In enzyme catalysis, the formation of the enzyme-substrate complex causes a reduction in the entropy of the system. If entropy reduces, the gibbs free energy will become less negative. How then is enzyme catalysis favorable? Are there any other factors that off-set this?
As @canadianer hinted, the answer is enthalpy.
ΔG = ΔH - TΔS
For binding to be a spontaneous process ΔG must be negative. If the binding of a ligand to a protein decreases system entropy then ΔS is negative and the entropy term -TΔS is positive. However ligand binding involves the formation of various protein-ligand interactions (hydrogen bonds, ionic interactions, van der Waals interactions) all of which make a negative contribution to the change in enthalpy, so ΔH is negative. If |ΔH| > |TΔS| then ΔG will be negative.
This simplified account ignores the parallel changes taking place due to the ligand 'dissociating' from the solvent.