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How can an exothermic reaction be accompanied by an increase in entropy?
DG = DH - TDS.
Here, DS is the entropy of the system, and not that including the surrounding, right? So, how can an exothermic reaction cause an increase in entropy of the system? Shouldn't the fact that energy is being given out by the system result in a decrease of thermal and positional entropy of the system? Could someone please help make this clear...?
I'd like to be a little more specific. How does LOSS of energy of the SYSTEM be accompanied by an INCREASE in entropy of the SYSTEM? The DS in the equation of DG is the entropy of the system only, right?
2 Answers
- 1 decade agoFavorite Answer
Well u have a case where System has a liberty of heat exchange with environment...so the Surrounding comes in play....and equation consist change in entropy of surroundings also......so ur query is absurd...(energy givent out by system ..to whom ???? == surroundings....so u cant ignore it.)
however in adiabatic cases.....This phenomenon dont apply.
