Question

The energy needed for the following process is 1.960 × 104 kJ/mol: Li(g) → Li3+(g) + 3e− If the first ionization energy of lithium is 5.20 × 102 kJ/mol, calculate the second ionization energy of lithium—that is, the energy required for the process: Li+(g) → Li2+(g) + e− A hydrogen-like ion is an ion containing only one electron. The energies of the electron in a hydrogen-like ion are given by En = −(2.18 × 10−18 J) (Z2)(1/n2) where En is the energy of the electron in the hydrogen-like ion, n is the principal quantum number, and Z is the atomic number of the element. Second ionization energy of lithium: × 10 kJ/mol (Enter your answer in scientific notation)

Answer 1

Answer:

Second Ionization energy, IE₂ = 7.27 x 10³ kJ/mol

Explanation:

Using En = -2.18 x 10¹⁸ J (Z²)(1/n²)

Z = atomic number; n = energy state of electron

Lithium has atomic number, Z = 3

Energy for n = 1:

E₁ = -2.18 x 10¹⁸ J (3²)(1/1²) = -1.962 x 10⁻¹⁷ J

When an electron is completely remove from an atom, n = ∞

Energy for n = ∞:

E∞ = -2.18 x 10¹⁸ J (3²)(1/∞²) = 0 J

Energy absorbed by the electron in order to be removed from the atom completely, ΔE = E∞ - E₁

ΔE = 0 - (-1.962 x 10⁻¹⁷ J) = 1.962 x 10⁻¹⁷ J = 1.962 x 10⁻²⁰ kJ

Ionization energy in kJ/mole of Li²⁺ ions, IE₃ is then calculated thus:

1 mole of Li²⁺ ions contains 6.02 x 10²³ ions

Ionization energy in kJ/mole of Li²⁺ ions = 1.962 x 10⁻²⁰ kJ x 6.02 x 10²³ = 1.185 x 10⁴ kJ/mol

Total energy, Etotal = IE₁ + IE₂ + IE₃

Second Ionization energy, IE₂ = Etotal - ( IE₁ + IE₃

IE₂ = 1.960 x 10⁴ kJ/mol - (520 kJ/mol + 1.185 x 10⁴ kJ/mol)

IE₂ = 7.27 x 10³ kJ/mol