Why does changing the temperature shift the equilibrium?
Suppose you have some reaction
$$\ce{A &lt;=&gt; B}.$$
The equilibrium constant for the reaction is $K$ and the Gibbs free energy change is $\Delta G$. The equilibrium constant is
$$K = \frac{[\ce{B}]}{[\ce{A}]}.$$
Increasing value of $K$ shifts the equilibrium towards the right, i.e. more $\ce{B}$, and reducing the value shifts it to the left, i.e. more $\ce{A}$. The equilibrium constant and the Gibbs free energy are related by
$$ K = \exp\left(-\frac{\Delta G}{RT}\right)$$
Putting in the expression you give for $\Delta G$ gives us
\begin{align}
 K &amp;= \exp\left(-\frac{\Delta H - T\Delta S}{RT}\right) \\
   &amp;= \exp\left(-\frac{\Delta H}{RT}\right)\, 
      \exp\left(\frac{\Delta S}{R}\right).
\end{align}
If we assume the variation of $\Delta H$ and $\Delta S$ is small we can ignore the entropy term because it's a constant, and we get
$$K \propto \exp\left(-\frac{\Delta H}{RT}\right).$$
If the reaction is endothermic $\Delta H$ is positive, so we have the exponential of a negative number and this is less than one. If we increase the temperature we decrease $\Delta H/(RT)$ and the exponential increases. So increasing the temperature makes the equilibrium coefficient bigger, i.e. it drives the endothermic reaction.
To understand why this happens consider what happens when we make the temperature very high.  If you do this the value of $\Delta H/(RT)$ approaches zero and the exponential approaches unity, i.e. the concentrations of $\ce{A}$ and $\ce{B}$ are the same. What we've done is make the thermal energy much greater than the activation energy in either direction, so the rates of the forward and backward reactions are the same.