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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.

Tony Stark
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    Your book is sloppy? – Zhe Sep 01 '20 at 13:51
  • @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 https://chem.libretexts.org/Bookshelves/Introductory_Chemistry/Book%3A_Introductory_Chemistry_(CK-12)/18%3A_Kinetics/18.04%3A_Potential_Energy_Diagrams – Tony Stark Sep 01 '20 at 13:55
  • Instead of using what appears to be the Arrhenius equation, trying looking at the Eyring equation instead: https://en.wikipedia.org/wiki/Eyring_equation – Zhe Sep 01 '20 at 15:34
  • @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. – Tony Stark Sep 01 '20 at 15:58
  • The Eyring equation is from modern transition state theory. The Arrhenius equation is a fine starting point, and shares some similar features. – Zhe Sep 01 '20 at 16:06
  • Related: https://chemistry.stackexchange.com/q/115923/72973 – Karsten Sep 08 '20 at 10:45

1 Answers1

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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
Safdar Faisal
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Mitradip Das
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  • Refer to NCERT Chemistry Class 12 Part 1 - Chemical Kinetics page 112 Graph 4.7 . It clearly mentions Potential Energy to Y-axis. – Tony Stark Sep 01 '20 at 06:52
  • Even here they are deliberately using Potential Energy on the Y-axis along with same Enthalpy representation https://chem.libretexts.org/Bookshelves/Introductory_Chemistry/Book%3A_Introductory_Chemistry_(CK-12)/18%3A_Kinetics/18.04%3A_Potential_Energy_Diagrams – Tony Stark Sep 01 '20 at 07:09
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    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. – Safdar Faisal Sep 01 '20 at 16:11
  • So are you saying that in the Y-axis of my diagram,it should contain Enthalpy? – Tony Stark Sep 02 '20 at 02:15
  • @TonyStark The Y axis is dependent on reaction conditions. You cannot expect to use Helmholtz free energy while studying reaction in $\mu$PT conditions. – Mitradip Das Sep 03 '20 at 06:30
  • Also, the Arrhenius activation energy is not equal to $\Delta^\neq H^\circ$. For a bimolecular reaction according to transition-state theory, $E_a =\Delta^\neq H^\circ + 2RT $, for a bimolecular reaction. – Antonio de Oliveira-Filho Sep 10 '20 at 20:01