2 Fe(s) + 6 HCl(aq) →2 FeCl3(aq) + 3 H2(g) ΔHa Fe2O3(s) + 6 HCl(aq) → 2 FeCI3(aq) + 3 H2O(l) ΔHb 2 H2(g) + O2(g) → 2 H2O(l) ΔHc Show how these equations must be summed together according to Hess's Law to determine ΔH for the combustion of iron (target equation shown below). Also show clearly how the ΔH values of each of the three reactions must be manipulated to determine the enthalpy of combustion of iron. 4 Fe(s) + 3 O2(g) → 2 Fe2O3(s) ΔH = ?

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

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

The combustion reaction of iron is:

[tex]4Fe(s)+3O_2(g)\rightarrow 2Fe_2O_3(s)[/tex] [tex]\Delta H=?[/tex]

The intermediate balanced chemical reaction will be,

(1) [tex]2Fe(s)+6HCl(aq)\rightarrow 2FeCl_3(aq)+3H_2(g)[/tex] ; [tex]\Delta H_a[/tex]

(2) [tex]Fe_2O_3(s)+6HCl(aq)\rightarrow 2FeCI_3(aq)+3H_2O(l)[/tex] ; [tex]\Delta H_b[/tex]

(3) [tex]2H_2(g)+O_2(g)\rightarrow 2H_2O(l)[/tex] ; [tex]\Delta H_c[/tex]

Now we are multiplying reaction 1 by 2, reaction 3 by 3 and reverse reaction of 2 by 2 and then adding all the equations, we get:

(1) [tex]4Fe(s)+12HCl(aq)\rightarrow 4FeCl_3(aq)+6H_2(g)[/tex] ; [tex](2\times \Delta H_a)[/tex]

(2) [tex]4FeCI_3(aq)+6H_2O(l)\rightarrow 2Fe_2O_3(s)+12HCl(aq)[/tex] ; [tex](-2\times \Delta H_b)[/tex]

(3) [tex]6H_2(g)+3O_2(g)\rightarrow 6H_2O(l)[/tex] ; [tex](3\times \Delta H_c)[/tex]

The expression for enthalpy of combustion of iron will be,

[tex]\Delta H=2\times \Delta H_a-2\times \Delta H_b+3\times \Delta H_c[/tex]

[tex]\Delta H=2\Delta H_a-2\Delta H_b+3\Delta H_c[/tex]

Therefore, the enthalpy of combustion of iron is, [tex]\Delta H=2\Delta H_a-2\Delta H_b+3\Delta H_c[/tex]