By using the bond energies given (picture attached) the change in enthalpy for the reaction will be -78 KJ
Bond energy of N≡N = 942 KJ/mol
Bond energy of H-H = 432 KJ/mol
Bond energy of N-H = 386 KJ/mol
Enthalpy change of the reaction = ?
Enthalpy change is the amount of heat evolved or absorbed during a chemical process.
The balanced chemical equation for the given reaction
N₂ + 3H₂ → 2NH₃
So,
number of moles of N≡N bonds = 1
number of moles of H-H bonds = 3
number of moles of N-H bonds = 2(3) = 6
In the first step bond breakage occurs between N≡N and H-H, as bond breakage absorbed heat so we use the plus sign here. In the second step bond formation releases energy so we use the minus sign here.
Calculate the enthalpy of the reaction
Enthalpy of reaction = (1 × BE of N≡N) + (3 × BE of H-H) - (6 × BE of N-H)
BE = bond energy
Put the bond energy values
Enthalphy of reaction = (1 × 942) KJ + (3 × 432) KJ - (6 × 386 ) KJ
Enthalphy of reaction = -78 KJ
You can also learn about enthalpy from the following question:
https://brainly.com/question/13996238
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