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

Step-by-step 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)* \Delta H_f^0_((Fe))+n_(Al_2O_3)* \Delta H_f^0_((Al_2O_3))]-[n_(Al)* \Delta H_f^0_(Al)+n_(Fe_2O_3)* \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* 0)+(1* -1675.5)]-[(2* 0)+(1* -824.2)]=-851.5kJ`

ΔH∘rxn for the thermite reaction is -851.5 kJ