Answer
$\Delta E=-3463\,kJ$
$\Delta H= -3452\,kJ$
Work Step by Step
$q=-3452\,kJ$ (heat is released and therefore q is negative)
$w=-11\,kJ$ (work is done by the system and therefore w is negative)
$\Delta E=q+w= -3452\,kJ-11\,kJ=-3463\,kJ$
At constant pressure, the change in enthalpy is just the heat released or absorbed.
$\Delta H=q_{p}=-3452\,kJ$
