1
$\begingroup$

My book gives the following curve:

enter image description here

It gives the following relation :$$ΔH= E_{\mathrm{activation,forward}}-E_{\mathrm{activation,backward}} \tag{1}$$

But I suspect that $ E_{\mathrm{activation,forward}}-E_{\mathrm{activation,backward}}$ corresponds to change in Internal Energy of the system ΔU.

or $$ΔU= E_{\mathrm{activation,forward}}-E_{\mathrm{activation,backward}} \tag{2}$$

However,at a constant pressure, heat of reaction or Enthalpy change $$ΔH = ΔU + PΔV \tag{3}$$

Statement (3) contradicts statements (1) and (2) or is valid only for $ΔV=0$


This source contains the same graph : https://chem.libretexts.org/Bookshelves/Introductory_Chemistry/Book%3A_Introductory_Chemistry_(CK-12)/18%3A_Kinetics/18.04%3A_Potential_Energy_Diagrams


What is going on here?

In case, you feel that the graph is given only to cover a particular kind of reaction, please provide relevant comments.

$\endgroup$
  • 3
    $\begingroup$ Your book is sloppy? $\endgroup$ – Zhe Sep 1 at 13:51
  • $\begingroup$ @Zhe It does contain some misprints but overall it is a very good book for undergraduates. Moreover I have seen similar graphs in other books and on internet like chem.libretexts.org/Bookshelves/Introductory_Chemistry/… $\endgroup$ – Tony Stark Sep 1 at 13:55
  • $\begingroup$ Instead of using what appears to be the Arrhenius equation, trying looking at the Eyring equation instead: en.wikipedia.org/wiki/Eyring_equation $\endgroup$ – Zhe Sep 1 at 15:34
  • $\begingroup$ @Zhe Sir I am just an undergrad and Erying equation seems a little beyond my understanding. Please explain what are you trying to convey here through this equation. $\endgroup$ – Tony Stark Sep 1 at 15:58
  • $\begingroup$ The Eyring equation is from modern transition state theory. The Arrhenius equation is a fine starting point, and shares some similar features. $\endgroup$ – Zhe Sep 1 at 16:06
5
$\begingroup$

The reason might be that while drawing the reaction energy profile, we forget to mention what energy we are mentioning in the Y-axis. The following conventions are generally used:

  • If reaction conditions are constant NVT, energy in Y-axis should represent internal energy.
  • If reaction conditions are constant NPT, energy in Y-axis should represent enthalpy.
  • If reaction conditions are constant $\mu$VT, energy in Y-axis should represent Helmholtz free energy.
  • If reaction conditions are constant $\mu$PT, energy in Y-axis should represent Gibbs free energy.
  • If you are looking at single molecule, the energy will be the total energy of the molecule (kinetic energy + potential energy).

In case of reactions, the last three are generally used.

Hence, the "Energy" Y-axis changes based on reaction conditions. In the question, it seems you might have got mixed up somehow.


Summary of the terms used:

  • N: No of molecules
  • $\mu$: Chemical potential
  • V: Volume
  • P: Pressure
  • T: Temperature
| improve this answer | |
$\endgroup$
  • $\begingroup$ Refer to NCERT Chemistry Class 12 Part 1 - Chemical Kinetics page 112 Graph 4.7 . It clearly mentions Potential Energy to Y-axis. $\endgroup$ – Tony Stark Sep 1 at 6:52
  • $\begingroup$ Even here they are deliberately using Potential Energy on the Y-axis along with same Enthalpy representation chem.libretexts.org/Bookshelves/Introductory_Chemistry/… $\endgroup$ – Tony Stark Sep 1 at 7:09
  • 1
    $\begingroup$ Please refrain from using a footnote that adds nothing to the answer. They can only reply in the comments and a greeting/footnote is unnecessary(adding very little value) in a Q&A forum. $\endgroup$ – Safdar Sep 1 at 16:11
  • $\begingroup$ So are you saying that in the Y-axis of my diagram,it should contain Enthalpy? $\endgroup$ – Tony Stark Sep 2 at 2:15
  • $\begingroup$ @TonyStark The Y axis is dependent on reaction conditions. You cannot expect to use Helmholtz free energy while studying reaction in $\mu$PT conditions. $\endgroup$ – Mitradip Das Sep 3 at 6:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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