H2 + F2--> 2HF deltaH = 518.0 kJ

Question

H2 ----> 2H deltaH = 436.8 kJ
F2 ---> 2F deltaH = 158.2 kJ

(a) calculate Delta H for H + > HF

(b) what is the standard enthalpy change of formation of F2

Help please!!! i think this is easier than i'm making it.
for (b) would i take 158.2/2 to get 79.1 kJ ?

Answer 1

For b, isn't delta Hf for elements in the free state zero?

For a, I would do this.
eqn1 H2 + F2 ==> 2HF -518.0 kJ(did you omit the minus sign)?
eqn 2 2F ==> F2 -158.2
eqn 3 2H ==> H2 -431.8
Now add eqn 1 to 2 and 3 and you should end up with
2H + 2F ==> 2HF which is just twice what you want. Add delta H values and take 1/2 the total.

Answer 2

equation for the reaction of eugenol with O3/zn+H

Answer 3

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