Gibbs free energy

[Sammy1024]

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?

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[Teleologist]

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.

 

[Sammy1024]

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.

 

[cjcarter]

You need to use Gibb's Free Energy equation: ΔG=ΔH−TΔS. Negative enthalpy and positive entropy = negative delta G.

 

[Teleologist]

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.

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.

 

[Sammy1024]

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.

 

[Czarcasm]

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.

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.

 

[Czarcasm]

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.

Was this an AAMC Question or TBR?

I need to stop reading Q&A posts. I feel like people keep posting AAMC specific questions and it's gonna influence my practice scores. I wish people would post warnings before hand.