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

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 molar enthalpy (in kJ/mol) for the reaction C₂H₄ (g) + 6 F₂ (g) → 2 CF₄ (g) + 4 HF (g).

Answer 1

Hess's Law:

Given:

Eq.1. H₂ (g) + F₂ (g) → 2 HF (g) ∆H° = -79.2 kJ/mol

Eq.2. C (s) + 2 F₂ (g) → CF₄ (g) ∆H° = 141.3 kJ/mol

Eq.3. 2 C(s) + 2 H₂ (g) → C₂H₄ (g) ∆H° = -97.6 kJ/mol

Asked for:

C₂H₄ (g) + 6 F₂ (g) → 2 CF₄ (g) + 4 HF (g) TARGET EQUATION

Procedure:

reverse Eq.3: C₂H₄ (g) → 2 C(s) + 2 H₂ (g) ∆H = 97.6 kJ/mol

copy Eq.2 x2: 2C (s) + 4 F₂ (g) → 2CF₄ (g) ∆H = 282.6 kJ/mol

copy Eq.1 x2: 2H₂ (g) + 2 F₂ (g) → 4 HF (g) ∆H = -158.5 kJ/mol

Add them up and combine/cancel like terms to get...

C₂H₄ (g) + 2C (s) + 4 F₂ (g) + 2H₂ (g) + 2 F₂ (g) → 2 C(s) + 2 H₂ (g) + 2CF₄ (g) + 4 HF (g)

C₂H₄ (g) + 6 F₂ (g) → 2CF₄ (g) + 4 HF (g) TARGET EQUATION

∆H = 97.6 + 282.6 - 158.5 = 221.7 kJ/mol

(as always, be sure to check the math)