These lessons are amazing! Wish I found your channel before I was halfway through the semester haha... I do have to take pchem 2 though so I think I'll still be using these a ton.
Derivative of S with respect to t at constant volume =Cp/T. But when we derived this relation in the previous videos we assumed reversible heat .So can we say that the Gibbs Helmholtz equation is valid for reversible processes and would turn into an inequality for an irreversible process ?
(First, I think you mean constant pressure, not constant volume) But to answer your main question: no! The Gibbs-Helmholtz equation is, in fact, always valid. That is the wonderful thing about using state variables instead of path-dependent variables. One you have managed to find the result (via any path), the answer must be the same (for all paths).
Not sure if this answers your question but this equation basically says that the rate of change of free energy, with respect to temperature, is negative with respect to temperature squared, not sure how this can be converted to systems in equilibrium but so long as the system is closed, this negative proportional relationship means that changing the temperature favours the reaction that reverses the direction of temperature change
No, not directly or easily. Formation energies are a difference between energies of products and reactants for a particular family of reactions. The Gibbs-Helmholtz equation describes how changes in G and H with temperature are related
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These lessons are amazing! Wish I found your channel before I was halfway through the semester haha... I do have to take pchem 2 though so I think I'll still be using these a ton.
Better late than never! I'm glad you found them, and that you find them helpful
Thank you so much, was so easy with your explanation
I appreciate that, thanks
Derivative of S with respect to t at constant volume =Cp/T. But when we derived this relation in the previous videos we assumed reversible heat .So can we say that the Gibbs Helmholtz equation is valid for reversible processes and would turn into an inequality for an irreversible process ?
(First, I think you mean constant pressure, not constant volume)
But to answer your main question: no! The Gibbs-Helmholtz equation is, in fact, always valid. That is the wonderful thing about using state variables instead of path-dependent variables. One you have managed to find the result (via any path), the answer must be the same (for all paths).
@@PhysicalChemistry Got it .Thank you .
I learned from other class that this equation can lead to Le Chatelier's principle about the change of T, but I still can't understand how.
Not sure if this answers your question but this equation basically says that the rate of change of free energy, with respect to temperature, is negative with respect to temperature squared, not sure how this can be converted to systems in equilibrium but so long as the system is closed, this negative proportional relationship means that changing the temperature favours the reaction that reverses the direction of temperature change
What a legend. You even write backwards so we can read your notes 👍
I hate to disappoint you, but I'm writing forwards. The computer is the one doing things backwards: ruclips.net/video/YmvJVkyJbLc/видео.html
@@PhysicalChemistry i did imagine some such arrangement lol
@@PhysicalChemistry😢always disappointed it’s ok sir
Sir this is too good
Can you derive Formation energy from Gibbs Helmholtz equation?
No, not directly or easily. Formation energies are a difference between energies of products and reactants for a particular family of reactions. The Gibbs-Helmholtz equation describes how changes in G and H with temperature are related
Thank you sir ❤️🙏
You are very welcome
Can you explain in simplest form this derivation
thx 🇲🇦
yw
oh i hate metallurgical thermodynamics...