homework - ∆Enthalpy of diatomic molecules

Question:

The reaction between hydrogen and oxygen to yield water vapor has $\Delta{}H^\circ = -484~\mathrm{kJ}$. How much $PV$ work is done, and what is the value of $\Delta{}E$ in kilojoules for the reaction of $0.50\:\mathrm{mol}$ of $\ce{H2}$ with $0.25\:\mathrm{mol}$ of $\ce{O2}$ at atmospheric pressure if the volume change is $-5.6\:\mathrm{L}$?

$$\ce{2H_{2(g)} + O_{2(g)} -> 2H2O_{(g)}}\ \ \ \ \ \Delta{}H^\circ=-484\:\mathrm{kJ}$$

I use the formula $\Delta E=\Delta H-P\Delta V$ to determine $\Delta E$. However, when determing the enthalpy, the solutions manual does this

$$\Delta H=\dfrac{-121\:\mathrm{kJ}}{0.50\:\mathrm{mol}~\ce{H2}}.$$

Where does the $-121$ come from? From my understanding, since there are two moles $\ce{H2}$, $\Delta H$ should be $-242\:\mathrm{kJ}$. Or do we take into account all four hydrogen atoms? That would give us $-121\:\mathrm{kJ}$ $\left(\dfrac{-484\:\mathrm{kJ}}{4}\right)=-121\:\mathrm{kJ}$