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The first IP of lithium is 5.41eV and electron affinity of Cl is -3.61eV. Calculate ΔH in KJmol^–1 for the reaction
Li_(g) + Cl_(g) → Li_(g)⁺ + Cl^–_(g)

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173.7

1737

17.3

1.73

answer is A.

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Detailed Solution

$Li (g) + Cl (g) \to {Li}^{+} (g) + {Cl}^{-} (g); ∆ H = IP + EA ∆ H = 5.41 - 3.61 = 1.8 ev . = 1.8 \times 1.602 \times 10^{- 19} \times 6.023 \times 10^{23} = 173.67 KJ$

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