Questions

The energy required for the following process is $1.96 \times 10^{4} kJ {mol}^{- 1}$ . ${Li}_{(g)} \to 3 e^{-} + L {i^{3 +}}_{(g)}$ . If the first ionization energy of lithium is $520 kJ {mol}^{- 1}$ , what is the second ionization energy?

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8369 kJ {mol}^{- 1}

7272 kJ {mol}^{- 1}

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6399 kJ {mol}^{- 1}

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detailed solution

Correct option is B

$\begin{aligned} Li (g) \to {Li}^{3 +} (g) + e; ΔH = 1.96 \times 10^{4} {kJmol}^{- 1} \dots . (i) \\ Li (g) \to {Li}^{+} (g) + e; {IE}_{1} = 520 kJ {mol}^{- 1} \dots . (ii) \\ {Li}^{+} (g) \to {Li}^{2 +} (g) + e; {IE}_{2} = {a kJmol}^{- 1} \dots . (iii) \\ {Li}^{2 +} (g) \to {Li}^{3 +} (g) + e; {IE}_{3} = b kJ {mol}^{- 1} \dots \dots (iv) \end{aligned}$

$\begin{aligned} b = E_{1} for {Li}^{2 +} \times N_{A} = E_{1} H \times Z^{2} \times N_{A} = 2.18 \times 10^{- 18} \times 3^{2} \times 6.023 \times 10^{23} {Jmol}^{- 1} \\ ∴ eqs (i), (ii), (iii), (iv) \\ 1.96 \times 10^{4} = 520 + a + 11.81 \times 10^{3} \\ a = 7270 kJ {mol}^{- 1} \end{aligned}$