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

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SUMMARY

The average S-F bond enthalpy in sulfur hexafluoride (SF6) can be calculated using the standard enthalpy of formation values for SF6(g), sulfur (S(g)), and fluorine (F(g)). The values are -1100 kJ/mol for SF6, 275 kJ/mol for S, and 80 kJ/mol for F. The correct calculation for the average S-F bond enthalpy is derived from the equation: average bond enthalpy = -ΔH of formation / number of bonds, resulting in an average bond enthalpy of 183.33 kJ/mol for the S-F bond.

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  • Understanding of standard enthalpy of formation values
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  • Ability to balance chemical equations
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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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