Answer:
The answer to your question is: ΔH = -849.5 KJ/mol
Explanation:
Data
ΔH = ?
H Fe2O3 = -825.5 KJ/mol
H Al2O3 = -1675 KJ/mol
H Al = 0 KJ/mol
H Fe = 0 KJ/ mol
Reaction
2Al(s) + Fe2O3(s) → 2Fe(s) + Al2O3(s)
ΔH = H products - H reactants
ΔH = H Al2O3 - H Fe2O3
ΔH = -1675 - (- 825.5)
ΔH = -1675 + 825.5
ΔH = -849.5 KJ/mol
Answer : The standard enthalpy of the reaction is, -849.5 kJ/mol
Explanation :
Enthalpy change : It is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. It is represented as [tex]\Delta H^o[/tex]
The equation used to calculate enthalpy change is of a reaction is:
[tex]\Delta H^o_{rxn}=\sum [n\times \Delta H^o_f(product)]-\sum [n\times \Delta H^o_f(reactant)][/tex]
The equilibrium reaction follows:
[tex]2Al(s)+Fe_2O_3(s)\rightleftharpoons 2Fe(s)+Al_2O_3(s)[/tex]
The equation for the enthalpy change of the above reaction is:
[tex]\Delta H^o_{rxn}=[(n_{(Al_2O_3)}\times \Delta H^o_f_{(Al_2O_3)})+(n_{(Fe)}\times \Delta H^o_f_{(Fe)})]-[(n_{(Fe_2O_3)}\times \Delta H^o_f_{(Fe_2O_3)})+(n_{(Al)}\times \Delta H^o_f_{(Al)})][/tex]
We are given:
[tex]\Delta H^o_f_{(Al_2O_3(s))}=-1675kJ/mol\\\Delta H^o_f_{(Al(s))}=0kJ/mol\\\Delta H^o_f_{(Fe_2O_3(s))}=-825.5kJ/mol\\\Delta H^o_f_{(Fe(s))}=0kJ/mol[/tex]
Putting values in above equation, we get:
[tex]\Delta H^o_{rxn}=[(1\times -1675)+(2\times 0)]-[(1\times -825.5)+(2\times 0)]=-849.5kJ/mol[/tex]
Therefore, the standard enthalpy of the reaction is, -849.5 kJ/mol
