# Gibb's free energy confusion!

$$\Delta G = \Delta H - T \Delta S$$

I don't understand this equation satisfactorily.
I have learned so far that endothermic reaction having a positive $$\Delta S_{system}$$ is spontaneous at higher temperature because $$T\Delta S$$ out-compete the enthalpy term and make the free energy change negative.
But I don't understand why $$T\Delta S _{sorrounding}$$ should not increase with temperature as $$T\Delta S _{system}$$ does.
Clearly $$\Delta H$$ =$$-T\Delta S _{sorrounding}$$
so we can rewrite the gibbs free energy equation
$$\Delta G= -T\Delta S _{sorrounding} - T\Delta S _{system}$$
$$\Delta G = -T ( \Delta S _{sorrounding} + \Delta S _{system})$$

if $$\Delta S _{sorrounding}$$ and $$\Delta S _{system}$$ both are constant then the sign of $$\Delta G$$ is independent of temperature. If $$\Delta G$$ is positive increasing temperature just make it more positive and If $$\Delta G$$ is negative increasing temperature just make it more negative. Temperature cannot change the sign.

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Borek
Mentor
When you use the equation to describe isolated system you don't have to worry about the rest of the Universe.

When you use the equation to describe isolated system you don't have to worry about the rest of the Universe.
But I need to know why?
Isn't Del H = -T*Del S (sorrounding)? Clearly it is a function of T, then why it would be unaltered by temperature {but T*Del S (system) is altered}?

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Mapes
Clearly $$\Delta$$ H = -T$$\Delta$$S$$^{ }_{sorrounding}$$
But that is a side point. The key issue is this: if you're writing $\Delta H=T\Delta S^\mathrm{surr}$, you're already assuming that the reaction is spontaneous; otherwise energy would not be transferred. And when you write $\Delta G= -T\Delta S^\mathrm{surr} - T\Delta S^\mathrm{sys}$, you're assuming that it's barely spontaneous, that $\Delta G=0$. Otherwise more energy would be present from the spontaneous forward reaction.