# Gibbs free energy

Discussion in 'MCAT Study Question Q&A' started by Sammy1024, May 15, 2014.

1. ### Sammy1024

701
58
Dec 12, 2013
I did a question where the react did decomp violently when it was heated.

And then the reaction was something liquid into a bunch of moles of gas.

I could tell that entropy would be super positive but I didnt know how to tell the gibbs free energy.

There was a graph saying enthalpy was negative for the forward reaction with it being something like -350 but how do you determine what gibbs free energy was?

I thought that since it said "when it was heated" meant that you had to put in energy for it to react to gibbs would be positive?

3. ### Teleologist 2+ Year Member

616
158
Jul 7, 2013
The delta G of the reaction is negative when the reaction is spontaneous - i.e. the reaction will actually occur with no need for an input of energy.

Because the reaction system was heated before the violent reaction commenced, this suggests to me there is an activation energy barrier. This suggests to me that the delta G for the reaction at the temperature it was at *before* heating is positive (non-spontaneous) because activation energy had to be overcome - i.e. energy had to be inputted.

4. ### Sammy1024

701
58
Dec 12, 2013
That's what I put in the answer but it said delta G was negative. The answer said that since it said decomp "violently" it means that it is exothermic, meaning delta H is negative and then when you plug it into the equation with a positive delta S, delta G is negative.

5. ### cjcarter 2+ Year Member

991
906
Jan 21, 2014
IL
You need to use Gibb's Free Energy equation: ΔG=ΔH−TΔS. Negative enthalpy and positive entropy = negative delta G.

6. ### Teleologist 2+ Year Member

616
158
Jul 7, 2013
Right, I would say this is a crappy question because the temperature isn't specified in the problem. In the explanation however it's pretty clear the author is referring to the reaction system at the temperature it was at *after* heating. Note how I emphasized temperature in my original question.

7. ### Sammy1024

701
58
Dec 12, 2013
I guess looking back at it, it makes sense. I just hate the wording on a bunch of questions whether it's AAMC or TBR or whatever.

8. ### CzarcasmHakuna matata, no worries. 2+ Year Member

966
405
Jun 22, 2013
Crypts of Lieberkühn
The relevant equation is: dG = dH - TdS. In this case, an exothermic reaction which also increases in entropy favors spontaneity. You can confirm that quantitatively by plugging in the values. If they didn't provide the entropy change for the reaction, look at the reaction occuring to gauge what effect it'll have on delta G. If a lot of gas is produced then it has a huge effect on delta G, but if the moles of gas is the same in both products and reactants, then you should expect deltaG to be very close to or near deltaH.

9. ### CzarcasmHakuna matata, no worries. 2+ Year Member

966
405
Jun 22, 2013
Crypts of Lieberkühn