Question

The energy needed for Li gives to Li+3 + 3e is
1.96 x 10^4 KJ mole' If the first ionisation
energy of Li is 520 KJ mole 'calculate second
ionisation energy for Li. Given IE, for
H = 2.18 x 10^-18 J atom"​

Answer 1

The energy needed for Li gives to Li³⁺ + 3e¯ is 1.96 × 10⁴ KJ/mol. the first ionisation energy of Li is 520 kJ/mol.

To find : the second ionisation energy for Li.

solution : as Li ⇒Li³⁺ + 3e¯, I.E = 1.96 × 10⁴ kJ/mol

Li ⇒Li⁺ + e¯, I.E₁ = 520 kJ/mol

Li⁺⇒Li⁺⁺ + e¯, I.E₂ = a kJ/mol

Li²⁺ ⇒Li³⁺ + e¯, I.E₃ = Bohr's energy for Li²⁺ (as Li²⁺ is Hydrogen like atom)

= 13.6 × (3)²/1²

= 13.6 × 9

= 122.4 eV/atom

we know, 1 eV/atom = 96.49 kJ/mol

so, I.E₃ = 122.4 × 96.49 = 11810.376 kJ/mol

now I.E = I.E₁ + I.E₂ + I.E₃

⇒1.96 × 10⁴ = 520 + a + 11810.376

⇒19600 - 11810.376 - 520 = a

⇒a = 7269.624 kJ/mol

Therefore the second ionisation energy for Li is 7269.624 kJ/mol

Answer 2

Answer:

7270

Explanation:

7269.624 can be approximately written as 7270