Answer
-1368 $\frac{KJ}{mol}$
Work Step by Step
H=U+PV ==> $\Delta_{r}$H=$\Delta_{r}$U+$\Delta$(PV)
As the enthalpy change of the reaction ($\Delta_{r}$H) is meaningful when the pressure P is constant, the formula for $\Delta_{r}$H above is expressed as: $\Delta_{r}$H=$\Delta_{r}$U+P$\Delta$V
Now, let's assume all gas components in the reaction system to be ideal gases. For ideal gases in a reaction system with constant P, P$\Delta$V=($\Delta$n)RT, where $\Delta$n is the change of moles of all gas components in the equation. In this reaction, $\Delta$n=n(CO2)+n(H2O)-n(O2)=2+3-3=2
So, $\Delta_{r}$H=$\Delta_{r}$U+($\Delta$n)RT
=(-1373$\times$$10^{3}$$\frac{J}{mol}$)+2$\times$(8.31$\frac{J}{mol·K}$)$\times$298K
$\approx$ -1368$\times$$10^{3}$$\frac{J}{mol}$= -1368 $\frac{KJ}{mol}$
