Question

36) Calculate the enthalpy of formation of anhydrous aluminium chloride, Al2Cl6 from the following data :

(i) 2Al(s) + 6HCl (aq) → Al2Cl6(aq) + 3H2(g);

△ r H ° = - 1004 . 0 k J

(ii) H2(g) + Cl2 (aq) → 2HCl(g);

△ rH ° = - 183 . 9 k J

(iii) HCl(g) + aq → HCl(aq);

△ r H ° = - 73 . 2 k J

(iv) Al2Cl6(s) + aq → Al2Cl6(aq);

△ r H ° = - 643 . 0 k J

Answer 1

Hey dear,

● 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...