Why is this step exothermic?

From Wade's "Organic Chemistry" [1, p. 1003-1004]:



> 
> **MECHANISM 21-7 Conversion of an Ester to an Amide (Ammonolysis of an Ester)**  
> 
> [![Mechanism 21-7](https://i.stack.imgur.com/Mrqne.png)](https://i.stack.imgur.com/Mrqne.png)
> [![Example](https://i.stack.imgur.com/FXAUP.png)](https://i.stack.imgur.com/FXAUP.png)
> 
> 
> 


Step 2, loss of a poor leaving group is justified by claiming the step is exothermic, hence the TS energy isn't very high (as would be the case w/ $\mathrm{S\_N2}$ since it's neither exo or endo)


Why is this step exothermic though? No neutralization of charge and no net additional bonds were formed (1 pi bond formed, 1 sigma bond broken) 


### Reference


1. Wade, L. G. Organic Chemistry; Pearson Education, **2012,**. ISBN 978-0-321-81139-4.


