Deriving the condtion for spontaneity using gibbs free energy While deriving the condition for spontaneity, $\Delta$G$\leqslant$0,
we start by saying that 
$\Delta S_{tot}$ $\ge$0 $\Rightarrow$ $\Delta S_{sys}$ + $\Delta S_{surr}$ $\ge0$   
If $Q$ is the heat transferred to the system from the surroundings, then $−Q$ is the heat lost by the surroundings, so that   $\Delta S_{ext}$ = - ${Q \over T}$, corresponds to the entropy change of the surroundings.
$\Delta S_{int}$ - ${Q \over T}$ $\ge 0 $,
But, isn't   $\Delta S_{int}$ =  ${Q \over T}$ , since $Q$ is the heat gained by the system?
Consequently, wouldn't we always get $\Delta S_{int}$ - ${Q \over T}$ = ${Q \over T}$- ${Q \over T}$=0 ?
 A: There is an asymmetry between the system and the surroundings. In fact, at constant temperature $T$, we are in a canonical or grand-canonical situation (depending on cases), where the "surrondings" is a heat reservoir, with Energy, Number of Particles,etc... much greater than the system. When we modify some external data (ex : volume), the set (sytem + heat reservoir) is evolving until it reaches a thermodynamic equilibrium, and this is done by a positive variation of the total entropy (sytem + heat reservoir) and a diminution of the (Helmholtz, Gibbs) free energy of the system.
The heat reservoir is so big, than we may consider it approximatively in thermodynamic equilibrium, during this evolution, with a temperature T,so, if we consider the heat gained by the heat reservoir $Q_R$, the variation of entropy of the heat reservoir  is approximately $\Delta S_R = \frac{Q_R}{T}$.
On the other hand, during the evolution, the system itself  cannot be considered in thermal equilibrium, so the notion of temperature is not well defined, so we cannot express a variation of entropy of the system using variation of heat and temperature like $\Delta S = \int dS$ with $dS = \frac{\delta Q}{T}$.
For instance, for simplicity, consider Helmotz free energy, and consider that no work is exchanged between the system and the reservoir. The conservation of energy is written : 
$$\Delta U + \Delta U_R=0 \tag{1}$$
where $U$ and $U_R$ are the internal energies of the system and the heat reservoir.
We have : 
$$ \Delta U_R = \Delta Q_R = T \Delta S_R\tag{2}$$
The first equality comes from the hypothesis of no work exchange, and the second equality comes from the approximated thermodynamic equilibrium of the heat reservoir.
The total entropy (system + heat reservoir) is increasing during the evolution, so we have : 
$$\Delta S + \Delta {S_R} \ge 0 \tag{3}$$
Now, from $(1)$ and $(2)$, we get : $\Delta {S_R} = - \frac{\Delta U}{T}$, so, finally : $\frac{1}{T} (T\Delta S - \Delta U) >0$, and noting the Helmoltz free energy $F = U- TS$, we may write : 
$$\Delta F \le 0 \tag{4}$$
