First and second ionization enthalpies of lit | vedclass

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(B) The total ionization energy required to convert $1 	ext{ mole}$ of $Li(g)$ to $Li^{2+}(g)$ is the sum of the first and second ionization enthalpi

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(B) The total ionization energy required to convert $1 	ext{ mole}$ of $Li(g)$ to $Li^{2+}(g)$ is the sum of the first and second ionization enthalpies: $IE_{total} = 520 + 7297 = 7817 	ext{ kJ mol}^{-1}$.Given mass of $Li = 3.5 	ext{ mg} = 3.5 	imes 10^{-3} 	ext{ g}$.Molar mass of $Li = 7 	ext{ g mol}^{-1}$.Number of moles of $Li = \frac{3.5 	imes 10^{-3} 	ext{ g}}{7 	ext{ g mol}^{-1}} = 0.5 	imes 10^{-3} 	ext{ mol}$.Total energy required = $	ext{Moles} 	imes IE_{total} = 0.5 	imes 10^{-3} 	ext{ mol} 	imes 7817 	ext{ kJ mol}^{-1} = 3.9085 	ext{ kJ}$.Rounding to the nearest integer, we get $4 	ext{ kJ}$.

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