## Gibbs Free Energy Problem on pg 39

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

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

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

Jasmine_Esparza_2A
Posts: 12
Joined: Wed Sep 21, 2016 2:58 pm
Been upvoted: 1 time

### Gibbs Free Energy Problem on pg 39

On page 39 in the course reader, there is an example problem Lavelle went over. It says that deltaG is equal to zero.

Why does the delta G equal to zero?

Angela To 2B
Posts: 20
Joined: Wed Sep 21, 2016 2:55 pm

### Re: Gibbs Free Energy Problem on pg 39

In the example, you are trying to find the temperature at which the reaction is spontaneous. That's when delta G is negative. We set delta G to zero in order to find the temperature at which delta G is zero. In this case, that temperature was 333K. This means that for the reaction to be spontaneous, or delta G negative, the temperature would have to be greater than 333K. In other words, we set delta G to find the temperature. I hope this helps!

nicoleclarke_lec1M
Posts: 25
Joined: Sat Jul 09, 2016 3:00 am

### Re: Gibbs Free Energy Problem on pg 39

In this example, (delta)G would equal zero because that is the temperature at which (delta)H=T*(delta)S. The reason we want this exact value is that any temperature above this value will be just enough to make the latter half of the equation (the T*deltaS part) a higher number than the deltaH value. So, if you are subtracting a bigger number from a smaller one, your answer (the deltaG) will be negative. This is essential in order to have a spontaneous reaction.

He also mentioned in class that this is considered the boiling point; at any phase change, the energy levels are the same so therefore there would be no change in energy (no deltaG).

Hope this helps.