Enthalpy change of formation of an element = zero ? Watch

Ari Ben Canaan
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Report Thread starter 7 years ago
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Given the following data:

C(s) + 2F2(g) ---> CF4(g); H1 = –680 kJ mol–1

F2(g) ---> 2F(g); H2 = +158 kJ mol–1

C(s) ---> C(g); H3 = +715 kJ mol–1


calculate the average bond enthalpy (in kJ mol–1) for the C––F bond.

So without thinking I solved this question as follows :

x - [-715-158*2] = -680

x = -1711

-1711/4 = -427.75 (correct answer as per markscheme)

Then I stopped and thought.... why did I use the enthalpy change of formation of Fluorine in this equation ?????

I was told that the enthalpy change of formation of elements in their standard state is zero...

What's the explanation ?
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charco
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You didn't - you used the bond enthalpy of fluorine.
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