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So for example, if H is endothermic, then the enthalpy term is positive, meaning energy is absorbed by system overall, meaning net bonds broken, meaning the equation leans towards nonspontaneity. If entropy is negative, then -T∆S is positive, meaning net bonds formed, meaning equation leans towards nonspontaneity. It appears that they oppose one another in regards to bond formation...

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    $\begingroup$ Endothermic does not necessarily mean that there is net bond breakage. For example, you can break one strong bond and form many weak ones, and the number of bonds will increase even though the reaction is endothermic. $\endgroup$
    – Andrew
    Commented Jun 12, 2022 at 16:46

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It is not about opposing terms but from experimental necessity. In the early days before the second law was understood it was assumed that the most efficient utilisation of chemical energy consisted of converting all the heat into work. But experimentally this proved not to be the case.

Lewis and Randall (Thermodynamics) put it very clearly

It is true that according to the first law, the external work performed must be equal to the loss of heat content of a system, unless some heat is given to or taken from the surroundings, but this is precisely the point first seen by Gibbs. When an isothermal reaction runs reversibly $T\Delta S$ is the heat absorbed from the surroundings, and if this is positive, the work done will be even greater than the heat of reaction.

Thus entropy and so the free energy had to be invented to understand the experimental data.

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  • $\begingroup$ discovered and explained not invented $\endgroup$
    – jimchmst
    Commented Jun 16, 2022 at 4:16
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The combined effect of $ \Delta H-T\Delta S $ is a measure of spontaneity. The individual terms (i.e., $ \Delta H$ and $T\Delta S $) are not the actual measure of spontaneity. But in general we say that exothermic reactions are spontaneous (it is when $T \Delta S $ term is negligible in comparison with $ \Delta H$) and sometimes if $\Delta S>0$ we say that the process is spontaneous (if $\Delta H$ term is negligible in comparison with $T\Delta S$ term).

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  • $\begingroup$ Itepends on the direction the system is displaced from equilibrium. the differences determine where the equilibrium lies $\endgroup$
    – jimchmst
    Commented Jun 16, 2022 at 5:21
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At equilibrium, $$\Delta G = 0 = \Delta H-T \Delta S$$ $$\Delta H = T\Delta S$$ This means that a system at equilibrium is stuck there unless matter [chemical potential] or energy is changed to perturb the equilibrium.

Ice + heat = water; add some heat; $T \Delta S$ increases; $\Delta H$ must now increase; this means that heat must be removed and ice melts to maintain equilibrium. Spontaneity direction depends on which side the reaction is displaced from equilibrium. Both directions are spontaneous. One way is energy driven, the other way is entropy driven.

Energy and entropy really do work as a team.

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  • $\begingroup$ Please visit this page, this page and this one on how to format your future posts better with MathJax and Markdown. $\endgroup$
    – andselisk
    Commented Jun 16, 2022 at 5:43

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