Gibbs Free Energy and Enthalpy

  • #1
When I studied chemistry in high school, I learnt 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!
 

Answers and Replies

  • #2
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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.
 
  • #3
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?
 
  • #4
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Looking at the enthalpy change determines whether the reaction is exothermic or endothermic.
 
  • #5
Looking at the enthalpy change determines whether the reaction is exothermic or endothermic.
Um... but the system apparently is absorbing heat from the ambient?
 
  • #7
The book then said that the difference (78kJ/mol.) comes from absorbing heat from the environment.
 
  • #9
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To get us started, rather than examining an electrolytic battery reaction, lets 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
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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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