Chapter 3: Problem 52
Let IP stand for ionization potential. The IP, and \(\mathrm{IP}_{2}\) of \(\mathrm{Mg}\) are 178 and \(348 \mathrm{kcal} \mathrm{mol}^{-1}\). The energy required for the following reaction is: \(\mathrm{Mg} \rightarrow \mathrm{Mg}^{2+}+2 \mathrm{e}^{-}\) (a) \(+178\) kcal (b) \(+526 \mathrm{kcal}\) (c) \(-170\) kcal (d) \(-526\) kcal
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