Kinetics Vs Thermodynamics : Apparent contradiction in the definition of Enthalpy in the two disciplines

Question

My book gives the following curve:

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 $ΔV=0$

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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

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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.

Answer 1

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.

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Summary of the terms used:

• N: No of molecules
• $\mu$: Chemical potential
• V: Volume
• P: Pressure
• T: Temperature