Question

Aluminum chloride can be formed from its elements:
(i) 2Al(s) + 3Cl2 (g) ⟶ 2AlCl3 (s) ΔH° = ? Use the reactions here to determine the ΔH° for reaction (i):
(ii) HCl(g) ⟶ HCl(aq) ΔH(ii) ° = −74.8 kJ
(iii) H2 (g) + Cl2 (g) ⟶ 2HCl(g) ΔH(iii) ° = −185 kJ
(iv) AlCl3 (aq) ⟶ AlCl3 (s) ΔH(iv) ° = +323 kJ/mol
(v) 2Al(s) + 6HCl(aq) ⟶ 2AlCl3 (aq) + 3H2 (g) ΔH(v) ° = −1049 kJ

Answer 1

Answer: The $\Deltas H^o_{formation}$ for the reaction is -1406.8 kJ.

Explanation:

Hess’s law of constant heat summation states that the amount of heat absorbed or evolved in a given chemical equation remains the same whether the process occurs in one step or several steps.

According to this law, the chemical equation is treated as ordinary algebraic expressions and can be added or subtracted to yield the required equation. This means that the enthalpy change of the overall reaction is equal to the sum of the enthalpy changes of the intermediate reactions.

The chemical reaction for the formation reaction of $AlCl_3$ is:

$2Al(s)+3Cl_2(g)\rightarrow 2AlCl_3 (s)$    $\Delta H^o_{formation}=?$

The intermediate balanced chemical reaction are:

(1) $HCl(g)\rightarrow HCl(aq.)$    $\Delta H_1=-74.8kJ$    ( ×  6)

(2) $H_2(g)+Cl_2(g)\rightarrow 2HCl(g)$    $\Delta H_2=-185kJ$     ( ×  3)

(3) $AlCl_3(aq.)\rightarrow AlCl_3(s)$    $\Delta H_3=+323kJ$     ( ×  2)

(4) $2Al(s)+6HCl(aq.)\rightarrow 2AlCl_3(aq.)+3H_2(g)$    $\Delta H_4=-1049kJ$

The expression for enthalpy of formation of $AlCl_3$ is,

$\Delta H^o_{formation}=[6\times \Delta H_1]+[3\times \Delta H_2]+[2\times \Delta H_3]+[1\times \Delta H_4]$

Putting values in above equation, we get:

$\Delta H^o_{formation}=[(-74.8\times 6)+(-185\times 3)+(323\times 2)+(-1049\times 1)]=-1406.8kJ$

Hence, the $\Deltas H^o_{formation}$ for the reaction is -1406.8 kJ.