## Gibbs Free Energy: Spontaneous versus Unstable?

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

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

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

Cowasjee_Sanaea_3E
Posts: 27
Joined: Fri Jul 22, 2016 3:00 am

### Gibbs Free Energy: Spontaneous versus Unstable?

In problem number 9.63 in the textbook it asks:
Determine which of the following compounds are stable with respect to decomposition into their elements under standard conditions at 25 degrees C.

The answer was that those with negative free energies were stable. However my understanding was that being unstable meant spontaneity and therefore the unstable compounds would be negative (as negative Gibbs free energy means spontaneous). Is this not the case because it is talking about decomposition?

Also, how come you don't have to do the full reaction of decomposition equation to find the free energy? And instead just look at the compounds delta G directly?

Kareem
Posts: 18
Joined: Wed Sep 21, 2016 3:00 pm

### Re: Gibbs Free Energy: Spontaneous versus Unstable?

The question is asking about stability in terms of the decomposition, meaning is it stable to decompose into its elements. The questions is basically asking if it will decompose spontaneously or not, so when it decomposes is it stable or not.

ntyshchenko
Posts: 21
Joined: Wed Sep 21, 2016 2:59 pm

### Re: Gibbs Free Energy: Spontaneous versus Unstable?

But if the Gibbs free energy of formation of PCl5 is -305 then doesn't that mean that the decomposition of PCl5 is +305 and thus the decomposition of it is thermodynamically unstable?