The following is known as the thermite reaction: 2Al(s)+Fe2O3(s)→Al2O3(s)+2Fe(s) This highly exothermic reaction is used for welding massive units, such as propellers for large ships. Using enthalpies of formation in Appendix C, calculate ΔH∘ for this reaction.

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Answer:

[tex]\Delta H_{rxn}=-847.6\ KJ/mol.[/tex]

Explanation:

The balanced chemical equation is:

[tex]2Al(s)+Fe_2O_3(s)-->Al_2O_3+2Fe(s)[/tex]

Now, standard values of [tex]\Delta H_f\ of\ all\ participants\ of\ reaction.[/tex]

[tex]\Delta H_f(Al_2O_3)=-1669.8\ KJ/mol\\\Delta H_f(Fe_2O_3)=-822.2\ KJ/mol\\\Delta H_f(Al(s))=0\ KJ/mol\\\Delta H_f(Fe(s))=0\ KJ/mol\\[/tex]

Now, We know [tex]\Delta H_{rxn}=\Delta H_{products}-\Delta H_{reactants}[/tex]

Putting all values of [tex]\Delta H_f[/tex] to above equation.

We get,

[tex]\Delta H_{rxn}=-847.6\ KJ/mol.[/tex]

Hence, this is the required solution.

-847.6 kJ/mol is heat of formation or heat change for the given thermite  reaction.

What is exothermic reaction?

Those reactions which releases heat during a chemical reaction is known as exothermic reaction.

Given chemical reaction is:

2Al(s) + Fe₂O₃(s) → Al₂O₃(s) + 2Fe(s) and heat of formation is calculated as:

ΔH⁰ = ΔH⁰ₓ - ΔH⁰ₙ, where

ΔH⁰ₓ = heat of formation of product

ΔH⁰ₙ = heat of formation of reactant

At standard condition heat of formation of reactants and products are listed below:

Al₂O₃ = -1669.8 kJ/mol

Fe₂O₃ = -822.2 kJ/mol

Fe & Al = 0

So, ΔH⁰ = -1669.8 - (-822.2) = -847.6 kJ/mol

Hence, heat of formation or heat change for the given reaction is -847.6 kJ/mol.

To know more about exothermic reaction, visit the link:

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