Insights Blog
-- Browse All Articles --
Physics Articles
Physics Tutorials
Physics Guides
Physics FAQ
Math Articles
Math Tutorials
Math Guides
Math FAQ
Education Articles
Education Guides
Bio/Chem Articles
Technology Guides
Computer Science Tutorials
Forums
Classical Physics
Quantum Physics
Quantum Interpretations
Special and General Relativity
Atomic and Condensed Matter
Nuclear and Particle Physics
Beyond the Standard Model
Cosmology
Astronomy and Astrophysics
Other Physics Topics
Trending
Featured Threads
Log in
Register
What's new
Search
Search
Search titles only
By:
Classical Physics
Quantum Physics
Quantum Interpretations
Special and General Relativity
Atomic and Condensed Matter
Nuclear and Particle Physics
Beyond the Standard Model
Cosmology
Astronomy and Astrophysics
Other Physics Topics
Menu
Log in
Register
Navigation
More options
Contact us
Close Menu
JavaScript is disabled. For a better experience, please enable JavaScript in your browser before proceeding.
You are using an out of date browser. It may not display this or other websites correctly.
You should upgrade or use an
alternative browser
.
Forums
Physics
Classical Physics
Gibbs Free energy vs Gibbs Free energy at standard state
Reply to thread
Message
[QUOTE="sgstudent, post: 4509428, member: 384987"] ΔG° is the measure of Gibbs Free energy change at 1 bar but no specified temperature and also the stoichiometric amounts depending on the equation of the chemical reaction. For example, if X ->2Y then the ΔG° would be equal to ΔG°f(2Y)-ΔG°f(X). While for ΔG it is a general term for Gibbs Free Energy that might not have the same properties as the ΔG° (1 bar, stoichiometric amounts) So I was wondering, for a equilibrium reaction the ΔG=0 as when it reaches the lowest point the gradient=0. At this point, there is a certain composition of reactants to products that would not be 0% reactants and 100% products as that would mean the reaction is irreversible and reaches a lowest point at 100% products. The graph for the reversible reaction would look like this: [url]http://postimg.org/image/r9vtp0fh3/[/url] So now in most questions, such as in phase changes questions they would give us the ΔH° and ΔS° and they then would want us to find the temperature at which equilibrium takes place. So using the formula ΔG°=ΔH°-TΔS° we would let ΔG° be 0 and solve for T. The part that i don't get is that ΔG° is a non-zero value and that we shouldn't be able to use ΔH° or ΔS° to find ΔG because either ΔH° or ΔS° represents 100% complete reaction. So I'm not sure why we are allowed to do those thing actually. Thanks in advance for the help :) [/QUOTE]
Insert quotes…
Post reply
Forums
Physics
Classical Physics
Gibbs Free energy vs Gibbs Free energy at standard state
Back
Top