Using the equations

Question

2 Fe (s) + 3 Cl₂ (g) → 2 FeCl₃ (s) ∆H° = -800.0 kJ/mol Si(s) + 2 Cl₂ (g) → SiCl₄ (s) ∆H° = -640.1 kJ/mol Determine the molar enthalpy (in kJ/mol) for the reaction 3 SiCl₄ (s) + 4 Fe (s) → 4 FeCl₃ (s) + 3 Si (s)

Answer 1

Hess's Law

3 SiCl₄ (s) + 4 Fe (s) → 4 FeCl₃ (s) + 3 Si (s) .. TARGET EQUATION

Given:

eq.1) 2 Fe (s) + 3 Cl₂ (g) → 2 FeCl₃ (s) ∆H° = -800.0 kJ/mol

eq.2) Si(s) + 2 Cl₂ (g) → SiCl₄ (s) ∆H° = -640.1 kJ/mol

Procedure:

reverse eq.2 and multiply by 3: 3SiCl₄(s) ==> 3Si(s) + 6Cl₂(g) .. ∆H = +2400 kJ/mol

copy eq.1 and multiply by 2: 4Fe(s) + 6Cl₂(g) ==> 4FeCl₃(s) .. ∆H = -1600 kJ/mol

Add and combine/cancel like terms to get:

3SiCl₄(s) + 4Fe(s) + 6Cl₂(g) ==> 3Si(s) + 6Cl₂(g) + 4FeCl₃(s)

3SiCl₄(s) + 4Fe(s) ==> 3Si(s) + 4FeCl₃(s) .. TARGET EQUATION

∆H = 2400 kJ/mol + -1600 kJ/mol = 800 kJ/mol