Gibbs free energy in standard state vs. equilibrium












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 $Delta G=Delta G^circ + RTln{Q}$. 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







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  • 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♦
    14 hours ago












  • $begingroup$
    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
    13 hours ago










  • $begingroup$
    I find that terribly confusing and wrong if it claims $K = 1$.
    $endgroup$
    – orthocresol♦
    13 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
    12 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
    11 hours ago


















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 $Delta G=Delta G^circ + RTln{Q}$. 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







share|improve this question







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♦
    14 hours ago












  • $begingroup$
    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
    13 hours ago










  • $begingroup$
    I find that terribly confusing and wrong if it claims $K = 1$.
    $endgroup$
    – orthocresol♦
    13 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
    12 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
    11 hours ago
















1












1








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 $Delta G=Delta G^circ + RTln{Q}$. 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







share|improve this question







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.







$endgroup$




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






share|improve this question







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







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




share|improve this question






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









asked 14 hours ago









user76122user76122

91




91




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user76122 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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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.






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Check out our Code of Conduct.








  • 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♦
    14 hours ago












  • $begingroup$
    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
    13 hours ago










  • $begingroup$
    I find that terribly confusing and wrong if it claims $K = 1$.
    $endgroup$
    – orthocresol♦
    13 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
    12 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
    11 hours ago
















  • 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♦
    14 hours ago












  • $begingroup$
    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
    13 hours ago










  • $begingroup$
    I find that terribly confusing and wrong if it claims $K = 1$.
    $endgroup$
    – orthocresol♦
    13 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
    12 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
    11 hours ago










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♦
14 hours ago






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














$begingroup$
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
13 hours ago




$begingroup$
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
13 hours ago












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




$begingroup$
I find that terribly confusing and wrong if it claims $K = 1$.
$endgroup$
– orthocresol♦
13 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
12 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
12 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
11 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
11 hours ago












2 Answers
2






active

oldest

votes


















4












$begingroup$

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 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









$endgroup$





















    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











    $endgroup$













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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

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      active

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      active

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      4












      $begingroup$

      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 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









      $endgroup$


















        4












        $begingroup$

        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 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









        $endgroup$
















          4












          4








          4





          $begingroup$

          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 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









          $endgroup$



          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 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












          share|improve this answer



          share|improve this answer










          answered 9 hours ago









          Night WriterNight Writer

          2,453223




          2,453223























              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











              $endgroup$


















                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











                $endgroup$
















                  1












                  1








                  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











                  $endgroup$



                  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.







                  share|improve this answer














                  share|improve this answer



                  share|improve this answer








                  edited 9 hours ago

























                  answered 9 hours ago









                  theoristtheorist

                  2388




                  2388






















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                      19世紀

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                      この記事は検証可能な参考文献や出典が全く示されていないか、不十分です。 出典を追加して記事の信頼性向上にご協力ください。 （ 2012年3月 ） 千年紀: 2千年紀 世紀: 18世紀 - 19世紀 - 20世紀 十年紀: 1800年代 1810年代 1820年代 1830年代 1840年代 1850年代 1860年代 1870年代 1880年代 1890年代 19世紀に君臨した大英帝国。 19世紀 （じゅうきゅうせいき）は、西暦1801年から西暦1900年までの100年間を指す世紀。 目次 1 19世紀の歴史 1.1 国民国家の成立 1.2 帝国主義の興隆 1.3 列強の植民地争奪戦 2 できごと 2.1 1800年代 2.2 1810年代 2.3 1820年代 2.4 1830年代 2.5 1840年代 2.6 1850年代 2.7 1860年代 2.8 1870年代 2.9 1880年代 2.10 1890年代 2.11 1900年代 3 文化 3.1 文学 3.2 音楽 3.3 思想 3.4 科学 3.5 技術 4 人物 4.1 ヨーロッパ 4.1.1 政治家・王族 4.1.1.1 フランス 4.1.1.2 オーストリア＝ハンガリー 4.1.1.3 ロシア 4.1.1.4 イギリス 4.1.1.5 ドイツ（プロイセンほかドイツ領邦を含む） 4.1.1.6 北欧 4.1.1.7 イタリア 4.1.1.8 スペイン 4.1.1.9 ベルギー 4.1.1.10 ギリシア 4.1.2 軍人 4.1.3 実業家 4.1.4 科学と技術 4.1.5 思想と哲学・人文諸学 4.1.6 宗教 4.1.7 文学 4.1.8 美術 4.1.9 音楽 4.1.10 社会事業家 4.1.11 探検家・旅行家 4.1.12 料...
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