Question

Helium can be excited to the ${1s}^1{2p}^1$ configuration by light of $58.44nm$. The lowest excited singlet state, with the configuration ${1s}^1,{2s}^1$ lies $4857c^{-1}$ below ${1s}^1{2p}^1$ state. What would be the average $He-H$ bond energy ($kJ{mol}^{-1}$) have to be in order that $He{H}_2$ could form non-endothermically from $He$ and $H_2$? Assume that the compound would form from the lowest excited singlet state of helium. Neglect any differences between $\Delta U$ and $\Delta H$. Take $\Delta h_f\left(H\right)=218.0Kj/mol$.

Answer 1

Formation of $He{H}_2$ requires energy equal to sum of
(i) Energy for excitation from ${1s}^2$ to ${1s}^1,{2s}^1$ to form $He$ singlet is equal to: [Energy needed for excitation from ${1s}^2$ to ${1s}^1{2p}^1$-energy level difference in between ${1s}^1{2s}^1$ and ${1s}^1{2p}^1$]
Thus $E_{He}singlet=\frac{hc}{{\lambda }_1}-\frac{hc}{{\lambda }_2}=3.40\times {10}^{-18}-9.66\times {10}^{-20}J$
$=3.30\times {10}^{-18}J/molecule$
where ${\lambda }_1=58.44\times {10}^{-9}m;\frac{1}{{\lambda }_2}=4857{cm}^{-1}$
(ii) Energy to produce two mole of $H$,i.e $2\times 218.0=436kJ/mol$
Thus, $E$ for 2 mol bonds of $He-H$
$=[3.30\times {10}^{-18}\times 6.023\times {10}^{23}+436\times {10}^3]J/mol$
$=2423.5kJ{mol}^{-1}$
$E_{He-H}=1211.8kJ{mol}^{-1}$