Question

Calculate enthalpy of hydration of anhydrous al2cl6 from

Answer 1

Answer:

Answer -

∆Hf(Al2Cl6) = -1351 kJ/mol

● Explaination -

To get equation for formation of anhydrous Al2Cl6, we'll have to -

Eqn.(i) + 3×Eqn.(ii) + 6×Eqn.(iii) - Eqn.(iv) :-

2Al(s) + 3Cl2(aq) → Al2Cl6

To calculate ∆Hf(Al2Cl6),

∆Hf(Al2Cl6) = ∆H1 + 3∆H2 + 6∆H3 - ∆H4

∆Hf(Al2Cl6) = -1004 + 3(-184) + 6(-73) - (-643)

∆Hf(Al2Cl6) = -1351 kJ/mol

Therefore, enthalpy of formation of anhydrous aluminum chloride is -1351 kJ/mol.

Hope this helps you...