universally favored (GC)

[P0W3RL1FT3R]

how is it that the universe favors disorder (positive entropy) yet also favors exothermic reactions (negative enthalpy)?

i understand it will produce a negative gibbs free energy, but conceptually i just dont understand it..

dont the two contradict themselves? i'm surely missing something..

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[tommyngu]

I'll give this one a shot. 😀
I believe your confusion lies in interpreting these values in terms of the system and surroundings.
The 2nd law of thermodynamics states that all spontaneous processes proceed such that the entropy of the system and surroundings (and therefore the universe) increases. This applies to isolated systems and the universe itself is considered an isolated system. ⌂S(universe) = ⌂S(system) + ⌂(surroundings) This is a law so we'll just have to accept it and go from there.
Exothermic reactions are also spontaneous with ⌂H < 0. Enthalpy is defined in terms of the system. Having a negative ⌂H, such as ⌂H = -10kJ means that 10kJ of heat energy was released from the system and so 10kJ of heat energy was absorbed by the surroundings which would be some other object in the universe.
⌂S(system) = ⌂S(whatever released energy) + ⌂S(whatever absorbed energy).
⌂S(whatever released energy) = ⌂q/T1.
⌂S(whatever absorbed energy) = ⌂q/T2.
So ⌂S(system) = [-⌂q/T1] + [⌂q/T2].
Since T1>T2 (heat was released from 1 and transferred to 2), your ⌂S(system) > 0 showing that it's spontaneous.
This increases the temperature of the surroundings, therefore the entropy of the surroundings increases. The entropy of the surroundings increases more than the decrease in entropy of the system so the overall ⌂S of the universe is positive and therefore spontaneous.

Simply put, exothermic reactions increase entropy of the whole universe. If you have a hot object and set it on a cool object, the hot object will transfer heat to the cooler object increasing its entropy more than the decrease in entropy from the hot object.

Sorry, I know this still might be confusing.

 

[P0W3RL1FT3R]

Oh, so the imagine below is the enthalpy and entropy of the system?
so.. condensation releases energy (negative enthalpy) and becomes more ordered (negative enthalpy) but the surroundings gain energy from the exothermic phase change- resulting in disorder, thus a positive entropy?

 

[tommyngu]

This image shows how entropy of the system changes as you go from phase to phase in the order of increasing entropy for the top and decreasing entropy for the bottom.
For the top diagram, going from solid to liquid (fusion), liquid to gas (vaporization), or solid to gas (sublimation) will increase the disorder of the system so ⌂S > 0. The enthalpy of these processes will be positive since energy is required /added to the system in order for the processes to go in this direction.
For the bottom diagram, going from gas to liquid (condensation), liquid to solid (freezing), or gas to solid (deposition) will decrease the disorder of the system so ⌂S < 0. The enthalpy of these processes will be negative because energy is released in these processes.
So yeah, entropy here is the entropy of the system in which the process/reaction takes place. And the enthalpy is the enthalpy of that process so I supposed you can think of it as the enthalpy of the system if that helps you.
Furthermore, to relate it to my previous post (and hopefully not confusing you more), the bottom diagram depicts exothermic processes in which ⌂S of the system is negative. But ⌂S of the universe would be positive making everything you've learned about enthalpy and entropy agree with what is spontaneous (⌂H<0 and ⌂S(universe)>0)

 

[P0W3RL1FT3R]

This image shows how entropy of the system changes as you go from phase to phase in the order of increasing entropy for the top and decreasing entropy for the bottom.
For the top diagram, going from solid to liquid (fusion), liquid to gas (vaporization), or solid to gas (sublimation) will increase the disorder of the system so ⌂S > 0. The enthalpy of these processes will be positive since energy is required /added to the system in order for the processes to go in this direction.
For the bottom diagram, going from gas to liquid (condensation), liquid to solid (freezing), or gas to solid (deposition) will decrease the disorder of the system so ⌂S < 0. The enthalpy of these processes will be negative because energy is released in these processes.
So yeah, entropy here is the entropy of the system in which the process/reaction takes place. And the enthalpy is the enthalpy of that process so I supposed you can think of it as the enthalpy of the system if that helps you.
Furthermore, to relate it to my previous post (and hopefully not confusing you more), the bottom diagram depicts exothermic processes in which ⌂S of the system is negative. But ⌂S of the universe would be positive making everything you've learned about enthalpy and entropy agree with what is spontaneous (⌂H<0 and ⌂S(universe)>0)

Thanks,
I was pretty confused but I think I understand now 👍