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Delta G at boiling point

Posted: Tue Feb 26, 2019 11:10 pm
by Fionna Shue 4L
Why is standard Gibbs free energy for the vaporization of water at 100 degrees Celsius 0?

Re: Delta G at boiling point

Posted: Wed Feb 27, 2019 11:46 am
by MaanasO 1A
This is assuming constant temperature and pressure. Sooooooooooooooooo....:

del(S) = q/T = del(H)/T
T * del(S) = T * del(H)/T = del(H)
del(G) = del(H) - T * del(S) = del(H) - del(H) = 0

Hope that helps!

Re: Delta G at boiling point

Posted: Wed Feb 27, 2019 1:22 pm
by Alexa Tabakian 1A
It is 0 because water vaporizes at 100 degrees, therefore, the system is technically in equilibrium because the question asks for the delta g of vaporization.