Gibbs Free Energy : What is it trying to say actually?
You can prove that the Gibbs free energy is equal to the maximum amount of work that a certain reaction (or any physical process in general) can perform, i.e. $\Delta G = w_{max}$. Such a proof can be found in any decent Physical Chemistry textbook (such as Atkins'). The remaining energy must be "lost" as heat to basically account for the observation that is essentially stated with the Clausius inequality $\Delta S \geq 0$ (i.e. the entropy of the universe is always increasing) . Of course the definition of entropy $\Delta S = \frac{q_{rev}}{T}$ and the Clausius inequality that is derived from it (and hence the Gibbs free energy) are inherently connected to Kelvins (and Clausius') statement of the second law. You can see from the definition (don't forget the change of entropy of the surroundings!) that (since $\Delta U = q + w$) the entropy is decreasing when heat is completely converted to work.

Ask yourself the following: if heat could be completely converted to work, then the collision of an inelastic ball with the earth (or any other object) would be reversible, have you ever seen a tennis ball suddenly jumping back from the ground?