Question

One reaction involved in the conversion of iron ore to the metal is FeO(s) + CO(g) → Fe(s) + CO2(g) Use Hess’s Law to calculate ΔHrxn given the following. 3 Fe2O3(s) + CO(g) → 2 Fe3O4(s) + CO2(g) ΔH = −47.0 k???? Fe2O3(s) + 3 CO(g) → 2 Fe(s) + 3 CO2(g) ΔH = −25.0 k???? Fe3O4(s) + CO(g) → 3 FeO(s) + CO2(g) ΔH = 19.0 k????

Answer 1

`\delta H_(rxn) = -66.0 \ kJ/mole`

Step-by-step explanation:

Given that:

`3FeO_3_((s))+CO_((g)) \to 2Fe_3O_4_((s)) +CO_(2(g)) \ \ \delta H = -47.0 \ kJ/mole -- equation (1) \\ \\ \\ Fe_2O_3_((s)) +3CO_((g)) \to 2FE_((s)) + 3CO_(2(g)) \ \ \delta H = -25.0 \ kJ/mole -- equation (2) \\ \\ \\ Fe_3O_4_((s)) + CO_((g)) \to 3FeO_((s)) + CO_(2(g)) \ \delta H = 19.0 \ kJ/mole -- equation (3)`

From equation (3) , multiplying (-1) with equation (3) and interchanging reactant with the product side; we have:

`3FeO_((s)) + CO_(2(g)) \to Fe_3O_4_((s)) + CO_((g)) \ \delta H = -19.0 \ kJ/mole -- equation (4)`

Multiplying (2) with equation (4) ; we have:

`6FeO_((s)) + 2CO_(2(g)) \to 2Fe_3O_4_((s)) + 2CO_((g)) \ \delta H = -38.0 \ kJ/mole -- equation (5)`

From equation (1) ; multiplying (-1) with equation (1); we have:

`2Fe_3O_4_((s)) +CO_(2(g)) \to 3FeO_3_((s))+CO_((g)) \ \ \delta H = 47.0 \ kJ/mole -- equation (6)`

From equation (2); multiplying (3) with equation (2); we have:

`3 Fe_2O_3_((s)) +9CO_((g)) \to 6FE_((s)) + 9CO_(2(g)) \ \ \delta H = -75.0 \ kJ/mole -- equation (7)`

Now; Adding up equation (5), (6) & (7) ; we get:

`6FeO_((s)) + 2CO_(2(g)) \to 2Fe_3O_4_((s)) + 2CO_((g)) \ \delta H = -38.0 \ kJ/mole -- equation (5)`

`2Fe_3O_4_((s)) +CO_(2(g)) \to 3FeO_3_((s))+CO_((g)) \ \ \delta H = 47.0 \ kJ/mole -- equation (6)`

`3 Fe_2O_3_((s)) +9CO_((g)) \to 6FE_((s)) + 9CO_(2(g)) \ \ \delta H = -75.0 \ kJ/mole -- equation (7)`

`FeO \ \ \ + \ \ \ CO \ \ \to \ \ \ \ Fe_((s)) + \ \ CO_(2(g)) \ \ \ \delta H = - 66.0 \ kJ/mole`

`\delta H_(rxn) = \delta H_1 + \delta H_2 + \delta H_3` (According to Hess Law)

`\delta H_(rxn) = (-38.0 + 47.0 + (-75.0)) \ kJ/mole`

`\delta H_(rxn) = -66.0 \ kJ/mole`