Trying to figure out heat of reaction A question in Thermodynamics of Natural Systems by G. M. Anderson is to find the "standard heat of reaction" of this reaction:
$$\ce{C8H16N2O3(aq) + H2O(l) -&gt; C6H13NO2(aq) + C2H5NO2(aq)}$$

As far as I understand it, it should be a rather simple problem of finding $\Delta_\mathrm r H$.
So it should be:

$\Delta_\mathrm r H=\Delta_\mathrm fH (\ce{C6}) + \Delta_\mathrm fH (\ce{C2}) - \Delta_\mathrm fH (\ce{C8}) - \Delta_\mathrm fH (\ce{H2O})$ (note that I shortened the organic formulae)

According to the data from the same textbook, it's:
$(-632077~\pu{J/mol}-513988~\pu{J/mol})-(-847929~\pu{J/mol}-285830~\pu{J/mol})=-12306~\pu{J/mol}$

However, the answer given is $−89475~\pu{J/mol}$.

Am I missing something? Do I need to add some terms of pressure or volume expansion? Is this not the enthalpy that I'm looking for?
There is a "readme" note from the author that says:


  The answers are not guaranteed, in the sense that not only
  might I make a mistake here and there, but I might have used different sources of
  data, or the data in a database may have changed. I have not been able to check all
  the answers, or to make them all consistent with the data in Appendix B, although I
  believe that most are.


So I think that this problem is an example of one of those situations.  

The book itself works out the problem on page 267 and gets the same answer as you: −12306 J/mol.