Answer :
Answer:
The bond dissociation enthalpy of O-F bond is 188 kJ/mol.
Explanation:
[tex]\Delta H_{O-F}=?[/tex]
[tex]\Delta H_{O-H}=467 kJ/mol[/tex]
[tex]\Delta H_{H-F}=565 kJ/mol[/tex]
[tex]\Delta H_{O-O}=498 kJ/mol[/tex]
[tex]OF_2(g) + H_2O(g) \longrightarrow O=O(g) + 2HF(g) \Delta H^o = -318 kJ[/tex]
Total bond energies of the reactant :
[tex]= 2 mol\times \Delta H_{O-F} + 2 mol\times \Delta H_{O-H}[/tex]
[tex]= 2 mol\times \Delta H_{O-F} + 2 mol\times 467 kJ/mol[/tex]
Total bond energies of the product:
[tex]= 1 mol\times \Delta H_{O-O} + 2 mol\times \Delta H_{H-F}[/tex]
[tex]= 1 mol\times 498 kJ/mol+ 2 mol\times 565 kJ/mol=1,628 kJ[/tex]
Standard enthalpy of reaction : [tex]\Delta H^o [/tex]
=Total bond energies of the reactants - Total bond energies of the products
[tex]-318 kJ=(2 mol\times \Delta H_{O-F} + 2 mol\times 467 kJ/mol)-1,628 kJ[/tex]
[tex]2 mol\times \Delta H_{O-F}=-318 kJ+1,628 kJ-934 kJ=376 kJ[/tex]
[tex]\Delta H_{O-F}=\frac{376 kJ}{2 mol}=188 kJ/mol[/tex]
The bond dissociation enthalpy of O-F bond is 188 kJ/mol.
Based on the bond energies of the reactants and products, the bond energy of the O–F bond is 188kJ
What is enthalpy change of a reaction?
The enthalpy change of a reaction is the energy evolved or absorbed when reactant molecules react to form products.
- Enthalpy change, ΔH° = Total bond energies of the reactants - Total bond energies of the products
From the equation of reaction:
OF2(g) + H2O(g) ⟶ O=O(g) + 2HF(g) ΔH° = –318 kJ
Using the bond energies of the reactants and products given:
2 (O-F) + 2 ×(O-H) - (O=O) + 2 (H–F) = –318 kJ
2(O-F) + 2 ×(467) - (498) + 2 (565) = –318 kJ
2(O-F) = 376 kJ
O-F m= 188 kJ
Therefore, the bond energy of the O–F bond is 188kJ
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