Gibbs Free Energy Problem on pg 39






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Jasmine_Esparza_2A
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Gibbs Free Energy Problem on pg 39

Postby Jasmine_Esparza_2A » Thu Jan 26, 2017 5:56 pm

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
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Joined: Wed Sep 21, 2016 2:55 pm

Re: Gibbs Free Energy Problem on pg 39

Postby Angela To 2B » Thu Jan 26, 2017 6:10 pm

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
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Joined: Sat Jul 09, 2016 3:00 am

Re: Gibbs Free Energy Problem on pg 39

Postby nicoleclarke_lec1M » Thu Jan 26, 2017 6:11 pm

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.


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