The first ionization enthalpy (IE\(_1\)) and second ionization enthalpy (IE\(_2\)) of Mg(g) are 178 and 348 kcal mol\(^{-1}\) respectively. The energy required for the reaction Mg(g) \(\rightarrow\) Mg\(^{2+}\)(g) + 2e\(^{-}\) (in kcal mol\(^{-1}\)) is:
Show Hint
- +170
- +526
- -170
- -526
The Correct Option is B
Solution and Explanation
The given reaction can be broken down into two steps: 1. Mg(g) \(\rightarrow\) Mg\(^{+}\)(g) + e\(^{-}\) (IE\(_1\)) 2. Mg\(^{+}\)(g) \(\rightarrow\) Mg\(^{2+}\)(g) + e\(^{-}\) (IE\(_2\)) The overall reaction is the sum of these two steps: Mg(g) \(\rightarrow\) Mg\(^{2+}\)(g) + 2e\(^{-}\) The energy required for the overall reaction is the sum of the ionization enthalpies:
Energy = IE\(_1\) + IE\(_2\) Given IE\(_1\) = 178 kcal mol\(^{-1}\) and IE\(_2\) = 348 kcal mol\(^{-1}\), we have: Energy = 178 + 348 = 526 kcal mol\(^{-1}\)
Final Answer: +526 kcal mol\(^{-1}\).
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