Consider the following data.
(i) 2Al(s) + 6HCl(aq) → Al$$_2$$Cl$$_6$$(aq) + 3H$$_2$$(g) + 1200 kJ/mol
(ii) H$$_2$$(g) + Cl$$_2$$(g) → 2HCl(g) + 164 kJ/mol.
(iii) HCl(g) + aq → HCl(aq) + 83 kJ/mol.
(iv) Al$$_2$$Cl$$_6$$(s) + aq → Al$$_2$$Cl$$_6$$(aq) + 663 kJ/mol
The enthalpy of formation of anhydrous solid Al$$_2$$Cl$$_6$$ is :

To determine the standard enthalpy of formation $$\Delta_f H^\ominus$$ of anhydrous solid $$Al_2Cl_6$$, we first identify the target reaction:

$$2Al(s) + 3Cl_2(g) \rightarrow Al_2Cl_6(s)$$

Using Hess's law, we combine the given thermochemical equations appropriately to obtain the desired reaction.

The given equations are:

$$2Al(s) + 6HCl(aq) \rightarrow Al_2Cl_6(aq) + 3H_2(g), \qquad \Delta H = -1200 \text{ kJ mol}^{-1}$$

$$H_2(g) + Cl_2(g) \rightarrow 2HCl(g), \qquad \Delta H = -164 \text{ kJ mol}^{-1}$$

$$HCl(g) + aq \rightarrow HCl(aq), \qquad \Delta H = -83 \text{ kJ mol}^{-1}$$

$$Al_2Cl_6(s) + aq \rightarrow Al_2Cl_6(aq), \qquad \Delta H = -663 \text{ kJ mol}^{-1}$$

Equation (i) is used without modification. Equation (ii) is multiplied by $$3$$ to convert $$3$$ moles of $$H_2$$ into $$6$$ moles of $$HCl(g)$$:

$$3H_2(g) + 3Cl_2(g) \rightarrow 6HCl(g), \qquad \Delta H = 3(-164) = -492 \text{ kJ mol}^{-1}$$

Equation (iii) is multiplied by $$6$$ to convert gaseous hydrochloric acid into aqueous hydrochloric acid:

$$6HCl(g) + aq \rightarrow 6HCl(aq), \qquad \Delta H = 6(-83) = -498 \text{ kJ mol}^{-1}$$

Equation (iv) is reversed so that aqueous $$Al_2Cl_6$$ is converted into solid $$Al_2Cl_6$$:

$$Al_2Cl_6(aq) \rightarrow Al_2Cl_6(s) + aq, \qquad \Delta H = +663 \text{ kJ mol}^{-1}$$

On adding these modified equations, all intermediate species cancel, leaving the required overall reaction:

$$2Al(s) + 3Cl_2(g) \rightarrow Al_2Cl_6(s)$$

Therefore, the standard enthalpy of formation is

$$\Delta_f H^\ominus = (-1200) + (-492) + (-498) + 663$$

$$\Delta_f H^\ominus = -2190 + 663$$

$$\Delta_f H^\ominus = -1527 \text{ kJ mol}^{-1}$$

Hence, the standard enthalpy of formation of anhydrous solid $$Al_2Cl_6$$ is $$-1527 \text{ kJ mol}^{-1}$$.