Answer: [tex]\Delta H[/tex] for the given reaction is -851.4 kJ/mol.
Explanation:
Enthalpy of the reaction is equal to the total sum of the standard enthalpies of the formation of products minus the total sum of the standard enthalpies of formation of reactants. It is represented by [tex]\Delta H_{reaction}[/tex]
Mathematically,
[tex]\Delta H_{rxn}=\Delta H_{products}-\Delta H_{reactants}[/tex]
For the given reaction:
[tex]2Al(s)+Fe_2O_3(s)\rightarrow Al_2O_3(s)+2Fe(s)[/tex]
The standard enthalpy of the elements present in their standard state is always zero.
[tex]\Delta H_{rxn}[/tex] for the reaction is calculated by:
[tex]\Delta H_{rxn}=[1(\Delta H_{Al_2O_3})+2(\Delta H_{Fe})]-[1(\Delta H_{Fe_2O_3})+2(\Delta H_{Al})][/tex]
We are given:
[tex]\Delta H_{Fe_2O_3}=-824.3kJ/mol\\\Delta H_{Fe}=0kJ/mol\\\Delta H_{Al_2O_3}=-1675.7kJ/mol\\\Delta H_{Al}=0kJ/mol[/tex]
Putting values in above equation, we get:
[tex]\Delta H_{rxn}=[1(-1675.7)+2(0)]-[1(-824.3)+2(0)]kJ/mol\\\\\Delta H_{rxn}=-851.4kJ/mol[/tex]
Hence, [tex]\Delta H[/tex] for the given reaction is -851.4 kJ/mol.
Answer:
ΔHrxn = [(1) x (-1675.7 kJ/mole) + (2) x (0 kJ/mole)] - [(1) x (-824.3 kJ/mole) + (2) x (0 kJ/mole)] = -1675.7 kJ/mole + 824.3 kJ/mole = -851.4 kJ/mole.
Explanation:
