Gibbs Free Energy and Enthelpy

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SUMMARY

The discussion centers on the relationship between Gibbs Free Energy (ΔG) and Enthalpy (ΔH) in thermodynamic reactions, particularly in battery systems. It is established that for a battery reaction with ΔG = -394 kJ/mol and ΔH = -316 kJ/mol, the difference of 78 kJ/mol indicates that the system absorbs heat from the environment. The key takeaway is that while ΔH < 0 suggests an exothermic reaction, the presence of electrical work complicates the determination of whether the process is endothermic or exothermic. The heat absorbed does not equate to ΔH due to the electrical work involved.

PREREQUISITES
  • Understanding of Gibbs Free Energy and Enthalpy concepts
  • Familiarity with thermodynamic equations, specifically ΔG = ΔH - TΔS
  • Knowledge of battery chemistry and electrochemical reactions
  • Basic principles of ideal gas reactions and standard heats of formation
NEXT STEPS
  • Study the relationship between Gibbs Free Energy and Enthalpy in various chemical reactions
  • Learn about the standard heats of formation and standard free energies of formation for ideal gases
  • Explore the concept of reversible reactions in thermodynamics
  • Investigate the role of electrical work in thermodynamic systems, particularly in batteries
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Chemistry students, thermodynamics researchers, and professionals in electrochemistry who seek to deepen their understanding of energy transformations in chemical reactions.

Mayan Fung
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When I studied chemistry in high school, I learned that if the change of enthalpy of a reaction ΔH > 0 , the reaction is endothermic, and if ΔH<0, it is exothermic.
However in thermodynamic class, I learnt:
$$ ΔG = ΔH - TΔS $$
For a reaction of a battery, the data reads
ΔG = -394kJ/mol. (which is also the electrical work by the battery), ΔH = -316kJ/mol.
The book then said that the difference (78kJ/mol.) comes from absorbing heat from the environment. So is this an endothermic reaction? However, ΔH<0. Is the system in fact absorbing heat or releasing heat?

Thanks!
 
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If the battery is maintained at constant temperature while it is doing the electrical work it will absorb that amount of heat from the environment and convert that heat to electrical work.
 
Bystander said:
If the battery is maintained at constant temperature while it is doing the electrical work it will absorb that amount of heat from the environment and convert that heat to electrical work.
Can I say I can't determine whether the process is endothermic or not only by looking at the enthalpy?
 
Looking at the enthalpy change determines whether the reaction is exothermic or endothermic.
 
Chestermiller said:
Looking at the enthalpy change determines whether the reaction is exothermic or endothermic.
Um... but the system apparently is absorbing heat from the ambient?
 
Chan Pok Fung said:
Um... but the system apparently is absorbing heat from the ambient?
Who says?
 
The book then said that the difference (78kJ/mol.) comes from absorbing heat from the environment.
 
Chan Pok Fung said:
The book then said that the difference (78kJ/mol.) comes from absorbing heat from the environment.
What book is this?
 
To get us started, rather than examining an electrolytic battery reaction, let's first start out simpler by considering a chemical reaction involving ideal gases. Please pick out an ideal gas reaction you would like to look, and look up the standard heats of formation and standard free energies of formation of the reactants and products at 25 C and 1 atm. We will use this to determine the standard heat of reaction and free energy of reaction, and will talk about how to carry out the reaction reversibly at 25 C and 1 atm.
 
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  • #10
It is from "An Introduction to Thermal Physics" by Daniel V. Schroeder. The following is a direct quote from the book:
$$ Pb + PbO_2 + 4H^+ + 2SO^{2-}_4 → 2PbSO_4 + 2H_2O $$
" ΔG for this reaction is -394kJ/mol, at standard pressure, temperature, and concentration of the solution. So the electrical work produced under these conditions, per mole of metallic lead, is 394kJ. Meanwhile, ΔH for this reaction is -316kJ/mol, so the energy that comes out of the chemicals is actually less than the work done, by 78kJ. This extra energy comes from heat, absorbed from the environment."
 
  • #11
I can now see what your problem is. The change in enthalpy is equal to the amount of heat Q added only if the work involved in the process is P-V work at constant pressure. In a battery system, there is electrical work involved. So Q is not equal to ##\Delta H##. In this system, if the process is carried out reversibly, the amount of heat added is ##Q = T\Delta S##
 
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  • #12
Thanks for your explanation!
 

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