ΔH° of the following reaction c) 2H₂(g) + O₂(g) → 2H₂O(g) is -834kJ.
What is Bond Enthalpy?
The minimum amount of energy required to break down or form the bonds in chemical reaction is known as bond enthalpy.
It is calculated as:
ΔHrxn=∑ΔH bonds broken(reactants)−∑ΔH bonds formed(products)
In order to calculate ΔHrxn for the given equation we need the following data :
Bond energies in kJ/mol
Now,
The given reaction is ,
2H₂(g) + O₂(g) → 2H₂O(g)
Here 2 mole of H₂ and 1 mole of O₂ breaks to form 2 moles of H₂O.
hence,
We know that,
ΔHrxn=∑ΔH bonds broken(reactants)−∑ΔH bonds formed(products)
Hence,
ΔHrxn= [ 2× B.E(H-H) + B.E (O-O)] - [4×B.E (O-H)]
=(2×436 +146)kJ - 4×463kJ
= 1018kJ-1852kJ
ΔHrxn= -834kJ
Thus from the above conclusion we can say that,ΔH° of the following reaction c) 2H₂(g) + O₂(g) → 2H₂O(g) is -834kJ.
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