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