thermodynamics - Why does the Gibbs Free Energy change determine spontaneity of a reaction?

Here is a doubt that arose in my mind while studying the chapter Thermodynamics of the 11th standard. (It might be silly, but I would love it if you try to understand what I mean to ask before down voting. Also, I want the doubt cleared.)

1. We have been told that if the value of $\Delta G$ or Gibbs' free energy change is negative or zero the process is spontaneous or in equilibrium, respectively. If otherwise, it is non-spontaneous.
   Why? What does $\Delta G$ exactly signify that if it is positive the process won't happen by itself. I mean, I know what $\Delta G$ is equal to, but why does spontaneity of the reaction depend on its sign?

Answer

By the second law of thermodynamics, your main goal is to ensure that $$\Delta S_{\mathrm{universe}} \geq 0$$

$$\Delta S_{\mathrm{universe}} = \Delta S_{\mathrm{sys}} + \Delta S_{\mathrm{external}}$$

$\Delta S_{\mathrm{sys}}$ is measurable on your system. $\Delta S_{\mathrm{external}}$ is a bit more challenging, since you're not going to count the microstates of the external system, so you compute indirectly by heat transfer to the surrounding. Thus,

$$\Delta S_{\mathrm{external}} = -\frac{\Delta H}{T}$$

Combine:

$$T\Delta S_{\mathrm{universe}} = T\Delta S_{\mathrm{sys}} - \Delta H$$

or

$$\Delta G = -T\Delta S_{\mathrm{universe}} = \Delta H -T\Delta S$$

And with the sign flip, you're looking for $\Delta G \leq 0$.

Importantly, I should point out that this is not a closed system in the sense that we assume constant-pressure and not constant-volume.

If you're looking at a process that is constant-volume, you need to replace $\Delta H$ (which has pressure-volume work compensation) with $\Delta U$ (which does not) and replace $\Delta G$ (Gibbs free energy, constant pressure) with $\Delta F$ (Helmholtz free energy, constant volume).