What is the relationship between temperature and Gibbs free energy?

[Ahmed Abdullah]

\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.
I know I must be missing something. Please help!

A^{ }_{A}

 

[Borek]

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

 

[Ahmed Abdullah]

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}?

 

[Mapes]

Clearly \Delta H = -T\DeltaS^{ }_{sorrounding}

Not clearly. That equation only holds if you have a very large heat bath surrounding the system with the same temperature T, so that all heat transfer is reversible. It's important to make this distinction.

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.

So you can no longer vary T and ask whether the reaction is spontaneous!