0
$\begingroup$

I am trying to make sense of these two graphs. They appear to have similar axes, yet display different curves.

Please let me know if this understanding is correct:

In the top image (x-axis = "Reaction coordinate"), this shows the free energy as we convert X equivalents of A to Y equivalent of B (with X and Y relating to the stoichiometric coefficients). So this is looking at an individual reaction essentially.

And for the bottom image (x-axis = "Reaction progress"), this shows the free energy versus the reaction quotient (Q) essentially.

A large part of my confusion is why the bottom image does not indicate activation energy yet the top one does. Does this have to do with Gibbs free energy being a state function. It is true that over the course of the reaction, reactants require some activation energy, but it is not represented in the bottom image since Gibbs free energy is a state function? Perhaps the top image shows the actual "path" taken, while the bottom image just indicates the overall change in Gibbs free energy as Q changes.

enter image description here

enter image description here

$\endgroup$
3
  • 1
    $\begingroup$ chemistry.stackexchange.com/questions/173557/… $\endgroup$
    – Mithoron
    Jul 6, 2023 at 16:55
  • 1
    $\begingroup$ chemistry.stackexchange.com/questions/456/… $\endgroup$
    – Mithoron
    Jul 6, 2023 at 16:55
  • 1
    $\begingroup$ @ait2001 the diagrams you are asking about have no relation between them. The first refer to the reaction coordinate (chemical kinetics) while the second refers to the reaction progress (chemical thermodynamics). You should be able to differentiate those concepts to understand the difference. $\endgroup$
    – PAEP
    Jul 6, 2023 at 16:57

1 Answer 1

2
$\begingroup$

The lower plot is OK, it shows that the Gibbs energy is a minimum at equilibrium, i.e. the slope of free energy vs extent of reaction $\xi$ is zero so $\displaystyle \Delta G_r=\left(\frac{\partial G}{\partial \xi}\right)_{T,p}=0$, then $\Delta G^\text{o}=-RT\ln(K_e)$

The upper plot is far less well defined and really only reminds us that there is a barrier between reactants and products, and the use of Gibbs free energy here really only shows that entropy is considered as well as enthalpy, but using $E_a$ is then confusing as it is an energy not free energy. The equilibrium between A and B is at the minimum of the curve in the lower plot. Recall, that there is an activation energy from B to A but is not shown in the top picture. The reaction coordinate is not defined in any real sense either, it is just used to indicate that bonds are broken and others formed.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.