Question

The energy needed for $Li\left(g\right)⟶{Li}^{3+}\left(g\right)+3e$ is $1.96\times {10}^{4}kJ$ ${mol}^{-1}$. If the first ionization energy of $Li$ is $520kJ$ ${mol}^{-1}$, the second ionization energy for $Li$ in $3635\times xkJ$ $mo{l}^{-1}$. Find $x$.
Given ${IE}_1$ for $H=2.18\times {10}^{-18}kJ$ $mo{l}^{-1}$.

Answer 1

Given

$Li\left(g\right)⟶{Li}^{3+}\left(g\right)+e;\Delta H=1.96\times {10}^{4}kJ{mol}^{-1}....\left(i\right)$

$Li\left(g\right)⟶{Li}^{+}\left(g\right)+e;{IE}_1=520kJ{mol}^{-1}....\left(ii\right)$

${Li}^{+}\left(g\right)⟶{Li}^{2+}\left(g\right)+e;{IE}_2=akJ{mol}^{-1}.....\left(iii\right)$

${Li}^{2+}\left(g\right)⟶{Li}^{3+}\left(g\right)+e;{IE}_3=bkJ{mol}^{-1}.......\left(iv\right)$

Also

$b=E_1$ for ${Li}^{2+}\times N_A=E_{1}H\times Z^2\times N_A=2.18\times {10}^{-18}\times 3^2\times 6.023\times {10}^{23}J{mol}^{-1}$

$=11.81\times {10}^{3}kJ{mol}^{-1}$

$\therefore$ eqs $\left(i\right),\left(ii\right),\left(iii\right),\left(iv\right)$

$1.96\times {10}^4=520+a+11.81\times {10}^3$

$a=7270kJ{mol}^{-1}$

So the total second ionization energy is $3635\times xkJmo{l}^{-1}$ $=7270kJ{mol}^{-1}$

$\therefore x=2$