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

each  column  of the  table  is called a group. elements in a group share similar properties. groups are  read  from  top  to  bottom. metal. metalloid. nonmetal. solid. liquid. gas. fe. hg. o.  2. 3. 4. 5. 6. 7. 8. 9.  2. beryllium. 9.012. 4.  sodium. 22.990.  11. 3.  magnesium. 24.305.  12.  potassium. 39.098.  19. 4.  calcium. 40.078.  20. i guess  i  

c is having 4 valence electrons and is bonded to three hydrogens and one sulphur atom, therefore, the number of bond pair is equal to four. as all the valence electrons are used in forming the bonds so, there is no lone pair left on carbon.

s is having 6 valence electrons and is bonded to one other sulfur atom and one carbon atom, therefore, the number of bond pair is equal to 2. as only two valence electrons are used in forming the bonds so, four electrons are left. therefore, two lone pairs are present on s.