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Ionisation Energy

The energy ne...

Question

The energy needed for $L i (g) ⟶ {L i}^{3 +} (g) + 3 e$ is $1.96 \times 10^{4} k J$ ${m o l}^{- 1}$ . If the first ionization energy of $L i$ is $520 k J$ ${m o l}^{- 1}$ , the second ionization energy for $L i$ in $3635 \times x k J$ $m o l^{- 1}$ . Find $x$ .
Given ${I E}_{1}$ for $H = 2.18 \times 10^{- 18} k J$ $m o l^{- 1}$ .

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Solution

Given
$L i (g) ⟶ {L i}^{3 +} (g) + e; Δ H = 1.96 \times 10^{4} k J {m o l}^{- 1} . . . . (i)$
$L i (g) ⟶ {L i}^{+} (g) + e; {I E}_{1} = 520 k J {m o l}^{- 1} . . . . (i i)$
${L i}^{+} (g) ⟶ {L i}^{2 +} (g) + e; {I E}_{2} = a k J {m o l}^{- 1} . . . . . (i i i)$
${L i}^{2 +} (g) ⟶ {L i}^{3 +} (g) + e; {I E}_{3} = b k J {m o l}^{- 1} . . . . . . . (i v)$
Also
$b = E_{1}$ for ${L i}^{2 +} \times N_{A} = E_{1} H \times Z^{2} \times N_{A} = 2.18 \times 10^{- 18} \times 3^{2} \times 6.023 \times 10^{23} J {m o l}^{- 1}$
$= 11.81 \times 10^{3} k J {m o l}^{- 1}$
$∴$ eqs $(i), (i i), (i i i), (i v)$
$1.96 \times 10^{4} = 520 + a + 11.81 \times 10^{3}$
$a = 7270 k J {m o l}^{- 1}$
So the total second ionization energy is $3635 \times x k J m o l^{- 1}$ $= 7270 k J {m o l}^{- 1}$
$∴ x = 2$

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