Calculating S-F Bond Enthalpy from Standard Enthalpy of Formation Values

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To calculate the average S-F bond enthalpy in SF6, the standard enthalpy of formation values for SF6, S, and F are used. The bond enthalpy is derived from the enthalpy of formation of SF6, which is -1100 kJ/mol, divided by the number of S-F bonds in the molecule, resulting in an average bond enthalpy of 183.33 kJ/mol. The discussion emphasizes that the enthalpy of formation is equivalent to the enthalpy of the reaction when SF6 is formed from its constituent atoms. Participants are encouraged to calculate the overall enthalpy change for the reaction to confirm their findings. This method effectively illustrates the relationship between bond enthalpy and standard enthalpy of formation.
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Homework Statement


The values of standard enthalpy of formation of SF6(g), S(g), F(g) are -1100, 275, 80 kJ/mol resp. What is the average S-F bond enthalpy in SF6?


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The Attempt at a Solution


I read that the bond enthalpy is negative of enthalpy of formation from constituent atoms.
So average bond entalpy of S-F bond is 1100/6= 183.33kJ/mol. Is this correct?
 
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The heat of a reaction is = (ΔH of Products - ΔH of Reactants).

Consider the balanced equation SF6(g) --> S(g) + 6F(g).

Try calculating the enthalpy for the entire reaction based on the information given to you, and go from there.
 
What is the difference between enthalpy of entire reaction and enthalpy of formation of SF6? I think the enthalpy of formation is equal to enthalpy of reaction as SF6 is formed from its constituent atoms.
 
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