Question

An element undergoes reaction as shown:
$X+2e-$, energy released $=30.87eV/atom$. If the energy released, is used to dissociate $4$g of $H_2$ molecules, equally into $H^{+}$ and $H^{\ast }$, where $H^{\ast }$ is excited state of H atoms, where the electrons travels in in orbit whose circumference equal to four times it's de Broglie's wavelength. Determine the least moles of X that would be required:
Given : I.E. of H =$13.6eV/atom$, bond energy of $H_2=4.526eV/molecule$

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is B $2$

Here, each molecule of $H_2$

1.dissociates (this requires bond dissociation enthalpy) into

2.$H^{+}$(for which we require ionization enthalpy) and

3.$H^{\ast }$(which is the excited H atom)

this atom is in the fourth state. Energy in this state $=13.6/n^2=13.6/16$

therefore energy supplied = 13.6 – (13.6/16) = (15/16) 13.6

And we have a total of 4g $H_2$molecules means $2N_a$molecules

Therefore energy required =

$=2\times 6.022\times {10}^{23}$(B.D.E + I.E. + energy for excitation)

$=2\times 6.022\times {10}^{23}(4.526+13.6+(15/16\left)13.6\right)$

$=(2\times 6.022\times {10}^{23}\times 30.876)eV$

This energy is supplies by atoms of X. Energy released per atom = 30.87eV /atom

so, atoms required $=2\times 6.022\times {10}^{23}=2$ moles.