Given that :
`O(g) + e^(-) to O^(-)(g) DeltaH = -142 kJ "mol"^(-1)`
`O(g) + 2e^(-) to O^(2-)(g) Delta H = + 712 kJ "mol"^(-1)`
what will be the `DeltaH` for the reaction:
`O^(-)(g) + e^(-) to O^(2-)(g)`

Explanation

To find the enthalpy change (ΔH) for the reaction:

\[ O^-(g) + e^-(g) \rightarrow O^{2-}(g) \]

we will use the given reactions and their corresponding enthalpy changes.

### Given Reactions:
1. \( O(g) + e^-(g) \rightarrow O^-(g) \) \(\Delta H = -142 \, \text{kJ/mol}\)
...