## Class Example

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

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

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

Alexandra Carpenter 1G
Posts: 32
Joined: Sat Jul 22, 2017 3:00 am

### Class Example

In class, we went over a problem where we needed to calculate the temperature needed to make the reaction spontaneous. I understand that delta G needs to be negative for the reaction to be considered as such. We got 333K by setting the delta G to 0, but would that be the final answer? Or would the answer be anything higher than 333K, because then that would make delta G negative?

Johann Park 2B
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Joined: Thu Jul 27, 2017 3:01 am
Been upvoted: 1 time

### Re: Class Example

Setting T = 333K is not the final answer; it's the "boiling point." But it is necessary to find that value to come up with the final answer: T > 333K favors the forward reaction and makes delta G negative.

Anna Okabe
Posts: 30
Joined: Fri Sep 29, 2017 7:06 am

### Re: Class Example

That would not be the final answer, because that is when liquid and gas phase both coexist. Therefore the answer is any temp >333K.

Liz White 1K
Posts: 35
Joined: Thu Jul 13, 2017 3:00 am

### Re: Class Example

With the information that we had, we were forced to solve for G = 0, which gives us the boiling point at which the liquid will become a gas. However, since it must be a gas, anything greater than 333K will be correct for this equation. Any higher temperature will result in G having a negative value, and when G is negative, the reaction is spontaneous.

Lorie Seuylemezian-2K
Posts: 62
Joined: Fri Sep 29, 2017 7:04 am

### Re: Class Example

Dr. Lavelle continued to explain that the final answer is anything higher than 333k will result in a spontaneous reaction in the forward direction.

Amanda Wu 2C
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Joined: Sat Jul 22, 2017 3:00 am

### Re: Class Example

Setting delta G to 0 merely gives the boiling point, in this case T=333 K. At this temperature, both the gas and liquid phases exist; therefore, the forward reaction isn't favored until T>333 K, causing delta G to be negative. If finding the boiling point by setting delta G to 0 complicates the calculation, I suppose you can just create an inequality: 0 > delta H - T(delta S) so that you can go directly to final answer by solving for T.

Nicole Anisgard Parra 2H
Posts: 39
Joined: Sat Jul 22, 2017 3:01 am

### Re: Class Example

Because we are solving for a single variable (T) and there are two unknowns (T and G), we set G=0 in order to find the boiling point at which the liquid and gas state can coexist. Once we find the T at which they coexist, we know that a T higher than the one we found will allow for only gases to exist, which tells us when the process is spontaneous (G will be less than zero, or negative, at that point)