Chemistry Test ...

Calculate \(\Delta H / k J\) for the following reaction using the listed standard enthalpy of reaction data.
\[
\begin{array}{l}
2 N_{2}(g)+5 O_{2}(g) \longrightarrow 2 N_{2} O_{5}(s) \\
N_{2}(g)+3 O_{2}(g)+H_{2}(g) \longrightarrow 2 H N O_{3}(a q) ; \Delta H / k J=-414.0 \\
N_{2} O_{5}(s)+H_{2} O(I) \longrightarrow 2 H N O_{3}(a q) ; \Delta H / k J=-86.0 \\
2 H_{2}(g)+O_{2}(g) \longrightarrow 2 H_{2} O(I) ; \Delta H / k J=-571.6
\end{array}
\]

Consider the following changes.
\[
\begin{array}{ll}
M(s) \rightarrow M(g) \ldots \ldots(i) & \\
M(s) \rightarrow M^{2+}(g)+2 e^{-} & \ldots \ldots(i i) \\
M(g) \rightarrow M^{+}(g)+e^{-} & \ldots \ldots(i i i) \\
M^{+}(g) \rightarrow M^{2+}(g)^{+} e^{-} & \ldots \ldots(i v) \\
M(g) \rightarrow M^{2+}(g)+2 e^{-} & \ldots \ldots(v)
\end{array}
\]
The second ionization energy of \(M\) can be calculated from the energy values associated with:

The formation of the oxide ion \(O^{2-}(g)\) requires first an exothermic and then an endothermic step as shown below.
\(O_{(g)}+e^{-}=O^{-}(g), \Delta H^{0}=-142 k J m o l^{-1}\)
\(O^{-}(g)+e^{-}=O^{2-}(g), \Delta H^{0}=+844 k J m o l^{-1}\)
This is because :