Question

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.

Answer 1

Answer:

$\Delta H_{rxn}=-847.6\ KJ/mol.$

Explanation:

The balanced chemical equation is:

$2Al(s)+Fe_2O_3(s)-->Al_2O_3+2Fe(s)$

Now, standard values of $\Delta H_f\ of\ all\ participants\ of\ reaction.$

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

Now, We know $\Delta H_{rxn}=\Delta H_{products}-\Delta H_{reactants}$

Putting all values of $\Delta H_f$ to above equation.

We get,

$\Delta H_{rxn}=-847.6\ KJ/mol.$

Hence, this is the required solution.