Question

Find ∆H for the reactions given below using Hess's Law. Show your work on a separate

sheet of paper.

1. Fe₂O3(s) + 2Al(s) →Al₂O3(s) + 2Fe(s)

a. 2Al(s) + 3/20₂(g) → Al₂O3(s); ∆H = -1675.7 kJ

b. 2Fe(s) + 3/20₂(g) →Fe₂O3(s); AH = -824.2 kJ

Answer 1

To find ∆H for the reaction Fe₂O₃(s) + 2Al(s) → Al₂O₃(s) + 2Fe(s) using Hess's Law, we need to use the given reactions (a) and (b) and manipulate them to obtain the desired overall reaction.

Given:

a. 2Al(s) + 3/2O₂(g) → Al₂O₃(s); ∆H = -1675.7 kJ

b. 2Fe(s) + 3/2O₂(g) → Fe₂O₃(s); ∆H = -824.2 kJ

To match the stoichiometry of the desired reaction, we need to multiply reaction (a) by 2 and reaction (b) by 2:

2a. 4Al(s) + 3O₂(g) → 2Al₂O₃(s); ∆H = 2*(-1675.7 kJ) = -3351.4 kJ

2b. 4Fe(s) + 3O₂(g) → 2Fe₂O₃(s); ∆H = 2*(-824.2 kJ) = -1648.4 kJ

Now, we can add these manipulated reactions to obtain the desired overall reaction:

4Al(s) + 3O₂(g) + 4Fe(s) + 3O₂(g) → 2Al₂O₃(s) + 2Fe₂O₃(s)

To find the ∆H for the overall reaction, we add the ∆H values of the manipulated reactions:

∆H = (-3351.4 kJ) + (-1648.4 kJ) = -4999.8 kJ

Therefore, the ∆H for the reaction Fe₂O₃(s) + 2Al(s) → Al₂O₃(s) + 2Fe(s) using Hess's Law is -4999.8 kJ.

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