Why would a reaction be nonspontaneous at higher temperatures?

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In the context of the Gibbs free energy equation, a reaction with a negative ΔH (exothermic) and negative ΔS can lead to a positive ΔG at higher temperatures, indicating nonspontaneity. This occurs because, at elevated temperatures, the TΔS term outweighs the negative ΔH, resulting in an overall positive ΔG. Such behavior aligns with the second law of thermodynamics, as the reaction would decrease the universe's total entropy. Consequently, if a reaction is nonspontaneous at high temperatures, its reverse reaction becomes spontaneous. The discussion emphasizes that while ΔG must be negative for spontaneity, all reactions can occur to some extent under specific conditions.
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Typically we think of a higher temperature speeding up the reaction rate and/or supplying the activation energy of a reaction. So why is it the case that some reactions are only spontaneous at lower temperatures?
Using the gibbs free energy equation ## \Delta G = \Delta H - T \Delta S ##, If I have a reaction where ##\Delta H## is negative (exothermic?) and ## \Delta S## is negative it makes ##\Delta G## positive at higher temperatures which means the reaction is nonspontaneous at higher temperatures. Why would this be the case?
 
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Jaccobtw said:
Summary: Typically we think of a higher temperature speeding up the reaction rate and/or supplying the activation energy of a reaction. So why is it the case that some reactions are only spontaneous at lower temperatures?

Using the gibbs free energy equation ## \Delta G = \Delta H - T \Delta S ##, If I have a reaction where ##\Delta H## is negative (exothermic?) and ## \Delta S## is negative it makes ##\Delta G## positive at higher temperatures which means the reaction is nonspontaneous at higher temperatures. Why would this be the case?
Did you show this with the van't Hoff equation? What does the equilibrium constant look like as a function of temperature. What is the value of K at 25C?
 
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Chestermiller said:
Did you show this with the van't Hoff equation? What does the equilibrium constant look like as a function of temperature. What is the value of K at 25C?
The way I understand it is if a reaction in nonspontaneous at a high temperature, then its opposite reaction is spontaneous at that same high temperature. So a reaction is still spontaneous if the parts are there.

As far as why can't an exothermic reaction that reduces entropy of its system (rather than surroundings) occur at a high temperature (## \Delta H ## is negative, ##T## is high, ##\Delta S## is negative) is because (my guess) it would break the second law of thermodynamics. An exothermic reaction at high temperatures barely increases the relative entropy of the surroundings while decreasing the entropy of the system more, leading to an overall decrease in entropy of the universe.
 
Jaccobtw said:
The way I understand it is if a reaction in nonspontaneous at a high temperature, then its opposite reaction is spontaneous at that same high temperature. So a reaction is still spontaneous if the parts are there.

As far as why can't an exothermic reaction that reduces entropy of its system (rather than surroundings) occur at a high temperature (## \Delta H ## is negative, ##T## is high, ##\Delta S## is negative) is because (my guess) it would break the second law of thermodynamics. An exothermic reaction at high temperatures barely increases the relative entropy of the surroundings while decreasing the entropy of the system more, leading to an overall decrease in entropy of the universe.
You realize that this thing about ##\Delta G^0## having to be negative in order for a reaction to be spontaneous is just a rule of thumb, right? If simply means that the equilibrium constant is > 1. All reactions will occur spontaneously to some extent if you mix pure reactants, and all reactions will occur spontaneously in reverse to some extent if you mix pure products.
 
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