# Homework Help: Gibb's free energy confusion!

1. Jan 7, 2009

### 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.

A$$^{ }_{A}$$

Last edited: Jan 7, 2009
2. Jan 7, 2009

### Staff: Mentor

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

3. Jan 7, 2009

### Ahmed Abdullah

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}?

Last edited: Jan 7, 2009
4. Jan 7, 2009

### Mapes

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!