Given the following thermochemical equations:
$(i)$ $H_{2(g)} + \frac{1}{2}O_{2(g)} \rightarrow H_2O_{(l)}; \Delta H = -285 \text{ kJ}$
$(ii)$ $N_2O_{5(g)} + H_2O_{(l)} \rightarrow 2HNO_{3(l)}; \Delta H = -76.6 \text{ kJ}$
$(iii)$ $N_{2(g)} + 3O_{2(g)} + H_{2(g)} \rightarrow 2HNO_{3(l)}; \Delta H = -348.2 \text{ kJ}$
Calculate the $\Delta H$ for the reaction: $2N_{2(g)} + 5O_{2(g)} \rightarrow 2N_2O_{5(g)}$. (in $\text{ kJ}$)

  • A
    $572$
  • B
    $419$
  • C
    $14.5$
  • D
    $26.8$

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$S_{(g)} + \frac{3}{2} O_{2(g)} \rightarrow SO_{3(g)} + 2x \ kcal$
$SO_{2(g)} + \frac{1}{2} O_{2(g)} \rightarrow SO_{3(g)} + y \ kcal$
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