Question

Calculate the enthalpy of reduction for the reaction: Fe2O3(s) + 3CO(g) → 2Fe(s) + 3CO2(g)given that the standard enthalpy of formation of Fe2O3(s), CO(g), and CO2(g) are -824kJ/mol, -110.5kJ/mol, and -393.5kJ/mol respectively. Assume all reactants and products are in their standard states.

Answer 1

To calculate the enthalpy of reduction for the given reaction, we can use the following formula:

ΔH reaction  = Σ ΔHf products  - Σ ΔHf reactants

where ΔH reaction  is the enthalpy of reduction, ΔHf products  is the sum of the standard enthalpies of formation of the products, and ΔHf reactants  is the sum of the standard enthalpies of formation of the reactants.

For the given reaction:

Fe2O3 s  + 3CO g  → 2Fe s  + 3CO2 g

The standard enthalpy of formation of Fe s  is 0 kJ/mol since it is in its elemental form.

Now, let's calculate the enthalpy of formation for the products and reactants:

Σ ΔHf products  =  2 × 0 kJ/mol  +  3 × -393.5 kJ/mol  = -1180.5 kJ/mol

Σ ΔHf reactants  =  -824 kJ/mol  +  3 × -110.5 kJ/mol  = -1155.5 kJ/mol

Now, we can calculate the enthalpy of reduction:

ΔH reaction  = Σ ΔHf products  - Σ ΔHf reactants  = -1180.5 kJ/mol -  -1155.5 kJ/mol  = -25 kJ/mol

Therefore, the enthalpy of reduction for the given reaction is -25 kJ/mol.