If the reaction is spontaneous, is it the ΔS surroundings that are always positive or ΔS universe?
I tried researching this on Google. However, the answers that I have found vary with different explanation.
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1$\begingroup$ Hint: What would Delta G say about that? Consider spontaneous but endothermic processes. $\endgroup$– PoutnikJan 8, 2022 at 11:21
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2$\begingroup$ It can be also said $$\Delta S_\mathrm{universe} \ge - \frac { \Delta G_\mathrm{r}} { T}$$ for constant T,p. $\endgroup$– PoutnikJan 8, 2022 at 12:07
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If a reaction is spontaneous, is decided by ΔG= ΔH-TΔS. The value needs to be negative. The second law of thermodynamics tells us that the entropy of the universe always stays the same or increases. Sometimes, for a chemical reaction with negative ΔG, ΔS is negative, but ΔH is as well, that means that the entropy of the system we are describing decreases. How is this compatible with the law I stated earlier? Well, the entropy of the system decreases, but the heat dissipated to the surroundings increases the surroundings’ entropy again, so that the second law of thermodynamics is valid. Otherwise, if a reaction has positive entropy change, it’s evident that is outweighs the (sometimes) positive enthalpy change. Hope that I could clear the doubts, if you have any questions feel free to ask.
