The given question is incomplete. The complete question is
The thermite reaction, in which powdered aluminum reacts with iron oxide, is highly exothermic: [tex]2Al(s)+Fe_2O_3(s)\rightarrow Al_2O_3(s)+2Fe(s)[/tex]. Use standard enthalpies of formation to find ΔH∘rxn for the thermite reaction. Express the heat of the reaction in kilojoules to four significant figures.
Answer: ΔH∘rxn for the thermite reaction is -851.5 kJ
Explanation:
The balanced chemical reaction is :
[tex]2Al(s)+Fe_2O_3(s)\rightarrow Al_2O_3(s)+2Fe(s)[/tex]
We have to calculate the enthalpy of reaction [tex](\Delta H^o)[/tex].
[tex]\Delta H^o=H_f_{product}-H_f_{reactant}[/tex]
[tex]\Delta H^o=[n_{Fe}\times \Delta H_f^0_{(Fe)}+n_{Al_2O_3}\times \Delta H_f^0_{(Al_2O_3)}]-[n_{Al}\times \Delta H_f^0_(Al)+n_{Fe_2O_3}\times \Delta H_f^0_{(Fe_2O_3)}][/tex]
where,
We are given:
[tex]\Delta H^o_f_{(Fe_2O_3(s))}=-824.2kJ/mol\\\Delta H^o_f_{(Al_2O_3(s))}=-1675.7kJ/mol\\\Delta H^o_f_{(Fe(s))}=0kJ/mol\\\Delta H^o_f_{(Al(s)}=0kJ/mol[/tex]
Putting values in above equation, we get:
[tex]\Delta H^o_{rxn}=[(2\times 0)+(1\times -1675.5)]-[(2\times 0)+(1\times -824.2)]=-851.5kJ[/tex]
ΔH∘rxn for the thermite reaction is -851.5 kJ
