Question

he thermite reaction, in which powdered aluminum reacts with iron(III) oxide, is highly exothermic. 2 Al(s) + Fe2O3(s) Al2O3(s) + 2 Fe(s) Use standard enthalpies of formation to find for the thermite reaction

Answer 1

The given question is incomplete. The complete question is

The thermite reaction, in which powdered aluminum reacts with iron oxide, is highly exothermic: $2Al(s)+Fe_2O_3(s)\rightarrow Al_2O_3(s)+2Fe(s)$. 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 :

$2Al(s)+Fe_2O_3(s)\rightarrow Al_2O_3(s)+2Fe(s)$

We have to calculate the enthalpy of reaction $(\Delta H^o)$.

$\Delta H^o=H_f_{product}-H_f_{reactant}$

$\Delta H^o=[n_{Fe}\times \Delta H_f^0_{(Fe)}+n_{Al_2O_3}\times \Delta H_f^0_{(Al_2O_3)}]-[n_{Al}\times \Delta H_f^0_(Al)+n_{Fe_2O_3}\times \Delta H_f^0_{(Fe_2O_3)}]$

where,

We are given:

$\Delta H^o_f_{(Fe_2O_3(s))}=-824.2kJ/mol\\\Delta H^o_f_{(Al_2O_3(s))}=-1675.7kJ/mol\\\Delta H^o_f_{(Fe(s))}=0kJ/mol\\\Delta H^o_f_{(Al(s)}=0kJ/mol$

Putting values in above equation, we get:

$\Delta H^o_{rxn}=[(2\times 0)+(1\times -1675.5)]-[(2\times 0)+(1\times -824.2)]=-851.5kJ$

ΔH∘rxn for the thermite reaction is -851.5 kJ