Determine the enthalpy for the reaction...

Question

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

Answer 1

This is another problem using Hess' Law.

Target equation = 3SiCl₄(s) + 4Fe(s) ==> 4FeCl₃(s) + 3Si(s)

eq. 1: Si(s) + 2Cl₂ ==> SiCl₄(s) ∆H = -640 kJ/mol

eq. 2: 2Fe(s) + 3Cl₂(g) ==> 2FeCl₃(s) ∆H = -800 kJ/mol

reverse eq. 1 and multiply it by 3: 3SiCl₄ ==> 3Si + 6Cl₂ ∆H = +640 x 3 = +1920 kJ

copy eq. 2 and multiply it by 2: 4Fe + 6Cl₂ ==> 4FeCl₃ ∆H = -800 x 2 = - 1600 kJ

_____________________________________________ add the two equations....

3SiCl₄ + 4Fe + 6Cl₂ ==> 3Si + 6Cl₂ + 4FeCl₃ and add the ∆H values +1920 - 1600 = +320 kJ

Cancel like terms on both sides of the equation to obtain the target equation....

3SiCl₄ + 4Fe ==> 3Si + 4FeCl₃ and ∆H = +320 kJ