve. Rather they are related to the slope of that curve.
$endgroup$
– Night Writer
Mar 25 at 19:25
1
$begingroup$
Thanks! By digging deeper, I found the following answer: chemistry.stackexchange.com/a/41864/76122 There it is claimed, that $Q=1$ has to lie on the curve, since $Q$ alters from $infty$ to zero.
$endgroup$
– user76122
Mar 28 at 9:42
|
show 3 more comments
$begingroup$
So, based on your answer and on the comments above, I think that my understanding increases. These are just two different point of views or two different points of the same equation: $Delta G=Delta G^circ + RTln{Q}$. My misunderstanding arises from the equation: $Delta G^circ = - RTln{K}$, since it seems to connect standard ($Delta G^circ$) state with equilibrium ($K$) and that's not the meaning of this formula, correct?
$endgroup$
– user76122
Mar 25 at 10:12
$begingroup$
@user76122 $K$ and $Delta G^circ$ are very much connected. One point that is confusing for a beginner is that $Delta G^circ$ does not describe the change in free energy from reactants to products at equilibrium. I'll add this to my answer.
$endgroup$
– Night Writer
Mar 25 at 11:42
$begingroup$
At equilibrium, there is no change in free energy neither a net conversion from reactants to products (or vice versa). In my understanding, if you consider the "hanging rope"-style curve of $G(xi)$, $Delta G^circ$ and $K$ are just two different "points" on that curve, describing different things. The connection is, that these points lie on the same curve and therefore, can be calculated from each other. This is what is shown by the equations. Did you mean it that way (or at least in a similar way)? I'd like to emphasize that I'm trying to understand and not to correct you :)
$endgroup$
– user76122
Mar 25 at 12:53
$begingroup$
@user76122 I would not interpret $Delta G^circ$ and $K$ as points on the progress curve. Rather they are related to the slope of that curve.
$endgroup$
– Night Writer
Mar 25 at 19:25
1
$begingroup$
Thanks! By digging deeper, I found the following answer: chemistry.stackexchange.com/a/41864/76122 There it is claimed, that $Q=1$ has to lie on the curve, since $Q$ alters from $infty$ to zero.
$endgroup$
– user76122
Mar 28 at 9:42
$begingroup$
So, based on your answer and on the comments above, I think that my understanding increases. These are just two different point of views or two different points of the same equation: $Delta G=Delta G^circ + RTln{Q}$. My misunderstanding arises from the equation: $Delta G^circ = - RTln{K}$, since it seems to connect standard ($Delta G^circ$) state with equilibrium ($K$) and that's not the meaning of this formula, correct?
$endgroup$
– user76122
Mar 25 at 10:12
$begingroup$
So, based on your answer and on the comments above, I think that my understanding increases. These are just two different point of views or two different points of the same equation: $Delta G=Delta G^circ + RTln{Q}$. My misunderstanding arises from the equation: $Delta G^circ = - RTln{K}$, since it seems to connect standard ($Delta G^circ$) state with equilibrium ($K$) and that's not the meaning of this formula, correct?
$endgroup$
– user76122
Mar 25 at 10:12
$begingroup$
@user76122 $K$ and $Delta G^circ$ are very much connected. One point that is confusing for a beginner is that $Delta G^circ$ does not describe the change in free energy from reactants to products at equilibrium. I'll add this to my answer.
$endgroup$
– Night Writer
Mar 25 at 11:42
$begingroup$
@user76122 $K$ and $Delta G^circ$ are very much connected. One point that is confusing for a beginner is that $Delta G^circ$ does not describe the change in free energy from reactants to products at equilibrium. I'll add this to my answer.
$endgroup$
– Night Writer
Mar 25 at 11:42
$begingroup$
At equilibrium, there is no change in free energy neither a net conversion from reactants to products (or vice versa). In my understanding, if you consider the "hanging rope"-style curve of $G(xi)$, $Delta G^circ$ and $K$ are just two different "points" on that curve, describing different things. The connection is, that these points lie on the same curve and therefore, can be calculated from each other. This is what is shown by the equations. Did you mean it that way (or at least in a similar way)? I'd like to emphasize that I'm trying to understand and not to correct you :)
$endgroup$
– user76122
Mar 25 at 12:53
$begingroup$
At equilibrium, there is no change in free energy neither a net conversion from reactants to products (or vice versa). In my understanding, if you consider the "hanging rope"-style curve of $G(xi)$, $Delta G^circ$ and $K$ are just two different "points" on that curve, describing different things. The connection is, that these points lie on the same curve and therefore, can be calculated from each other. This is what is shown by the equations. Did you mean it that way (or at least in a similar way)? I'd like to emphasize that I'm trying to understand and not to correct you :)
$endgroup$
– user76122
Mar 25 at 12:53
$begingroup$
@user76122 I would not interpret $Delta G^circ$ and $K$ as points on the progress curve. Rather they are related to the slope of that curve.
$endgroup$
– Night Writer
Mar 25 at 19:25
$begingroup$
@user76122 I would not interpret $Delta G^circ$ and $K$ as points on the progress curve. Rather they are related to the slope of that curve.
$endgroup$
– Night Writer
Mar 25 at 19:25
1
1
$begingroup$
Thanks! By digging deeper, I found the following answer: chemistry.stackexchange.com/a/41864/76122 There it is claimed, that $Q=1$ has to lie on the curve, since $Q$ alters from $infty$ to zero.
$endgroup$
– user76122
Mar 28 at 9:42
$begingroup$
Thanks! By digging deeper, I found the following answer: chemistry.stackexchange.com/a/41864/76122 There it is claimed, that $Q=1$ has to lie on the curve, since $Q$ alters from $infty$ to zero.
$endgroup$
– user76122
Mar 28 at 9:42
|
show 3 more comments
$begingroup$
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.
answered Mar 24 at 20:20
theoristtheorist
2488
$endgroup$
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.
answered Mar 24 at 20:20
theoristtheorist
2488
$endgroup$
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.
answered Mar 24 at 20:20
theoristtheorist
2488
$endgroup$
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.
answered Mar 24 at 20:20
theoristtheorist
2488
edited Mar 24 at 20:29
answered Mar 24 at 20:20
theoristtheorist
2488
answered Mar 24 at 20:20
theoristtheorist
2488
answered Mar 24 at 20:20
theoristtheorist
2488
2488
add a comment |
add a comment |
Thanks for contributing an answer to Chemistry Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fchemistry.stackexchange.com%2fquestions%2f111475%2fgibbs-free-energy-in-standard-state-vs-equilibrium%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
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♦
Mar 24 at 15:48
$begingroup$
ok, but the formulae say: At standard state $Delta G^circ = -RTln{K}$ 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
Mar 24 at 16:14
$begingroup$
I find that terribly confusing and wrong if it claims $K = 1$.
$endgroup$
– orthocresol♦
Mar 24 at 16:20
$begingroup$
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
Mar 24 at 17:36
$begingroup$
@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
Mar 24 at 18:24