## 9.65

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

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

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

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Lucian1F
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### 9.65

Can someone help me figure out how to do 9.65? It asks "Which of the following compounds become less stable with respect to the elements as the temperature is raised: a. PCl5 (g) b. HCN (g) c. NO (g) d. SO2 (g)"

PranithaPrasad
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### Re: 9.65

As I understand it, we can calculate the standard entropies of formation for each compound and whichever is lowest (in this case, it would be PCl5) would be the one that is less stable.

Britney Alvey 1B
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### Re: 9.65

Write a balanced chemical equation for each compound and calculate the standard entropies of formation. If the standard entropy of formation is negative, then the compound will be less stable at a higher temperature. If the standard entropy of formation is positive, the compound will be more stable at a higher temperature.

Tiffany Dao 1A
Posts: 32
Joined: Fri Sep 29, 2017 7:05 am

### Re: 9.65

Since everything naturally goes towards disorder or an increased entropy, that is why the most stable would be when entropy is highest. It takes more energy to decrease entropy, therefore meaning that any negative entropy of formation is unstable.

Christy Zhao 1H
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### Re: 9.65

The compound becomes more unstable when the change in entropy is negative and temperature increases. So, after the standard entropies of formation are calculated, PCl5 would be the compound that is less stable since it is negative.

Curtis Tam 1J
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### Re: 9.65

I get how to calculate entropy and enthalpy to see how temperature affects gibbs free energy but doesn't the book have to specify which direction it's going in. In problem 63, it uses the same compounds and asks us to calculate based on its decomposition but in this problem it doesn't specify, and the signs of the enthalpy and entropy values depend on the direction of the reaction.

Timothy Kim 1B
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Joined: Fri Sep 29, 2017 7:04 am

### Re: 9.65

Would a spontaneous reaction (negative free energy) be more stable?

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