## Gibbs free energy dependence on enthalpy and entropy

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

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

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

Christina_F_3F
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Joined: Fri Jul 22, 2016 3:00 am

### Gibbs free energy dependence on enthalpy and entropy

In class, we discussed how a reaction with a negative enthalpy change (aka an exothermic reaction) and a positive entropy change (increase in "randomness" or "disorder") will cause the Gibbs free energy to be negative and thus the reaction will be spontaneous at all temperatures. For what kind of reactions does this exist? To me, the idea seems a bit of a contradiction; exothermic reactions are typically associated with bond making, which would suggest that the change in entropy would be negative. Likewise, if there is an increase in entropy in a reaction, you would expect bonds to be broken, not made. Basically, I'm just wondering if reactions that have both a negative delta H and a positive delta S exist, or if they are just a theoretical concept in the calculation of Gibbs free energy meant to deepen understanding.

Chem_Mod
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### Re: Gibbs free energy dependence on enthalpy and entropy

This is a really interesting point you have brought up about the conceptual picture we have for bonds being broken/formed. To best answer your question it might be helpful to think of any generic chemical reaction. When proceeding from reactants to products, regardless of the spontaneity, we can imagine breaking the bonds of the reactants, and then forming the new bonds for the products. This is a general scheme that we see over and over in chemistry, and it happens all of the time.

Now for the subtlety. Depending on the particular types and number of bonds that are being broken compared to the ones that are being formed, a reaction may either be exothermic or endothermic, and the same is true for comparing the total standard entropy of all of the products formed when the new bonds are forged, compared to the total standard entropy of the reactants before their bonds were broken. That difference will determine if the standard entropy change for the reaction is positive or negative

So, in most traditional reactions, bonds are going to be broken, and bonds are going to be formed, but its the relative change in enthalpy for the stability of those bonds, that will determine if it is exothermic, and the relative change in entropy for the sum of all the products, compared to the sum of all of the reactants that will determine the total change in entropy when the reaction proceeds. Putting this all together, some reactions are going to be spontaneous at any temperature if they are exothermic, and the products also yield an increase in entropy.

Hope that helps

Calderon_Humberto_3E
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### Re: Gibbs free energy dependence on enthalpy and entropy

So when G is minimum why can't the system no longer change?

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