## 9.63

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

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

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

Rakhi Ratanjee 1D
Posts: 59
Joined: Fri Sep 29, 2017 7:04 am

### 9.63

This question asks to determine which compounds are stable in terms of decomposition into elements under standard conditions. The answer referenced the Gibbs free energy of formation and stated that if the Gibbs free energy is positive then the reaction is not favorable and if it is negative, then it is favorable. Therefore, A and D were stable. However, doesn't using the Gibbs free energy of formation imply that the reaction is in creating the compound, not decomposing it? Don't we have to flip the sign of the Gibbs free energy value to find the energy of the reverse reaction?

Joanne Guan 1B
Posts: 30
Joined: Sat Jul 22, 2017 3:01 am

### Re: 9.63

The book gives definitions for thermodynamically stable/unstable compounds. A stable compound has a negative standard Gibbs free energy, meaning the reaction goes toward the compound. An unstable compound has a positive standard Gibbs free energy, meaning the reaction goes toward the elements. The question is asking if the compounds are stable or not. Stability refers to the thermodynamic tendency of a compound decomposing into its elements.

zanekoch1A
Posts: 25
Joined: Thu Jul 13, 2017 3:00 am

### Re: 9.63

Doesn't the sign of gibbs that makes the reaction more stable depend on which side of the reaction the compound is written? If it is on the reactants then a positive gibbs would imply stability and if it is on the products a negative.