First and second ionization enthalpies of lithium are 520 kJ mol^ -1 and 7297 kJ mol^ -1 respectively. Energy required to convert 3.5 mg lithium (g) into Li ^ 2+ (g) [ Li(g) Li ^ 2+ (g) ] is _______ kJ mol^ -1 . (nearest integer) [Molar mass of Li = 7 g mol^ -1 ]
A. A
The total energy required to convert 1 mole of Li(g) to Li ^ 2+ (g) is the sum of the first and second ionization enthalpies: E = IE_1 + IE_2 = 520 + 7297 = 7817 kJ mol ^ -1 Given mass of lithium = 3.5 mg = 3.5 10^ -3 g Number of moles of lithium = 3.5 10^ -3 g 7 g mol ^ -1 = 5 10^ -4 mol Energy required for 5 10^ -4 mol of Li is: Energy = 5 10^ -4 mol 7817 kJ mol ^ -1 = 3.9085 kJ Rounding to the nearest integer, we get 4 KJ. Answer: 4
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