ammonia reacts with oxygen to produce nitrogen and water?
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= -1267.2 kJ
The equation for the reaction is;
4NH3 + 3O2 → 2N2 + 6 H2O
ΔHrxn = ∑(ΔHf,products) - ∑(ΔHf. reactants)
ΔHf of NH3 =-45.9 kJ/mol , while ΔHf of H2O = -241.8 kJ/mol
Therefore;
ΔHrxn = 6(-241.8 kJ/mol) - 4(45.9 kJ/mol)
= -1450.8 - (-183.6)
= -1450.8 + 183.6)
= -1267.2 kJ
Answer:
- 1276.2 kJ.
Explanation:
ΔHrxn = ∑(n ΔHproducts) - ∑(n ΔHreactants)
∴ ΔHrxn = [(6 x ΔHf H₂O) + (2 x ΔHf N₂)] - [(4 x ΔHf NH₃) + (3 x ΔHf O₂)],
∴ ΔHrxn = [(6 x - 241.8 kJ/mol) + (2 x 0)] - [(4 x -45.9 kJ/mol) + (3 x 0)] = -1450.8 kJ/mol + 183.6 kJ/mol = - 1276.2 kJ.