Why is my conclusion inconsistent with the van't Hoff equation? Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)What kind of equilibrium constant we use for Gibbs free energy and Van't Hoff equation?What is the name for the equation ΔG = ΔG° + RT ln Q?Finding the thermodynamics of protein unfolding from temperature and absorbance using fluorescence spectroscopy?What's the source of “2.303” in Van't Hoff equation?Derivation of van 't Hoff equation for temperature dependence of equilibrium constantHow to derive Van't Hoff equation for Henry's constantVan't Hoff Equation with changing EnthalpyHow did Williard Gibbs come up with the Gibbs equation?Density calculation with cubic equation of stateWhy might copper have a lower heat capacity than lithium according to the Shomate Equation?

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Why is my conclusion inconsistent with the van't Hoff equation?



Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)What kind of equilibrium constant we use for Gibbs free energy and Van't Hoff equation?What is the name for the equation ΔG = ΔG° + RT ln Q?Finding the thermodynamics of protein unfolding from temperature and absorbance using fluorescence spectroscopy?What's the source of “2.303” in Van't Hoff equation?Derivation of van 't Hoff equation for temperature dependence of equilibrium constantHow to derive Van't Hoff equation for Henry's constantVan't Hoff Equation with changing EnthalpyHow did Williard Gibbs come up with the Gibbs equation?Density calculation with cubic equation of stateWhy might copper have a lower heat capacity than lithium according to the Shomate Equation?










3












$begingroup$


Let's say I hypothesize that a graph of $ln K$ vs. $1/T$ has a slope of $-∆G^circ/R$ and a $y$-intercept of $0$. I prove it simply:



$$∆G^circ = -RTln K quadtoquad ln K = -frac∆G^circRT$$



This matches the linear form $y = mx + b$. Thus, plotting $ln K$ vs. $1/T$ would have a slope $m = -∆G^circ/R$ and a $y$-intercept $b = 0$.



However, I understand that a van't Hoff plot defines a graph of $ln K$ vs. $1/T$ to have a slope of $-ΔH^circ/R$ and a $y$-intercept of $∆S^circ/R$. It is clear from the relation $∆G^circ = ∆H^circ - TΔS^circ$ that my final equation is thermodynamically equivalent to the van't Hoff equation. I do not disagree that



$$ln K = -frac∆H^circRT + frac∆S^circR,$$



but if I were to experimentally measure temperature and calculate the equilibrium constant temperature, why should I expect the y-intercept to be $∆S^circ/R$ as defined by van't Hoff rather than $0$ as I defined above? Why should I expect the slope to be $-ΔH^circ/R$ instead of $-ΔG^circ/R$? What makes the van't Hoff equation match experimentally determined values over the equation $ln K = -∆G^circ/(RT)$?










share|improve this question









New contributor




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







$endgroup$
















    3












    $begingroup$


    Let's say I hypothesize that a graph of $ln K$ vs. $1/T$ has a slope of $-∆G^circ/R$ and a $y$-intercept of $0$. I prove it simply:



    $$∆G^circ = -RTln K quadtoquad ln K = -frac∆G^circRT$$



    This matches the linear form $y = mx + b$. Thus, plotting $ln K$ vs. $1/T$ would have a slope $m = -∆G^circ/R$ and a $y$-intercept $b = 0$.



    However, I understand that a van't Hoff plot defines a graph of $ln K$ vs. $1/T$ to have a slope of $-ΔH^circ/R$ and a $y$-intercept of $∆S^circ/R$. It is clear from the relation $∆G^circ = ∆H^circ - TΔS^circ$ that my final equation is thermodynamically equivalent to the van't Hoff equation. I do not disagree that



    $$ln K = -frac∆H^circRT + frac∆S^circR,$$



    but if I were to experimentally measure temperature and calculate the equilibrium constant temperature, why should I expect the y-intercept to be $∆S^circ/R$ as defined by van't Hoff rather than $0$ as I defined above? Why should I expect the slope to be $-ΔH^circ/R$ instead of $-ΔG^circ/R$? What makes the van't Hoff equation match experimentally determined values over the equation $ln K = -∆G^circ/(RT)$?










    share|improve this question









    New contributor




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







    $endgroup$














      3












      3








      3





      $begingroup$


      Let's say I hypothesize that a graph of $ln K$ vs. $1/T$ has a slope of $-∆G^circ/R$ and a $y$-intercept of $0$. I prove it simply:



      $$∆G^circ = -RTln K quadtoquad ln K = -frac∆G^circRT$$



      This matches the linear form $y = mx + b$. Thus, plotting $ln K$ vs. $1/T$ would have a slope $m = -∆G^circ/R$ and a $y$-intercept $b = 0$.



      However, I understand that a van't Hoff plot defines a graph of $ln K$ vs. $1/T$ to have a slope of $-ΔH^circ/R$ and a $y$-intercept of $∆S^circ/R$. It is clear from the relation $∆G^circ = ∆H^circ - TΔS^circ$ that my final equation is thermodynamically equivalent to the van't Hoff equation. I do not disagree that



      $$ln K = -frac∆H^circRT + frac∆S^circR,$$



      but if I were to experimentally measure temperature and calculate the equilibrium constant temperature, why should I expect the y-intercept to be $∆S^circ/R$ as defined by van't Hoff rather than $0$ as I defined above? Why should I expect the slope to be $-ΔH^circ/R$ instead of $-ΔG^circ/R$? What makes the van't Hoff equation match experimentally determined values over the equation $ln K = -∆G^circ/(RT)$?










      share|improve this question









      New contributor




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







      $endgroup$




      Let's say I hypothesize that a graph of $ln K$ vs. $1/T$ has a slope of $-∆G^circ/R$ and a $y$-intercept of $0$. I prove it simply:



      $$∆G^circ = -RTln K quadtoquad ln K = -frac∆G^circRT$$



      This matches the linear form $y = mx + b$. Thus, plotting $ln K$ vs. $1/T$ would have a slope $m = -∆G^circ/R$ and a $y$-intercept $b = 0$.



      However, I understand that a van't Hoff plot defines a graph of $ln K$ vs. $1/T$ to have a slope of $-ΔH^circ/R$ and a $y$-intercept of $∆S^circ/R$. It is clear from the relation $∆G^circ = ∆H^circ - TΔS^circ$ that my final equation is thermodynamically equivalent to the van't Hoff equation. I do not disagree that



      $$ln K = -frac∆H^circRT + frac∆S^circR,$$



      but if I were to experimentally measure temperature and calculate the equilibrium constant temperature, why should I expect the y-intercept to be $∆S^circ/R$ as defined by van't Hoff rather than $0$ as I defined above? Why should I expect the slope to be $-ΔH^circ/R$ instead of $-ΔG^circ/R$? What makes the van't Hoff equation match experimentally determined values over the equation $ln K = -∆G^circ/(RT)$?







      thermodynamics free-energy






      share|improve this question









      New contributor




      Mateen Kasim 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




      Mateen Kasim 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








      edited 7 hours ago









      Karsten Theis

      4,604542




      4,604542






      New contributor




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









      asked 10 hours ago









      Mateen KasimMateen Kasim

      213




      213




      New contributor




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      Check out our Code of Conduct.





      New contributor





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






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




















          2 Answers
          2






          active

          oldest

          votes


















          4












          $begingroup$

          In the linear form $y = mx + b$, both $m$ and $b$ are constants, i.e. they don't depend on $x$. On the other hand, $Delta G^circ$ definitely depends on the temperature (and consequently on its inverse $1/T$). So if you plot a function $$f(x) = m x$$ where $m$ is not a constant but a function dependent on $x$, you might get something unexpected. In your case, $x$ is $1/T$ and $$m = -fracDelta HR + fracT Delta SR$$



          The $y$-intercept corresponds to an infinitely high temperature where $-fracDelta HR times frac1T$ tends to zero and $fracT Delta SR times frac1T$ cancels to be just $fracDelta SR$.






          share|improve this answer











          $endgroup$












          • $begingroup$
            Isn't ∆Gº a constant defined at 298 K? The relation ∆Gº=-RTlnK gives the standard change in free energy (i.e. 298 K, 100 kPa, 1M), so wouldn't this value be constant for a given reaction? Plus, if we were talking about non-standard values, doesn't change in enthalpy depend on temperature as well?
            $endgroup$
            – Mateen Kasim
            5 hours ago






          • 1




            $begingroup$
            @MateenKasim You are varying the temperature. The Gibbs energy and the equilibrium constant are significantly temperature-dependent. The enthalpy is temperature-dependent too, but to a much lesser degree.
            $endgroup$
            – Karsten Theis
            4 hours ago










          • $begingroup$
            If you want to see the rigorous treatment, look at the answer posted by @Chet_Miller.
            $endgroup$
            – Karsten Theis
            18 mins ago



















          3












          $begingroup$

          The fact of the matter is that the differential version of your equation



          $$fracmathrmdlnKmathrmdleft(frac1Tright) = -fracDelta G^circR$$



          is not exact (because $Delta G^circ$ is a function of $T$) while the form of the van't Hoff equation involving differentials



          $$fracmathrmdlnKmathrmdleft(frac1Tright) = -fracDelta H^circR$$



          is exact. Moreover, the derivation of the van't Hoff equation properly takes into account the fact that, in varying temperature $T$, the initial and final states for $Delta G^circ$ are constrained to be at 1 bar. So, in the van't Hoff development, the temperature derivative of $Delta G^circ$ is exactly given by



          $$fracDelta G^circmathrmdT = -Delta S^circ$$



          This important constraint is not even addressed in your approach.



          Finally, do you have a reference where it asserts that the intercept at $(1/T) to 0$ is supposed to be $Delta S^circ/R$?






          share|improve this answer











          $endgroup$













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            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            4












            $begingroup$

            In the linear form $y = mx + b$, both $m$ and $b$ are constants, i.e. they don't depend on $x$. On the other hand, $Delta G^circ$ definitely depends on the temperature (and consequently on its inverse $1/T$). So if you plot a function $$f(x) = m x$$ where $m$ is not a constant but a function dependent on $x$, you might get something unexpected. In your case, $x$ is $1/T$ and $$m = -fracDelta HR + fracT Delta SR$$



            The $y$-intercept corresponds to an infinitely high temperature where $-fracDelta HR times frac1T$ tends to zero and $fracT Delta SR times frac1T$ cancels to be just $fracDelta SR$.






            share|improve this answer











            $endgroup$












            • $begingroup$
              Isn't ∆Gº a constant defined at 298 K? The relation ∆Gº=-RTlnK gives the standard change in free energy (i.e. 298 K, 100 kPa, 1M), so wouldn't this value be constant for a given reaction? Plus, if we were talking about non-standard values, doesn't change in enthalpy depend on temperature as well?
              $endgroup$
              – Mateen Kasim
              5 hours ago






            • 1




              $begingroup$
              @MateenKasim You are varying the temperature. The Gibbs energy and the equilibrium constant are significantly temperature-dependent. The enthalpy is temperature-dependent too, but to a much lesser degree.
              $endgroup$
              – Karsten Theis
              4 hours ago










            • $begingroup$
              If you want to see the rigorous treatment, look at the answer posted by @Chet_Miller.
              $endgroup$
              – Karsten Theis
              18 mins ago
















            4












            $begingroup$

            In the linear form $y = mx + b$, both $m$ and $b$ are constants, i.e. they don't depend on $x$. On the other hand, $Delta G^circ$ definitely depends on the temperature (and consequently on its inverse $1/T$). So if you plot a function $$f(x) = m x$$ where $m$ is not a constant but a function dependent on $x$, you might get something unexpected. In your case, $x$ is $1/T$ and $$m = -fracDelta HR + fracT Delta SR$$



            The $y$-intercept corresponds to an infinitely high temperature where $-fracDelta HR times frac1T$ tends to zero and $fracT Delta SR times frac1T$ cancels to be just $fracDelta SR$.






            share|improve this answer











            $endgroup$












            • $begingroup$
              Isn't ∆Gº a constant defined at 298 K? The relation ∆Gº=-RTlnK gives the standard change in free energy (i.e. 298 K, 100 kPa, 1M), so wouldn't this value be constant for a given reaction? Plus, if we were talking about non-standard values, doesn't change in enthalpy depend on temperature as well?
              $endgroup$
              – Mateen Kasim
              5 hours ago






            • 1




              $begingroup$
              @MateenKasim You are varying the temperature. The Gibbs energy and the equilibrium constant are significantly temperature-dependent. The enthalpy is temperature-dependent too, but to a much lesser degree.
              $endgroup$
              – Karsten Theis
              4 hours ago










            • $begingroup$
              If you want to see the rigorous treatment, look at the answer posted by @Chet_Miller.
              $endgroup$
              – Karsten Theis
              18 mins ago














            4












            4








            4





            $begingroup$

            In the linear form $y = mx + b$, both $m$ and $b$ are constants, i.e. they don't depend on $x$. On the other hand, $Delta G^circ$ definitely depends on the temperature (and consequently on its inverse $1/T$). So if you plot a function $$f(x) = m x$$ where $m$ is not a constant but a function dependent on $x$, you might get something unexpected. In your case, $x$ is $1/T$ and $$m = -fracDelta HR + fracT Delta SR$$



            The $y$-intercept corresponds to an infinitely high temperature where $-fracDelta HR times frac1T$ tends to zero and $fracT Delta SR times frac1T$ cancels to be just $fracDelta SR$.






            share|improve this answer











            $endgroup$



            In the linear form $y = mx + b$, both $m$ and $b$ are constants, i.e. they don't depend on $x$. On the other hand, $Delta G^circ$ definitely depends on the temperature (and consequently on its inverse $1/T$). So if you plot a function $$f(x) = m x$$ where $m$ is not a constant but a function dependent on $x$, you might get something unexpected. In your case, $x$ is $1/T$ and $$m = -fracDelta HR + fracT Delta SR$$



            The $y$-intercept corresponds to an infinitely high temperature where $-fracDelta HR times frac1T$ tends to zero and $fracT Delta SR times frac1T$ cancels to be just $fracDelta SR$.







            share|improve this answer














            share|improve this answer



            share|improve this answer








            edited 7 hours ago

























            answered 9 hours ago









            Karsten TheisKarsten Theis

            4,604542




            4,604542











            • $begingroup$
              Isn't ∆Gº a constant defined at 298 K? The relation ∆Gº=-RTlnK gives the standard change in free energy (i.e. 298 K, 100 kPa, 1M), so wouldn't this value be constant for a given reaction? Plus, if we were talking about non-standard values, doesn't change in enthalpy depend on temperature as well?
              $endgroup$
              – Mateen Kasim
              5 hours ago






            • 1




              $begingroup$
              @MateenKasim You are varying the temperature. The Gibbs energy and the equilibrium constant are significantly temperature-dependent. The enthalpy is temperature-dependent too, but to a much lesser degree.
              $endgroup$
              – Karsten Theis
              4 hours ago










            • $begingroup$
              If you want to see the rigorous treatment, look at the answer posted by @Chet_Miller.
              $endgroup$
              – Karsten Theis
              18 mins ago

















            • $begingroup$
              Isn't ∆Gº a constant defined at 298 K? The relation ∆Gº=-RTlnK gives the standard change in free energy (i.e. 298 K, 100 kPa, 1M), so wouldn't this value be constant for a given reaction? Plus, if we were talking about non-standard values, doesn't change in enthalpy depend on temperature as well?
              $endgroup$
              – Mateen Kasim
              5 hours ago






            • 1




              $begingroup$
              @MateenKasim You are varying the temperature. The Gibbs energy and the equilibrium constant are significantly temperature-dependent. The enthalpy is temperature-dependent too, but to a much lesser degree.
              $endgroup$
              – Karsten Theis
              4 hours ago










            • $begingroup$
              If you want to see the rigorous treatment, look at the answer posted by @Chet_Miller.
              $endgroup$
              – Karsten Theis
              18 mins ago
















            $begingroup$
            Isn't ∆Gº a constant defined at 298 K? The relation ∆Gº=-RTlnK gives the standard change in free energy (i.e. 298 K, 100 kPa, 1M), so wouldn't this value be constant for a given reaction? Plus, if we were talking about non-standard values, doesn't change in enthalpy depend on temperature as well?
            $endgroup$
            – Mateen Kasim
            5 hours ago




            $begingroup$
            Isn't ∆Gº a constant defined at 298 K? The relation ∆Gº=-RTlnK gives the standard change in free energy (i.e. 298 K, 100 kPa, 1M), so wouldn't this value be constant for a given reaction? Plus, if we were talking about non-standard values, doesn't change in enthalpy depend on temperature as well?
            $endgroup$
            – Mateen Kasim
            5 hours ago




            1




            1




            $begingroup$
            @MateenKasim You are varying the temperature. The Gibbs energy and the equilibrium constant are significantly temperature-dependent. The enthalpy is temperature-dependent too, but to a much lesser degree.
            $endgroup$
            – Karsten Theis
            4 hours ago




            $begingroup$
            @MateenKasim You are varying the temperature. The Gibbs energy and the equilibrium constant are significantly temperature-dependent. The enthalpy is temperature-dependent too, but to a much lesser degree.
            $endgroup$
            – Karsten Theis
            4 hours ago












            $begingroup$
            If you want to see the rigorous treatment, look at the answer posted by @Chet_Miller.
            $endgroup$
            – Karsten Theis
            18 mins ago





            $begingroup$
            If you want to see the rigorous treatment, look at the answer posted by @Chet_Miller.
            $endgroup$
            – Karsten Theis
            18 mins ago












            3












            $begingroup$

            The fact of the matter is that the differential version of your equation



            $$fracmathrmdlnKmathrmdleft(frac1Tright) = -fracDelta G^circR$$



            is not exact (because $Delta G^circ$ is a function of $T$) while the form of the van't Hoff equation involving differentials



            $$fracmathrmdlnKmathrmdleft(frac1Tright) = -fracDelta H^circR$$



            is exact. Moreover, the derivation of the van't Hoff equation properly takes into account the fact that, in varying temperature $T$, the initial and final states for $Delta G^circ$ are constrained to be at 1 bar. So, in the van't Hoff development, the temperature derivative of $Delta G^circ$ is exactly given by



            $$fracDelta G^circmathrmdT = -Delta S^circ$$



            This important constraint is not even addressed in your approach.



            Finally, do you have a reference where it asserts that the intercept at $(1/T) to 0$ is supposed to be $Delta S^circ/R$?






            share|improve this answer











            $endgroup$

















              3












              $begingroup$

              The fact of the matter is that the differential version of your equation



              $$fracmathrmdlnKmathrmdleft(frac1Tright) = -fracDelta G^circR$$



              is not exact (because $Delta G^circ$ is a function of $T$) while the form of the van't Hoff equation involving differentials



              $$fracmathrmdlnKmathrmdleft(frac1Tright) = -fracDelta H^circR$$



              is exact. Moreover, the derivation of the van't Hoff equation properly takes into account the fact that, in varying temperature $T$, the initial and final states for $Delta G^circ$ are constrained to be at 1 bar. So, in the van't Hoff development, the temperature derivative of $Delta G^circ$ is exactly given by



              $$fracDelta G^circmathrmdT = -Delta S^circ$$



              This important constraint is not even addressed in your approach.



              Finally, do you have a reference where it asserts that the intercept at $(1/T) to 0$ is supposed to be $Delta S^circ/R$?






              share|improve this answer











              $endgroup$















                3












                3








                3





                $begingroup$

                The fact of the matter is that the differential version of your equation



                $$fracmathrmdlnKmathrmdleft(frac1Tright) = -fracDelta G^circR$$



                is not exact (because $Delta G^circ$ is a function of $T$) while the form of the van't Hoff equation involving differentials



                $$fracmathrmdlnKmathrmdleft(frac1Tright) = -fracDelta H^circR$$



                is exact. Moreover, the derivation of the van't Hoff equation properly takes into account the fact that, in varying temperature $T$, the initial and final states for $Delta G^circ$ are constrained to be at 1 bar. So, in the van't Hoff development, the temperature derivative of $Delta G^circ$ is exactly given by



                $$fracDelta G^circmathrmdT = -Delta S^circ$$



                This important constraint is not even addressed in your approach.



                Finally, do you have a reference where it asserts that the intercept at $(1/T) to 0$ is supposed to be $Delta S^circ/R$?






                share|improve this answer











                $endgroup$



                The fact of the matter is that the differential version of your equation



                $$fracmathrmdlnKmathrmdleft(frac1Tright) = -fracDelta G^circR$$



                is not exact (because $Delta G^circ$ is a function of $T$) while the form of the van't Hoff equation involving differentials



                $$fracmathrmdlnKmathrmdleft(frac1Tright) = -fracDelta H^circR$$



                is exact. Moreover, the derivation of the van't Hoff equation properly takes into account the fact that, in varying temperature $T$, the initial and final states for $Delta G^circ$ are constrained to be at 1 bar. So, in the van't Hoff development, the temperature derivative of $Delta G^circ$ is exactly given by



                $$fracDelta G^circmathrmdT = -Delta S^circ$$



                This important constraint is not even addressed in your approach.



                Finally, do you have a reference where it asserts that the intercept at $(1/T) to 0$ is supposed to be $Delta S^circ/R$?







                share|improve this answer














                share|improve this answer



                share|improve this answer








                edited 4 hours ago









                andselisk

                19.6k665127




                19.6k665127










                answered 4 hours ago









                Chet MillerChet Miller

                6,6411613




                6,6411613




















                    Mateen Kasim is a new contributor. Be nice, and check out our Code of Conduct.









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