Question

consider the following thermochemical information.
2 al(s) + 3/2 o₂(g) → al₂o₃(s)
δh = -1601 kj/mol
3 fe(s) + 3/2 o₂(g) → fe₂o₃(s)
δh = -821 kj/mol
2 al(s) + fe₂o₃(s) → 2 fe(s) + al₂o₃(s)
calculate the enthalpy change, in kj/mol, for the reaction

Explanation:

Step1: Identify the target reaction

We want to find $\Delta H$ for \(2\mathrm{Al}(s)+\mathrm{Fe}_{2}\mathrm{O}_{3}(s)\to 2\mathrm{Fe}(s)+\mathrm{Al}_{2}\mathrm{O}_{3}(s)\).

Step2: Use Hess's Law

Hess's Law states that the enthalpy change of a reaction is the same whether the reaction takes place in one step or in a series of steps.
We know:

1. \(2\mathrm{Al}(s)+\frac{3}{2}\mathrm{O}_{2}(g)\to\mathrm{Al}_{2}\mathrm{O}_{3}(s)\), \(\Delta H_{1}=- 1601\mathrm{kJ/mol}\)
2. \(3\mathrm{Fe}(s)+\frac{3}{2}\mathrm{O}_{2}(g)\to\mathrm{Fe}_{2}\mathrm{O}_{3}(s)\), \(\Delta H_{2}=-821\mathrm{kJ/mol}\)

If we reverse the second reaction (\(\mathrm{Fe}_{2}\mathrm{O}_{3}(s)\to3\mathrm{Fe}(s)+\frac{3}{2}\mathrm{O}_{2}(g)\), \(\Delta H_{2}^{'} = 821\mathrm{kJ/mol}\)) and add it to the first reaction:

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By Hess's Law, \(\Delta H=\Delta H_{1}+\Delta H_{2}^{'}\)

Substitute \(\Delta H_{1}=-1601\mathrm{kJ/mol}\) and \(\Delta H_{2}^{'}=821\mathrm{kJ/mol}\)

\(\Delta H=-1601 + 821\)

Step3: Calculate the enthalpy change

\(\Delta H=- 780\mathrm{kJ/mol}\)

Answer:

\(-780\mathrm{kJ/mol}\)