I really need help with this fast!

Question

Using the equations H₂ (g) + F₂ (g) → 2 HF (g) ∆H° = -79.2 kJ/mol C (s) + 2 F₂ (g) → CF₄ (g) ∆H° = 141.3 kJ/mol 2 C(s) + 2 H₂ (g) → C₂H₄ (g) ∆H° = -97.6 kJ/mol Determine the enthalpy (in kJ/mol) for the reaction C₂H₄ (g) + 6 F₂ (g) → 2 CF₄ (g) + 4 HF (g).

Answer 1

This will be a situation where you need to "add" the chemical equations together to form the final result.

So if I have the chemical equations

1) A+B -> C ∆H° =10

2) D+E -> B ∆H° =20

3) G+A -> D ∆H° =30

and I want to show that

C+G -> 2D+E

My first step would be to reverse reaction 1) so as to make sure I have C as a reactant, so

1) C -> A+B ∆H° =-10

2) D+E -> B ∆H° =20

3) G+A -> D ∆H° =30

Notice how we switch the enthalpy of 1), since the reverse of a reaction where you put energy in, is a reaction where energy is released. Now if I add all of these up, you will notice that A is both a reactant and a product, meaning that you can remove it from the "added" equation, because if you break down, and recreate a substance, there is no net energy gained or lost.

Since, in my final equation, I don't want B to be present either, we should also reverse equation 2), so that B is a reactant and a product just like A.

1) C -> A+B ∆H° =-10

2) B -> D+E ∆H° =-20

3) G+A -> D ∆H° =30

Now, if we add these equations up, and remove substances on both sides of the equation, we get:

C+B ->2D+E

To get the total enthalpy, now just add the enthalpy of each reaction

-10 + (-20) + (30)= 0

Hope this helps