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

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, ΔHr = -551.7 kJ____________(1)
Multiply equation (iii) by 6.
6HCl + aq ----> 6HCl(aq), ​ΔHr = -439.2 kJ____________(2)
Add eq (i), (1) and (2), we get,
3Cl2(aq) + 2Al(s) ----> ​Al2Cl6(aq)​, ​ΔHr = ​-1994.9 kJ______________(3)
Reverse equation (iv)
Al2Cl6(aq) ----> Al2Cl6(s) + aq, ​​ΔHr = 643.0 kJ_________(4)
Add (3) and (4), we get:
2Al(s) + 3Cl2(aq) ----> Al2Cl6(s), ​ ​​ΔHr = -1351.9 kJ
Heat of formation = -1351.9  kJ

Regards

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