36) Calculate the enthalpy of formation of anhydrous aluminium chloride, Al₂Cl₆ from the following data :
(i)  2Al(s) + 6HCl (aq) $\to$  Al₂Cl₆(aq) + 3H₂(g);
                                              ${△}_{r}H°=-1004.0 kJ$
(ii) H₂(g) + Cl₂ (aq)  $\to$ 2HCl(g);
                                             $△$ _rH$°$ = $-183.9 kJ$
(iii) HCl(g) + aq $\to$ HCl(aq);
                                        ${△}_{r}H°=-73.2 kJ$
(iv) Al₂Cl₆(s)  + aq $\to$ Al₂Cl₆(aq);
                                            ${△}_{r}H°=-643.0 kJ$

Dear Student,

The enthalpy of formation of Al2Cl6(s) can be calculated from following reaction:
2Al(s) + 3Cl2(aq) ----> Al2Cl6(s)
There are 3 moles of Cl2, so multiply eq (ii) given in question by 3.
3H2(g) + 3Cl2(aq) ----> 6HCl, $\Delta$H_r = -551.7 kJ____________(1)
Multiply equation (iii) by 6.
6HCl + aq ----> 6HCl(aq), ​$\Delta$H_r = -439.2 kJ____________(2)
Add eq (i), (1) and (2), we get,
3Cl2(aq) + 2Al(s) ----> ​Al2Cl6(aq)​, ​$\Delta$H_r = ​-1994.9 kJ______________(3)
Reverse equation (iv)
Al2Cl6(aq) ----> Al2Cl6(s) + aq, ​​$\Delta$H_r = 643.0 kJ_________(4)
Add (3) and (4), we get:
2Al(s) + 3Cl2(aq) ----> Al2Cl6(s), ​ ​​$\Delta$H_r = -1351.9 kJ
Heat of formation = -1351.9  kJ

Regards