If the sign for delta G is negative (spontaneous process) and the sign for delta S is positive (more disorder) for both dissolving processes, how could one be endothermic (positive delta H) and one be exothermic (negative delta H)

Question

Aileen Beasley

Chemistry

15 October, 20:59

If the sign for delta G is negative (spontaneous process) and the sign for delta S is positive (more disorder) for both dissolving processes, how could one be endothermic (positive delta H) and one be exothermic (negative delta H)

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Answers (1)

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Kassandra Reilly · Answer 1

Answer: From your question,

One could be exothermic which means that the final enthalpy will be less than the initial enthalpy. H = Hf-Hi (Hf
In Endothermic reaction, the entropy is lowered by absorbing energy in the surronding. By so doing, the surronding losses energy and the reaction is not spontaneous.

H is positive and S (entropy) is positive.

Explanation:

Exothermic reaction is the reaction where heat is released In the surronding which lead to increase in the surronding Temperature.

Endothermic reaction is the reaction that absorb heat from the surronding and decrease the surronding Temperature.