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 Determine the enthalpy for the reaction C (s) + 4 HF (g) → CF₄ (g) + 2 H₂ (g).

Answer 1

Hess' Law:

Target equation = C(s) + 4HF(g) ==> CF₄(g) + 2H₂(g)

eq.1: H₂(g) + F₂(g) ==> 2HF(g) ∆H = -79.2 kJ/mol

eq.2: C(s) + 2F₂(g) ==> CF4(g) ∆H = 141.3 kJ/mol

Reverse eq.1 and multiply it by 2: 4HF(g) ==> 2H₂(g) + 2F₂(g) ∆H = 2x+79.2 = 158.4 kJ/mole

Copy eq. 2 as it is......................C(s) + 2F₂(g) ==> CF₄(g) ∆H = 141.3 kJ

Add both equations

__________________________________________________

4HF(g) + C(s) + 2F₂(g) ===> 2H₂(g) + 2F₂(g) + CF₄(g) now cancel like terms to get the target equation....

4HF(g) + C(s) ===> 2H₂(g) + CF₄(g) and add the ∆H values 158.4 + 141.3 = 299.7 kJ/mole