Calculate enthalpy change if delta h is 22.9 fpt nh3
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The balanced chemical equation for formation of NH3 would be as follows:
12N2(g) + 32H2(g)-->NH3(g)
The energy required to break required bonds for the formation of product is calculated as:
1/2 mole of N2 will require 712 kJ·mol-1/2 = 356 kJ/mol
3/2 mole of H2 will require 3× 436 kJ·mol-1/2 = 654 kJ/mol
Total energy require to break the bonds = (356 + 654) kJ/mol = 1010 kJ/mol
total energy of reaction = (total energy of bonds broken) - (total energy of bonds formed)
-46.3 kJ/mol =1010) kJ/mol - 3(N-H)
3(N-H) = 1010 kJ/mol + 46.3kJ/mol = 1056.3 kJ/mol
(N-H) = 1056.3 kJ/mol /3 = 352.1kJ/mol
Hope this helps you ☺️☺️⭐✨✨⭐✌️✌️
The balanced chemical equation for formation of NH3 would be as follows:
12N2(g) + 32H2(g)-->NH3(g)
The energy required to break required bonds for the formation of product is calculated as:
1/2 mole of N2 will require 712 kJ·mol-1/2 = 356 kJ/mol
3/2 mole of H2 will require 3× 436 kJ·mol-1/2 = 654 kJ/mol
Total energy require to break the bonds = (356 + 654) kJ/mol = 1010 kJ/mol
total energy of reaction = (total energy of bonds broken) - (total energy of bonds formed)
-46.3 kJ/mol =1010) kJ/mol - 3(N-H)
3(N-H) = 1010 kJ/mol + 46.3kJ/mol = 1056.3 kJ/mol
(N-H) = 1056.3 kJ/mol /3 = 352.1kJ/mol
Hope this helps you ☺️☺️⭐✨✨⭐✌️✌️
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