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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 Li2+(g) [ Li(g) Li2+(g)] is kJ. (nearest integer) [Molar mass of Li = 7 g mol-1]
4
Moles of Li = 3.5 10-3 g7 g mol-1 = 5 10-4 mol
Li(g) Li2+(g) + 2e- needs iH1 + iH2 per mole
= 520 + 7297 = 7817 kJ mol-1
∴ Energy required = 5 10-4 7817 = 3.91 kJ ≈ 4 kJ
Answer: 4
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