Enthalpy (Chemistry)

Question

Find the ΔH for the reaction below, given the following reactions and subsequent ΔH values:

FeO (s) + CO (g) --> Fe (s) + CO2 (g)      = ?

 

Fe2O3 (s) + 3 CO (g) --> 2 Fe (s) + 3 CO2 (g)         = -23 kJ

3 Fe2O3 (s) + CO (g) --> 2 Fe3O4 (s) + CO2 (g)    = -39 kJ

Fe3O4 (s) + CO (g) --> 3 FeO (s) + CO2 (g)            = + 18 kJ

Answer 1

Use Hess' Law

FeO (s) + CO (g) --> Fe (s) + CO2 (g)  TARGET EQUATION

Given:

eq.1: Fe2O3 (s) + 3 CO (g) --> 2 Fe (s) + 3 CO2 (g)      = -23 kJ

eq.2: 3 Fe2O3 (s) + CO (g) --> 2 Fe3O4 (s) + CO2 (g)    = -39 kJ

eq.3: Fe3O4 (s) + CO (g) --> 3 FeO (s) + CO2 (g)        = + 18 kJ

Procedure:

eq.3 reverse & ÷3: FeO(s) + 1/3CO2(g) ==> 1/3Fe3O4(s) + 1/3CO ... ∆H = 1/3(-18 kJ) = -6 kJ

eq.1÷2: 1/2Fe2O3(s) + 3/2CO(g) ==> Fe(s) + 3/2CO2(g) ... ∆H = 1/2(-23 kJ) = -11.5 kJ

eq.2 reverse & ÷6: 1/3Fe3O4(s) + 1/6CO2(g) ==> 1/2Fe2O3(s) + 1/6CO(g) ...∆H = (1/6)(+39 kJ) = 6.5 kJ

Add them up to get:

FeO(s) + 1/3CO2(g) + 1/2Fe2O3(s) + 3/2CO(g) + 1/3Fe3O4(s) + 1/6CO2(g) => 1/3Fe3O4(s) + 1/3CO + Fe(s) + 3/2CO2(g) + 1/2Fe2O3(s) + 1/6CO(g)

Combine and cancel like terms to get:

FeO(2) + CO(g) ==> Fe(s) + CO2(g) ... TARGET EQUATION

∆H = -6 kJ + -11.5 kJ + 6.5 kJ

∆H = -11 kJ