## Delta G

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

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

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

Sophie 1I
Posts: 51
Joined: Fri Sep 29, 2017 7:04 am

### Delta G

In a sample of 1 mol of gas initially at 1atm and 298K thats heated at constant pressure to 350K, and is then compressed isothermally it its initial volume and finally cooled to 298K at constant volume, why is Delta G 0?

Evelyn L 1H
Posts: 67
Joined: Fri Sep 29, 2017 7:05 am

### Re: Delta G

Since G is a state function, only the final and initial states matter, not the path to get there. Since the final and initial states are the same, then delta G is 0.

Ramya Lakkaraju 1B
Posts: 67
Joined: Fri Sep 29, 2017 7:03 am

### Re: Delta G

This is also true for enthalpy and entropy, also state functions. If the initial state and final states are the same, then the change in that quantity is zero.

Brigitte Phung 1F
Posts: 50
Joined: Thu Jul 27, 2017 3:00 am
Been upvoted: 1 time

### Re: Delta G

Delta G is not pathway dependent, so it is only necessary to examine the initial and final states! On the other hand, q, w, and E are not state functions so their pathways are significant in calculations.