## Delta G at boiling point

$\Delta G^{\circ}= \Delta H^{\circ} - T \Delta S^{\circ}$

$\Delta G^{\circ}= -RT\ln K$

$\Delta G^{\circ}= \sum \Delta G_{f}^{\circ}(products) - \sum \Delta G_{f}^{\circ}(reactants)$

Fionna Shue 4L
Posts: 59
Joined: Fri Sep 28, 2018 12:18 am

### Delta G at boiling point

Why is standard Gibbs free energy for the vaporization of water at 100 degrees Celsius 0?

MaanasO 1A
Posts: 72
Joined: Fri Sep 28, 2018 12:26 am

### Re: Delta G at boiling point

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!

Alexa Tabakian 1A
Posts: 38
Joined: Fri Sep 28, 2018 12:20 am

### Re: Delta G at boiling point

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.