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 ]

Step-by-step solution

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 = IE₁ + IE₂ = 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