## Delta G at Boiling Point

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

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

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

donnanguyen1d
Posts: 64
Joined: Fri Sep 29, 2017 7:04 am

### Delta G at Boiling Point

why is gibbs free energy zero at 100 boiling point (number 9.91)

Tasnia Haider 1E
Posts: 55
Joined: Sat Jul 22, 2017 3:01 am

### Re: Delta G at Boiling Point

At boiling point, the change in temperature is zero so change in enthalpy is also zero (infer that pressure is constant so W=0). Since change in enthalpy is zero, we also want change in entropy to be zero. This happens when W=0.

Shanmitha Arun 1L
Posts: 53
Joined: Fri Sep 29, 2017 7:04 am
Been upvoted: 1 time

### Re: Delta G at Boiling Point

At boiling point, temperature and pressure are constant. Therefore, the change in enthalpy would be zero.

Pooja Nair 1C
Posts: 55
Joined: Thu Jul 13, 2017 3:00 am

### Re: Delta G at Boiling Point

If you remember the graph of phase transitions, as a phase transition is about to occur, energy is still supplied to the system but temperature doesn't change immediately. Thus, at the point of a phase transition, such as the boiling point, there is no change in temperature. As delta G is a factor of change in temperature, delta G = 0 at the boiling point.

Return to “Gibbs Free Energy Concepts and Calculations”

### Who is online

Users browsing this forum: No registered users and 2 guests