Answer:
The answer to your question is ΔH rxn = - 849.5 kJ/mol
Explanation:
Reaction
2Al(s) + Fe₂O₃ (s) ⇒ 2Fe (s) + Al₂O₃ (s)
Enthalpy of reaction
ΔHrxn = ∑ H products - ∑ H reactants
ΔHrxn = ∑( H (Al₂O₃) + H (Fe) - ∑ ( H Fe₂O₃ + H Al)
ΔHrxn = [-1675 + 0] - [ -825.5 + 0]
ΔHrxn = -1675 + 825.5
ΔHrxn = - 849.5 kJ/mol
The standard enthalpy of reaction can be defined as the reaction change that takes place when there has been transfer of energy in the reactions.
The standard enthalpy of reaction ([tex]\rm \bold{\Delta H_r_x_n}[/tex]) can be given by:
[tex]\rm \bold{\Delta H_r_x_n}[/tex] = Enthalpy of formation of products - Enthalpy of formation of reactants
The enthalpy of formation of reactants :
= 2 [tex]\times[/tex] [tex]\rm \Delta H_f\;Al[/tex] + 1 [tex]\rm \times\;\Delta H_f\;Fe_2O_3[/tex]
Since Al is in a standard state, the enthalpy of formation is 0.
= 0 + (-825.5 kJ/mol)
= -825.5 kJ/mol
The enthalpy of formation of products:
= 2 [tex]\times[/tex] [tex]\rm \Delta H_f\;Fe[/tex] + 1 [tex]\rm \times\;\Delta H_f\;Al_2O_3[/tex]
Since Fe is in a standard state, the enthalpy of formation is 0.
= 0 + (-1675 kJ/mol.)
= -1675 kJ/mol
[tex]\rm \bold{\Delta H_r_x_n}[/tex] = -1675 kJ/mol - (-825.5 kJ/mol)
[tex]\rm \bold{\Delta H_r_x_n}[/tex] = -1675 kJ/mol + 825.5 kJ/mol
[tex]\rm \bold{\Delta H_r_x_n}[/tex] = -849.5 kJ/mol.
The standard enthalpy for the given reaction has been -849.5 kJ/mol.
For more information about the standard enthalpy, refer to the link:
https://brainly.com/question/10583725
