## Homework 6th Edition 9.63

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

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

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

Tamera Scott 1G
Posts: 65
Joined: Fri Sep 28, 2018 12:27 am

### Homework 6th Edition 9.63

For 9.63, how would you determine which are stable with respect to the decomposition into their elements?

Grace Kim 1J
Posts: 60
Joined: Fri Sep 28, 2018 12:18 am
Been upvoted: 1 time

### Re: Homework 6th Edition 9.63

The question asks for the stability of the compounds. If the formation energy is positive, the compound form has more energy and is thus more unstable. If the formation energy is negative, the compound is thermodynamically stable.

Hope this helped!

Vincent Li 4L
Posts: 48
Joined: Fri Sep 28, 2018 12:19 am

### Re: Homework 6th Edition 9.63

The idea is that the lower the energy of the substance, the more stable it is as a compound. In cases where the formation is negative from the elements to the substance, this means that energy was lost as the elements reacted to create the compound, or that the compound has lower energy and is therefore stable. When the formation energy is positive, then energy was gained, and that means the compound is then less stable.

Jeremiah Hutauruk
Posts: 67
Joined: Fri Sep 28, 2018 12:28 am

### Re: Homework 6th Edition 9.63

I think it is because more complex compounds have more bonds, thus more energy to break. And by having more bonds, it needs more energy to keep its form and more energy means more entropy, thus making it unstable.