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Question

A hypothetical element X undergoes a reaction as shown:
X+2eX2, energy released = 30.87 eV/atom. If the energy released is used to dissociate 4 g of H2 molecule equally into H+ and H , where H is the excited state of H atoms where the electron travels in orbit whose circumference is equal to four times its de-broglie's wavelength. Determine the least moles of X that would be required:
Given: I.E of H = 13.6 eV/atom , bond energy of H2 = 4.526 eV/molecule.

A
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B
2
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C
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D
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Solution

The correct option is B 2
Number of moles of H2 = 42 = 2 mol
Energy released for x mol = -30.87 eV/atom ×NA×x eV
Bond energy for 2 mol = 4.526×NA×2 eV
Ionisation enthalpy = 2×13.6×NA eV
2 mol of hydrogen is excited from the ground state to the excited state which is given as the fourth orbit.
So, En=13.6Z2n2
= ΔEn for 2 mol = 2×NA×13.6(1142)
Balancing the energy released and absorbed,
4.526×NA×2+(2×13.6×NA)+2×NA×13.6(1142)+(30.87×NA×x eV)
On solving, x = 2 mol

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