Using the equations 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 enthalpy (in kJ/mol) for the reaction
3 SiCl₄ (s) + 4 Fe (s) → 4 FeCl₃ (s) + 3 Si (s)

The enthalpy for the formation of 4 moles of has been 320.3 kJ/mol.

The enthalpy of formation of 1 mole at standard temperature and pressure has been termed the standard enthalpy of formation.

The reaction enthalpy to be calculated has reactants of 3 moles of , and 4 moles of Fe.

The enthalpy for the formation of has been -640.1 kJ/mol.

The enthalpy for the dissociation of has been the reverse of the formation reaction = +640.1 kJ/mol.

The enthalpy for the dissociation of 3 moles of has been = 3 640.1 kJ/mol.

The enthalpy for the dissociation of 3 moles of has been = 1920.3 kJ/mol.

The enthalpy for the formation of 2 moles of = -800 kJ/mol

The enthalpy for the formation of 4 moles of = -1600 kJ/mol.

The final reaction has 3 moles of  , and 4 moles of .

Thus, the final enthalpy of reaction will be: 1920 kJ/mol + (-1600 kJ/mol)

The final enthalpy of reaction will be = 320.3 kJ/mol.

For more information about the enthalpy of reaction, refer to the link:

link

The enthalpy : 320.3 kJ/mol

The change in enthalpy in the formation of 1 mole of the elements is called enthalpy of formation  

The enthalpy of formation measured in standard conditions (25 ° C, 1 atm) is called the standard enthalpy of formation (ΔHf °)  

Based on the principle of Hess's Law,

the change in enthalpy of a reaction will be the same even though it is through several stages or ways  

Reaction

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

4Fe (s) + 6Cl₂ (g) → 4FeCl₃ (s) ∆H° = -1600.0 kJ/mol

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

Reverse

SiCl₄ (s) → Si(s) + 2 Cl₂ (g) ∆H° = 640.1 kJ/mol x 3

3SiCl₄ (s) → 3Si(s) + 6Cl₂ (g) ∆H° = 1920.3 kJ/mol

+

3 SiCl₄ (s) + 4 Fe (s) → 4 FeCl₃ (s) + 3 Si (s) ∆H° = 320.3 kJ/mol

6.4 cm (max) - 25 cm (min)