Respuesta :
The reaction;
O(g) +O2(g)→O3(g), ΔH = sum of bond enthalpy of reactants-sum of food enthalpy of products.
ΔH = ( bond enthalpy of O(g)+bond enthalpy of O2 (g) - bond enthalpy of O3(g)
-107.2 kJ/mol = O+487.7kJ/mol =O+487.7 kJ/mol +487.7kJ/mol =594.9 kJ/mol
Bond enthalpy (BE) of O3(g) is equals to 2× bond enthalpy of O3(g) because, O3(g) has two types of bonds from its lewis structure (0-0=0).
∴2BE of O3(g) = 594.9kJ/mol
Average bond enthalpy = 594.9kJ/mol/2
=297.45kJ/mol
∴ Averange bond enthalpy of O3(g) is 297.45kJ/mol.
O(g) +O2(g)→O3(g), ΔH = sum of bond enthalpy of reactants-sum of food enthalpy of products.
ΔH = ( bond enthalpy of O(g)+bond enthalpy of O2 (g) - bond enthalpy of O3(g)
-107.2 kJ/mol = O+487.7kJ/mol =O+487.7 kJ/mol +487.7kJ/mol =594.9 kJ/mol
Bond enthalpy (BE) of O3(g) is equals to 2× bond enthalpy of O3(g) because, O3(g) has two types of bonds from its lewis structure (0-0=0).
∴2BE of O3(g) = 594.9kJ/mol
Average bond enthalpy = 594.9kJ/mol/2
=297.45kJ/mol
∴ Averange bond enthalpy of O3(g) is 297.45kJ/mol.
The average bond enthalpy in ozone is 303.0 KJ/mol .
Given here,
The change of enthalpy - 107.2 kJ/mo
BE of O2 - 498.7 kJ/mol
The reaction
[tex]\rm \bold{ O(g) +O_2(g)\rightarrow O_3(g)}[/tex]
the change of enthalpy of reaction
H= (BE reactants) - (BE products)
-107.2KJ/mol= 498.7 kJ/mol- (bond enthalpy of products (BE)
The bond enthalpy of Ozone 2 times the bonding enthaply of products Because there's 2 kinds of bonds in O3
605.9 KJ/mol = 2BE
BEproducts = 303.0 KJ/mol
Hence, we can conclude that the average bond enthalpy in ozone is 303.0 KJ/mol .
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