Question

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:

• A. +170
• B. +526
• C. -170
• D. -526

Hint:

Ionization enthalpy is the energy required to remove an electron from a gaseous atom or ion. The total energy required to remove multiple electrons is the sum of the individual ionization enthalpies.

Answer 1

The Correct Option is B

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\(^{-1}\).