Based on bond energies, what will be the heat of reaction for the formation of 2 moles of hydrogen fluoride from one mole each of hydrogen gas and fluorine gas?

ΔH₍reaction₎ = ΔH₍ hydrogen gas bond-breakage₎ + ΔH₍fluorine gas bond-breakage₎ + 2.ΔH₍hydrogen fluoride bond-formation₎ .

Explanation:

Reaction:

H₂ + F₂ --> 2.HF      ΔH₍reaction₎

In this reaction, bonds are broken in the hydrogen gas and fluorine gas molecules which then lead to the new bond formation, that is, bond between hydrogen and fluorine atoms which explains the formation of hydrogen fluoride.

ΔH₍reaction₎ = ΔH₍bond-breakage₎ + ΔH₍bond-formation₎

ΔH₍bond-breakage₎ is just ΔH₍ hydrogen gas bond-breakage₎ + ΔH₍fluorine gas bond-breakage₎.

ΔH₍bond-formation₎ is 2.ΔH₍hydrogen fluoride bond-formation₎ .Since the ΔH₍bond-formation₎ is in the units kJ/mol, a factor of 2 comes in the picture

Therefore,

ΔH₍reaction₎ = ΔH₍ hydrogen gas bond-breakage₎ + ΔH₍fluorine gas bond-breakage₎ + 2.ΔH₍hydrogen fluoride bond-formation₎ .

Note: Care should be taken while assigning opposite signs to ΔH₍bond-breakage₎ and ΔH₍bond-formation₎.

considering the question as per the attachment

here we can see that the radon nuclide has 222 as the atomic mass number and 86 as the atomic number.

when it undergoes nuclear fission it gives polonium which has atomic mass number as 218 and atomic number will be 84 along with a helium nucleus which has 4 atomic mass number and 2 as its atomic number.

222 = 218 + 4 (atomic mass number)

86 = ? + 2; therefore ? = 86-2 = 84.

the atomic mass number of polonium will be 84.