Consider this mechanism for acyl substitution by nucleophilic addition-elimination:
To start with, the nucleophile having a lone pair of electrons attacks carbon (because of oxygen being more electronegative?) and forms the conformation/arrangement of atoms as shown in the image.
But (there is an equilibrium sign) doesn't this indicate that the atoms are going back to the initial state against their original tendency which was responsible for the formation of the product in the first place? For example, in second step, what would make a nucleophile attached to a positive carbon site, leave to form back the reactants of step one?
Which force would make nucleophile again leave the carbon?
Edit (25/5/17): Above reaction mechanism seems to be supported by "isotope labeling experiments". See Nucleophilic acyl substitution -- Wikipedia.
Edit (26/5/17): For more clarification: (These comments were present below the deleted answer of Berry Holmes)
Are the intermediates given in any mechanism experimentally verified by any of the techniques (though they may be having short life span?)? I will search for it, if you know it, it will be helpful.
I have found one such experiment: http://pubs.acs.org/doi/abs/10.1021/ac60221a003?journalCode=ancham ; I am still seeing the reasoning behind every action of atoms in the reaction using even your answer, I will say you after completion of that process.
Let us consider only nucleophillic addition, i.e. first two steps; nucleophile in the second step is the only product, there may be energy for it due to temperature, this kinetic energy can be used to collide with either of the reactants in the first step. If it collides with nuclephile containing compound :Nu-H, Nu+ in second step may form bond with :Nu-H from by being attached with carbon of second step, this seems to not form the first reactants back again.
If we consider another possibility of collision of second reactant with the acyl compound, the attack of Nu+ of second step goes to oxygen of acyl compound, here I am now not seeing any easy way of the initial products being formed again. Even if energy profile diagram of the above reaction has less energy of activation for the back reaction, which could be easily attainable from the initial given energy for forward reaction, I am not seeing any way of this energy being used to form back the reactants.
There seems to be one possibility for the compound in second step to produce the reactants back. The negative charge in second step of O- might produce tendency for its lone pair to form double bond with carbon at the center, this needs to make -(Nu+-H) group leave the central carbon. We need to now explore on how -(Nu+-H) group is a better leaving group than other groups, and even on the time taken by oxygen's lone pair to form double bond, to know on whether this happens before H of -(Nu+-H) attaches to L.
Image Source: Solomons, Fryhle and Snyder's "Organic Chemistry" [International Student Version (India) 11th edition (Page No. 784)].