Question

Calculate the standard enthalpy change for the reaction

2 Al_(s) + Fe₂O_3(s)                                           2 Fe(s) + Al₂O_3(s)  

Given that

2 Al_(s) +   3/2 O_{2 (g)}                             Al₂O_3(s)     ΔH rxn = -1669.8 kJ/mol

2 Fe _(s)     +    3/2 O_{2 (g)}                              Fe₂O_3(s)    ΔH rxn = -822.2 kJ/mol

Answer 1

We must use Hess Law, which states that we can change the HRxn if we modify by multiplication and inversion ( negative sign) the HRxn values

so

First...

we requrie 2Al(s) in the left, so leave equation (1) as it is

we need Fe(s) in the right side, so we must invert (2) this implies a change in in HRXn to -HRxn

so

2Al(s) +   3/2 O2 (g) --> Al2O3(s) ΔH rxn = -1669.8 kJ/mol

Fe2O3(s) --> 2 Fe (s)     +    3/2 O2 (g)   ΔH rxn = -(-822.2) kJ/mol

add all:

2Al(s) + 3/2 O2 (g) + Fe2O3(s) --> Al2O3(s) + 2 Fe (s)     +    3/2 O2 (g) HRxn = -1669.8 + 822.2

cancel common terms

2Al(s) + Fe2O3(s) --> Al2O3(s) + 2 Fe (s) HRxn = -1669.8 + 822.2 = -847.6kJ

which is what we wanted, so

HRxn = -847.6kJ/mol