Question

The Energy needed For $Li\left(g\right)\to {\left(Li\right)}^{3+}\left(g\right)+3e^{-}$ , is $19600kJmol{e}^{-1}$ . The first ionisation energy of $Li\left(g\right)=520{\left(kJmole\right)}^{-1}$ . Calculate $I{E}_2$ For $Li\left(g\right)$ .
(ionisation energy of $H=13.6eV$ )

• A. $\text{75.3 eV/species}$
• B. $\text{25.30 eV/species}$
• C. $\text{30.45 eV/species}$
• D. $\text{62.40 eV/species}$

Answer 1

Answer: (A)

$Li(g)\to L{i}^{3+}(g)+3e;\Delta H=1.96\times {\left(10\right)}^{4}kJmo{l}^{-1}\dots .(i)$
$Li(g)\to L{i}^{+}(g)+e;I{E}_1=520kJmo{l}^{-1}\dots .(ii)$
$L{i}^{+}\left(g\right)\to L{i}^{2+}\left(g\right)+e;I{E}_2=akJmo{l}^{-1}\dots ..\left(iii\right)$
$L{i}^{2+}(g)\to L{i}^{3+}(g)+e;I{E}_3=bkJmo{l}^{-1}\dots ..(iv)$
Also,
$b=E_{1}forL{i}^{2+}\times N_A$
$=H\times Z^2\times N_A=2.18\times 10^{-18}\times 3^2\times 6.023\times 10^{23}Jmo{l}^{-1}=11.81\times 10^{3}kJmo{l}^{-1}$
$\therefore eqs\left(i\right),\left(ii\right),\left(iii\right),\left(iv\right)$
$1.96\times 10^4=520+a+11.81\times 10^3$
$a=7270{\text{kJmol}}^{-1}$
$a=75.3eV/species$
$\left[1\text{ ev/atom }=96.5\text{kJ/mol}\right]$