The reaction between hydrogen and oxygen to yield water vapor has $\Delta H^\circ = \pu{- 484 kJ}$. How much $pV$ work is done, and what is the value of $\Delta E$ in kilojoules for the reaction of $\pu{0.50 mol}$ of $\ce{H2}$ with $\pu{0.25 mol}$ of $\ce{O2}$ at atmospheric pressure if the volume change is $-\pu{5.6L}?$
$$\ce{2H2(g) + O2(g) -> 2H2O(g)} \qquad \Delta H^\circ = \pu{- 484 kJ}$$
I use the formula $\Delta E=\Delta H - p\Delta V$ to determine $\Delta E.$ However, when determining the enthalpy, the solutions manual does this:
$$\Delta H = \frac{\pu{-121 kJ}}{\pu{0.50 mol}~\ce{H2}}$$
Where does the $-121$ come from? From my understanding, since there are two moles $\ce{H2}$, $\Delta H$ should be $-\pu{242 kJ}$. Or do we take into account all four hydrogen atoms? That would give us $\pu{-484 kJ}/4 = -\pu{121 kJ}.$